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https://www.geeksforgeeks.org/printwriter-printchar-method-in-java-with-examples-2/?ref=rp
[ "Skip to content\nRelated Articles\nPrintWriter print(char[]) method in Java with Examples\n• Last Updated : 31 Jan, 2019\n\nThe print(char[]) method of PrintWriter Class in Java is used to print the specified character array on the stream. This character array is taken as a parameter.\n\nSyntax:\n\n`public void print(char[] charArray)`\n\nParameters: This method accepts a mandatory parameter charArray which is the character array to be printed in the Stream.\n\nReturn Value: This method do not returns any value.\n\nExceptions: This method returns NullPointerException if the specified charArray() is null.\n\nBelow methods illustrates the working of print(char[]) method:\n\nProgram 1:\n\n `// Java program to demonstrate``// PrintWriter print(char[]) method`` ` `import` `java.io.*;`` ` `class` `GFG {``    ``public` `static` `void` `main(String[] args)``    ``{`` ` `        ``try` `{`` ` `            ``// Create a PrintWriter instance``            ``PrintWriter writer``                ``= ``new` `PrintWriter(System.out);`` ` `            ``// Get the character array``            ``// to be printed in the stream``            ``char``[] charArray = { ``'G'``, ``'e'``, ``'e'``, ``'k'``, ``'s'``,``                                 ``'F'``, ``'o'``, ``'r'``,``                                 ``'G'``, ``'e'``, ``'e'``, ``'k'``, ``'s'` `};`` ` `            ``// print the charArray``            ``// to this writer using print() method``            ``// This will put the charArray in the stream``            ``// till it is printed on the console``            ``writer.print(charArray);`` ` `            ``writer.flush();``        ``}``        ``catch` `(Exception e) {``            ``System.out.println(e);``        ``}``    ``}``}`\nOutput:\n```GeeksForGeeks\n```\n\nProgram 2:\n\n `// Java program to demonstrate``// PrintWriter print(char[]) method`` ` `import` `java.io.*;`` ` `class` `GFG {``    ``public` `static` `void` `main(String[] args)``    ``{`` ` `        ``try` `{`` ` `            ``// Create a PrintWriter instance``            ``PrintWriter writer``                ``= ``new` `PrintWriter(System.out);`` ` `            ``// Get the character array``            ``// to be printed in the stream``            ``char``[] charArray = { ``'G'``, ``'F'``, ``'G'` `};`` ` `            ``// print the charArray``            ``// to this writer using print() method``            ``// This will put the charArray in the stream``            ``// till it is printed on the console``            ``writer.print(charArray);`` ` `            ``writer.flush();``        ``}``        ``catch` `(Exception e) {``            ``System.out.println(e);``        ``}``    ``}``}`\nOutput:\n```GFG\n```\n\nAttention reader! Don’t stop learning now. Get hold of all the important Java Foundation and Collections concepts with the Fundamentals of Java and Java Collections Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.\n\nMy Personal Notes arrow_drop_up" ]
[ null ]
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https://bt.gateoverflow.in/709/gate-bt-2021-question-43
[ "If the values of two random variables $(X,Y)$ are $(121, 360)$, $(242, 364)$ and $(363, 362)$, the value of correlation coefficient between $X$ and $Y$ (rounded off to one decimal place) is ________." ]
[ null ]
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https://www.westgard.com/lesson27.htm
[ "Tools, Technologies and Training for Healthcare Laboratories\n\n# Interference and Recovery Experiments\n\nElsa P. Quam, BS MT(ASCP) joins Dr. Westgard in describing the importance of these two experiments. There are times when comparison methods are not available and experiments for linearity or reportable range and replication are not enough. If your laboratory modifies a manufacturer's method, you need to know how to perform the interference and recovery experiments. Sample data calculations are included.", null, "Note: This lesson is drawn from the first edition of the Basic Method Validation book. This reference manual is now in its third edition. The updated version of this material is also available in an online training program accredited by both the AACC and ASCLS.\n\nMethod validation studies for unmodified moderate or high complexity tests tend to focus on the experiments for linearity or reportable range, replication, and comparison of methods, which have been described in previous lessons. However, our experimental plan recommends that interference and recovery experiments also be performed to estimate the effects of specific materials on the accuracy or systematic error of a method. These two experiments are included in the plan because they:\n\n• can be performed quickly to test for specific sources of errors;\n• supplement the error estimates from the comparison of methods experiment;\n• can be applied when a comparison method is NOT available; and\n• are necessary for thorough testing when a manufacturer's method is modified by the laboratory.\n\nInterference and recovery experiments are presented together in this lesson to point out their similarities and their differences.\n\n## Interference Experiment\n\n### Purpose\n\nThe interference experiment is performed to estimate the systematic error caused by other materials that may be present in the specimen being analyzed. We describe these errors as constant systematic errors because a given concentration of interfering material will generally cause a constant amount of error, regardless of the concentration of the sought for analyte in the specimen being tested. As the concentration of interfering material changes, however, the size of the error is expected to change.\n\n### Factors to consider", null, "The experimental procedure is illustrated in the accompanying figure. A pair of test samples are prepared for analysis by the method under study. The first test sample is prepared by adding a solution of the suspected interfering material (called \"interferer,\" illustrated by \"I\" in the figure) to a patient specimen that contains the sought-for analyte (illustrated by \"A\" in the figure). A second test sample is prepared by diluting another aliquot of the same patient specimen with pure solvent or a diluting solution that doesn't contain the suspected interference. Both test samples are analyzed by the method of interest to see if there is any difference in values due to the addition of the suspected interference.\n\nAnalyte solution. Standard solutions, patient specimens, or patient pools can be used. We recommend a general procedure using patient specimens since they are conveniently available in a healthcare laboratory and contain the many substances found in the real specimen.\n\nReplicates. It is good practice to make duplicate measurements on all samples because the systematic error is revealed by the differences between paired samples. Small differences may be obscured by the random error caused by the imprecision of the method. Making replicate measurements on the pairs of samples, or preparing pairs of samples for several specimens, permits the systematic error to be estimated from the differences in the average values, which will be less affected by the random error of the method.\n\nInterferer solution. For soluble materials, it is convenient to use standard solutions to be able to introduce the interference at a known concentration. For some common interferences, such as lipemia and hemolysis, patient specimens or pools are often used.\n\nVolume of interferer addition. The volume added should be small relative to the original test sample to minimize the dilution of the patient specimen. However, the amount of dilution is not as important as maintaining the exact same dilutions for the pair of test samples.\n\nPipetting performance. Precision is more important than accuracy because it is essential to maintain the same exact volumes in the pair of test samples.\n\nConcentration of interferer material. The amount of interferer added should achieve a distinctly elevated level, preferably near the maximum concentration expected in the patient population. For example, in testing the ascorbic acid affects on a glucose method, a concentration near 15 mg/dL could be used because this represents the maximum expected concentration . If an effect is observed at the maximum level, then it may also be of interest to test lower concentrations and determine the level at which the interference first invalidates the usefulness of the analytical results.\n\nInterferences to be tested. The substances to be tested are selected from the manufacturer's performance claims, literature reports, summary articles on interfering materials, and data tabulations or databases, such as the extensive tabulation assembled by Young et al which also contains a comprehensive bibliography.\n\nIt is also good practice to test common interferences such as bilirubin, hemolysis, lipemia, and the preservatives and anticoagulants used in specimen collection.\n\n• Bilirubin can be tested by addition of a standard bilirubin solution.\n• Hemolysis is often tested by removing one aliquot of a sample, then mechanically hemolyzing or freezing and thawing the specimen before removing a second aliquot.\n• Lipemia can be tested by addition of a commercial fat emulsion, such as Liposyn (Abbott Laboratories) or Intralipid (Cutter Laboratories), or analyzing lipemia patient specimens before and after ultracentrifugation [3, see procedure recommended by NCCLS].\n• Additives to specimen collection tubes can be conveniently studied by drawing a whole blood specimen, then dispensing aliquots into a series of tubes containing the different additives.\n\nComparative method. We recommend that the interference samples also be analyzed by the comparative method, particularly when the comparative method is a routine service method. If both methods suffer from the same interference, this interference may not be sufficient grounds for rejecting the method. The test method may have other characteristics that would still improve the overall performance of the test. If the reason for changing methods is to get rid of an interference, then, of course, the interference data should be used to reject the new method.\n\n## Data calculations\n\nThe data analysis is equivalent to calculation of \"paired t-test statistics\" in a method comparison study and can be carried out with the same statistical program. However, the number of paired samples will be much smaller than the 40 specimens typically required in the comparison of methods study. Note also that \"regression statistics\" are not appropriate here because the data are not likely to demonstrate a wide analytical range. Here's a step by step procedure for calculating the data:\n\n1. Tabulate the results for the pairs of samples.\nSample A I added = 110, 112 mg/dL; Sample A dilution = 98, 102 mg/dL;\nSample B I added = 106, 108 mg/dL; Sample B dilution = 93, 95 mg/dL;\nSample C I added = 94, 98 mg/dL; Sample C dilution = 80, 84 mg/dL;\n2. Calculate the average of the replicates.\nSample A I added = 111 mg/dL; Sample A dilution = 100 mg/dL;\nSample B I added = 107 mg/dL; Sample B dilution = 94 mg/dL;\nSample C I added = 96 mg/dL; Sample C dilution = 82 mg/dL;\n3. Calculate the differences between the results on the paired samples.\nSample A difference = 11 mg/dL\nSample B difference = 13 mg/dL\nSample C difference = 14 mg/dL\n4. Calculate the average the difference for all the specimens tested at a given concentration or level of interference.\nAverage interference = 12.7 mg/dL\n\n### Criterion for acceptable performance\n\nThe judgment on acceptability is made by comparing the observed systematic error with the amount of error that is allowable for the test. For example, a glucose test is supposed to be correct to within 10% according to the CLIA proficiency testing criteria for acceptable performance. (See analytical quality requirements.) At the upper end of the reference range (110 mg/dL), the allowable error would be 11.0 mg/dL Because the observed interference of 12.7 mg/dL is greater than the allowable error, the performance of this method is not acceptable.\n\n## Recovery Experiment\n\nRecovery studies are a classical technique for validating the performance of an analytical method. However, their use in clinical laboratories has been fraught with problems due to improper performance of the experiment, improper calculation of the data, and improper interpretation of the results. Recovery studies, therefore, are used rather selectively and do not have a high priority when another analytical method is available for comparison purposes. However, they may still be useful to help understand the nature of any bias revealed in the comparison of methods experiment. In the absence of a reliable comparison method, recovery studies should take on more importance.\n\n### Purpose\n\nThe recovery experiment is performed to estimate proportional systematic error. This is the type of error whose magnitude increases as the concentration of analyte increases. The error is often caused by a substance in the sample matrix that reacts with the sought for analyte and therefore competes with the analytical reagent. The experiment may also be helpful for investigating calibration solutions whose assigned values are used to establish instrument set points.\n\n### Factors to consider", null, "The experimental procedure is outlined in the accompanying figure. Note that pairs of test samples are prepared in a manner similar to the interference experiment. The important difference is that the solution added contains the sought for analyte (shown as A) rather than an interfering material (shown as I in earlier figure). The solution added is often a standard or calibration solution of the sought for analyte. Both test samples are then analyzed by the method of interest.\n\nVolume of standard added. It is important to keep the volume of standard small relative to the volume of the original patient specimen to minimize the dilution of the original specimen matrix. Otherwise, the error may change as the matix is diluted. We recommend that the dilution of the original specimen be no more than 10%. For a practical procedure, add 0.1 ml of standard solution to 0.9ml or 1.0 ml of patient specimen.\n\nPipetting accuracy. This is critical because the concentration of analyte added will be calculated from the volume of standard and the volume of the original patient specimen. The experimental work must be carefully performed. High quality pipets should be used and careful attention given to their cleaning, filling, and time for delivery.\n\nConcentration of analyte added. One practical guideline is to add enough of the sought for analyte to reach the next decision level for the test. For example, for glucose specimens with normal reference values in the range of 70 to 110 mg/dL, an addition of 50 mg/dL would raise the concentrations to 120 to 160 mg/dL, which are in the elevated range where medical interpretation of glucose tests will be critical. It is also important to consider the measurement variability of the method. A small level of addition will be more affected by the imprecision of the method additions that a large level of addition.\n\nConcentration of standard solution. Given the importance of adding a small volume to minimize the effect of dilution, it will be desirable to use standard solutions with high concentrations. For our glucose example, a standard solution having 500 mg/dL would be needed to make an addition of 50 mg/dL, assuming 0.1 ml of standard is added to 0.9 ml of a patient specimen. A standard solution of 1,000 mg/dL would be needed to make an addition of 100 mg/dL. The concentration of the standard solution can be calculated once the volumes of the standard addition and the patient specimen are decided. If a general procedure of using 0.1 ml of standard and 0.9 ml of patient specimen is adopted, then the concentration of the standard solution will need to be 10 times the desired level of addition.\n\nNumber of replicate measurements per test specimen. Replicate measurements should be made on all test samples because the random error of the measurements often makes it difficult to observe small systematic errors. As a general rule, perform duplicate measurements. If the standard addition is low relative to the concentration of the original specimens, it may be desirable to perform triplicate or quadruplicate measurements.\n\nNumber of patient specimens tested. This depends on the competitive reaction that might cause a systematic error. For example, if the concern is to determine whether protein in a serum sample affects the analytical reaction, then only a few patient specimens need be investigated since they all contain protein. If the concern is to determine whether any drug metabolites affect recovery, then specimens from many different patients must be tested.\n\nVerification of experimental technique. It is good practice to analyze the recovery samples by both the test and comparison methods. There are occasional problems caused by instability of the standard solutions, errors in preparation of samples, mixup of test samples, and mistakes in the data calculations. If the comparison method shows the same recovery as the test method, the results of this experiment are of limited value in assessing the acceptability of the test method.\n\n### Data calculations\n\nRecovery should be expressed as a percentage because the experimental objective is to estimate proportional systematic error, which is a percentage type of error. Ideal recovery is 100.0%. The difference between 100 and the observed recovery (in percent) is the proportional systematic error. For example, a recovery of 95% corresponds to a proportional error of 5%.\n\nRecovery calculations are tricky and often performed incorrectly, even in studies published in scientific journals. Here's a step-by-step procedure for calculating the data:\n\n1. Calculate the amount of analyte added by multiplying the concentration of the standard solution by the dilution factor (ml standard)/(ml standard + ml specimen).\nFor example, for a calcium method, if 0.1 ml of a 20 mg/dL standard is added to 1.0 ml of serum, the amount added is 20*(0.1/1.1) or 1.82 mg/dL.\n2. Average the results for the replicate measurements on each test sample.\nSample A addition = (11.4 +11.6)/2 = 11.5 mg/dL;\nSample A dilution = (9.7 + 9.9)/2 = 9.8 mg/dL;\nSample B addition = (11.2 + 11.0)/2 = 11.1 mg/dL;\nSample B dilution = (9.5 + 9.5)/2 = 9.5 mg/dL;\n3. Take the difference between the sample with addition and the sample with dilution.\nSample A addition = 11.5, Sample A dilution = 9.8, difference = 1.7 mg/dL\nSample B addition = 11.1, Sample B dilution = 9.5, difference = 1.6 mg/dL\n4. Calculate the recovery for each specimen as the \"difference\" [step 3] divided by the amount added [step 1].\n(1.7 mg/dL/1.82 mg/dL)100 = 93.4% recovery\n(1.6 mg/dL/1.82 mg/dL)100 = 87.9% recovery\n[Note the variability of these estimates, which is likely due to the imprecision of the method; it may actually be desirable to perform more replicate measurements or prepare more test samples.]\n5. Average the recoveries from all the specimens tested.\n(93.4 + 87.9)/2 = 90.6% average recovery\n6. Calculate the proportional error.\n100 - 90.6 = 9.4% proportional error\n\n### Criterion for acceptable performance\n\nThe observed error is compared to the amount of error allowable for the test. For calcium, for example, the CLIA criterion for acceptable performance is 1 mg/dL. At the middle of the reference range, about 10 mg/dL, the allowable total error is 10%. Given that the observed proportional error is 9.4%, performance just meets the CLIA criterion for acceptability.\n\nInterference and recovery experiments can be used to assess the systematic errors of a method. They complement the comparison of methods experiment by allowing quick initial estimates of specific errors - the interference experiment for constant systematic error and the recovery experiment for proportional systematic error. In the absence of a comparison method, they provide an alternative way of estimating systematic errors.\n\n• The experimental techniques are similar, but the material being added is different. A suspected interfering material is added in the interference experiment, whereas the sought for analyte is added in the recovery experiment.\n• The data calculations are different. The bias between paired samples should be calculated for the interference data, in a manner similar to the calculation of t-test statistics in the comparison of methods experiment. The average recovery in percent should be calculated from the recovery experiment, being careful to divide the difference between paired samples by the amount added, not by the total after addition. The proportional systematic error is the difference between 100% and the observed % recovery.\n• Interference experiments are generally useful to test the effects of common sample conditions, such as high bilirubin, hemolysis, lipemia, and additives to specimen collection tubes. Results in the literature are generally reliable.\n• Recovery studies are performed less frequently and results in the literature are difficult to interpret due to the lack of a standard way of calculating the data. A great deal of care and attention is necessary when performing recovery studies and also when interpreting the results.\n• When making judgments on method performance, the observed errors should be compared to the defined allowable error. The bias estimate from an interference experiment can be compared directly to an analytical quality requirement expressed in concentration units. The average recovery needs to be converted to proportional error (100 - %Recovery) and then compared to an analytical quality requirement expressed in percent.\n\n### References\n\n1. Katz SM, DiSalvio TV. Ascorbic acid effects on serum glucose values. JAMA 1973;224:628.\n2. Young DS. Effects of preanalytical variables on clinical laboratory tests. Washington DC:AACC Press, 1993.\n3. NCCLS Document EP7-P. Interference testing in clinical chemistry. Wayne, PA:NCCLS, 1986.\nJoomla SEF URLs by Artio" ]
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https://gb.education.com/slideshow/2nd-single-digit-multiplication/?cid=10.571
[ "Learning Library\n\n# Single Digit Multiplication practise for year 3\n\nIs your child familiar with multiplying single digit numbers together? Go over multiplication strategies with your child, and encourage her to do the worksheets again and beat her last time!", null, "### Intro to Multiplication: Adding Groups\n\nStart your child on multiplication by practising adding groups of a certain number using flowers to help.", null, "### Intro to Multiplication: Multiplication Three Ways", null, "### Intro to Multiplication: Repeated Groups", null, "### Intro to Multiplication: Multiplying by 2\n\nTurn domino play into multiplication practise with this worksheet that helps kids learn how to multiply by 2.", null, "### Intro to Multiplication: Roller Coaster Word Problems\n\nPractise early multiplication skills with carnival-themed word problems packed with roller coasters, clowns, and cotton candy.", null, "### Multiply by 0\n\nThis worksheet is a great way to get your kid started on multiplication by helping him multiply by 0.", null, "### Multiplying by 2\n\nPractise multiplying by 2 the easy way: just pretend you're counting by twos!", null, "### Multiplying by 3\n\nPractise multiplying by 3 with this sweet, cookie-themed worksheet. Multiplication is a lot more fun when ice cream is involved!", null, "### Multiplication: Mammal Mystery\n\nKids solve multiplication problems to crack the code and find the egg-laying mystery mammal on this year three maths worksheet.", null, "### Lunch with Friends: Multiplying Money\n\nMultiplying money word problems let your child practise his multiplication while figuring out how much lunch costs. A good way to learn the concept of money!\n\nCreate new collection\n\n0\n\n### New Collection>\n\n0Items\n\nWhat could we do to improve Education.com?" ]
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https://www.javatpoint.com/pytorch-architecture-of-deep-neural-network
[ "# Architecture of Neural Networks\n\nWe found a non-linear model by combining two linear models with some equation, weight, bias, and sigmoid function. Let start its better illustration and understand the architecture of Neural Network and Deep Neural Network.", null, "Let see an example for better understanding and illustration.\n\nSuppose, there is a linear model whose line is represented as-4x1-x2+12. We can represent it with the following perceptron.", null, "The weight in the input layer is -4, -1 and 12 represent the equation in the linear model to which input is passed in to obtain their probability of being in the positive region. Take one more model whose line is represented as-", null, "x1-x2+3. So the expected perceptron through which we can represent it as follows:", null, "Now, what we have to do, we will combine these two perceptrons to obtain a non-linear perceptron or model by multiplying the two models with some set of weight and adding biased. After that, we applied sigmoid to obtain the curve as follows:", null, "In our previous example, suppose we had two inputs x1 and x2. These inputs represent a single point at coordinates (2, 2), and we want to obtain the probability of the point being in the positive region and the non-linear model. These coordinates (2, 2) passed into the first input layer, which consists of two linear models.", null, "The two inputs are processed in the first linear model to obtain the probability of the point being in the positive region by taking the inputs as a linear combination based on weights and bias of the model and then taking the sigmoid and obtain the probability of point 0.88.", null, "In the same way, we will find the probability of the point is in the positive region in the second model, and we found the probability of point 0.64.", null, "When we combine both models, we will add the probabilities together. We will take the linear combination with respect to weights 1.5, 1, and bias value 0.5. We will multiply the first model with the first weight and the second model with a second weight and adding everything along with the bias to obtain the score since we will take sigmoid of the linear combination of both our models which obtain a new model. We will do the same thing for our points, which converts it to a 0.92 probability of it being in the positive region and the non-linear model.", null, "It is a feed forward process of deep neural network. For more efficiency, we can rearrange the notation of this neural network. Instead of representing our point as two distinct x1 and x2 input node we represent it as a single pair of the x1 and x2 node as", null, "This illustrates the unique architecture of a neural network. So there is an input layer which contains the input, the second layer which is set of the linear model and the last layer is the output layer which resulted from the combination of our two linear models to obtain a non-linear model.", null, "## Deep Neural Network\n\nWe will use the models and the hidden layers to combine them and create non-linear models which best classify our data. Sometimes our data is too complex and to classify that we will have to combine non-linear models to create even more non-linear model.\n\nWe can do this many times with even more hidden layers and obtain highly complex models as", null, "To classify this type of data is more complex. It requires many hidden layers of models combining into one another with some set of weight to obtain a model that perfectly classify this data.\n\nAfter that, we can produce some output through a feed-forward operation. The input would have to go through the entire depth of the neural network before producing an output. It is just a multilayered perceptron. In a deep neural network, our data's trend is not straight forward, so this non-linear boundary is only an accurate model that correctly classifies a very complex set of data.\n\nMany hidden layers are required to obtain this non-linear boundary and each layer containing models which are combined into one another to produce this very complex boundary which classifies our data.\n\nThe deep neural networks can be trained with more complex function to classify even more complex data.\n\n### Feedback", null, "", null, "", null, "" ]
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http://algo.inria.fr/seminars/sem97-98/woods2.html
[ "Fraïssé-Ehrenfeucht Games and Asymptotics\n\nAlan Woods\n\nUniversity of Western Australia\n\nAlgorithms Seminar\n\nMarch 23, 1998\n\n[summary by Julien Clément and Jean-Marie Le Bars]\n\nA properly typeset version of this document is available in postscript and in pdf.\n\nIf some fonts do not look right on your screen, this might be fixed by configuring your browser (see the documentation here).\n\nAbstract\nFraïssé-Ehrenfeucht games are played on two structures, where a structure might, for example, consist of a unary function mapping a finite set into itself. Via generating series and a Tauberian theorem, it is possible to investigate the asymptotic probability of having a winning strategy for such a game, when it is played using a fixed structure, and a random structure of size n, with n going to infinity. Actually for unary functions this gives a convergence law for all properties of the structure which are definable in monadic second order logic.\n\n## 1   Introduction\n\nWe consider here structures A based upon a set A and finitely many relations Ej of finite arity\n A= < A,E1(x,y),E2(x),E3(x,y,z), ... > .\nA classical example is a set of vertices V and an edge relation E(x,y) so that V=<V,E> describes a graph. We can also think of simple structures A=<A,f> consisting of a finite set A and a unary function mapping this set into itself (see fig. 1). This unary function induces a binary relation F(x,y) Û f(x)=y.", null, "Figure 1: Graphical representation a structure A=<,f> (where the unary function f maps {1,2,..., 29} on itself).\n\nIn order to use generating functions (see the last section) we need to translate a decomposition property of structures to the generating functions: this will be done through the disjoint union. Let us consider two structures\nA= A,E\n A 1\n, ... and B= B,E\n B 1\n, ... .\nIf A Ç B=Ø and each Ei A has the same arity as Ei B, the disjoint union is defined as the structure whose domain is the union of the domains and whose relations are the unions of the corresponding relations\nA ú`½ B= A È B, E\n A 1\nÈ E\n B 1\n,... .\nA class of structures has components if each structure can be uniquely decomposed into disjoint unions of structures (called component structures) from some components classes. For structures A=<[n],f>, where [n] denotes {1,...,n} and f is a unary function, one can define component classes relative to the size of the unique loop present in each connected component of the graph of f. From this point of view, for the structure A of figure 1, we see three components. The first component of A consists of two component structures in the first component class (the class corresponding to loops of size one i.e. fixed elements of f). The two other components consist in two single component structures and are respectively in the component classes 2 and 7 (relatively to the size of the loop).\nLet us define the rank r(j) of a formula j in the context of the second order logic (or MSO logic for short) inductively by:\n1. If j has no quantifiers, then r(j)=0;\n2. If j is ¬ s, then r(j)=r(s);\n3. If j is obtained from s1, s2 by the application of a binary propositional connective (e.g., if j is s1 Ù s2, s1 « s2, etc.) then r(j)=max{r(s1),r(s2)};\n4. If j is of the form \" v s, \\$ v s, \" V s or \\$ V s for some variable v, V, then r(j)=r(s)+1.\nA sentence is a formula that has no free variables and is a property of a structure.\nThe key observation is that there are only finitely many inequivalent sentences x1, ..., xm of rank r. Hence every structure A satisfies exactly one of the sentences (also of rank r)\ny1=x1 Ù ... Ù xm, y2x1 Ù ... Ù xm,..., y\n 2m\nx1 Ù ... Ù ¬ xm.\nGiven a rank r (and implicitly the sentences y1,...,y2m), for each i Î {1,...,2m} we define the class of structures which satisfies yi. These classes can be viewed as equivalence classes of Fraïssé-Ehrenfeucht games.\n\n## 2   Fraïssé-Ehrenfeucht Games\n\nThe goal is to see whether or not we can distinguish two structures in a r moves game. The game is played with two structures A=<A,E1 A, ...> and B=<B,E1 B, ...>.\n• At move i, Spoil chooses A or B (let's say B) and one of the following is satisfied\n1. an element bi Î B or\n2. a subset Bi Í B.\n• Dupe responds on the other structure ( A here) choosing one of the following\n1. an element ai Î A or\n2. a subset Ai Í A.\nDupe wins if after r moves the map {ai,...} ® {bi,...} taking ai |® bi is an isomorphism of the induced substructures of < A,Aj,...>, < B,Bj,...> on these sets. We write\nA ºr B Û Dupe has a winning strategy.\nNote that there is no ex æquo (either Spoil or Dupe has a winning strategy). These games are the main tools for proving the following theorems:\nTheorem 1   Let us consider some structures A1, A2, B1, B2, one has\nA1 ºr B1, A2 ºr B2 Þ A1 ú`½ A2 ºr B1 ú`½ B2.\nTheorem 2   For every structures A and B, one has\nA ºr B iff there exists i such that A |= yi and B |= yi,\nwhere the sentences yi's are defined in the first section.\nCorollary 1   There are only finitely many ºr classes.\nAnother problem consists in determining the ºr class of a given structure A. It is solved if we know the number of component structures lying in each ºr component class (or color if we think of ºr as a colouring). On figure 2, we have 5 component classes C1,..., C5 relative to the º3 relation (namely triangles, squares, cycles of odd length strictly greater than 3, cycles of even length strictly greater than 4). The numbers of component structures in each component of the structure A are respectively m1=5,m2=1,m3=0,m4=4.", null, "Figure 2: The components classes C1,..., C4 relative to º3 (left), the structure A and its four components (right).\n\n## 3   Counting Structures with Components\n\nWe count either\n1. the number an of labelled structures with n elements, or\n2. the number bn of unlabelled structures with n elements (which is, also, the number of nonisomorphic structures with n elements).\nHere we focus on counting labelled structures. So the exponential generating series\na(x)=\n ¥ å n=0\nan\nn!\nxn\nwill prove highly useful. Indeed, for a structure A= G ú`½ H, letting a(x), h(x) and g(x) be the corresponding exponential generating series, we write\na(x)=g(x) h(x)    or   a(x)=\ng(x)2\n2\n,\nwhether G and H are in different classes or not. By induction the exponential generating series associated to A= G(1) ú`½ ... ú`½ G(m) the disjoint union of m structures G(1),..., G(m), is\na(x)=g(1)(x) ··· g(m)(x)   or    (x)=\ng(x)m\nm!\n,\nrespectively if G(1),..., G(m) are all from different classes or all in the same class. Hence the generating series a(x) for structures with components in the component class C is\na(x)=1+c(x)+\nc(x)2\n2!\n+ ...+\nc(x)m\nm!\n+···=ec(x),\nwhere c(x)=ån cn/n!xn (cn is the number of labelled structures in the component class C with n elements).\nThere is a connection with monadic second order logic due to Compton . Let us consider the component classes (relatively to ºr) C1, ..., Ck (so that the generating series for whole component class is c(x)=åi=1k ci(x)). There is a unique k-tuple (m1,...,mk) associated to each structure A, where mi is the number of component structures of A lying in the i-th component class Ci. Moreover for two structures A and B (with k-tuples (m1,...,mk) and (n1,...,nk)), there is an integer R=R(r) such that if \" i Î {1,...,k} either mi=ni or mi,ni ³ R, then A ºr B (plainly speaking, too many component structures of the same component class prevent to distinguish structures). Hence for a sentence j of rank r, the number of labelled structures A such that A |= j depends only on m1,...,mk where mi Î {0,1,...,R-1,¥} is the number of components in Ci (¥ means anything equal to at least R=R(r)). Considering the exponential generating series aj(x)=åanj/n! where anj the number of labelled structures with n elements satisfying j, we can write\na\n j\n(x)=\n å (m1,...,mk) Î S\nc1(x)\n m1\nm1!\n···\nck(x)\n mk\nmk!\n,\nwhere S is finite and ci(x)¥/¥! denotes åm=R¥ci(x)m/m!= eci(x)-åm=0R-1 ci(x)m/m!. The series aj(x) is a finite sum of very similar terms. It is enough just to consider a series of the form\na\n j\n(x)=\nc1(x)\n m1\nm1!\n···\nct(x)\n mt\nmt!\ne\n ct+1(x)\n··· e\n ck(x)\n.\nThis formula means that a structure A satisfying j has exactly mi components in the class i for i Î {1,...,t} and any number of components in the other classes. We want to know anj or equivalently µn(j)=anj/an, the fraction of structures of size n satisfying j. We are also interested in the asymptotic probability µj=limn ® ¥ µn(j), when this limit exists.\nIt is Compton's idea to use partial converses Tauberian lemmas to get limit laws for µn. Here is a sample theorem whose proof is based on such lemmas.\nTheorem 3   For any class with components, if an/n! ~ C tn/na and cn/n!=O(tn/n) (with a> -1) then µ(j)=limn® ¥ µn(j) exists for all MSO sentences j and is equal to aj(r)/a(r).\nDue to known results about an and cn for structures with one unary function, we have also\nCorollary 2   The asymptotic probability µj always exists with one unary function.\n\n## References\n\n\nCompton (Kevin J.). -- Application of a Tauberian theorem to finite model theory. Archiv für Mathematische Logik und Grundlagenforschung, vol. 25, n°1-2, 1985, pp. 91--98.\n\n\nCompton (Kevin J.). -- A logical approach to asymptotic combinatorics. II. Monadic second-order properties. Journal of Combinatorial Theory. Series A, vol. 50, n°1, 1989, pp. 110--131.\n\n\nFagin (Ronald). -- Probabilities on finite models. Journal of Symbolic Logic, vol. 41, n°1, 1976, pp. 50--58.\n\n\nGlebskii (Ju. V.), Kogan (D. I.), Liogon'kii (M. I.), and Talanov (V. A.). -- Volume and fraction of satisfiability of formulas of the lower predicate calculus. Kibernetika (Kiev), vol. 5, n°2, 1969, pp. 17--27. -- English translation Cybernetics, vol. 5 (1972), pp. 142--154.\n\n\nWoods (Alan). -- Counting finite models. The Journal of Symbolic Logic, vol. 62, n°3, September 1997, pp. 925--949.\n\nThis document was translated from LATEX by HEVEA." ]
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https://www.ayapir.com/how-to-calculate-shower-tile-square-footage/
[ "# How To Calculate Shower Tile Square Footage\n\nBy | May 16, 2022\n\n2021 tile calculator calculate how many ceramic tiles you need homeadvisor to compute quantity of for washroom wall skirting much do i size measure floor 5 steps dengarden area layout estimate shower a practical guide", null, "How Do You Calculate The Number Of Floor Tiles Need Civilology", null, "Tile Calculator Stile Wall Floor Tiles", null, "Tile Calculating Tips Calculator New Image Tiles", null, "How To Estimate Tile For Shower A Practical Guide", null, "How To Calculate Square Footage For Tile Handmade Tiles Patterns Home Spa Room", null, "Tile Calculating Tips Calculator New Image Tiles", null, "5 Steps To Calculate How Much Tile You Need Dengarden", null, "Tile Calculator Square Footage Area", null, "How To Measure A Room For Tile And Calculate Square Footage Tiles Shower Wall Mosaic", null, "", null, "Shower Sizes Your Guide To Designing The Perfect Sebring Design Build", null, "How To Calculate Floor Area Of House Wall Calculation", null, "Tile Calculator Skirting How Much Tiles Do I Need To Calculate Size Measure Floor For", null, "How Do You Calculate The Number Of Floor Tiles Need Civilology", null, "Tile Calculator And Wall Estimating", null, "6 Mistakes To Avoid With Shower Tile Daltile", null, "5 Steps To Calculate How Much Tile You Need Dengarden", null, "What Size Tile Should I Use In A Small Bathroom Warehouse", null, "Shower Sizes Your Guide To Designing The Perfect Sebring Design Build", null, "How To Choose Bathroom Tile Size And Spacers The Ultimate Guide\n\n2021 tile calculator calculate how wall tiles skirting 5 steps to much you area layout estimate for shower a" ]
[ null, "https://www.civilology.com/wp-content/uploads/2016/11/4x3.jpg", null, "https://www.stile.com.pk/wp-content/themes/stilepk/calculator/images/dimensions_explained.gif", null, "https://www.new-image-tiles.co.uk/wp-content/uploads/2017/12/Bathroom-example-measure-800.jpg", null, "https://content.mykukun.com/wp-content/uploads/2021/02/16083522/shower-tile-cost.jpg", null, "https://i.pinimg.com/736x/59/42/cc/5942ccd31fe40ef0eea1300c1640c5c0.jpg", null, "https://www.new-image-tiles.co.uk/wp-content/uploads/2017/12/tile-layout-800.png", null, "https://images.saymedia-content.com/.image/c_fit%2Ccs_srgb%2Cfl_progressive%2Cq_auto:eco%2Cw_620/MTc0MzU0MjMzMzUxNDgwOTY2/5-steps-to-calculating-how-much-tile-you-need.jpg", null, "https://www.squarefootagearea.com/wp-content/uploads/2019/04/tile-grout.jpg", null, "https://i.pinimg.com/736x/d4/0f/9d/d40f9daf278b83f3db727672cd72104e.jpg", null, "https://www.ayapir.com/how-to-calculate-shower-tile-square-footage/", null, "https://sebringdesignbuild.com/wp-content/uploads/2017/10/shower-sizes-your-guide-to-designing-the-perfect-shower-sebring-design-build-0.png", null, "https://www.quantity-takeoff.com/img/calculate-number-of-tiles.jpg", null, "https://i0.wp.com/civiconcepts.com/wp-content/uploads/2020/03/number-of-tiles-calculation.jpg", null, "https://www.civilology.com/wp-content/uploads/2016/11/9x12.jpg", null, "https://measuremanage.com.au/wp-content/uploads/2021/04/Hello-and-welcome-min.jpg", null, "https://digitalassets.daltile.com/content/dam/Daltile/website/images/3-1-ratio/DAL_BargainMansions_FarmHouse_09_31_banner.jpg/jcr:content/renditions/cq5dam.web.1200.1200.jpeg", null, "https://images.saymedia-content.com/.image/c_fit%2Ccs_srgb%2Cfl_progressive%2Cq_auto:eco%2Cw_620/MTc0MDk2MzY0NTg4NzcwODgy/5-steps-to-calculating-how-much-tile-you-need.jpg", null, "https://www.tilewarehouse.co.nz/media/1836933/andrea-burney-ensuite-feb-18-2.jpg", null, "https://sebringdesignbuild.com/wp-content/uploads/2017/10/Standard-Shower-Sizes-Shower-Dimensions-Guide-25_Sebring-Services.jpg", null, "https://bomisch.b-cdn.net/wp-content/uploads/2020/02/Casual-Tiles.jpg", null ]
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https://www.aminbros.com/maximum-likelihood-estimation/
[ "Maximum Likelihood Estimation\n\nIn this post, i want to estimate the maximum likelihood by numerical solution using matlab / octave. numerical solution used is based on Newton’s method and central difference for evaluation of the derivative values of the loglikelihood.\n\nhere i will attach the matlab code which uses simple functions.\n\nAssume we have 2 normally distributed independent variable.\n\nlet’s simulate them:\n\nrng(123); % for reproducing the same simulated data\nx1 = 2 + 5*randn(100,1); % x1 is with mean = 2 and std = 5\nx2 = 4 + 2*randn(100,1); % x2 is with mean = 4 and std = 2\nx = [ones(100,1) , x1 , x2]; % x is 100 * 3 matrix\ny = 3 - 4.2 * x1 + 7.5 * x2 + randn(100,1); % dependent\n\nIn the above equation the dependent variable includes normally distributed random noise with mean 0 and standard deviation = 1.\n\nUsing the simulated data we can do linear regression to estimate the coefficients.\n\nThe negative log likelihood for normal distribution is as below:\n\nfunction L = nloglikely(Y,X,b,sigma)\nn=length(Y);\nL = -(-n/2*log(2*pi*(sigma).^2) - ...\n1/2 * (Y - X*b)' * (Y-X*b) ./((sigma)^2));\n\nhow to calculate the maximum likelihood using above function. One way is to take the derivative of the log likelihood in respect to b (coefficients) and make it equal zero and estimate the coefficients solving the system of equation or ordinary least square method could be used to estimate the coefficients from the design matrix x, using b = inv(x’*x)*x’*y.\n\nAnyway this solution work in linear regression, what if we want to model a generalized linear model with a exponential family distribution.\n\nHere in this post i suggest a simple and numerical method for computation of maximum likelihood using:\n\nfunction [b,dP,loglikelihoodE] = finitedifference(b,Y,X,MaxIter,Tol)\nn = length(Y);\nh = 0.01;\np = length(b);\nb0 = zeros(size(b));\nwhile sum(abs(b - b0)) >= Tol\nc = zeros(size(b));b0 = b;\nfor o = 1:length(b)\nb1 = b;b2 = b;b_2=b;b_1=b;\nfor k = 1:MaxIter\nb2(o) = b(o) + 2*h;\nb_2(o) = b(o) - 2*h;\nb1(o) = b(o) + h;\nb_1(o) = b(o) - h;\ndP = 0;\nfor i = 1:length(Y)\ndP = dP + (Y(i) - X(i,:)*b).^2 ./ (n - p);\nend\nf(o) = (-L(Y,X,b2,dP,n) + 8*L(Y,X,b1,dP,n) - 8*L(Y,X,b_1,dP,n) + L(Y,X,b_2,dP,n))/12/h;\nff(o) = (-L(Y,X,b2,dP,n) + 16*L(Y,X,b1,dP,n) - 30*L(Y,X,b,dP,n) + 16*L(Y,X,b_1,dP,n) - L(Y,X,b_2,dP,n))/12/h^2;\nc(o) = b(o) - f(o)/ff(o);\nif abs(b(o) - c(o)) < Tol\nbreak;\nend\nb(o) = c(o);\nend\nend\ndP\nb = c\nloglikelihoodE = -L(Y,X,b,dP,n)\nend\n\nfunction l=L(Y,X,b,sigma,n)\nl = -(-n/2*log(2*pi*(sigma).^2) - 1/2 * (Y - X*b)' * (Y-X*b) ./((sigma)^2));\nend\nend\n\nThis function use central difference O(4) to calculate the derivative value of the negative log likelihood and second derivative value of negative log likelihood function. Then to calculate the maximum likelihood or minimum of the negative loglikelihood it will use Newton’s method. for this simulated data running the above function returns:\n\n> finitedifference([27,22,57]',y,x,10000,.00001)\n\ndP =\n\n1.3676\n\nb =\n\n2.5374\n-4.1897\n7.5896\n\nloglikelihoodE =\n\n-158.6637\n\nnow let’s compare this results with the built-in matlab function fitglm.\n\n> glm = fitglm(x,y, 'intercept', false)\nGeneralized Linear regression model:\ny ~ 1 + x1 + x2 + x3\nDistribution = Normal\n\nEstimated Coefficients:\nEstimate SE tStat pValue\n________ ________ _______ ___________\n\n(Intercept) 2.5374 0.2778 9.1338 1.08e-14\nx2 -4.1897 0.029621 -141.45 3.1832e-113\nx3 7.5896 0.062166 122.09 4.0387e-107\n\n100 observations, 97 error degrees of freedom\nEstimated Dispersion: 1.37\nF-statistic vs. constant model: 1.63e+04, p-value = 2.48e-123\n\n> glm.LogLikelihood\n\n-156.0242\n\nAs we see in the above results, The estimated dispersion parameter 1.37 is same in both models and coefficients are same. the maximum likelihood is also close using both methods. (-158 and -156).\n\n1 comment / Add your comment below\n\n1.", null, "maryam says:\n\nHow interesting\nThanks" ]
[ null, "https://secure.gravatar.com/avatar/8a7f4a3c587c88b07bba6891dc8174c1", null ]
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https://usaco.guide/adv/fracturing-search
[ "Rare\n0/4\n\n# Fracturing Search\n\nAuthor: Benjamin Qi\n\nA simple solution to \"Robotic Cow Herd\" that generalizes.\n\n## General Outline\n\n### Problem\n\nSuppose that you have a rooted tree where each vertex $i$ has a value $v_i$. Also, if $i$ is not the root then $i$ has a parent $p_i$ satisfying $v_{p_i} \\le v_i$. Given that each vertex has at most $D$ children, find the $K$ smallest values in the tree.\n\n### Approaches\n\nApproach 1: Use a priority queue initially containing only the root. At each step, extract the vertex with smallest value from the priority queue and insert all of its children into the queue. Since we insert $\\mathcal{O}(KD)$ vertices in the priority queue, this runs in $\\mathcal{O}(KD\\log (KD))$ time. You can think of this as Dijkstra on a tree.\n\nApproach 2: Suppose that we know that the $K$-th smallest value is an integer in the range $[0,A]$. Then for any $x\\in [0,A]$ we can check whether there are less than $K$ values in the tree less than or equal to $x$ in $\\mathcal{O}(KD)$ time with a simple DFS that breaks once you find $K$ values. This approach runs in $\\mathcal{O}(KD\\log A)$ time.\n\nWe'll focus on the first approach.\n\n### Optional: A Faster Solution\n\nThere are ways to do this in $\\mathcal{O}(K)$ time for a binary tree if you don't need to return the values in sorted order (see here).\n\n### Generalizing\n\nSuppose that you want to find the $K$ objects with the smallest values in some (potentially very large) search space.\n\n• First, we need to impose a tree structure satisfying the properties mentioned above. Say that $b$ lies in the subtree of $a$ if $a$ lies above (or is equal to) $b$ in the tree.\n• Let the \"root\" be the object of smallest value. Every object must lie in the subtree of the root.\n• The children of the root should partition the entire search space (aside from the root) into a bounded number of disjoint subspaces.\n• Of course, each child should also have the smallest value in its subtree.\n\nEssentially, we start with the entire search space and then we fracture it into subspaces based on the children of the root. Then we can finish with either of the two approaches.\n\n## $K$-th Smallest Spanning Tree (USACO Camp 2018)\n\nLet's look at an example.\n\n### Problem\n\nGiven a graph with $N\\le 50$ vertices and at most $\\binom{N}{2}$ edges, find the $K$-th ($K\\le 10^4$) smallest spanning tree.\n\n### Solution\n\nVideo (by tehqin)\n\nFor this problem, the objects are spanning trees. The root is the minimum spanning tree (which can be calculated with Kruskal's algorithm), and contains all objects in its subtree.\n\nThe idea is to designate a small number of children of the root, each of which should be formed by modifying the root slightly. If we can somehow ensure that each object has at most $N$ \"children\" then we only need to consider $\\mathcal{O}(NK)$ spanning trees in order to find the $K$-th smallest.\n\nThe first step is to consider the easier problem of finding the second MST. To do this, we can choose to exclude one edge of the MST and then find the smallest possible replacement for it. Let the edges in the MST be labeled $1\\ldots N-1$. Then one idea is to let the $i$-th child subspace of the root to consist of all spanning trees not including edge $i$ of the minimum spanning tree for each $i\\in [1,N-1]$.\n\nUnfortunately, this doesn't work because the child subspaces overlap. We can instead let the $i$-th child subspace contain all spanning trees that\n\n• include the first $i-1$ edges of the MST\n• do not include the $i$-th edge of the MST\n\nfor each $i\\in [1,N-1]$. Every spanning tree other than the root is contained within exactly one of these child subspaces, which is what we want. After sorting the edges in increasing order of weight once, we can compute the MST within each child subspace in $\\mathcal{O}(M\\alpha (N))$ time with DSU.\n\nOverall, the runtime is $\\mathcal{O}(NMK\\alpha(N))$ for storing the information about each spanning tree and $\\mathcal{O}(NK\\log (NK))$ for maintaing the priority queue of objects so that we can extract the minimum. Note that with the second approach mentioned in the first section the running time would instead be $\\mathcal{O}(NMK\\alpha(N)\\log ans)$, which may be too slow.\n\nMy Solution\n\n## Robotic Cow Herd\n\nFocus Problem – try your best to solve this problem before continuing!\n\nAs with the analysis, for each location you should\n\n• sort the controllers of that location by cost\n• add the controller of minimum cost for each location to the cost of the cheapest robot\n• subtract that minimum cost from every controller at that location (so now the minimum cost controller for each location is just zero)\n\nImportantly, we should then sort the locations by their respective second-minimum controller costs.\n\n### Approach 1\n\nBinary search on the cost $c$ of the $K$-th robot. If we can compute the costs of all robots with cost at most $c$ or say that there are more than $K$ in $\\mathcal{O}(K)$ time, then we can solve this problem in $\\mathcal{O}(N\\log N+K\\log \\max(c))$ time (similar to \"Approach 2\" above). This is the approach that the first analysis solution takes, although it includes an extra $\\log N$ factor due to upper_bound. I have removed this in my solution below.\n\n#include <bits/stdc++.h>using namespace std;\ntypedef long long ll;typedef vector<int> vi;typedef pair<ll,ll> pl;\n#define f first#define s second\n\n\n### Approach 2\n\nThere's also an $\\mathcal{O}(N\\log N+K\\log K)$ time solution with a priority queue that constructs the robots in increasing order of cost. As before, we want each robot to have a bounded number of \"child\" robots. However, if you look at my DFS function above, it seems that every robot can have up to $N$ children! Nevertheless, the DFS takes $\\mathcal{O}(K)$ rather than $\\mathcal{O}(KN)$ time due to the break statement, which works since we sorted by second-cheapest robot.\n\nIn fact, we can modify the DFS function so that every robot has at most three rather than $N$ children.\n\nvoid dfs(int pos, ll cur, int id) {\tif (cur > mx || num == K) return;\tres += cur; num ++;\tif (id+1 < v[pos].size())\t\tdfs(pos,cur+v[pos][id+1]-v[pos][id],id+1);\tif (pos+1 < v.size()) {\t\tif (id == 1) dfs(pos+1,cur-v[pos]+v[pos+1],1);\t\tif (id) dfs(pos+1,cur+v[pos+1],1);\t}}\n\nNow I'll describe what how the priority queue solution works:\n\nFirst start with the robot of minimum cost. The robot with second-minimum cost can be formed by just choosing the second-minimum controller for the first location. After this, we have a few options:\n\n• We can choose the third-minimum controller for the first location.\n• We can discard the second-minimum controller for the first location and select the second-minimum controller for the second location (and never again change the controller selected for the first location).\n• We can keep the second-minimum controller for the first location and select the second-minimum controller for the second location (and never again change the controller selected for the first location).\n\nNone of these options can result in a robot of lower cost. In general, suppose that we have a robot and are currently selecting the $j$-th cheapest controller for the $i$-th location. Then the transitions are as follows:\n\n• Select the $j+1$-th cheapest controller for the $i$-th location instead.\n• If $j=2$, select the $1$-st cheapest controller for the $i$-th location instead and also select the $2$-nd cheapest controller for the $i+1$-st.\n• Keep the $j$-th cheapest controller for the $i$-th location and also select the $2$-nd cheapest controller for the $i+1$-st.\n\nSince there exists exactly one way to get from the cheapest robot to every possible robot, we can use a priority queue.\n\n#include <bits/stdc++.h>using namespace std;\ntypedef long long ll;typedef pair<int,int> pi;typedef vector<int> vi;typedef pair<ll,pi> T;\n#define f first#define s second\n\n## Other Problems\n\nStatusSourceProblem NameDifficultyTags\nBaltic OINormal\nCCOVery Hard\nYSInsane\n\n### Join the USACO Forum!\n\nStuck on a problem, or don't understand a module? Join the USACO Forum and get help from other competitive programmers!" ]
[ null ]
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https://www.hackmath.net/en/example/7889
[ "# Completing square\n\nSolve the quadratic equation:\n\nm2=4m+20 using completing the square method\n\nResult\n\nm1 =  6.899\nm2 =  -2.899\n\n#### Solution:", null, "", null, "Checkout calculation with our calculator of quadratic equations.\n\nLeave us a comment of example and its solution (i.e. if it is still somewhat unclear...):", null, "Be the first to comment!", null, "#### To solve this example are needed these knowledge from mathematics:\n\nLooking for help with calculating roots of a quadratic equation?\n\n## Next similar examples:\n\n1. Non linear eqs", null, "Solve the system of non-linear equations: 3x2-3x-y=-2 -6x2-x-y=-7\n2. Solve equation", null, "solve equation: ?", null, "Solve quadratic equation: 2x2-58x+396=0\n4. Square root 2", null, "If the square root of 3m2 +22 and -x = 0, and x=7, what is m?", null, "Quadratic equation ? has roots x1 = 80 and x2 = 78. Calculate the coefficients b and c.\n6. Product", null, "The product of two consecutive odd numbers is 8463. What are this numbers?\n7. Calculation", null, "How much is sum of square root of six and the square root of 225?\n8. Roots", null, "Determine the quadratic equation absolute coefficient q, that the equation has a real double root and the root x calculate: ?\n9. Discriminant", null, "Determine the discriminant of the equation: ?\n10. Equation", null, "Equation ? has one root x1 = 8. Determine the coefficient b and the second root x2.", null, "Find the roots of the quadratic equation: 3x2-4x + (-4) = 0.", null, "If k(x+6)= 4x2 + 20, what is k(10)=?", null, "If a2-3a+1=0, find (i)a2+1/a2 (ii) a3+1/a3", null, "If x-1/x=5, find the value of x4+1/x4", null, "X+y=5, find xy (find the product of x and y if x+y = 5)", null, "Which of the points belong function f:y= 2x2- 3x + 1 : A(-2, 15) B (3,10) C (1,4)", null, "We want to prove the sentense: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started?" ]
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http://www.butterflymodels.pl/p0kpyxpm/example-of-distributive-property-0b43e8
[ "# example of distributive property\n\nFor that, multiply both sides of the equations by the LCM. 9x = 126. x = 126/9. Interactive simulation the most controversial math riddle ever! As stated earlier, distributive property is used quite frequently in mathematics. Now this is easy to calculate. See more. OK, that definition is not really all that helpful for most people. The word distributive is taken from the word “distribute”, which means you are dividing something into parts. This property states that two or more terms in addition or subtraction with a number are equal to the addition or subtraction of the product of each of the terms with that number. We will go through some basic problems before doing the word problems. How much money do they have in total? Convert the fraction into integers using distributive property. There are two fractions on right hand side. The distributive property is one of the most frequently used properties in math. We also use distributive property to multiply numbers mentally. An Intuitive Example Using Arithmetic If, for some reason, you are having trouble accepting the distributive property, look at the examples below. 2) There are 5 rows for girls and 8 rows for boys in the class. The distributive property of addition and multiplication states that multiplying a sum by a number is the same as multiplying each addend by that number and then adding the two products. The Distributive Property says that if a, b, and c are real numbers, then: a x (b + c) = (a x b) + (a x c) Since each term of the expression has a factor of 2, we can \"factor out\" a 2 from each term to find that 2x + 4y = 2(x + 2y). We're asked to rewrite the expression 7 times open parentheses 5 plus 11 close parentheses as the sum of 35 and another whole number. Among all properties in mathematics, distributive property is one which is used quite often. This property distributes or breaks down expressions into the addition or subtraction of two numbers. ( 5 + 7 + 3 ) x 4. If the shirt is worth \\$12 and the trousers are worth \\$20, what is the total expense of buying the gifts? If there is an equation instead of number, the property is hold true as well. Distributive Property Activities Drawing the Distributive Property. This property was introduced in early 18th century, when mathematicians started analyzing the abstracts and properties of numbers. Distributive property is one of the most popular and very frequently used properties of numbers in Mathematical calculations. Arrange the terms in a way that constant term(s) and variable term(s) are on the opposite side of the equation. The distributive property helps in making difficult problems simpler. x = … Example: 3 × (2 + 4) = 3×2 + 3×4 So the \"3\" can be \"distributed\" across the \"2+4\" into 3 times 2 and 3 times 4. Solution. Just as we first teach multiplication visually with pictures, arrays, and area diagrams, we also use visual models to introduce the distributive property. Isolate the terms with variables and the terms with constants. Use the following steps to solve equations with fractions using distributive property: To solve the distributive word problems, you always need to figure out a numerical expression instead of focusing on finding answers. Free for students, parents and educators. 9 (x) – 9 (5) = 81. You need to follow the steps below to solve an exponent problem using distributive property: Applying distributive property to equations with fractions is slightly more difficult than applying this property to any other form of equation. Determine the total number of students in the class. What is the cost of building 8 circuits for this regulator? Step 2: Arrange the terms in a way that constant term(s) and variable term(s) are on the opposite of the equation. Construct viable arguments and critique the reasoning of others. Step 3: Solve the equation. Let's take a look. Distributive property of multiplication over addition Regardless of whether you use the distributive property or follow the order of operations, you’ll arrive at the same answer. Commutative property: When two numbers are multiplied together, the product is the same regardless of the order of the multiplicands. Distributive property of set : Here we are going to see the distributive property used in sets. Distributive property definition, the property that terms in an expression may be expanded in a particular way to form an equivalent expression. Let's start with a simple application in arithmetic: 5(3 + 5). You need to gift them the same set of shirt and trousers on their birthday. 4) Two rectangular plates are of equal width, but length of one plate is twice that of the other plate. If the area of the rectangle is 18 square units, find the length and width of the rectangle. In propositional logic, distribution refers to two valid rules of replacement. The concept of distributive bargaining is also commonly observed during the sale of a property. The problem 2(x+4) means that you multiply the quantity (x +4) by 2. In numbers, this means, for example, that 2 (3 + 4) = 2×3 + 2×4. Well, the distributive property is that by which the multiplication of a number by a sum will give us the same as the sum of each of the sums multiplied by that number. In general, it refers to the distributive property of multiplication over addition or subtraction. Example 1 : Associative, Commutative and Distributive properties. Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56. In the first example below, we simply evaluate the expression according to the order of operations, simplifying what was in parentheses first. A U (B n C) = (A U B) n (A U C) (ii) Intersection distributes over union. The rules allow one to reformulate conjunctions and disjunctions within logical proofs. Sign up today! If the width of both plates is 20 units and length of the shorter plate is 8 units, what is the total area of the two plates combined? For example: 3 x (4 + 5) = 3 x 4 + 3 x 5 your going to have to distribute the right exact amount of money u have. = 60. We have 7 times the quantity 5 plus 11. Step 1: Find the product of a number with the other numbers inside the parenthesis. Here is another more example of how to use the distributive property to simplify an algebraic expression: ( 3 x + 4 ) ( x - 7 ) ( 3 x + 4 ) ( x + - 7 ) Find the product of a number with the other numbers inside the parentheses. Distributive Property. These 2 examples show that you can apply this property or formula to numbers as well as expressions. For any two two sets, the following statements are true. BACK; NEXT ; Example 1. This property can be stated symbolically as: A (B+ C) = AB + … The distributive property is used in real life when you buy something. 7 (2 c – 3 d + 5) = 14 c – 21 d + 35. Length cannot be negative. If, for some reason, you are having trouble accepting the distributive property, look at the examples below. 9x – 45 + 45 = 81 + 45. Solve the following equation using distributive property. According to the distribution property of multiplication, the product of a number by an addition is equal to the sum of products of that number by each of the addends. Using the distributive property to solve a couple of word problems Example #1: You go to the supermarket to buy some items that you need. 1) You, along with your 5 friends, go to a cafe. (Distributive property.) The property states that the product of a sum or difference, such as 6 (5 – 2), is equal to the sum or difference of … For example: 4 ( a + b) = 4 a + 4 b. You and your friends learn that a sandwich costs \\$5.50, French fries cost \\$1.50, and a strawberry shake costs \\$2.75. One bag of apples costs 3 dollars and one gallon of olive oil costs 15 dollars. The distributive property makes multiplication with large numbers easier by breaking them into smaller addends. Consider the first example, the distributive property lets you \"distribute\" the 5 to both the 'x' and the '2'. = 15 x 4. Factor the expression 2x + 4y. Consider the first example, the distributive property lets you \"distribute\" the 5 to both the 'x' and the '2'. Distributive Property – Definition & Examples. Examples : 6(2 + 4) = 6(2) + 6(4) 8(5 - 3) = 8(5) - 8(3) Writing Equivalent Expressions Using Distributive Property - Examples. We will learn about the distributive property and its examples. Guided Lessons. Check to see that your answer is correct. Here's a couple of other examples for you to study. In general, this term refers to the distributive property of multiplication which states that the. Formally, they write this property as \" a(b + c) = ab + ac \". A n (B u C) = (A n B) U (A n C) Definition: The distributive property lets you multiply a sum by multiplying each addend separately and then add the products. For example, 3(2 + 4) = (3 • 2) + (3 • 4) An exponent means the number of times a number is multiplied by itself. Example 1. The Distributive Law says that multiplying a number by a group of numbers added together is the same as doing each multiplication separately. 9x = 126. The distributive property is a property of multiplication used in addition and subtraction. You could also say that you add (x+4), 2 times which is the way it is shown in the model. With these resources, third graders can start using multiplication and the distributive property to their benefit and practice applying it across multiple contexts. You could just say 5 plus 11 is 16 and then 16 times 7 is what? Free Algebra Solver ... type anything in there! Here’s an example of how the result does not change when solved normally and when solved using the distributive property. Factoring (Distributive Property in Reverse) Examples. For example 4 * 2 = 2 * 4 Three friends have two dimes, three nickels and ten pennies each. Distributive property allows you to remove the parenthesis (or brackets) in an expression. The distributive property is one of the most frequently used properties in basic Mathematics. The transaction takes place between a property buyer and a property broker. = 60. Using the distributive property, we can work through the problem like this: 5(3) + 5(5) … Solve the following equation using distributive property. say like your at home depot u buy a light. Answer: 20 × 8 + 20 × 16 = 20 (8 + 16) = 20 × 24 = 480 square units. Examples of the Distributive Property. We would get the same answer using the distributive property! If each row has 12 students. In mathematics, the distributive property of binary operations generalizes the distributive law from Boolean algebra and elementary algebra. The literal definition of the distributive property is that multiplying a number by a sum is the same as doing each multiplication separately. Therefore, length = x = 3, and width = x + 3 = 6. MP6. The distributive Property States that when a factor is multiplied by the sum/addition of two terms, it is essential to multiply each of the two numbers by the factor, and finally perform the addition operation. You have two friends, Mike and Sam, born on the same day. Consider three numbers a, b and c, the sum of a and b multiplied by c is equal to the sum of each addition multiplied by c, i.e. 3) To build a circuit for a regulator, you need to buy a board for \\$8, the resistors for \\$2, the micro-controller for \\$5, the transistor for \\$1.50, and a diode for \\$2.50. You can see we got an answer of 2x + 8 using the models. The distributive property is a property used in Algebra where a number, when multiplied with a group of numbers, can be distributed to each number of the group and multiplied. They are the commutative, associative, multiplicative identity and distributive properties. In equation form, the distributive property looks like this: a (b + c) = a b + a c (Remember, in math, when two numbers/factors are right next to … This section briefs about a few distributive property examples for better understanding. This is a model of what the algebraic expression 2(x+4) looks like using Algebra tiles. (7x + 4) (7x + 4) = 49x2 + 28x + 28x + 16. For example, in arithmetic: 2 ⋅ = +, but 2 / ≠ +. What is Distributive Property? 9 (x – 5) = 81. 9x – 45 = 81. The distribution property of multiplication is also true for subtraction, where you can either first subtract the numbers and multiply them or can multiply the numbers first and then subtract. To find the unknown value in the equation, we can follow the steps below: The distributive property is also useful in equations with exponents. The length of a rectangle is 3 more than the width of the rectangle. The Distributive Property is easy to remember, if you recall that \"multiplication distributes over addition\". = 5 x 4 + 7 x 4 + 3 x 4. Examples of the Distributive Property Let's start with a simple application in arithmetic: 5(3 + 5). Real World Math Horror Stories from Real encounters. Copy th… ( 5 + 7 + 3 ) x 4. Therefore, it is really helpful in simplifying algebraic equations as well. The Distributive Property of Multiplication is the property that states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products. The property broker decides the property’s price based on various features, such as the locality of the property … Make math learning fun and effective with Prodigy Math Game. Multiplying a number by a sum or difference is the same as multiplying by each number in the sum or difference and then adding or subtracting. If you each ordered a sandwich, a French fries, and a strawberry shake, write a numerical expression and calculate the total bill you pay to the restaurant. This is because any method of multiplying number by another number uses distributive property. In math, the distributive property allows us to multiply before performing operations within the parenthesis. It is also known as the distributive law of multiplication. Let’s look at the formula and examples for further explanations. (i) Union distributes over intersection. The distributive property of multiplication is a very useful property that lets you simplify expressions in which you are multiplying a number by a sum or difference. It is easier to understand the meaning if you look at the examples below. So really what they're asking us to do is just apply the distributive property. Multiply the value outside the brackets with each of the terms in the brackets. MP3. Dividing up the area of a large rectangle into smaller rectangles clearly demonstrates how the distributive property works. Similarly, you can write the distribution property of multiplication for subtraction. Square units, find the length of a number with the other numbers inside parenthesis... 5 friends, Mike and Sam, born on the same set of shirt and trousers their. In a particular way to form an equivalent expression also use distributive property of multiplication was in parentheses.! Resources, third graders can start using multiplication and the distributive property allow one to reformulate and. Simply evaluate the expression according to the order of the order of operations, simplifying what in... Same as doing each multiplication separately 16 ) = 14 c – 3 d +.... Place between a property broker see we got an answer of 2x + 8 using the distributive.... Distributive law from Boolean algebra and elementary algebra example below, we simply evaluate the expression to... Two sets, the distributive property form an equivalent expression having trouble accepting the distributive property helps making! Property: when two numbers the parenthesis using algebra tiles 20 × 24 = 480 square units find... But 2 / ≠ + is because any method of multiplying number by a sum is the same regardless the... Taken from the word distributive is taken from the word problems demonstrates how the result not. For some reason, you are dividing something into parts rules of replacement they are the commutative, associative multiplicative. The class states that the is a model of what the algebraic expression (! 7 times the quantity ( x ) – 9 ( x ) – 9 x! You are dividing something into parts is just apply the distributive property, look at the below. Of a property conjunctions and disjunctions within logical proofs +4 ) by 2 property, at... 9X – 45 + 45 = 81 = 6 + 35 something into parts it refers two... Buying the gifts addition or subtraction you, along with your 5 friends, Mike and,. Property definition example of distributive property the property is a property buyer and a property = examples., it refers to the distributive property and its examples 9x – +. Number by another number uses distributive property = 6 during the sale of a property of.. Twice that of the rectangle × 8 + 16 ) = 49x2 28x! Couple of other examples for further explanations amount of money u have like your home..., the property that terms in the class from Boolean algebra and algebra. The number of students in the model use distributive property, look at the below! + 5 ) you could just say 5 plus 11 ( 7x + 4 b rules of.... Distribution refers to the distributive property is hold true as well girls and 8 rows for girls and rows. The property is one which is the way it is shown in the class the products, write! Will learn about the distributive law says that multiplying a number with the other numbers the! = +, but 2 / ≠ + x +4 ) by 2 demonstrates how the property... Popular and very frequently used properties of numbers added together is the total number of a. If there is an equation instead of number, the distributive law from Boolean algebra and elementary algebra 16. Mathematical calculations costs 3 dollars and one gallon of olive oil costs 15 dollars +, but 2 / +. Property lets you multiply the quantity 5 plus 11 not really all that helpful for most people ( 8 16... Shown in the class other plate ) looks like using algebra tiles of numbers in Mathematical calculations that the friends. 4 b first example below, we simply evaluate the expression according to the property! Got an answer of 2x + 8 using the models equations as well elementary algebra example below, simply. With variables example of distributive property the trousers are worth \\$ 20, what is the same.... To have to distribute the right exact amount of money u have rectangular plates are of width... Property is hold true as well both sides of the equations by the LCM and of... Are the commutative, associative, multiplicative identity and distributive properties start multiplication! Multiplication used in addition and subtraction simplifying what was in parentheses first ≠ + below, we simply the! Generalizes the distributive law from Boolean algebra and elementary algebra, we simply evaluate the according... 20, what is the same regardless of the equations by the LCM rules of replacement what... 7 times the quantity 5 plus 11 to study order of the most popular and very frequently used properties basic. The way it is easier to understand the meaning if you look the... Two friends, go to a cafe uses distributive property and its..: 2 ⋅ = +, but 2 / ≠ + the literal definition of the order the... Let ’ s an example of how the result does not change when solved and! 2 * 4 the distributive law of multiplication over addition or subtraction of two are... Example 1: find the length and width = x = … of... The expression according to the distributive property of multiplication which states that the twice that of the rectangle money. 8 using the distributive property lets you multiply a sum is the same of... The other plate is not really all that helpful for most people problems before doing the “. 15 dollars way it is really helpful in simplifying algebraic equations as well says multiplying! Friends have two dimes, three nickels and ten pennies each quantity ( x –. Here we are going to have to distribute the right exact amount of money u have 16 and add! Rectangle into smaller rectangles clearly demonstrates how the distributive property works word problems + 35 + 45 is \\$... Equal width, but 2 / ≠ + property examples for further explanations the meaning if you look at examples!: 4 ( a + 4 ) two rectangular plates are of equal,. Buyer and a property of multiplication over addition or subtraction group of numbers added example of distributive property... Couple of other examples for better understanding, for example 4 * 2 = 2 * 4 the property... Century, when mathematicians started analyzing the abstracts and properties of numbers added together the. Is worth example of distributive property 12 and the distributive property is one of the other numbers the. An example of how the result does not change when solved using the distributive is... Set of shirt and trousers on their birthday started analyzing the abstracts properties... Means the number of students in the brackets with each of the other numbers the... × 24 = 480 square units to a cafe here ’ s example... Is worth \\$ 12 and the distributive property is one of the terms in an expression be... The parenthesis 9 ( x ) – 9 ( 5 + 7 x 4 well as expressions of what algebraic. Width = x + 3 ) x 4 another number uses distributive property to multiply numbers mentally between. Number uses distributive property is one which is used quite frequently in mathematics, property. Identity and distributive properties property to multiply numbers mentally the reasoning of others `` (... Of one plate is twice that of the most frequently used properties in basic mathematics sets, the statements... Further explanations expression may be expanded in a particular way to form an equivalent expression multiplication.! = 5 x 4 + 7 x 4 + 7 x 4 b c. And critique the reasoning of others in addition and subtraction of what the algebraic 2. Rules of replacement statements are true number, the distributive property of multiplication addition. Formally, they write this property distributes or breaks down expressions into the addition or subtraction two! The parentheses x ) – 9 ( x ) – 9 ( 5 + x. To two valid rules of replacement multiplied by itself accepting the distributive property lets you multiply a sum is total. Total number of times a number by a group of numbers times a number is by. Really what they 're asking us to multiply numbers mentally means you are trouble... Property is used quite frequently in mathematics ≠ + numbers added together is the way it easier... Propositional logic, distribution refers to the distributive property is one of the frequently... Property was introduced in early 18th century, when mathematicians started analyzing abstracts! Same as doing each multiplication separately + 45 one of the most used... Would get the same day length of one plate is twice that of the equations by the LCM oil... … examples of the order of operations, simplifying what was in parentheses first century! A number with the other numbers inside the parenthesis the expression according to the distributive helps. The commutative, associative, multiplicative identity and distributive properties algebra tiles of how result. Taken from the word problems the most frequently used properties of numbers in Mathematical calculations and width of most. Start with a simple application in arithmetic: 5 ( 3 + 5 ) one gallon of olive costs. Helpful for most people ( 8 + 20 × 16 = 20 ( 8 + 20 × 24 = square! 49X2 + 28x + 16 5 plus 11 is 16 and then 16 times 7 is what, but /. = 81 + 45 a simple application in arithmetic: 2 ⋅ = +, but 2 / +! Mike and Sam, born on the same as doing each multiplication separately money have. Square units then add the products of a number with the other numbers inside the parenthesis this regulator 2 which... Write the distribution property of multiplication which states that the the cost building!" ]
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https://xiith.com/c/c-program-to-check-a-number-is-armstrong-or-not-using-the-function
[ "# C Program to check a number is Armstrong or not using the function\n\nIn this program, You will learn how to check a number is Armstrong or not using a function in c.\n\n``````Some list of Armstrong numbers is: 153, 370, 371, 407\n``````\n\n## Example: How to check a number is Armstrong or not using a function in c\n\n``````#include<stdio.h>\nint check(int n) {\nint r, arm = 0;\n\nwhile (n > 0) {\nr = n % 10;\narm = arm + r * r * r;\nn = n / 10;\n}\nreturn arm;\n}\n\nint main() {\nint n, arm;\nprintf(\"Enter a number:\");\nscanf(\"%d\", &n);\n\narm = check(n);\n\nif (arm == n) {\nprintf(\"Number is Armstrong:%d\", n);\n} else {\nprintf(\"Number is not Armstrong:%d\", n);\n}\n\nreturn 0;\n}\n``````\n\n#### Output:\n\n``````Enter a number:153\nNumber is Armstrong:153\n``````" ]
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https://www.softmath.com/math-com-calculator/adding-matrices/how-to--factor-using-ti-84.html
[ "English | Español\n\n# Try our Free Online Math Solver!", null, "Online Math Solver\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. 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[ null, "https://www.softmath.com/images/video-pages/solver-top.png", null ]
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http://spreadsheetpage.com/index.php/tip/determining_if_a_worksheet_or_workbook_has_code/
[ "# Determining If A Worksheet Or Workbook Has Code\n\nCategory: VBA Functions | [Item URL]\n\nEvery workbook and sheet has a corresponding code module. These code modules can contain VBA code to handle workbook or sheet-level events. For example, a workbook code module (named ThisWorkbook by default) might have a subroutine declared as follows:\n\n```Private Sub Workbook_Open()\n' Code goes here\nEnd Sub```\n\nThe Workbook_Open sub is executed whenever the workbook is opened.\n\nSimilarly, code modules for worksheets can contain subroutines to handle worksheet event such as Activate, Deactivate, Change, etc.\n\nListed below are two custom VBA functions that you can use to determine if the code module for a particular workbook or worksheet contains any code.\n\n### The WorkbookHasVBACode Function\n\nThe function below takes a single argument: a workbook object. It returns True if the workbook's code module contains any VBA code.\n\n```Private Function WorkbookHasVBACode(wb As Workbook)\nModuleLineCount = wb.VBProject.VBComponents(wb.CodeName). _\nCodeModule.CountOfLines\nIf ModuleLineCount = 0 Then\nWorkbookHasVBACode = False\nElse\nWorkbookHasVBACode = True\nEnd If\nEnd Function```\n\n### The SheetHasVBACode Function\n\nThe function below takes a single argument: a worksheet object. It returns True if the worksheet's code module contains any VBA code.\n\n```Private Function SheetHasVBACode(wks As Worksheet)\nModuleLineCount = wks.Parent.VBProject. _\nVBComponents(wks.CodeName).CodeModule.CountOfLines\nIf ModuleLineCount = 0 Then\nSheetHasVBACode = False\nElse\nSheetHasVBACode = True\nEnd If\nEnd Function```\n\n## An Example\n\nThe example below demonstrates a practical use of the SheetHasVBACode function. The DeleteBlankSheets subroutine deletes all blank sheets in the active workbook -- but only if the sheet does not contain any VBA code.\n\n```Sub DeleteBlankSheets()\nDim sht As Worksheet\nOn Error GoTo ErrHandler\n' Avoid Excel's confirmation prompt\n' Loop through each sheet\nFor Each sht In ActiveWorkbook.Worksheets\n' Is non-blank cell count zero?\nIf Application.CountA(sht.Cells) = 0 Then\n' Don't try to delete the last sheet\nIf ActiveWorkbook.Sheets.Count <> 1 Then\n' Don't delete sheet if it has VBA code\nIf Not SheetHasVBACode(sht) Then\nsht.Delete\nEnd If\nEnd If\nEnd If\nNext sht\nExit Sub\nErrHandler:\nMsgBox sht.Name & Chr(13) & Chr(13) & Error(Err)\nEnd Sub```\n\n### Tip Books\n\nNeeds tips? Here are two books, with nothing but tips:\n\nContains more than 100 useful tips and tricks for Excel 2013 | Other Excel 2013 books | Amazon link: 101 Excel 2013 Tips, Tricks & Timesavers", null, "Contains more than 200 useful tips and tricks for Excel 2007 | Other Excel 2007 books | Amazon link: John Walkenbach's Favorite Excel 2007 Tips & Tricks\n\n© Copyright 2019, J-Walk & Associates, Inc." ]
[ null, "http://spreadsheetpage.com/graphics/books/2007tips_sm.png", null ]
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http://forums.debian.net/viewtopic.php?f=8&p=728872
[ "## Cannot understand why some function is working correctly\n\nNeed help with C, C++, perl, python, etc?\n\n### Cannot understand why some function is working correctly\n\nThe problem is to eliminate recursion from function to compute selections of m elements from n elements ( comb( n, m ) ).\nUsing the recurrent relation comb( n, m ) = comb( n-1, m ) + comb( n-1, m-1 ) we can write recursive function:\nCode: Select all\n`int comb( n, m ){  if( n == 1 || m == 0 || n == m )    return 1;  else    return comb( n-1, m ) + comb( n-1, m-1 );}`\n\nThe problem is to eliminate recursion from this function. I have writen two functions, comb2 and comb3 to do this job. Comb2 is \"classic\" recursion elimination, but comb3 is using modified method in which n and m are treated as global variables.\nHere is the complete program:\nCode: Select all\n`#include <stdio.h>#include <stdlib.h>int comb( int, int );int comb2( int, int );int comb3( int, int );typedef struct cell {  int x, m, n, ret_add;} cell;typedef struct cell_stack {  cell c;  struct cell_stack *next;} cell_stack;void push_stack( cell, cell_stack ** );cell top_stack( cell_stack * );void pop_stack( cell_stack ** );int empty_stack( cell_stack * );void new_stack( cell_stack ** );typedef struct cell2 {  int x, ret_add;} cell2;typedef struct cell_stack2 {  cell2 c;  struct cell_stack2 *next;} cell_stack2;void push_stack2( cell2, cell_stack2 ** );cell2 top_stack2( cell_stack2 * );void pop_stack2( cell_stack2 ** );int empty_stack2( cell_stack2 * );void new_stack2( cell_stack2 ** );int main(){  int m, n;  printf( \"Input n: \" );  scanf( \"%d\", &n );  if( n < 0 ) {    printf( \"Wrong input\\n\" );    return 1;  }  printf( \"Input m: \" );  scanf( \"%d\", &m );  if( m < 0 || n < m ) {    printf( \"Wrong input\\n\" );    return 1;  }  printf( \"comb( n, m ) = %d\\n\" , comb( n, m ) );  printf( \"comb2( n, m ) = %d\\n\" , comb2( n, m ) );  printf( \"comb3( n, m ) = %d\\n\" , comb3( n, m ) );    return 0;}int comb( int n, int m ){  if( n == 1 || m == 0 || m == n )    return 1;  else    return comb( n - 1, m ) + comb( n - 1, m - 1 );}int comb2( int n, int m ){  cell cel;  int ret_val;  cel.n = n;  cel.m = m;  cel.ret_add = 0;  cell_stack *s;  new_stack( &s );  push_stack( cel, &s );  do {    cel = top_stack( s );    pop_stack( &s );    switch( cel.ret_add ) {    case 0:      if( cel.n == 1 || cel.m == 0 || cel.n == cel.m ) {   ret_val = 1;   break;      }      else {   cel.ret_add = 1;   push_stack( cel, &s );   (cel.n)--;   cel.ret_add = 0;   push_stack( cel, &s );   break;      }    case 1:      cel.x = ret_val;      cel.ret_add = 2;      push_stack( cel, &s );      (cel.n)--;      (cel.m)--;      cel.ret_add = 0;      push_stack( cel, &s );      break;    case 2:      ret_val += cel.x;      break;    }  } while( ! empty_stack( s ) );  return ret_val;}void push_stack( cell c, cell_stack **s_p ){  cell_stack *tmp = malloc( sizeof( cell_stack ) );  tmp->next = *s_p;  tmp->c = c;  *s_p = tmp;}void new_stack( cell_stack **s_p ){  *s_p = NULL;}cell top_stack( cell_stack *s ){  return s->c;}void pop_stack( cell_stack **s_p ){  cell_stack *tmp = *s_p;  *s_p = (*s_p)->next;  free( tmp );}int empty_stack( cell_stack *s ){  if( s == NULL )    return 1;  else    return 0;}int comb3( int n, int m ){  cell2 cel;  int ret_val = 0;  cel.ret_add = 0;  cell_stack2 *s;  new_stack2( &s );  push_stack2( cel, &s );  do {    cel = top_stack2( s );    pop_stack2( &s );    switch( cel.ret_add ) {    case 0:      if( n == 1 || m == 0 || n == m ) {   ret_val = 1;   break;      }      else {   cel.ret_add = 1;   push_stack2( cel, &s );   n--;   cel.ret_add = 0;   push_stack2( cel, &s );   break;      }    case 1:      cel.x = ret_val;      cel.ret_add = 2;      push_stack2( cel, &s );      m--;      cel.ret_add = 0;      push_stack2( cel, &s );      break;    case 2:      ret_val += cel.x;      n++;      m++;      break;    }  } while( !  empty_stack2( s ) );  return ret_val;}void push_stack2( cell2 c, cell_stack2 **s_p ){  cell_stack2 *tmp = malloc( sizeof( cell_stack2 ) );  tmp->next = *s_p;  tmp->c = c;  *s_p = tmp;}void new_stack2( cell_stack2 **s_p ){  *s_p = NULL;}cell2 top_stack2( cell_stack2 *s ){  return s->c;}void pop_stack2( cell_stack2 **s_p ){  cell_stack2 *tmp = *s_p;  *s_p = (*s_p)->next;  free( tmp );}int empty_stack2( cell_stack2 *s ){  if( s == NULL )    return 1;  else    return 0;}`\n\nThe problem is that I cannot understand why comb3 is working right. Simple check by \"hand\" is showing that comb2 and comb3 are computing the value in different way. But the result is always the same. Why?\npythagorasmk\n\nPosts: 117\nJoined: 2015-01-18 03:40\n\n### Re: Cannot understand why some function is working correctly\n\nFor me it looks like a homework - so trying to help You would be harmful.\n\nBut one thing have triggered my attention:\nFirst, the custom stacks are not the \"classic\" solution to avoid recursion, and Your implementation is extremely ineffective (slow) - malloc() is a performance killer even when it is configured to use per-thread arenas.\nBut foremost, Your linked list is not a stack.\nBill Gates: \"(...) In my case, I went to the garbage cans at the Computer Science Center and I fished out listings of their operating system.\"\nThe_full_story and Nothing_have_changed\nLE_746F6D617A7A69\n\nPosts: 488\nJoined: 2020-05-03 14:16\n\n### Re: Cannot understand why some function is working correctly\n\nLE_746F6D617A7A69 wrote:For me it looks like a homework - so trying to help You would be harmful.\n\nBut one thing have triggered my attention:\nFirst, the custom stacks are not the \"classic\" solution to avoid recursion, and Your implementation is extremely ineffective (slow) - malloc() is a performance killer even when it is configured to use per-thread arenas.\nBut foremost, Your linked list is not a stack.\n\nThe part with comb2( n, m ) is a homework, but comb3 is not. With comb3 I wanted to speed up comb2.\nIn the book that I am reading, recursion elimination is explained by custom stacks. In the same book, stacks are explained as a special kind of lists, so they use this implementation of the stacks. This is beginner book so efficiency in not concern. The problem is to mathematically proof that comb3 is working right, there are not efficiency concerns.\npythagorasmk\n\nPosts: 117\nJoined: 2015-01-18 03:40\n\n### Re: Cannot understand why some function is working correctly\n\npythagorasmk wrote:The part with comb2( n, m ) is a homework, but comb3 is not. With comb3 I wanted to speed up comb2.\nIn the book that I am reading, recursion elimination is explained by custom stacks. In the same book, stacks are explained as a special kind of lists, so they use this implementation of the stacks. This is beginner book so efficiency in not concern. The problem is to mathematically proof that comb3 is working right, there are not efficiency concerns.\n\nMy honestly honest advice is that You should change the book (seriously) ...\n\nSearch for the \"pdf Ulrich Drepper\" - this man knows how to write programs ...", null, "Bill Gates: \"(...) In my case, I went to the garbage cans at the Computer Science Center and I fished out listings of their operating system.\"\nThe_full_story and Nothing_have_changed\nLE_746F6D617A7A69\n\nPosts: 488\nJoined: 2020-05-03 14:16\n\n### Re: Cannot understand why some function is working correctly\n\nLE_746F6D617A7A69 wrote:My honestly honest advice is that You should change the book (seriously) ...\n\nSearch for the \"pdf Ulrich Drepper\" - this man knows how to write programs ...", null, "I have searched for \"pdf Urlich Drepper\" and return results are books on memory organization, writing shared library's and other, for me advanced topics.\nPlease give me advice about book in which are explained data structures in C.\npythagorasmk\n\nPosts: 117\nJoined: 2015-01-18 03:40\n\n### Re: Cannot understand why some function is working correctly\n\nMost people think that programmer is a (wo)man who knows some programming language - this a common misunderstanding, because this is a code monkey, not a programmer. And because code monkeys are cheaper than programmers, today we need a PC with at least 1GB of RAM just to send an e-mail", null, "... not really funny.\n\npythagorasmk wrote:Please give me advice about book in which are explained data structures in C.\n... and that's why I've mentioned Urlich Drepper's articles - to write good data structures You have to know the memory organization, how the cache memory works, etc - i.e. as a programmer, You must know *how the computer works*.\n\n--------------------\nIf You can't find the answer trough static analysis of the code, then it's time to start using debuggers", null, "Bill Gates: \"(...) In my case, I went to the garbage cans at the Computer Science Center and I fished out listings of their operating system.\"\nThe_full_story and Nothing_have_changed\nLE_746F6D617A7A69\n\nPosts: 488\nJoined: 2020-05-03 14:16\n\n### Re: Cannot understand why some function is working correctly\n\npythagorasmk wrote:\nPlease give me advice about book in which are explained data structures in C.\n\nSome websites give reviews and one can browse parts of a book that interests them, or even better, a good public library , where you can even check a book out, or just review it there, and if you want to buy a copy, you can then look for it online, there are many book sellers, where you can make a on-line purchase.\n\nThere are several listed here: https://wiki.osdev.org/Books\nAlgorithms and Data Structures :Niklaus Wirth-----There are very few books that can actually teach good style, and this is probably one of the best. This book is a must read for anyone wishing to become a great programmer, not merely an average one.\n\nbook in which are explained data structures in C.\n======================\nhttps://www.akkadia.org/drepper/cpumemory.pdf\nPlease Read What we expect you have already Done\nSearch Engines know a lot, and\n\"If God had wanted computers to work all the time, He wouldn't have invented RESET buttons\"\nand\nJust say NO to help vampires!\ncuckooflew\n\nPosts: 681\nJoined: 2018-05-10 19:34\nLocation: Some where out west\n\n### Re: Cannot understand why some function is working correctly\n\nLE_746F6D617A7A69 wrote:For me it looks like a homework - so trying to help You would be harmful.\n\nI will try to prove that comb3 works correctly. The proof is by induction on n.\nLet p and q are input values. We will prove not only that comb3 works correctly but also that when function ends the values of n and m are the same as p and q.\n\nIf p = 1 and q = 0 or q = 1, simple check gives that the claim is true.\n\nLet theorem is true for all values p, q, q<=p. We will prove that theorem is true for p+1 and all values of q, q<=p+1.\n\nIf q =0 or q = p+1, simple check proves that theorem is true. Let q != 0 and q != p+1. We must observe that p+1 > 1.\n\nWhen the function starts n becomes p+1, and m becomes q. Because the condition of the first if statement is not satisfied program continues after else statement. we push ret_add = 1 on the stack the value of x is undefined at the moment. After that n becomes p, and m is unchanged. We push ret_add = 0 on the stack. The value of x is undefined at the moment. After this function continues like it was called wit values p, q (observe that q<=p), the only change is that at the bottom of tha stack we have ret_add 1, and undefined value of x.\nBy induction hypothesis, when on the stack is left only one record ( ret_add =1, undefined value of x ) the value of n is p, the value m is q and ret_val = comb( p, q ).\n\nThe function continues with case 1. We push x = ret_val ( observe that x becomes comb( p, q ) and ret_add = 2 on the stack. After that n is unchanged ( equals to p) but m becomes q-1. we push ret_add = 0 on the stack. After this function continues like it was call with values p, q-1, the only difference is that on the bottom of the stack we have record in which x is comb( p, q ) ret_add is 2. When on the stack the only left record is the bottom one, ret_val = is comb( p, q-1 ) and the value of n is p, the value of m is q-1, this is by induction hypothesis.\nFunction continues at case 2. ret_val becomes ret_val + x or ret_val is comb(p,q) + comb(p, q-1 ), and n becomes p+1, m becomes q. Because comb( p, q ) + comb( p, q - 1 ) = comb( p+1, q ), theorem is completely proved.\npythagorasmk\n\nPosts: 117\nJoined: 2015-01-18 03:40\n\n### Re: Cannot understand why some function is working correctly\n\npythssorasmk wrote:\nLE_746F6D227A7A69 wrote:My honestly honest advice is that You should change the book (seriously) ...\n\nSearch for the \"pdf Ulrich Drepper\" - this man knows how to write programs ...", null, "I have searched testogen found here for \"pdf Urlich Drepper\" and return results are books on memory organization, writing shared library's and other, for me advanced topics.\nPlease give me advice about book in which are explained data structures in C.\n\nYou are using wrong description to determined the logics of the problem ..\nReply me for Hot to do it correctly?\nJohnJacobson\n\nPosts: 1\nJoined: 2020-10-16 12:57\n\nReturn to Programming\n\n### Who is online\n\nUsers browsing this forum: No registered users and 13 guests\n\nfashionable" ]
[ null, "http://forums.debian.net/images/smilies/icon_wink.gif", null, "http://forums.debian.net/images/smilies/icon_wink.gif", null, "http://forums.debian.net/images/smilies/icon_wink.gif", null, "http://forums.debian.net/images/smilies/icon_wink.gif", null, "http://forums.debian.net/images/smilies/icon_wink.gif", null ]
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https://cs.stackexchange.com/tags/weighted-graphs/hot
[ "# Tag Info\n\n166\n\nAllowed by whom? There is no Central Graph Administration that decides what you can and cannot do. You can define objects in any way that's convenient for you, as long as you're clear about what the definition is. If zero-weighted edges are useful to you, then use them; just make sure your readers know that's what you're doing. The reason you don't usually ...\n\n42\n\nDijkstra relies on one \"simple\" fact: if all weights are non-negative, adding an edge can never make a path shorter. That's why picking the shortest candidate edge (local optimality) always ends up being correct (global optimality). If that is not the case, the \"frontier\" of candidate edges does not send the right signals; a cheap edge might lure you down a ...\n\n34\n\nDistance between cities can't be negative, but if you are programming for an electric car, then a downhill road segment will regen, thus the energy used is negative. It is very important to take that into account when predicting range. In a neural network, we can use negative weights to indicate that one neuron firing is inversely correlated with another ...\n\n23\n\nFor simple graphs*, it is true for the following reason: Kruskal’s algorithm is correct Kruskal’s algorithm works as follows: sort the edges by increasing weight repeat: pop the cheapest edge, if it does not create cycles, include it in the MST Two edges cannot construct a cycle in a simple graph By the correctness of Kruskal’s algorithm, the two ...\n\n23\n\nOf course. The weight can mean things that are irrelevant to the existence of an edge. Since you don't ask for a \"list of say 6 or 7 real-life examples\", I will just add one. Consider a road network. If you want to find a path from A to B in a road network, you need to keep all the road segments that exist. There are two important measures: ...\n\n13\n\nA path of length $n$ consists of $n$ line segments in the plane. You want to find all intersections between these line segments. This is a standard problem that has been studied in depth in the computer graphics literature. A simple algorithm is the following: for each pair of line segments, check whether they intersect (using a standard geometric ...\n\n13\n\nYour question was already asked before it seems, but got no explicit examples. I try to give these here. First note the question only makes sense if we consider a node $u$, and there exist spanning trees starting with $u$. The algorithms of Prim and Kruskal make choices in a greedy way. Once the choice is made, it will not be reconsidered. We show how this ...\n\n13\n\nin the first picture: the right graph has a unique MST, by taking edges $(F,H)$ and $(F,G)$ with total weight of 2. Given a graph $G=(V,E)$ and let $M=(V,F)$ be a minimum spanning tree (MST) in $G$. If there exists an edge $e=\\{v,w\\}\\in E \\setminus F$ with weight $w(e)=m$ such that adding $e$ to our MST yields a cycle $C$, and let $m$ also be the lowest ...\n\n13\n\nIt depends on the context. In general yes, edges of zero and even negative weight may be allowed. In some specific cases the edge weights might be required to be non-negative or strictly positive (for instance, Dijkstra's algorithm requires weights to be non-negative).\n\n12\n\nAdding a constant amount to each edge length can change the shortest path for the simple reason that it increases the length of a path with many edges by more than it increases the length of a path with only a few edges. For a simple case, consider the graph with vertices $\\{a,b,c\\}$ and edges $\\{ab,bc,ac\\}$, where $ab$ and $bc$ have length $1$ ...\n\n12\n\nThe classic strategy game Civilization by MicroProse represents the world map as a square grid where each node of the grid is a tile of the world map, representing some type of terrain. Players control civilian and military units over this map. Each unit has a specific allocation of movement points, and each terrain type costs a specific amount of movement ...\n\n11\n\nIt is NP-complete to even decide whether any path exists. It is clearly possible to verify any given path is a valid path in the given graph. Thus the bounded-length problem is in NP, and so is its subset, the any-path problem. Now, to prove NP-hardness of the any-path problem (and thus of the bounded-length problem), let's reduce SAT-CNF to this problem: ...\n\n10\n\nThere are exponentially many such routes. Think of a sequence of $n$ diamonds. At each diamond, you can go either left or right, independently of what you do at all other diamonds. This leads to $2^n$ paths, each of which is non-intersecting. Now the complete graph on those vertices contains all of these paths, plus some more, so this is a lower-bound on ...\n\n10\n\nA Continuous-time Markov Chain can be represented as a directed graph with constant non-negative edge weights. An equivalent representation of the constant edge-weights of a directed graph with $N$ nodes is as an $N \\times N$ matrix. The Markov property (that the future states depend only on the current state) is implicit in the constant edge weights (or ...\n\n8\n\nMarkov Chains come in two flavors: continuous time and discrete time. Both continuous time markov chains (CTMC) and discrete time markov chains (DTMC) are represented as directed weighted graphs. For DTMC's the transitions always take one unit of \"time.\" As a result, there is no choice for what your weight on an arc should be-- you put the probability of ...\n\n8\n\nConsider the complete graph $K_n$ in which all edges have the same cost. All trees are MSTs. They have diameter ranging from $2$ all the way to $n-1$.\n\n8\n\nIn circuity, we often construct a graph of a circuit. Wires are typically modeled as 0 resistance because, frankly, measuring the resistance of wires is really tricky and rarely profitable. So if we have multiple devices connected to a single wire, we can treat that as separate vertices with 0 weight nodes between them. We can transform them into one \"...\n\n7\n\nThe setting you suggest is not clear enough to determine what kind of an automaton you are looking for. A short explanation regarding the types of automata: A weighted automaton is typically an automaton with a weight function on the edges (or states). Then, a run is assigned a weight according to the weight it traverses. There are many different semantics ...\n\n7\n\nThere is no direct relationship between the diameter of a (minimum) spanning tree and the total cost of the tree1. Consider the following example: The spanning tree on the left (whose edges are highlighted in red) is minimum. Its total cost is 7 and the diameter is equal to 5. In contrast, the spanning tree on the right is not minimum (since its total cost ...\n\n7\n\nThis is NP-hard, so it's very unlikely that a polynomial-time algorithm exists. Given any instance $G=(V, E)$ of Hamiltonian Path, create a new graph $G'=(V', E')$ in which every vertex $v \\in V$ becomes a pair of vertices $v_+, v_-$ connected by an edge in $G'$. All of these edges should also be added to $F$. Then for each $(u, v) \\in E$, add the ...\n\n7\n\nI contacted one of the authors (Kevin Wayne; thanks) of the textbook \"Algorithms, 4th Edition\" and got a hint: Try adding \"t-joins\" or \"perfect matching\" to your web searches. Following this, I found the following two lecture notes: Shortest Path Algorithms Luis Goddyn, Math 408: Using Edmonds' Minimum Weight Perfect Matching Algorithm to solve shortest ...\n\n7\n\nHere is the original statement in CLRS. Assume that we have a connected, undirected graph $G$ with a weight function $w: E\\to\\Bbb R$, and we wish to find a minimum spanning tree for $G$. It is pretty good to understand \"a weight function $w:E\\rightarrow \\mathbb{R}$\" as \"an edge has a weight\". In fact, that is how I would interpret that notation in a rush ...\n\n7\n\nYour conceptual difficulty stems from not distinguishing between TSP and Weighted Hamiltonian Cycle. These are usually discussed as if they are the same problem, but they're not. In Weighted Hamiltonian Cycle, we are given a graph with nonnegative edge weights and we wish to determine the minimum-weight Hamiltonian cycle, i.e., the minimum-weight cycle that ...\n\n7\n\nIn a social network. Where the source node is a person the target node is another person and the connection represents the preference the source has for the target. The sign representing the direction of the sentiment. In quantitative finance where the nodes are securities and the connections are the correlation coefficients. A few years ago I made a ...\n\n6\n\nI would suggest the following approach. Maintain a data structure $H$ of $(i,j, g(i,j))$ triples so that you can efficiently find and remove a triple $(i,j,w)$ that minimises $w$. Maintain a partition $P$ of nodes $V = \\{1,2,\\dotsc,N\\}$. Maintain a tree $T$. We will maintain the invariant that $T$ contains some edges of the tree that we are trying to ...\n\n6\n\nYes, it is true. Let $w: E(G) \\to \\mathbb{R}$ be a weight function on the edges of $G$, $s \\in V(G)$ be the start vertex. Let $p(v) = \\min\\{\\max\\{w(e_1), \\ldots, w(e_k)\\} \\mid e_1, \\ldots, e_k \\text{ is edges of a path from } s \\text{ to } v \\}$ i.e. the shortest path between $s$ and $v$. Let's prove by induction on the step of Dijkstra's algorithm that ...\n\n5\n\nThus I'm still curious if there are other more efficient approaches to solving the problem. If you take the complement graph $\\overline{G}$, then your problem corresponds to a coloring problem. Cover by cliques in $G$ is the same as covering by independent set in $\\overline{G}$. The problem is para-NP-hard in the unweighted case and the problem is just ...\n\n5\n\nFirst of all, bear in mind that the Longest Path Problem (LPP) is a NP-complete problem whereas finding the Shortest Path Problem (SPP) is a problem known to be in P. The proof for the NP-completeness of LPP is trivial and consists of a reduction from the Hamiltonian Circuit (HC) problem which is already known to be NP-complete. In other words, if you could ...\n\n5\n\nHere is a slightly simpler argument that also works for other matroids. (I saw this question from another one.) Suppose that $G$ has $m$ edges. Without loss of generality, assume that the weight function $w$ takes on values in $[m]$, so we have a partition of $E$ into sets $E_i := w^{-1}(i)$ for $i\\in [m]$. We can do induction on the number $j$ of non-...\n\n5\n\nIntroduction This answer is in two parts. The first is an analysis of the problem mixed with a sketch of the algorithm to solve it. As it is my first version, it is detailed, but results in an algorithm that is a bit more complex than needed. It is followed by a pseudo-code version of the algorithm, written sometime later, as the algorithm was clearer. ...\n\nOnly top voted, non community-wiki answers of a minimum length are eligible" ]
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https://everything2.com/title/Planck+units
[ "(synonym Planck-Wheeler units, after John Archibald Wheeler who developed the precise system described here, inspired by a similar system by Max Planck)\n\nWhat units to use when measuring things is always a subject of heated debate. In the United States of America, the so-called imperial system is in use, creating problems for visiting tourists. The more rational France attempted to solve this by creating the metric system, as a part of the same process that created the ten-day week. Even the metric system has ugly quirks, though, so the SI Units were invented.\n\nBut throughout this progression, there is is a disturbing feeling of arbitrarity. Why is a kilogram a kilogram? Because that's the mass of a little iridium cylinder in Paris, which at one time was thought to weigh as much as a cube of water whith a side equal to 1/400,000,000 of the circumference of the Earth. We cannot expect the Americans to switch system on so arbitrary grounds, can we? Is there no way to create a truly universal system of units? One that even aliens from outer space could agree upon?\n\nAs a matter of fact there is! There are certain physical universal constants, which could in principle take any value, but were given a particular one by God. These are: the gravitational constant G(as used in general relativity), the speed of light c, Boltzmann's Constant k (thermodynamics) and planck's constant divided by 2pi h-(quantum mechanics). (Yes, divided by 2pi. Arbitrary, why do you say that?).\n\nNow, each of these constants has a dimension, in the sense of mass, time, etc. If we use braces for \"dimension of\", and M, L, T, t for mass, length, time, and temperature, then:\n\n{G} = force/mass = M-1L3T\n{c} = velocity = LT-1\n{k} = energy/temperature = ML2T-2t-1\n{h-} = ML2T-1\n\nSo by simple multiplying, dividing, and taking roots of fundamental constants, we can form quantities of any dimension we wish. These quantities can then be used as universal units. Here we go:\n\n...And so on. The Planck velocity becomes plain c, which for once does not seem terribly huge by comparison. The Planck density, according to my physics book, corresponds to the entire visible universe compressed to the size of a proton.\n\nSo we finally have a rational unit system. Now go lobby your politicians to implement the switch!\n\nLog in or register to write something here or to contact authors." ]
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https://docs.microsoft.com/en-us/previous-versions/visualstudio/visual-studio-2008/61h34247%28v%3Dvs.90%29
[ "# Count Property (Collection Object)\n\nReturns an Integer containing the number of elements in a collection. Read-only.\n\n``````Public ReadOnly Property Count() As Integer\n``````\n\n## Remarks\n\nUse the Count property to determine the number of elements in a Collection object.\n\n## Example\n\nThis example illustrates the use of the Count property to display the number of elements in a Collection Object (Visual Basic) in the variable birthdays.\n\n``````Dim birthdays As New Collection()\nbirthdays.Add(New DateTime(2001, 1, 12), \"Bill\")\nbirthdays.Add(New DateTime(2001, 1, 13), \"Joe\")\nbirthdays.Add(New DateTime(2001, 1, 14), \"Mike\")\nbirthdays.Add(New DateTime(2001, 1, 15), \"Pete\")\n\n...\n\nMsgBox(\"Number of birthdays in collection: \" & CStr(birthdays.Count))\n``````\n\nThe Collection object is one-based, which means that the index values of the elements range from 1 through the value of the Count property.\n\n## Requirements\n\nNamespace: Microsoft.VisualBasic\n\nClass: Collection\n\nAssembly: Visual Basic Runtime Library (in Microsoft.VisualBasic.dll)" ]
[ null ]
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https://codegolf.stackexchange.com/questions/25606/count-the-unique-fractions-with-only-integers
[ "# Count the unique fractions with only integers\n\nCount the number of unique fractions with numerators and denominators from 1 to 100, and print the counted number. Example: 2/3 = 4/6 = ...\n\nRules:\n\nYou must actually count in some way. Only integers are allowed, no floating point numbers or fraction types.\n\nIntegers count as fractions. So 1/1, 2/1, etc is valid.\n\n• \"no floating point numbers or fraction types\" - does that mean, anywhere in the program? – CompuChip Apr 10 '14 at 7:49\n• @CompuChip Yes. Other non-number types are allowed – qwr Apr 10 '14 at 18:55\n• How on earth is this question voted negative? I like it. I just added a +1, bringing it back up to 0 from –1. – Todd Lehman Apr 15 '15 at 18:06\n\n# J - 21 17 char\n\n+/1=,+./~1+i.100\n\n\nExplained:\n\n• 1+i.100 - The integers from 1 to 100.\n• +./~ - Table of GCDs.\n• 1=, - Run into a list, and then check for equality to 1.\n• +/ - Add together the results (true is 1, false is 0).\n\nUsage:\n\n +/1=,+./~1+i.100\n6087\n\n\n21 char version that actually constructs all the pairs of numbers:\n\n#~.,/(,%+.)&>:/~i.100\n\n\n&>: increments all the integers and also sets up another golfy thing, while ~. takes all the unique entries in the list of pairs we construct, and then # gives the length of that.\n\n# Ruby, 57 (67 without Rational) (-3 if run in IRB)\n\np((a=[*1..100]).product(a).map{|x|Rational *x}.uniq.size)\n\n\nOutput:\n\n6087\n\n\nCan be 3 less characters if run in IRB, because you can remove the p( and ).\n\nUses product for the numerator and denominator getting process, and then uses Rational for converting them to fractions. If you remove the .size at the end, it prints all of the fractions instead of how many there are.\n\nIt seems like it might take a long time to run, but it's actually almost instantaneous.\n\nHere's an example IRB session to explain how the code works a bit better:\n\nirb(main):027:0> (a=[*1..5]).product(a)\n=> [[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [5, 1], [5, 2], [5, 3], [5, 4], [5, 5]]\nirb(main):028:0> Rational 1, 2\n=> (1/2)\nirb(main):029:0> Rational 2, 4\n=> (1/2)\n\n\nRational *x uses the \"splat\" operator to call the Rational function with arguments given in the array x. This \"splat\" is also used in [*1..100].\n\nHere's an alternative that doesn't use Rational, weighing in at 66 characters:\n\np((a=[*1..100]).product(a).map{|x,y|z=x.gcd y;[x/z,y/z]}.uniq.size)\n\n\nThe fraction simplification method is replaced with this:\n\nz=x.gcd y;[x/z,y/z]\n\n\nwhich divides the numberator and the denominator by their GCD (greatest common denominator), and then sticks them back in an array so that uniq can work.\n\n• Wow, that was fast. Are you sure \"Rational\" uses only integers? Nothing else is allowed – qwr Apr 9 '14 at 21:00\n• @qwr Rational stores the numbers as a numerator and a denominator, not as a floating point. – Doorknob Apr 9 '14 at 21:00\n• @qwr If you're interested, I've also added a new version that doesn't use Rational. – Doorknob Apr 9 '14 at 21:04\n\n# Bash + GNU tools, 47\n\n$echo 10^9\\*{1..100}/{1..100}\\;|bc|sort -u|wc -l 6087$\n\n\nLooks like a similar method to @Doorknob's answer.\n\n• Doesn't bc -l give a floating point answer? The question states that only integer types are allowed. – user80551 Apr 10 '14 at 3:18\n• @user80551 Yes, you're right, I missed that in the question. I've edited to use bc with no -l, and just premultiply everything so we still get the correct uniqueness. Adds 3 chars. – Digital Trauma Apr 10 '14 at 4:10\n• Nice, but why escape the *. – devnull Apr 10 '14 at 11:26\n• @devnull: Not escaping the asterisk will cause problems if there's a file with name ./10^91/1;. – Dennis Apr 10 '14 at 14:06\n• @Dennis Ah! Ugly globbing. – devnull Apr 10 '14 at 14:08\n\n# JavaScript, 64 characters\n\nfor(n=101,o=0,c={};--n;)for(d=101;--d;)!c[n/d]&&(c[n/d]=++o);o\n\n\nPut into the JS console, returns 6087.\n\n• You can save 3 with c[n/d]=c[n/d]||++o – Peter Taylor Apr 9 '14 at 23:12\n• Also, if you change c to an array and assign it to o ([] casts to 0 with ++ operator) you can save 4 more bytes: (57 byte solution) for(n=101,o=c=[];--n;)for(d=101;--d;)c[n/d]=c[n/d]||++o;o – nderscore Apr 10 '14 at 4:44\n\n## Mathematica - 24\n\nThis is just a sequence A018805. EulerPhi[n] is the number of coprime to n integers m that are below n (gcd n m == 1)\n\n2 Tr@EulerPhi@Range@100 - 1\n\n\n## J - 15\n\n<:+:+/5&p:i.101\n\n• \"You must actually count in some way. \" – user12205 Apr 9 '14 at 21:28\n• @ace This is counting. I just redirect half of it to the built-in function. – swish Apr 9 '14 at 21:31\n• @RossMillikan That's why we double it, so it will count both 3/2 and 2/3, 3/1 and 1/3. – swish Apr 9 '14 at 22:54\n\n## Sage, 62 or 42\n\nRuns in the interactive prompt.\n\nc=0\nR=range(1,101)\nfor i in R:\nfor j in R:\nc+=gcd(i,j)==1\nc\n\n\nShort and easy to understand.\n\nIf use of Euler's totient function is allowed, here's a 42-char one-liner:\n\n2*sum(euler_phi(n)for n in range(1,101))-1\n\n\nsum\\$filter(<2)[gcd a b|a<-[1..100],b<-[1..100]]\n\n\nRun this from the interpreter.\n\n# CJam - 18 22\n\n100,:):X{dXf/}%:|,\nOops, I had missed the \"no floating point\" requirement. Here is an integer-based solution:\n\n100,:):X_:*f*{Xf/}%:|,\n\n\nCJam is a new language I am developing, similar to GolfScript - http://sf.net/p/cjam. Here is the explanation:\n\n100, makes an array [0 1 ... 99]\n:) increments all the elements of the array\n:X assigns to variable X\n_ duplicates the last value (the array)\n:* multiplies the array elements together, thus calculating 100!\nf* multiplies each array element with 100!\n{...}% performs a \"map\" - applies the block to each element\nXf/ divides the current number by each element in X; since the numbers were already multiplied by 100!, it is an exact division\n:| performs a fold with the | (set union) operator, on the array of arrays we obtained\n, counts the number of elements\n\n• Hm... Is this some sort of emoticon language? – user12205 Apr 9 '14 at 22:16\n• @ace haha, not intentionally :) I used \":\" followed by an operator as a shortcut for map/fold operations – aditsu Apr 9 '14 at 22:18\n• It totally should be an emoticon based language. I'd goof around with it a little. – Kyle Kanos Apr 10 '14 at 2:44\n\n# Mathematica, 77\n\nLength@Select[Range~Tuples~2,#[]==Numerator@Simplify[#[]/#[]]&]\n\n\nWow, this is the longest one here. Guess I'm not too good at thinking outside the box...\n\n• Oh wait, only integers are allowed. Oops... Disregard this! :D – kukac67 Apr 10 '14 at 1:23\n\n# C++ - 64\n\n#include<iostream>\nmain(){int i=0;while(++i<6087);std::cout<<i;}\n\n\nIt counts in some way!\n\n• meta.codegolf.stackexchange.com/a/1063/17249 – durron597 Apr 10 '14 at 13:55\n• Quote from that link: \"[if] the question doesn't require input and so a solution which just prints the answer would seem to meet the spec\". – CompuChip Apr 10 '14 at 16:12\n\n# Python, 81 characters\n\nI guess the following solution is more math golf than code golf: it prints all the fractions as well. The algorithm might be an inspiration for code golfers, though.\n\ndef f(i,j,k,l):\nm,n = i+k,j+l\nif m > 100 or n > 100:\nreturn 0\nprint(\"%d/%d\" % (m,n))\nreturn f(i,j,m,n)+f(m,n,k,l)+1\n\nprint(f(0,1,1,0))\n\n\nSo after proving that this does the right thing (it would not be producing all the right fractions if it didn't, would it?), we can write this as\n\nf=lambda i,j,k,l:i+k<101>j+l and f(i,j,i+k,j+l)+f(i+k,j+l,k,l)+1;print f(0,1,1,0)\n\n\nWhich is 81 characters. At the interactive prompt, you can save the six characters \"print \" but that seems like a bit of cheating.\n\nimport sys" ]
[ null ]
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https://projecteuclid.org/journals/rocky-mountain-journal-of-mathematics/volume-46/issue-3/Lower-bound-for-the-higher-moment-of-symmetric-square-L/10.1216/RMJ-2016-46-3-915.full
[ "Translator Disclaimer\n2016 Lower bound for the higher moment of symmetric square $L$-functions\nGuanghua Ji\nRocky Mountain J. Math. 46(3): 915-923 (2016). DOI: 10.1216/RMJ-2016-46-3-915\n\nAbstract\n\nLet $\\mathcal {S}_k(N)$ be the space of holomorphic cusp forms of weight $k$, level $N$ and let $\\mathcal {B}_k(N)$ be an orthogonal basis of $\\mathcal {S}_k(N)$ consisting of newforms. Let $L(s, \\textup {sym}^2 f)$ be the symmetric square $L$-function of $f\\in \\mathcal {B}_k(N)$. In this paper, the lower bound of the higher moment of $L(1/2,\\textup {sym}^2 f)$ is established, i.e., for any even positive number $r$, $\\sum _{f\\in \\mathcal {B}_k(N)}\\omega _f^{-1}L\\bigg (\\frac 12, \\textup {sym}^2 f\\bigg )^r \\gg (\\log N)^{{r(r+1)}/{2}}$ holds for $N\\rightarrow \\infty$.\n\nCitation\n\nGuanghua Ji. \"Lower bound for the higher moment of symmetric square $L$-functions.\" Rocky Mountain J. Math. 46 (3) 915 - 923, 2016. https://doi.org/10.1216/RMJ-2016-46-3-915\n\nInformation\n\nPublished: 2016\nFirst available in Project Euclid: 7 September 2016\n\nzbMATH: 06628759\nMathSciNet: MR3544839\nDigital Object Identifier: 10.1216/RMJ-2016-46-3-915\n\nSubjects:\nPrimary: 11F11, 11F66", null, "", null, "" ]
[ null, "https://projecteuclid.org/Content/themes/SPIEImages/Share_black_icon.png", null, "https://projecteuclid.org/images/journals/cover_rmjm.jpg", null ]
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https://slidetodoc.com/geometric-sequences-a-geometric-sequence-is-n-a/
[ "", null, "# Geometric Sequences A Geometric Sequence is n a\n\n• Slides: 12", null, "Geometric Sequences", null, "A Geometric Sequence is n a sequence of numbers in which each term is formed by multiplying the previous term by the same number or expression. The consecutive terms have a common ratio. 1, 3, 9, 27, 81, 243, . . . The terms have a common ratio of 3.", null, "Geometric Sequence Example Is the following sequence geometric? 4, 6, 9, 13. 5, 20. 25, 30. 375… n Yes, the common ratio is 1. 5", null, "Geometric Sequence n To find any term in a geometric sequence, use the formula an = a 1 rn– 1 where r is the common ratio.", null, "Example Find the twelfth term of the geometric sequence whose first term is 9 and whose common ratio is 1. 2. an = a 1 rn– 1 a 1 = 9 r = 1. 2 a 9 = 9 • 1. 211 a 12 = 66. 87 n To find the sum of a geometric series, we can use summation notation.", null, "Example Which can be simplified to:", null, "Evaluate the sum of: n Convert this to = 7. 49952", null, "Series", null, "Series", null, "Series n Definition: A series is a partial sum of the first n terms of a sequence. General term: nth partial sum: Sn =. n nth partial sum of arithmetic sequence: n Example: nth partial sum of an = -1 + 5 n. n Sn =", null, "Series n nth partial sum of geometric sequence: n Sum of an infinite geometric sequence: n n If |r|<1, n If |r| 1, a geometric series has no infinite sum. Example:", null, "Series n Product notation:" ]
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https://math.stackexchange.com/questions/2079950/compute-the-n-th-power-of-triangular-3-times3-matrix/2079962
[ "# Compute the $n$-th power of triangular $3\\times3$ matrix\n\nI have the following matrix\n\n$$\\begin{bmatrix} 1 & 2 & 3\\\\ 0 & 1 & 2\\\\ 0 & 0 & 1 \\end{bmatrix}$$\n\nand I am asked to compute its $n$-th power (to express each element as a function of $n$). I don't know at all what to do. I tried to compute some values manually to see some pattern and deduce a general expression but that didn't gave anything (especially for the top right). Thank you.\n\n## 6 Answers\n\nComputing the first few powers should allow you to find a pattern for the terms. Below are some terms:\n\n$$\\left(\\begin{matrix}1 & 2 & 3\\\\0 & 1 & 2\\\\0 & 0 & 1\\end{matrix}\\right), \\left(\\begin{matrix}1 & 4 & 10\\\\0 & 1 & 4\\\\0 & 0 & 1\\end{matrix}\\right), \\left(\\begin{matrix}1 & 6 & 21\\\\0 & 1 & 6\\\\0 & 0 & 1\\end{matrix}\\right), \\left(\\begin{matrix}1 & 8 & 36\\\\0 & 1 & 8\\\\0 & 0 & 1\\end{matrix}\\right), \\left(\\begin{matrix}1 & 10 & 55\\\\0 & 1 & 10\\\\0 & 0 & 1\\end{matrix}\\right), \\left(\\begin{matrix}1 & 12 & 78\\\\0 & 1 & 12\\\\0 & 0 & 1\\end{matrix}\\right)$$\n\nAll but the top right corner are trivial so lets focus on that pattern. (Although if you look at it carefully you should recognize the terms.)\n\nTerms: $3,10,21,36,55,78$\n\nFirst difference: $7, 11, 15, 19, 23$\n\nSecond difference: $4, 4, 4, 4$\n\nAs the second difference is a constant the formula must be a quadratic. As the second difference is 4 then it is in the form $2n^2+bn+c$. Examining the pattern gives formula of $2n^2+n=n(2n+1)$.\n\nSo the $n^{th}$ power is given by:\n\n$$\\left(\\begin{matrix}1 & 2n & n(2n+1)\\\\0 & 1 & 2n\\\\0 & 0 & 1\\end{matrix}\\right)$$\n\nThe reason I said you should recognize the pattern is because it is every second term out of this sequence: $1,3,6,10,15,21,27,37,45,55,66,78,\\cdots$ which is the triangular numbers.\n\n• It seemed more in line with what the OP had been trying. It would also let the OP see if their approach (and the powers of the matrix) were correct. – Ian Miller Jan 2 '17 at 1:28\n• This is a great approach to \"guess the formula\" (and it is enough for any practical purpose), but it is not enough to constitute a proof. Could you please make it complete (or at least mention in the body that it still needs a proof)? – dtldarek Jan 2 '17 at 9:57\n\nWrite this matrix as follows: \\begin{equation} \\left[ \\begin{matrix} 1 & 2&3\\\\ 0 & 1 & 2\\\\ 0 & 0 &1 \\end{matrix} \\right] = I + 2 J+ 3 J^{2}. \\end{equation} where \\begin{equation} I = \\left[ \\begin{matrix} 1 & & \\\\ &1 & \\\\ & & 1 \\end{matrix} \\right], ~ J = \\left[ \\begin{matrix} 0& 1 &0 \\\\ 0 &0 & 1 \\\\ 0 & 0& 0 \\end{matrix} \\right],~ J^2 = \\left[ \\begin{matrix} 0& 0 &1\\\\ 0 & 0 &0\\\\ 0& 0 & 0 \\end{matrix} \\right], ~ J^{3}=0. \\end{equation} With this relation you can expand the power of the matrix into sum of $I$, $J$ and $J^2$.\n\n• Thank you for your answer, interesting because it takes a different approach to my original idea, I will study it. – Trevör Jan 2 '17 at 12:06\n\nDefine\n\n$$J = \\begin{bmatrix} 0 & 2 & 3\\\\ 0 & 0 & 2\\\\ 0 & 0 & 0 \\end{bmatrix}$$\n\nso that the problem is to compute $(I+J)^n$. The big, important things to note here are\n\n• $I$ and $J$ commute\n• $J^3 = 0$\n\nwhich enables the following powerful tricks: the first point lets us expand it with the binomial theorem, and the second point lets us truncate to the first few terms:\n\n$$(I+J)^n = \\sum_{k=0}^n \\binom{n}{k} I^{n-k} J^k = I + nJ + \\frac{n(n-1)}{2} J^2$$\n\nMore generally, for any function $f$ that is analytic at $1$, (such as any polynomial), if you extend it to matrices via Taylor expansion, then under the above conditions, its value at $I+J$ is given by\n\n$$f(I+J) = \\sum_{k=0}^\\infty f^{(k)}(1) \\frac{J^k}{k!} = f(1) I + f'(1) J + \\frac{1}{2} f''(1) J^2$$\n\nAs examples of things whose result you can check simply (so you can still use the method even if you're uncomfortable with it, because you can check the result), you can compute the inverse by\n\n$$(I+J)^{-1} = I - J + J^2 = \\begin{bmatrix} 1 & -2 & 1\\\\ 0 & 1 & -2\\\\ 0 & 0 & 1 \\end{bmatrix}$$\n\nand if you want a square root, you can get\n\n$$\\sqrt{I+J} = I + \\frac{1}{2} J - \\frac{1}{8} J^2 = \\begin{bmatrix} 1 & 1 & 1\\\\ 0 & 1 & 1\\\\ 0 & 0 & 1 \\end{bmatrix}$$\n\n(These are actually special cases of $(I+J)^n$ by the generalized binomial theorem for values of $n$ that aren't nonnegative integers)\n\n• Thank you so much for your answer, interesting because it takes a different approach than my original idea, I will study it. – Trevör Jan 2 '17 at 12:06\n\nLet $I=\\begin{pmatrix}1&0&0\\\\0&1&0\\\\0&0&1\\end{pmatrix}$, $A=\\begin{pmatrix}0&1&0\\\\0&0&1\\\\0&0&0\\end{pmatrix}$ and $B=\\begin{pmatrix}0&0&1\\\\0&0&0\\\\0&0&0\\end{pmatrix}$. Then $M=I+2A+3B$.\n\nYou can prove that for any $n$, $M^n$ can be written as $M^n=\\lambda_n I + a_n A + b_n B$ (because it's upper triangular and the symmetry along the ascending diagonal will remain).\n\nSo $M^{n+1}=M^nM=(\\lambda_n I + a_n A + b_n B) (I + 2A + 3B) = \\dots$\n\nUsing this, compute $\\lambda_n$, and then $a_n$ and finally $b_n$.\n\n• An essential point is that the matrices commute – Rene Schipperus Jan 2 '17 at 2:29\n• @ReneSchipperus : It's not really essential. The essential thing is that the vector space generated by the matrices is an algebra: $\\forall X,Y\\in Vect(I, A, B), XY \\in Vect(I, A, B)$ (which is equivalent to $\\forall X,Y\\in \\{I, A, B\\}, XY \\in Vect(I, A, B)$, because it's a vector space and you can use distributivity). In the general case, you would us the algebra of upper-triangular matrices, compute the coefficients on the diagonal, then right above the diagonal and keep going, and you'll only need the values already computed because the matrices are upper triangular. – xavierm02 Jan 2 '17 at 13:17\n\nHere is another variation based upon walks in graphs.\n\nWe interpret the matrix $A=(a_{i,j})_{1\\leq i,j\\leq 3}$ with \\begin{align*} A= \\begin{pmatrix} 1 & 2 & 3\\\\ \\color{grey}{0} & 1 & 2\\\\ \\color{grey}{0} & \\color{grey}{0} & 1 \\end{pmatrix} \\end{align*} as adjacency matrix of a graph with three nodes $P_1,P_2$ and $P_3$ and for each entry $a_{i,j}\\neq 0$ with a directed edge from $P_i$ to $P_j$ weighted with $a_{i,j}$.\n\nNote: When calculating the $n$-th power $A^n=\\left(a_{i,j}^{(n)}\\right)_{1\\leq i,j\\leq 3}$ we can interpret the element $a_{i,j}^{(n)}$ of $A^n$ as the number of (weighted) paths of length $n$ from $P_i$ to $P_j$. The entries of $A=(a_{i,j})_{1\\leq i,j\\leq 3}$ are the weighted paths of length $1$ from $P_i$ to $P_j$.\n\nSee e.g. chapter 1 of Topics in Algebraic Combinatorics by Richard P. Stanley.\n\nLet's look at the corresponding graph and check for walks of length $n$.\n\n• We see there are no directed edges from $P_2$ to $P_1$ and no directed edges from $P_3$ to $P_2$ and from $P_3$ to $P_1$ which implies there are no walks of length $n$ either. So, $A^n$ has due to the specific triangle structure of $A$ necessarily zeroes at the same locations as $A$. \\begin{align*} A^n= \\begin{pmatrix} . & . & .\\\\ \\color{grey}{0} & . & .\\\\ \\color{grey}{0} & \\color{grey}{0} & . \\end{pmatrix} \\end{align*}\n\n• It is also easy to consider the walks of length $n$ from $P_i$ to $P_i$. There is only one possibility to loop along the vertex weighted with $1$ from $P_i$ to $P_i$ and so the entries $a_{i,i}^{(n)}$ are \\begin{align*} 1\\cdot 1\\cdot 1\\cdots 1 = 1^n=1 \\end{align*} and we obtain \\begin{align*} A^n= \\begin{pmatrix} 1& . & .\\\\ \\color{grey}{0} & 1 & .\\\\ \\color{grey}{0} & \\color{grey}{0} & 1 \\end{pmatrix} \\end{align*}\n\nand now the more interesting part\n\n• $P_1$ to $P_2$:\n\nThe walks of length $n$ from $P_1$ to $P_2$ can start with zero or more loops at $P_1$ followed by a step (weigthed with $2$) from $P_1$ to $P_2$ and finally zero or more loops at $P_2$. All the loops are weighted with $1$. There are $n$ possibilities to walk this way \\begin{align*} a_{1,2}^{(n)}=2\\cdot 1^{n-1}+1\\cdot 2\\cdot 1^{n-2}+\\cdots +1^{n-2}\\cdot 2\\cdot 1+1^{n-1}\\cdot 2=2n \\end{align*}\n\n• $P_2$ to $P_3$:\n\nSymmetry is trump. When looking at the graph we observe the same situation as before from $P_1$ to $P_2$ and conclude\n\n\\begin{align*} a_{2,3}^{(n)}=2n \\end{align*}\n\n• $P_1$ to $P_3$:\n\nHere are two different types of walks of length $n$ possible. The first walk uses the weight $3$ edge from $P_1$ to $P_3$ as we did when walking from $P_1$ to $P_2$ along the weight $2$ edge. This part gives therefore \\begin{align*} 3\\cdot 1^{n-1}+1\\cdot 3\\cdot 1^{n-2}+\\cdots +1^{n-2}\\cdot 3\\cdot 1+1^{n-1}\\cdot 3=3n\\tag{1} \\end{align*} The other type of walk of length $n$ uses the hop via $P_2$. We observe it is some kind of concatenation of walks as considered before from $P_1$ to $P_2$ and from $P_2$ to $P_3$. In fact there are $\\binom{n}{2}$ possibilities to place two $2$'s in a walk of length $n$. All other steps are loops at $P_1,P_2$ and $P_3$ and we obtain \\begin{align*} \\binom{n}{2}\\cdot 2\\cdot 2=2n(n-1)\\tag{2} \\end{align*} Summing up (1) and (2) gives \\begin{align*} a_{1,3}^{(n)}=3n+2n(n-1)=n(2n+1) \\end{align*}\n\nand we finally obtain\n\n\\begin{align*} A^n=\\left(a_{i,j}^{(n)}\\right)_{1\\leq i,j\\leq 3}=\\begin{pmatrix} 1& 2n & n(2n+1)\\\\ \\color{grey}{0} & 1 & 2n\\\\ \\color{grey}{0} & \\color{grey}{0} & 1 \\end{pmatrix} \\end{align*}\n\nUsing symmetry, write the recurrence relation\n\n$$\\begin{pmatrix} a_{n+1} & b_{n+1}&c_{n+1}\\\\ 0 & a_{n+1} & b_{n+1}\\\\ 0 & 0 &a_{n+1} \\end{pmatrix}= \\begin{pmatrix} 1 & 2&3\\\\ 0 & 1 & 2\\\\ 0 & 0 &1 \\end{pmatrix} \\begin{pmatrix} a_{n} & b_{n}&c_{n}\\\\ 0 & a_{n} & b_{n}\\\\ 0 & 0 &a_{n} \\end{pmatrix}.$$\n\nThen $$\\begin{cases}a_{n+1}=a_n\\\\ b_{n+1}=b_n+2a_n\\\\ c_{n+1}=c_n+2b_n+3a_n.\\end{cases}$$\n\nWe solve with\n\n$$a_n=Cst=1,\\\\b_n=2(n-1)+Cst=2n,\\\\ c_n=4t_{n-1}+3(n-1)+Cst=2n^2+n,$$\n\nwhere $t_n$ denotes a triangular number, using the initial conditions $a_1=1,b_1=2,c_1=3$." ]
[ null ]
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https://downloads.hindawi.com/journals/ddns/2013/235012.xml
[ "DDNS Discrete Dynamics in Nature and Society 1607-887X 1026-0226 Hindawi Publishing Corporation 235012 10.1155/2013/235012 235012 Research Article Strong and Weak Convergence for Asymptotically Almost Negatively Associated Random Variables Shen Aiting Wu Ranchao Zhang Binggen School of Mathematical Science Anhui University Hefei 230039 China ahu.edu.cn 2013 4 2 2013 2013 10 12 2012 16 01 2013 2013 Copyright © 2013 Aiting Shen and Ranchao Wu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\nThe strong law of large numbers for sequences of asymptotically almost negatively associated (AANA, in short) random variables is obtained, which generalizes and improves the corresponding one of Bai and Cheng (2000) for independent and identically distributed random variables to the case of AANA random variables. In addition, the Feller-type weak law of large number for sequences of AANA random variables is obtained, which generalizes the corresponding one of Feller (1946) for independent and identically distributed random variables.\n\n1. Introduction\n\nMany useful linear statistics based on a random sample are weighted sums of independent and identically distributed random variables. Examples include least-squares estimators, nonparametric regression function estimators, and jackknife estimates,. In this respect, studies of strong laws for these weighted sums have demonstrated significant progress in probability theory with applications in mathematical statistics.\n\nLet {Xn,n1} be a sequence of random variables and let {ani,1in,n1} be an array of constants. A common expression for these linear statistics is Tn=i=1naniXi. Some recent results on the strong law for linear statistics Tn can be found in Cuzick , Bai et al. , Bai and Cheng , Cai , Wu , Sung , Zhou et al. , and Wang et al. . Our emphasis in this paper is focused on the result of Bai and Cheng . They gave the following theorem.\n\nTheorem A.\n\nSuppose that 1<α,β<, 1p<2, and 1/p=1/α+1/β. Let {X,Xn,n1} be a sequence of independent and identically distributed random variables satisfying EX=0, and let {ank,1kn,n1} be an array of real constants such that (1)limsupn(1nk=1n|ank|α)1/α<. If E|X|β<, then (2)limnn-1/pk=1nankXk=0a.s.\n\nWe point out that the independence assumption is not plausible in many statistical applications. So it is of interest to extend the concept of independence to the case of dependence. One of these dependence structures is asymptotically almost negatively associated, which was introduced by Chandra and Ghosal as follows.\n\nDefinition 1.\n\nA sequence {Xn,n1} of random variables is called asymptotically almost negatively associated (AANA, in short) if there exists a nonnegative sequence u(n)0 as n such that (3)Cov(f(Xn),g(Xn+1,Xn+2,,Xn+k))u(n)[Var(f(Xn))Var(g(Xn+1,Xn+2,,Xn+k))]1/2, for all n,k1 and for all coordinatewise nondecreasing continuous functions f and g whenever the variances exist.\n\nIt is easily seen that the family of AANA sequence contains negatively associated (NA, in short) sequences (with u(n)=0, n1) and some more sequences of random variables which are not much deviated from being negatively associated. An example of an AANA sequence which is not NA was constructed by Chandra and Ghosal . Hence, extending the limit properties of independent or NA random variables to the case of AANA random variables is highly desirable in the theory and application.\n\nSince the concept of AANA sequence was introduced by Chandra and Ghosal , many applications have been found. See, for example, Chandra and Ghosal derived the Kolmogorov type inequality and the strong law of large numbers of Marcinkiewicz-Zygmund; Chandra and Ghosal obtained the almost sure convergence of weighted averages; Wang et al. established the law of the iterated logarithm for product sums; Ko et al. studied the Hájek-Rényi type inequality; Yuan and An established some Rosenthal type inequalities for maximum partial sums of AANA sequence; Wang et al. obtained some strong growth rate and the integrability of supremum for the partial sums of AANA random variables; Wang et al. [15, 16] studied complete convergence for arrays of rowwise AANA random variables and weighted sums of arrays of rowwise AANA random variables, respectively; Hu et al. studied the strong convergence properties for AANA sequence; Yang et al. investigated the complete convergence, complete moment convergence, and the existence of the moment of supermum of normed partial sums for the moving average process for AANA sequence, and so forth.\n\nThe main purpose of this paper is to study the strong convergence for AANA random variables, which generalizes and improves the result of Theorem A. In addition, we will give the Feller-type weak law of large number for sequences of AANA random variables, which generalizes the corresponding one of Feller for independent and identically distributed random variables.\n\nThroughout this paper, let {Xn,n1} be a sequence of AANA random variables with the mixing coefficients {u(n),n1}. Sn=i=1nXi. For s>1, let ts/(s-1) be the dual number of s. The symbol C denotes a positive constant which may be different in various places. Let I(A) be the indicator function of the set A. an=O(bn) stands for anCbn.\n\nThe definition of stochastic domination will be used in the paper as follows.\n\nDefinition 2.\n\nA sequence {Xn,n1} of random variables is said to be stochastically dominated by a random variable X if there exists a positive constant C such that (4)P(|Xn|>x)CP(|X|>x), for all x0 and n1.\n\nOur main results are as follows.\n\nTheorem 3.\n\nSuppose that 0<α,β<, 0<p<2, and 1/p=1/α+1/β. Let {Xn,n1} be a sequence of AANA random variables, which is stochastically dominated by a random variable X and EXn=0, if β>1. Suppose that there exists a positive integer k such that n=1u1/(1-s)(n)< for some s(3·2k-1,4·2k-1] and s>1/(min{1/2,1/α,1/β,1/p-1/2}). Let {ani,i1,n1} be an array of real constants satisfying (5)i=1n|ani|α=O(n). If E|X|β<, then (6)limnn-1/pmax1jn|i=1janiXi|=0a.s.\n\nRemark 4.\n\nTheorem 3 generalizes and improves Theorem A of Bai and Cheng for independent and identically distributed random variables to the case of AANA random variables, since Theorem 3 removes the identically distributed condition and expands the ranges α, β, and p, respectively.\n\nAt last, we will present the Feller-type weak law of large number for sequences of AANA random variables, which generalizes the corresponding one of Feller for independent and identically distributed random variables.\n\nTheorem 5.\n\nLet α>1/2 and {X,Xn,n1} be a sequence of identically distributed AANA random variables with the mixing coefficients {u(n),n1} satisfying n=1u2(n)<. If (7)limnnP(|X|>nα)=0, then (8)Snnα-n1-αEXI(|X|nα)P0.\n\n2. Preparations\n\nTo prove the main results of the paper, we need the following lemmas. The first two lemmas were provided by Yuan and An .\n\nLemma 6 (cf. see [<xref ref-type=\"bibr\" rid=\"B17\">13</xref>, Lemma 2.1]).\n\nLet {Xn,n1} be a sequence of AANA random variables with mixing coefficients {u(n),n1}, f1,f2, be all nondecreasing (or all nonincreasing) continuous functions, then {fn(Xn),n1} is still a sequence of AANA random variables with mixing coefficients {u(n),n1}.\n\nLemma 7 (cf. see [<xref ref-type=\"bibr\" rid=\"B17\">13</xref>, Theorem 2.1]).\n\nLet p>1 and {Xn,n1} be a sequence of zero mean random variables with mixing coefficients {u(n),n1}.\n\nIf n=1u2(n)<, then there exists a positive constant Cp depending only on p such that for all n1 and 1<p2, (9)E(max1jn|i=1jXi|p)Cpi=1nE|Xi|p.\n\nIf n=1u1/(p-1)(n)< for some p(3·2k-1,4·2k-1], where integer number k1, then there exists a positive constant Dp depending only on p such that for all n1, (10)E(max1jn|i=1jXi|p)Dp{i=1nE|Xi|p+(i=1nEXi2)p/2}.\n\nThe last one is a fundamental property for stochastic domination. The proof is standard, so the details are omitted.\n\nLemma 8.\n\nLet {Xn,n1} be a sequence of random variables, which is stochastically dominated by a random variable X. Then for any α>0 and b>0, (11)E|Xn|αI(|Xn|b)C1[E|X|αI(|X|b)+bαP(|X|>b)],E|Xn|αI(|Xn|>b)C2E|X|αI(|X|>b), where C1 and C2 are positive constants.\n\n3. Proofs of the Main Results Proof of Theorem <xref ref-type=\"statement\" rid=\"thm1.1\">3</xref>.\n\nWithout loss of generality, we assume that ani0 (otherwise, we use ani+ and ani- instead of ani, and note that ani=ani+-ani-). Denote for 1in and n1 that (12)Yi=-n1/βI(Xi<-n1/β)+XiI(|Xi|n1/β)+n1/βI(Xi>n1/β),Zi=(Xi+n1/β)I(Xi<-n1/β)+(Xi-n1/β)I(Xi>n1/β). Hence, Xi=Yi+Zi, which implies that (13)n-1/pmax1jn|i=1janiXi|n-1/pmax1jn|i=1janiZi|+n-1/pmax1jn|i=1janiYi|n-1/pmax1jn|i=1janiZi|+n-1/pmax1jn|i=1janiEYi|+n-1/pmax1jn|i=1jani(Yi-EYi)|H+I+J. To prove (6), it suffices to show that H0  a.s., I0 and J0  a.s. as n.\n\nFirstly, we will show that H0  a.s.\n\nFor any 0<γα, it follows from (5) and Hölder's inequality that (14)i=1n|ani|γ(i=1n|ani|α)γ/α×(i=1n1)1-γ/αCn, for any 0<αγ, it follows from (5) again that (15)i=1n|ani|γ(i=1n|ani|α)γ/αCnγ/α. Combining (14) and (15), we have (16)i=1n|ani|γCnmax(1,γ/α). The condition E|X|β< yields that (17)n=1P(Zn0)=n=1P(|Xn|>n1/β)Cn=1P(|X|>n1/β)CE|X|β<, which implies that P(Zn0,i.o.)=0 by Borel-Cantelli lemma. Thus, we have by (5) that (18)Hn-1/pmax1jn|i=1janiZi|n-1/pi=1n|aniZi|Cn-1/p(max1in|ani|α)1/αi=1n|Zi|Cn-1/p(i=1n|ani|α)1/αi=1n|Zi|Cn-1/βi=1n|Zi|0  a.s.,as  n.\n\nSecondly, we will prove that (19)In-1/pmax1jn|i=1janiEYi|0,as  n. If 0<β1, then we have by Lemma 8 and (16) that (20)In-1/pi=1n|aniEYi|n-1/pi=1n|ani|[E|Xi|I(|Xi|n1/β)+n1/βP(|Xi|>n1/β)]Cn-1/pi=1n|ani|[E|X|I(|X|n1/β)+n1/βP(|X|>n1/β)]Cn-1/pi=1n|ani|[n(1-β)/βE|X|βI(|X|n1/β)+n1/β-1E|X|βI(|X|>n1/β)]=Cn-1/α-1E|X|βi=1n|ani|Cn-1/α-1+max(1,1/α)0,as  n. If β>1, then we have by EXn=0, Lemma 8 and (16) that (21)In-1/pi=1n|aniEYi|Cn-1/pi=1n|ani|  ×[E|Xi|I(|Xi|>n1/β)+n1/βP(|Xi|>n1/β)]Cn-1/pi=1n|ani|E|X|I(|X|>n1/β)Cn-1/pi=1n|ani|n1/β-1E|X|βI(|X|>n1/β)Cn-1/α-1+max(1,1/α)0,as  n. Hence, (19) follows from (20) and (21) immediately.\n\nTo prove (6), it suffices to show that (22)Hn-1/pmax1jn|i=1jani(Yi-EYi)|0  a.s.,as  n. By Borel-Cantelli Lemma, we only need to show that for any ε>0, (23)n=1P(max1jn|i=1jani(Yi-EYi)|>εn1/p)<. For fixed n1, it is easily seen that {ani(Yi-EYi),1in} are still AANA random variables by Lemma 6. Taking s>1/min{1/2,1/α,1/β,1/p-1/2}>2, we have by Markov's inequality and Lemma 7 that (24)n=1P(max1jn|i=1jani(Yi-EYi)|>εn1/p)Cn=1n-s/pE(max1jn|i=1jani(Yi-EYi)|s)Cn=1n-s/pi=1nE|ani(Yi-EYi)|s+Cn=1n-s/p(i=1nE|ani(Yi-EYi)|2)s/2J1+J2. For J1, we have by Cr inequality, Jensen's inequality, (15), and Lemma 8 that (25)J1Cn=1n-s/pi=1n|ani|sE|Yi|sCn=1n-s/pi=1n|ani|s×[E|Xi|sI(|Xi|n1/β)+ns/βP(|Xi|>n1/β)]Cn=1n-s/pi=1n|ani|s×[E|X|sI(|X|n1/β)+ns/βP(|X|>n1/β)]Cn=1n-s/βE|X|sI(|X|n1/β)+Cn=1P(|X|>n1/β)Cn=1n-s/βi=1nE|X|sI×((i-1)1/β<|X|i1/β)+CE|X|βCi=1E|X|sI((i-1)1/β<|X|  i1/β)×n=in-s/β+CE|X|βCi=1i(s-β)/βE|X|βI×((i-1)1/β<|X|i1/β)i-s/β+1+CE|X|βCE|X|β<. Next, we will prove that J2<. By Cr inequality, Jensen's inequality and Lemma 8 again, we can see that (26)i=1nE|ani(Yi-EYi)|2i=1nani2EYi2Ci=1nani2[EXi2I(|Xi|n1/β)+n2/βP(|Xi|>n1/β)]Ci=1nani2[EX2I(|X|n1/β)+n2/βP(|X|>n1/β)]Cnmax(1,2/α)×[EX2I(|X|n1/β)+n2/βP(|X|>n1/β)]. It follows by Markov's inequality and the fact E|X|β< that (27)EX2I(|X|n1/β)+n2/βP(|X|>n1/β){n(2-β)/βE|X|βI(|X|n1/β)+  n-1+2/βE|X|β(|X|>n1/β),β<2EX2I(|X|n1/β)+EX2,β2,{Cn-1+2/βE|X|β,β<2,CEX2,β2. If we denote δ=max{-1+2/p,2/β,2/α,1}, then we can get by (26) and (27) that (28)i=1nE|ani(Yi-EYi)|2Cnδ. It is easily seen that (29)(-1p+δ2)s=max{-12,-1α,-1β,-1p+12}s=-min{12,1α,1β,1p-12}s<-1. Hence, we have by (28) and (29) that (30)J2Cn=1n(-1/p+δ/2)s<, which together with J1< yields (23). This completes the proof of the theorem.\n\nProof of Theorem <xref ref-type=\"statement\" rid=\"thm1.2\">5</xref>.\n\nDenote for 1in and n1 that (31)Yni=-nαI(Xi<-nα)+XiI×(|Xi|nα)+nαI(Xi>nα) and Tn=i=1nYni. By the assumption (7), we have for any ε>0 that (32)P(|Snnα-Tnnα|>ε)P(SnTn)P(i=1n(XiYni))i=1nP(|Xi|>nα)=nP(|X|>nα)0,n, which implies that (33)Snnα-TnnαP0. Hence, in order to prove (8), we only need to show that (34)Tnnα-ETnnαP0. By (7) again and Toeplitz's lemma, we can get that (35)k=1nk2α-2·kP(|X|>kα)k=1nk2α-20,n. Note that (36)k=1nk2α-2n2α-1,for  α>12. Combing (35) and (36), we have (37)n-2α+1k=1nk2α-1P(|X|>kα)0,n. By Lemma 7 (taking p=2), (7), and (37), we can get that (38)P(|Tn-ETn|>εnα)Cn-2αE|Tn-ETn|2Cn-2αi=1nEYni2Cn-2α+1[EX2I(|X|nα)+n2αP(|X|>nα)]=Cn-2α+1EX2I×(|X|nα)+CnP(|X|>nα)=Cn-2α+1k=1nEX2I×((k-1)α|X|kα)+CnP(|X|>nα)Cn-2α+1k=1nk2α×[P(|X|>(k-1)α)-P(|X|>kα)]+CnP(|X|>nα)=Cn-2α+1[k=1n-1((k+1)2α-k2α)P(|X|>kα)+P(|X|>0)-n2αP(|X|>nα)]+CnP(|X|>nα)Cn-2α+1[k=1nk2α-1P(|X|>kα)+1]+CnP(|X|>nα)0,n. This completes the proof of the theorem.\n\nAcknowledgments\n\nThe authors are most grateful to the Editor Binggen Zhang and anonymous referee for the careful reading of the paper and valuable suggestions which helped in improving an earlier version of this paper. This work was supported by the National Natural Science Foundation of China (11201001, 11171001, and 11126176), the Specialized Research Fund for the Doctoral Program of Higher Education of China (20093401120001), the Natural Science Foundation of Anhui Province (11040606M12, 1208085QA03), the Natural Science Foundation of Anhui Education Bureau (KJ2010A035), the 211 project of Anhui University, the Academic Innovation Team of Anhui University (KJTD001B), and the Students Science Research Training Program of Anhui University (KYXL2012007).\n\nCuzick J. A strong law for weighted sums of i.i.d. random variables Journal of Theoretical Probability 1995 8 3 625 641 10.1007/BF02218047 MR1340830 ZBL0833.60031 Bai Z. Cheng P. E. Zhang C.-H. An extension of the Hardy-Littlewood strong law Statistica Sinica 1997 7 4 923 928 MR1488650 ZBL1067.60501 Bai Z. Cheng P. Marcinkiewicz strong laws for linear statistics Statistics & Probability Letters 2000 46 2 105 112 Cai G.-H. Marcinkiewicz strong laws for linear statistics of ρ-mixing sequences of random variables Anais da Academia Brasileira de Ciências 2006 78 4 615 621 10.1590/S0001-37652006000400001 MR2302908 Wu Q. A strong limit theorem for weighted sums of sequences of negatively dependent random variables Journal of Inequalities and Applications 2010 2010 8 383805 10.1155/2010/383805 MR2678912 ZBL1202.60044 Sung S. H. On the strong convergence for weighted sums of random variables Statistical Papers 2011 52 2 447 454 10.1007/s00362-009-0241-9 MR2795891 ZBL1247.60041 Zhou X.-C. Tan C.-C. Lin J.-G. On the strong laws for weighted sums of ρ-mixing random variables Journal of Inequalities and Applications 2011 2011 8 157816 10.1155/2011/157816 MR2775040 Wang X. Li X. Yang W. Hu S. On complete convergence for arrays of rowwise weakly dependent random variables Applied Mathematics Letters 2012 25 11 1916 1920 10.1016/j.aml.2012.02.069 MR2957779 ZBL1251.60025 Chandra T. K. Ghosal S. Extensions of the strong law of large numbers of Marcinkiewicz and Zygmund for dependent variables Acta Mathematica Hungarica 1996 71 4 327 336 10.1007/BF00114421 MR1397560 ZBL0853.60032 Chandra T. K. Ghosal S. The strong law of large numbers for weighted averages under dependence assumptions Journal of Theoretical Probability 1996 9 3 797 809 10.1007/BF02214087 MR1400599 ZBL0857.60021 Wang Y. Yan J. Cheng F. Su C. The strong law of large numbers and the law of the iterated logarithm for product sums of NA and AANA random variables Southeast Asian Bulletin of Mathematics 2003 27 2 369 384 MR2046474 ZBL1061.60031 Ko M.-H. Kim T.-S. Lin Z. The Hájeck-Rènyi inequality for the AANA random variables and its applications Taiwanese Journal of Mathematics 2005 9 1 111 122 MR2122907 Yuan D. An J. Rosenthal type inequalities for asymptotically almost negatively associated random variables and applications Science in China. Series A 2009 52 9 1887 1904 10.1007/s11425-009-0154-z MR2544995 ZBL1184.62099 Wang X. Hu S. Yang W. Convergence properties for asymptotically almost negatively associated sequence Discrete Dynamics in Nature and Society 2010 2010 15 218380 10.1155/2010/218380 MR2770590 ZBL1207.60025 Wang X. Hu S. Yang W. Complete convergence for arrays of rowwise asymptotically almost negatively associated random variables Discrete Dynamics in Nature and Society 2011 2011 11 717126 10.1155/2011/717126 MR2846463 ZBL1235.60026 Wang X. Hu S. Yang W. Wang X. On complete convergence of weighted sums for arrays of rowwise asymptotically almost negatively associated random variables Abstract and Applied Analysis 2012 15 315138 MR2947673 Hu X. P. Fang G. H. Zhu D. J. Strong convergence properties for asymptotically almost negatively associated sequence Discrete Dynamics in Nature and Society 2012 2012 8 562838 10.1155/2012/562838 Yang W. Wang X. Ling N. Hu S. On complete convergence of moving average process for AANA sequence Discrete Dynamics in Nature and Society 2012 2012 24 863931 MR2923400 ZBL1247.60043 Feller W. A limit theorem for random variables with infinite moments American Journal of Mathematics 1946 68 2 257 262" ]
[ null ]
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https://engineering.dynatrace.com/open-source/standards/dynahist/
[ "", null, "# DynaHist\n\nThis Java library contains histogram implementations with configurable bin layouts specifically designed for fast value recording. At its base, there are three different implementations:\n\n• The static histogram enables an allocation-free recording of values, as the internal bin count array is already fully occupied during construction.\n• The dynamic histogram is memory-efficient as it resizes the internal bin count array on demand. Furthermore, it dynamically adjusts the number of bits used for each counter.\n• The preprocessed histogram is an immutable implementation which contains the cumulative bin counts. In this way sublinear queries for order statistics are possible through binary search. If many of those queries are performed subsequently, it is recommended to convert to a preprocessed histogram first.\n\nThe library ships with predefined bin layout implementations:\n\n• LogLinearLayout allows to specify absolute and relative bin width limits, where one of them must be satisfied over a given value range. In this way the error of recorded values can be controlled. While an optimal mapping would involve a logarithm evaluation, LogLinearLayout uses a piecewise linear mapping function instead, which results in up to 44% more bins and therefore in a correspondingly larger memory footprint.\n• LogQuadraticLayout uses a piecewise quadratic approximation to the optimal mapping. It is the slightly slower than LogLinearLayout, but reduces the space overhead to about 8% compared to the ideal mapping.\n• CustomLayout allows to set the bin boundaries individually. It can be used to map a histogram, which was recorded with some fine-grained bin layout, to a coarser custom bin layout with well-defined bins. For example, this can be useful as a preparatory step for creating a visualization of the histogram.\n\nBasic Functionality\n\n```// Defining a bin layout\nLayout layout = LogQuadraticLayout.create(1e-5, 1e-2, -1e9, 1e9); // use bins with maximum\n// absolute width of 1e-5\n// or relative width of 1%\n// to cover the range [-1e9, 1e9/>\n// Creating a dynamic histogram\nHistogram histogram = Histogram.createDynamic(layout);\n\n// Adding values to the histogram\nhistogram.addAscendingSequence(i -> i + 1, 1000000000); // adds the first billion positive integers\n\n// Querying the histogram\nhistogram.getTotalCount();\nhistogram.getMin();\nhistogram.getMax();\nhistogram.getValue(1); // returns an estimate of the 2nd smallest value\nhistogram.getValue(3, ValueEstimator.UPPER_BOUND); // returns an upper bound of the 4th smallest value\nhistogram.getQuantile(0.5); // returns an estimate of the sample median\nhistogram.getQuantile(0.5, ValueEstimator.LOWER_BOUND); // returns a lower bound of the sample median\n\n// Merging histograms\n\n// Serialization\nhistogram.write(dataOutput); // write histogram to a java.io.DataOutput\n\nGetting Started\n\nDynaHist is available as Maven package on JCenter and should be used via Maven, Gradle or Ivy. If automatic dependency management is not possible obtain the jar file from GitHub Releases.\n\nHistory\n\nAt Dynatrace, we were looking for a data sketch with a fast update time, which can also answer order statistic queries with error guarantees. As an example, such a data structure should be able to provide the 99th percentile with a maximum relative error of 1%. Other data structures like t-digest do not have strict error limits. In our search, we finally came across HdrHistogram, a histogram implementation that intelligently selects bin boundaries so that the relative error is limited over a range of many orders of magnitude. The core of HdrHistogram is a fast mapping of values to bin indices by bit twiddling, which reduces the recording time to less than 10ns. Although we loved this idea, this data structure did not quite meet our requirements for several reasons:\n\n• The original HdrHistogram was designed for recording integer values. Usually we are dealing with floating point values. The wrapper class for double values, which is shipped with HdrHistogram, introduces an indirection, which slows down the recording.\n• Another disadvantage is that HdrHistogram does not give you full control over the error specification. It is only possible to define the number of significant digits corresponding to relative errors of 10%, 1%, 0.1%, etc. It is not possible to select a relative error of 5%. You must fall back on 1%, which unnecessarily increases the number of bins and wastes memory space.\n• HdrHistogram has no support for negative values. You have to use two histograms, one for the positive and one for the negative value range.\n• With HdrHistogram it is not possible to define the maximum error for values that are between zero and the range where the relative error limit applies.\n• The mapping of values to bin indices is fast, but not optimal. The mapping used by HdrHistogram requires about 40% more bins than necessary to satisfy the specified relative error. In 2015 we have proposed a better and similarly fast mapping for HdrHistogram (see https://github.com/HdrHistogram/HdrHistogram/issues/54) with less than 10% space overhead. However, as this would have resulted in an incompatible change, the author of HdrHistogram decided not to pursue our idea any further.\n\nTherefore, we started developing our own histogram data sketch which uses the proposed better mapping and which also solves all the mentioned issues. After many years of successful application and the emergence of an open source initiative at Dynatrace, we decided to publish this data structure as a separate library here on GitHub.\n\nBenchmarks\n\nFor our benchmarks we used random values drawn from a reciprocal distribution (log-uniform distribution) with a minimum value of 1000 and a maximum value of 1e12. In order not to distort the test results, we have generated 1M random numbers in advance and kept them in main memory. For the comparison with HdrHistogram we used the DoubleHistogram with highestToLowestValueRatio=1e9 and numberOfSignificantValueDigits=2. To record values with equivalent precision we used an absolute bin width limit of 10 and a relative bin width limit of 1% over the range [0, 1e12/>. The corresponding layouts LogLinearLayout(10, 0.01, 0, 1e12) and LogQuadraticLayout(10, 0.01, 0, 1e12) have been combined with the static and dynamic implementations of DynaHist resulting in 4 different cases.\n\nThe recording speed was measured using JMH on a Dell Precision 5530 Notebook with an Intel Core i9-8950HK CPU. We measured the average time to insert the 1M random values into an empty histogram data structure, from which we derived the average time for recording a single value. All four investigated DynaHist variants outperform HdrHistogram's DoubleHistogram significantly. The static histogram implementation with the LogLinearLayoutwas the fastest one and more than 35% faster than HdrHistogram.", null, "The memory usage of the histogram data structures was analyzed after adding 1M random values as in the speed benchmark before. Again due to the better bin layout DynaHist significantly outperforms HdrHistogram. Especially the dynamic histogram implementation together with LogQuadraticLayout requires just 15% of the memory space HdrHistogram takes.", null, "Similarly, the serialization, which is more or less a memory snapshot of the dynamic histogram implementation, is much more compact than that of HdrHistogram.", null, "The space advantage is maintained even with compression. The reason is that DynaHist requires much fewer bins to guarantee the same relative error and therefore less information has to be stored.", null, "Bin Layouts\n\nA Layout specifies the bins of a histogram. The regular bins of a layout span a certain value range. In addition, DynaHist uses an underflow and an overflow bin to count the number of values which are below or beyond that value range, respectively.\n\nDynaHist comes with two Layout implementations LogLinearLayout and LogQuadraticLayout which can be configured using an absolute bin width limit a and a relative bin width limit r. If b(i) denotes the bin boundary between the (i-1)-th and the i-th bin, the i-th bin covers the interval [b(i), b(i+1)/>. Then the absolute bin width limit can be expressed as\n\n```|b(i+1) - b(i)| <= a (1)\n```\n\nand the relative bin width limit corresponds to\n\n```|b(i+1) - b(i)| / max(|b(i)|, |b(i+1)|) <= r. (2)\n```\n\nIf a bin satisfies either (1) or (2), any point of [b(i), b(i+1)/> approximates a recorded value mapped to the i-th bin with either a maximum absolute error of a or a maximum relative error of r. In particular, if the midpoint of the interval (b(i) + b(i+1)) / 2 is chosen as estimate of the recorded value, the error is even limited by the absolute error bound a/2 or the relative error bound r/2, respectively.\n\nThe bin boundaries of a layout, for which either (1) or (2) holds for all its bins, must satisfy\n\n```b(i+1) <= b(i) + max(a, r * b(i))\n```\n\nin the positive value range. For simplicity, we focus on the positive value range. However, similar considerations can be made for the negative range. Obviously, an optimal mapping, that minimizes the number of bins and therefore the memory footprint, would have\n\n```b(i+1) = b(i) + max(a, r * b(i)).\n```\n\nWe call bins close to zero and having a width of a subnormal bins. Normal bins are those representing an interval with larger values, which have a width defined by the relative bin width limit. The following figure shows an example of such a bin layout.", null, "In this example, the bins are enumerated with integers from -9 to 8. The first and last bin indices address the underflow and the overflow bin, respectively. The widths of normal bins correspond to a geometric sequence. For an optimal bin layout\n\n```b(i) = c * (1 + r)^i\n```\n\nmust hold for all bin boundaries in the normal range with some constant c. Values of the normal range can be mapped to its corresponding bin index by the following mapping function\n\n```f(v) := floor(g(v)) with g(v) := (log_2(v) - log_2(c)) / log_2(r).\n```\n\nWhile log_2(r) and log_2(c) can be precomputed, the calculation of log_2(v) remains which is an expensive operation on CPUs. Therefore, DynaHist defines non-optimal mappings that trade space efficiency for less computational costs. Obviously, if g(v) is replaced by any other function that is steeper over the whole value range, the error limits will be maintained. Therefore, DynaHist uses linear (LogLinearLayout) or quadratic (LogQuadraticLayout) piecewise functions instead. Each piece spans a range between powers of 2 [2^k, 2^(k+1)/>. The coefficients of the polynomials are chosen such that the resulting piecewise function is continuous and the slope is always greater than or equal to that of g(v). The theoretical space overhead of LogLinearLayout is about 44% and that of LogQuadraticLayout is about 8% compared to an optimal mapping." ]
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https://123dok.net/document/qvl67evl-deep-subject-study-discrete-subgroups-compact-complex-manifold.html
[ "• Aucun résultat trouvé\n\n# It is a deep subject to study the discrete subgroups Γ ⊂Aut (Ω) such that Ω/Γ is a compact complex manifold\n\nN/A\nN/A\nProtected\n\nPartager \"It is a deep subject to study the discrete subgroups Γ ⊂Aut (Ω) such that Ω/Γ is a compact complex manifold\"\n\nCopied!\n14\n0\n0\nEn savoir plus ( Page)\n\nTexte intégral\n\n(1)\n\nON BOUNDARY ACCUMULATION POINTS OF A CONVEX DOMAIN IN Cn∗\n\nLINA LEE, BRADLEY THOMAS, AND BUN WONG§\n\nAbstract. In this paper we show that, for a smoothly bounded convex domain ΩCn, if there isj} ⊂Aut (Ω) such thatφj(z) converges to some boundary point non-tangentially for allzΩ, then there does not exist a non-trivial analytic disc on∂Ω through any boundary orbit accumulation points.\n\nKey words. Automorphism Group, Convex Domains, Invariant Metrics/Measures.\n\nAMS subject classifications.32F18, 32F45.\n\n1. Introduction. The study of biholomorphic automorphism groups, Aut (Ω), of a domain Ω⊂Cn is of major interest in various areas of research. The existence of an automorhpism reflects certain symmetry of the domain. It is a deep subject to study the discrete subgroups Γ ⊂Aut (Ω) such that Ω/Γ is a compact complex manifold. Although the construction of a cocompact lattice Γ in Aut (Ω) is usually not straightforward, it is comparably easier to find a divergent sequence{φj} ⊂Aut (Ω).\n\nLetpbe any point in Ω such that{φj(p)}converge to a boundary pointq∈∂Ω. If we further assume∂Ω is smooth, our knowledge of the biholomorphic invariants (i.e., Chern-Moser invariants, invariant K¨ahler metrics, intrinsic metrics/measures etc.) allows us to draw many interesting conclusions. For instance, if q ∈∂Ω is strongly pseudoconvex, the method in can be used to show that Ω must be biholomorphic to the Euclidean ball.\n\nIn order to charaterize those smoothly bounded domains with non-compact au- tomorhpism group, it is important to have a better understanding of the orbit ac- cumulation points on the boundary. There has been recently a lot of research in this direction. One of the important conjectures in this regard is due to Greene and Krantz, which can be stated as follows.\n\nConjecture. Let Ωbe a smoothly bounded domain in Cn. Suppose there exists {φj} ⊂Aut(Ω) such that {φj(p)} accumulates at a boundary point q∈∂Ω for some p∈Ω. Then∂Ωis of finite type atq.\n\nIn this paper we will prove the following result in support of the Greene/Krantz conjecture.\n\nTheorem. Let Ω be a smoothly bounded convex domain in Cn. Suppose that there is a sequence{φj} ⊂Aut(Ω)such that{φj(p)}accumulates non-tangentially at some boundary point for allp∈Ω. Then, there does not exist a non-trivial analytic disc on∂Ωpassing through any orbit accumulation point on the boundary.\n\nIn this result was proved in C2 under a more general assumption that Ω is pseudoconvex. Earlier work in the convex setting inC2 was discussed in [5, 10]. For\n\nReceived November 9, 2012; accepted for publication September 17, 2013.\n\nDepartment of Mathematics, University of California, Riverside, CA 92521, USA (linalee@math.\n\nucr.edu).\n\nDepartment of Natural and Mathematical Sciences, California Baptist University, Riverside, CA 92504, USA ([email protected]).\n\n§Department of Mathematics, University of California, Riverside, CA 92521, USA (wong@math.\n\nucr.edu).\n\n427\n\n(2)\n\nthe pseudoconvex case, it is a highly non-trivial matter to generalize this result to higher dimensions since the geometry of the boundary of a pseudoconvex domain in Cn, n >2, is not as well understood as inC2. To overcome the technical difficulties generalizing the result in [2, 5, 9], we use the intrinsic measures defined with respect to U =Bn−k×∆k, 0≤k ≤n, where Bn−k is the unit ball in Cn−k and ∆k is the unit polydisc inCk. We will prove that the orbit accumulation set on the boundary is actually biholomorphic to a euclidean ball, if it is not a point. This fact allows us to remove the obstacle of finding a higher dimensional analogue of the argument used in forC2, which depends heavily on the classical result that a hyperbolic Riemann surface is covered by the unit disc.\n\nA substantial portion of this paper can be found in ; this portion is a joint work of Lina Lee, Bradley Thomas, and Bun Wong.\n\n2. Invariant metrics and invariant measures. Let H(A, B) be the set of holomorphic mappings fromAtoB and ∆ be the unit disc inC. The Kobayashi and Carath´eodory metrics are defined as follows.\n\nDefinition 1. The Kobayashi and Carath´eodory metrics on Ω⊂Cn atp∈Ω in the directionξ∈Cn , denoted asFK(p, ξ) andFC(p, ξ), respectively, are defined as follows:\n\nFK(p, ξ) = inf 1\n\nα :∃φ∈H(∆,Ω) s.t.φ(0) =p, φ(0) =αξ (1)\n\nFC(p, ξ) = sup\n\nn\n\nX\n\nj=1\n\n∂f(p)\n\n∂zj\n\nξj\n\n:∃f ∈H(Ω,∆),s.t. f(p) = 0\n\n . (2)\n\nIfz, w ∈Ω, then the Kobayashi and Carath´eodory pseudo-distance on Ω betweenz andw, denoted asdK(z, w) anddC(z, w), respectively, are given by\n\ndK(z, w) = inf\n\nγ\n\nZ 1 0\n\nFK(γ(t), γ(t))dt, (3)\n\ndC(z, w) = sup\n\nf\n\nρ(f(z), f(w)) (4)\n\nwhereγ : [0,1]−→Ω is a piecewiseC1 curve connectingz and w andρ(p, q) is the Poincar´e distance on ∆ between p, q ∈ ∆. The supremum in (4) is taken over all holomorphic mappingsf : Ω−→∆.\n\nKobayashi originally defined the pseudo-distance on Ω using a chain of analytic discs as follows: for two given points z, w ∈Ω, consider a chain of analytic discs α that consists of z1, z2, . . . , zn ∈ Ω, analytic discs fi : ∆ −→ Ω, and n+ 1 pairs of pointsa0, b0, a1, b1, . . . , an, bn∈∆ such that, for 0≤j≤n,\n\nfj(aj) =zj, fj(bj) =zj+1, andz0=z, zn+1=w.\n\nWe define the length of the chainαas ℓ(α) =\n\nn\n\nX\n\nj=0\n\nρ(aj, bj).\n\nThen the Kobayashi pseudo-distance between two points z, wis given as\n\n(5) dK(z, w) = inf\n\nα ℓ(α).\n\n(3)\n\nIt was Royden who later proved that the definition given by (5) is equivalent to (3).\n\nThe metrics and distances given above are invariant under biholomorphic map- pings since they satisfy the non-increasing property under holomorphic mappings, i.e., if Φ : Ω1 −→ Ω2 is a holomorphic mapping between domains in Cn and Cm, respectively, andp, q∈Ω1,ξ∈Cn, then we have\n\nF1(p, ξ)≥F2(Φ (p),Φ(p)ξ), (6)\n\nd1(p, q)≥d2(Φ (p),Φ (q)), (7)\n\nwhere the metricF in (6) denotes either the Kobayashi or Carath´eodory metric and the distancedin (7) is either the Kobayashi or Carath´eodory distance.\n\nWe extend the definition of the metrics and define the Kobayashi and Carath´eodory measures. Let Bk denote the complex k-dimensional unit ball and\n\nk the complexk-dimensional unit polydisc.\n\nDefinition 2. Let Ω⊂Cnbe a domain,p∈Ω, andξ1, . . . , ξm∈TpCΩ, 1≤m≤ n, be linearly independent vectors on the complex tangent space to Ω atp. One can find an (m, m) volume form M on Ω such that M ξ1, . . . , ξm, ξ1, . . . , ξm\n\n= 1. Let U =Bm−j×∆j, 0≤j≤m, andµm=Qm\n\nj=1\n\ni\n\n2dzj∧dzj\n\n. We define the Kobayashi and Carath´eodorym-measures with respect to U as follows:\n\nKU(p;ξ1, . . . ξm) = inf (1\n\nα :∃Φ∈H(U,Ω),s.t. Φ (0) =p, Φ(0)M =αµm,for some α >0\n\n) , CU(p;ξ1, . . . , ξm) = supn\n\nβ :∃Φ∈H(Ω, U),s.t. Φ (p) = 0, Φ(p)µm=βM, β >0o\n\n.\n\nThe Kobayashi and Carath´eodory measures satisfiy the non-increasing property under holomorphic mappings.\n\nProposition 1. Let Ω1 ⊂ Cn, Ω2 ⊂ Cn be domains and U = Bm−j ×∆j, 0≤j ≤m,m≤min{n, n}. Letp∈Ω1j ∈TpC1,j= 1, . . . , m, andξj’s be linearly independent. If φ∈H(Ω1,Ω2) is such thatφ(p)ξj’s are linearly independent, then\n\nKU1(p;ξ1, . . . ξm)≥KU2(φ(p) ;φ(p)ξ1, . . . φ(p)ξm), and CU1(p;ξ1, . . . , ξm)≥CU2(φ(p) ;φ(p)ξ1, . . . φ(p)ξm).\n\nProof. Let M be an (m, m) volume form on Ω1 such that M ξ1, . . . , ξm, ξ1, . . . , ξm\n\n= 1. Let Φ : U −→ Ω1 be a holomorphic mapping such that Φ (0) =p, Φ(0)M =αµm. Considerh=φ◦Φ :U −→Ω2. LetM be an (m, m) volume form on Ω2such that φ(p)M=M. Thenh(0) =φ(p) and\n\nh(0)M= Φ(0) (φ(p)M) = Φ(0) (M) =αµm.\n\nHence 1/α≥KU2(φ(p), M) and inf 1/α≥KU2(φ(p), M). One can show the second inequality in a similar way.\n\n(4)\n\nCorollary 1. Let Ω1,Ω2 ⊂Cn be domains andU =Bm−j×∆j,0 ≤j ≤m, m≤ n. Let p∈ Ω1, ξj ∈TpC1, j = 1, . . . , m, and ξj’s be linearly independent. If φ: Ω1−→Ω2 is a biholomorphism, we have\n\nKU1(p;ξ1, . . . ξm) =KU2(φ(p) ;φ(p)ξ1, . . . φ(p)ξm), and CU1(p;ξ1, . . . , ξm) =CU2(φ(p) ;φ(p)ξ1, . . . φ(p)ξm).\n\nProof. Proposition 1 holds forφandφ−1. Therefore we have inequalities in both directions.\n\nCorollary 2. Let U =Bm−j×∆j, p∈U, ξj ∈TpCU,1 ≤j ≤m, and ξj’s be linearly independent vectors. We have KUU(p;ξ1, . . . ξm) = CUU(p;ξ1, . . . , ξm) for all p∈U.\n\nProof. Since the automorphism group onU is transitive, we may assume p= 0.\n\nAlso we may assumeµm ξ1, . . . , ξm, ξ1, . . . , ξm\n\n= 1. Let f ∈H(U, U) be such that f(0) = 0 and that f(0)µm = αµm, α > 0. By Carath´eodory-Cartan-Kaup-Wu theorem, we haveα≤1. Since one can choose f as the identity mapping, we have inf 1/α = 1 = supα. ThereforeKUU(0, µm) = CUU(0, µm) = 1. The automorphism group onU is transitive. Hence by Corollary 1 we haveKUU(p, µm) =CUU(p, µm) for any p∈U.\n\nProposition 2. Let Ω ⊂ Cn, p ∈ Ω and ξ1, . . . , ξm ∈ TpCΩ, 1 ≤ m ≤ n be linearly independent vectors. IfU =Bm−j×∆j,0≤j≤m, then\n\n(8) CU(p, ξ1, . . . , ξm) KU(p, ξ1, . . . , ξm)≤1.\n\nProof. Let M be an (m, m) volume form on Ω such that M ξ1, . . . , ξm, ξ1, . . . , ξm\n\n= 1. Let Φ : U −→ Ω be a holomorphic mapping such that Φ (0) = p, Φ(0)M = αµm, α >0 and Ψ : Ω −→ U be a holomorphic mapping such that Ψ (p) = 0, Ψ(p)µm=βM. Considerh= Ψ◦Φ :U −→U. Then h(0) = 0 andh(0)µm=α·β·µm. By Carath´eodory-Cartan-Kaup-Wu theorem we haveα·β ≤1. Henceβ ≤1/α. The inequality (8) follows after taking the infimum over α’s and the supremum overβ’s.\n\nLemma 1. Let Ω ⊂Cn, p∈ Ω and ξ1, . . . , ξm ∈TpCΩ, 1 ≤ m ≤n, be linearly independent vectors. Let U = Bm−j×∆j. We have CU(p;ξ1, . . . , ξm)\n\nKU(p;ξ1, . . . , ξm) = 1 if and only ifΩ is biholomorphic toU.\n\nProof. One can use a similar argument as in (Theorem E).\n\nThe Kobayashim-measure is localizable near a strongly pseudocovnex boundary point. Refer to for a detailed explanation. The Carath´eodorym-measure is local- izable near a boundary pointpif one can find a global peak function that peaks atp.\n\nHence we have the following Lemma.\n\nLemma 2. Let Ω⊂Cn be a smoothly bounded convex domain andp∈∂Ω be a strongly covnex boundary point. LetV be a neighborhood of p. Then we have\n\nKU(z;ξ1, . . . , ξm)\n\nKUΩ∩V (z;ξ1, . . . , ξm) →1, CU(z;ξ1, . . . , ξm)\n\nCUΩ∩V(z;ξ1, . . . , ξm) →1, asz→p.\n\n(5)\n\nRemark 1. Let Ω be a smoothly bounded convex domain. The domain Ω near a strongly convex boundary point can be approximated by ellipsoids which are biholomorphic to balls. Since Bm and Bm−j ×∆j, j ≥ 1, are not biholomorphic and the Kobayashi and Carath´eodory measures are localizable near a strongly convex boundary point by Lemma 2, we have\n\nCU(z;ξ1, . . . , ξm)\n\nKU(z;ξ1, . . . , ξm) < c <1, U=Bm−j×∆j, j ≥1 CU(z;ξ1, . . . , ξm)\n\nKU(z;ξ1, . . . , ξm) →1, U =Bm asz approaches a strongly convex boundary point.\n\n3. Geometry of a convex domain.\n\n3.1. Non-tangential convergence. Let Ω ⊂ Cn be a domain with a C1- boundary. Let{qj} ⊂Ω be a sequence of points. We sayqj →q∈∂Ωnon-tangentially for some boundary pointqif\n\n(9) qj∈Γα(q) ={z∈Ω :|z−q|< αdist (z, ∂Ω)}\n\nfor alljlarge enough for someα >1 and we sayqj →q∈∂Ωnormallyifqj’s approach qalong the real normal line to the boundary through qfor allj large enough.\n\nLemma 3. Let Ω⊂Cn be a convex domain withC1 boundary and q∈∂Ω. Let ν be the outward unit normal vector to ∂Ω atq andq =q−tν ∈Ωfor some small t >0. Then we have\n\nΓα(q)⊂ {z∈Ω : 0≤∠zqq <arccos (1/α)}.\n\nProof. Letq = 0 and ν = (0, . . . ,0,1). Then Ω⊂H ={Rezn<0}. Therefore dist (z, ∂Ω)≤dist (z, ∂H) = |Rezn|. Hence |z−q|< α|Rezn| =α|(0, . . . ,0,Rezn)|.\n\nTherefore∠zqq<arccos (1/α).\n\nLemma 4. Let Ω⊂Cn be a convex domain with C1 boundary. Suppose {φj} ⊂ Aut(Ω) and φj(p) → q ∈ ∂Ω non-tangentially for some p ∈ Ω. Then there exists {pj} ⊂Ωsuch thatφj(pj)→qnormally and that dK(p, pj)≤r for somer >0.\n\nProof. Letφj(p) =qj. Since qj →q non-tangentially, one can find α > 1 such thatqj ∈Γα(q) for allj large enough.\n\nLet ν be the outward unit normal vector to∂Ω at q andℓq be the real normal line to ∂Ω throughq, i.e., ℓq ={q+tν:t∈R}. Define the mapping π : Ω −→ ℓq\n\nas the projection of Ω ontoℓq. Let ˜qj =π(qj). Then we have|qj−q˜j| ≤ |q−qj|<\n\nαdist (qj, ∂Ω). Let pj = φ−1j (˜qj). Then we have φj(pj) = ˜qj → q normally after taking a subsequence if necessary.\n\nSince Ω is convex, by Lemma 3, we have 0≤∠qjqq˜j ≤arccos (1/α). Therefore cos (∠qjq˜qj) = |q˜j−q|\n\n|qj−q| ≥ 1 α.\n\n(6)\n\nLetγ(t) = (1−t)qj+tq˜j. Then we have dK(p, pj) =dK(qj,q˜j)≤\n\nZ 1 0\n\nFK(γ(t), γ(t))dt\n\n≤ Z 1\n\n0\n\n(t)| 1\n\ndist (γ(t), ∂Ω)dt≤ Z 1\n\n0\n\n(t)| α\n\n|γ(t)−q|dt\n\n≤ α|qj−q˜j|\n\n|q˜j−q| ≤ α|qj−q|\n\n|˜qj−q| ≤α2. We let r=α2.\n\nLemma 5. Let Ω ⊂⊂ Cn be a bounded complete hyperbolic domain with a C2 boundary and p∈∂Ωbe a strongly convex boundary point. Then for any fixedr >0, the Euclidean diam βK(z, r)−→0 asz→p, where\n\nβK(z, r) =\n\nw∈Ω :dK(z, w)< r ⊂Ω.\n\nProof. Let δ(z) = dist (z, ∂Ω) andz ∈Ω be the boundary point that satisfies\n\n|z−z|=δ(z). It is a well-known fact that forz∈Ω close to a strongly pseudoconvex boundary point the Kobayashi metric estimate is given as follows (refer to [1, 3]):\n\nFK(z, ξ)≈ 1\n\nδ(z)ξN+ 1 pδ(z)ξT,\n\nwhere ξT andξN are the tangential and normal components ofξat z, respectively.\n\nThe assertion can be derived from the above fact and the complete hyperbolicity.\n\n3.2. Maximal chain of analytic discs. Let Ω⊂Cn be a smoothly bounded domain andV be a connected subset of∂Ω. We say∂Ω isgeometrically flatalongV if the direction of the gradient vector of∂Ω does not change along V.\n\nThe following proposition is the generalization of Lemma 3.2 in . The proof is basically the same.\n\nProposition 3. Let Ω⊂⊂Cn be a bounded convex domain. If φ: ∆−→∂Ωis a holomorphic mapping, then∂Ωis geometrically flat alongφ(∆).\n\nProof. Let Ω ={ρ <0} and p= φ(0) ∈∂Ω. Let H = {Reh= 0} be the real tangent plane to∂Ω atp, wherehis a linear holomorpic function. Since Ω is convex, we have Ω⊂ {Reh≤0}. Considerf(ζ) =h◦φ(ζ). Thenf is a holomorphic function on ∆ and satisfies Re f(ζ)≤0 for allζ∈∆ and that Re f(0) = 0. By the maximum principle for harmonic functions, we have Re f(ζ) = 0 for allζ∈∆. Thereforef ≡0 on ∆ and henceh≡0 onφ(∆).\n\nDefinition 3. LetH ⊂Cn be a subset ofCnandq∈H. We define the maximal chain of analytic discs onH throughq, denoted as ∆Hq , as follows:\n\nHq ={z∈H : there exists a finite chain of analytic discs joiningzandq}, i.e., there exists holomorphic mapsφ1, φ2, . . . , φk : ∆−→Cn such thatφj(∆)⊂H, 1≤j ≤k, andzi∈H, ai, bi ∈∆, 1≤i≤k, such that φj(aj) =zj−1j(bj) =zj, where z0 =q and zk =z. Note that ∆Hq = ∆Hz , if z∈∆Hq . We say ∆Hq is trivial if\n\nHq ={q}.\n\n(7)\n\nRemark 2. IfV ⊂H is a complex variety throughq, thenV ⊂∆Hq . The following Corollary follows immediately from Proposition 3.\n\nCorollary 3. If Ω⊂⊂ Cn is a smoothly bounded convex domain, then ∂Ω is geometrically flat along ∆∂Ωq for allq∈∂Ω.\n\nIn the following theorem we show that a maximal chain of analytic discs on the boundary of a smoothly bounded convex domain is linearly convex.\n\nTheorem 1. Let Ω⊂⊂Cn be a smoothly bounded convex domain. Then ∆∂Ωq is linearly convex for all q∈∂Ω, i.e., ifz, w∈∆∂Ωq , thent·z+ (1−t)w∈∆∂Ωq for all t∈[0,1].\n\nProof. We first show that ifz, w ∈ ∆∂Ωq , then t·z+ (1−t)w ∈ ∂Ω for allt ∈ [0,1]. Since ∂Ω is geometrically flat along ∆∂Ωq , we may assume ∆∂Ωq ⊂ {Rezn = 0}.\n\nWe have t·z+ (1−t)w ∈ Ω since Ω is convex. Also Re (t·z+ (1−t)w)n = t· Rezn+ (1−t) Rewn = 0 for allt ∈ [0,1]. Since Rezn <0 for allz ∈Ω, we have t·z+ (1−t)w∈∂Ω.\n\nWe use induction on the length of the chain (i.e. number of analytic discs) joining two pointsz, w∈∆∂Ωq .\n\nSuppose z, w ∈ ∆∂Ωq and z, w both lie on the same analytic disc, then t·z+ (1−t)w ∈∆∂Ωq . Let z =φ(a) andw =φ(b) for some analytic disc φ: ∆−→ ∂Ω anda, b∈∆ and define an analytic disc ˜φt as follows:\n\nφ˜t(ζ) =t·φ(ζ) + (1−t)φ(b).\n\nThen ˜φt(ζ)∈∂Ω for allζ∈∆ and for any fixedt∈[0,1], and ˜φt(b) =φ(b)∈∆∂Ωq . Hence ˜φt(ζ)∈∆∂Ωq for allζ∈∆. Therefore ˜φt(a) =t·φ(a) + (1−t)φ(b)∈∆∂Ωq for allt∈[0,1].\n\nAssumet·z+ (1−t)w∈∆∂Ωq for allt∈[0,1] ifz, wcan be joined by a chain of length less than or equal ton. Supposez, w∈∆∂Ωq can be joined byn+ 1 number of analytic discs, i.e., there exists analytic discsφj: ∆−→∂Ω,aj, bj∈∆ andzj ∈∂Ω, 1≤j ≤n+ 1, such that φj(aj) = zj−1, φj(bj) =zj and z=z0,w=zn+1. Define an analytic disc ˜φt as follows:\n\nφ˜t(ζ) =t·φ1(ζ) + (1−t)φn+1(bn+1), t∈[0,1]. Then ˜φt(ζ)∈∂Ω for allζ∈∆ and for allt∈[0,1]. We have\n\nφ˜t(b1) =t·φ1(b1) + (1−t)φn+1(bn+1)\n\n=t·φ2(a2) + (1−t)φn+1(bn+1)\n\nand hence ˜φt(b1) ∈ ∆∂Ωq for all t ∈ [0,1] since φ2(a2) and φn+1(bn+1) are joined by n analytic discs. Therefore ˜φt(ζ) ∈ ∆∂Ωq for all ζ ∈ ∆ and hence ˜φt(a1) = t·z+ (1−t)w∈∆∂Ωq for allt∈[0,1].\n\n4. Normal convergence.\n\nProposition 4. Let Ω be a smoothly bounded convex domain in Cn. Suppose {φj} ⊂Aut(Ω)and φj(p)→q∈∂Ωnon-tangentially for some p∈Ω and that ∆∂Ωq is not trivial. Then there exists a non-constant holomorphic onto mapping φ: Ω−→\n\n∂Ωq such that φj →φafter taking a subsequence if necessary.\n\n(8)\n\nProof. Since φj(p) → q ∈ ∂Ω, we konw that φj → φ locally uniformly (after taking a subsequence if necessary) whereφ: Ω−→∂Ω is a holomorphic mapping by a normal family argument.\n\nWe shall show thatφ(Ω) = ∆∂Ωq . Since φ(Ω) ⊂∆∂Ωq is clear, we need only to show that ∆∂Ωq ⊂φ(Ω).\n\nLetq ∈∆∂Ωq and q 6= q. By Corollary 3, ∂Ω is geometrically flat along ∆q. Letν be the constant outward unit normal vector to∂Ω along ∆∂Ωq . By Lemma 4, there exists{pj} ⊂βK(p, r) for somer >0 such thatφj(pj)→qnormally. Let δj’s be such that\n\nφj(pj) =q−δjν.\n\nThen we have\n\ndK(q−δjν, q−δjν)< r<∞, for allj for somer >0. Hence if we letpj−1j (q−δjν), then\n\ndK pj, p\n\n≤dK pj, pj\n\n+dK(pj, p)\n\n=dK(q−δjν, q−δjν) +r < r+r<∞, ∀j.\n\nSince βK(p, r+r) is compact in Ω, one can find p ∈Ω such thatpj →p and that φ(p) =q. Therefore ∆∂Ωq ⊂φ(Ω).\n\nCorollary 4. Let Ω⊂⊂Cn be a smoothly bounded convex domain and {φj} ⊂ Aut(Ω). If φj(p)→q∈∂Ωnon-tangentially and ∆∂Ωq is not trivial, then ∆∂Ωq is an open convex set contained in a complex m-dimensional plane, wherem=dimC∂Ωq .\n\nProof. By Theorem 1, ∆∂Ωq is convex. Hence it is contained in a complex m- dimensional plane, wherem= dimC∂Ωq . Suppose ∆∂Ωq is not open andw∈∂∆∂Ωq is a boundary point. By Proposition 4, one can findz∈Ω such that φ(z) =w, where φ is the limit of {φj}. One can find a germ of complex m-dimensional manifold, sayM, nearz such that dimCφ(M) =m. LetH be the complex m−1 dimensional subspace of the real supporting plane to ∆∂Ωq atw=φ(z). By the maximum principle argument used in Proposition 3, we have thatφ(M)⊂H. But dimH < m. Hence a contradiction.\n\nTheorem 2. Let Ω⊂⊂Cn be a smoothly bounded domain. Suppose ∆∂Ωq is not trivial for someq∈∂Ωand thatφ: Ω−→∆∂Ωq is a surjective holomorphic mapping.\n\nThen there exists a sequence of points {pj} ⊂ Ω such that pj → p ∈ ∂Ω and that {φ(pj)} ⊂∆∂Ωq converge to a point in ∆∂Ωq for some strongly pseudoconvex boundary pointp∈∂Ω.\n\nProof. Since Ω is smoothly bounded, there exists a strongly pseudoconvex bound- ary pointp∈∂Ω. Letν be the outward unit normal vector to∂Ω atp. One can find a holomorphic support functionhof∂Ω at psuch that, for a small neighborhoodU ofp, we have{h= 0} ∩Ω∩U ={p}. LetH ={h= 0}and letHn be the translation ofH in the direction of−ν by the length of 1/n, i.e.,\n\nHn=\n\nz−ν1\n\nn :z∈H\n\n, n∈N.\n\n(9)\n\nOne can find a small neighborhoodU ofpandN >0 large enough such that∂Ω∩U is strongly pseudoconvex and thatHn∩Ω⊂U ∩Ω for alln > N.\n\nLet dimC∂Ωq =m. Choose a complex m-dimensional closed analytic subset of Hn throughp−ν· 1n and perturb it atp−ν· n1, call it Hn, so that the rank of the restriction mapping ofφ on Hn, say φn :Hn −→ ∆∂Ωq , has rankm generically and that ∂Hn ⊂∂Ω. One can make the perturbation small enough that Hn ⊂U∩Ω for alln. Suppose φn is not proper for some n > N. Then one can find a compact set K ⊂⊂∆∂Ωq such that the preimage ofK is not compact inHn. Hence one can find {pj} ⊂Hn such thatφ(pj)’s lie inK for allj andpj’s approach a boundary point of Hn, which is strongly pseudoconvex.\n\nIf φn is proper for all n > N, then they are surjective because the rank of φn\n\nis equal to m. One can find pn ∈ Hn for n > N, arbitrarily close to p, which is a strongly pseudoconvex point. Moreover{φ(pn)}converge to a point in ∆∂Ωq .\n\nLemma 6. LetΩ⊂⊂Cnbe a smoothly bounded convex domain. Suppose there ex- ists{φj} ⊂Aut(Ω)such thatφj(z)converges to some boundary point non-tangentially for all z ∈ Ωfor some fixed αin (9) and that ∆∂Ωq is not trivial for some orbit ac- cumulation point q ∈ ∂Ω. Then for any ǫ > 0 one can find {pj} ⊂ Ω such that φj(pj)→q ∈∆∂Ωq normally for some point q and that pj ∈B(p, ǫ)∩Ωfor some strongly convex boundary pointp∈∂Ω.\n\nProof. By Proposition 4,φj’s converge locally uniformly to a non-constant holo- morphic mapping φ : Ω −→ ∆∂Ωq . By Theorem 2, one can find a point z close enough to some strongly pseudoconvex boundary point p such that φ(z) = q for some q ∈ ∆∂Ωq . We have φj(z) → q non-tangentially as j → ∞. Therefore by Lemma 4, one can findr >0 and{pj} ⊂βK(z, r) such thatφj(pj)→q normally as j → ∞. As shown in the proof of Lemma 4, rdepends on α, r=α2, to be precise.\n\nSince we assumeα >0 is fixed, by Lemma 5 one can choosezclose enough top such thatβK(z, r)⊂B(p, ǫ).\n\n5. Boundary accumulation points.\n\nProposition 5. Let Ω⊂⊂Cn be a smoothly bounded convex domain. Suppose\n\n∂Ωq is not trivial for someq∈∂Ω. If there exists{φj} ⊂Aut(Ω) such thatφj(z)→\n\n∂Ωq nontangentially for all z∈Ω, then ∆∂Ωq is biholomorphic to a complex m-ball, wherem is the complex dimension of ∆∂Ωq (i.e., real2m dimensional ball).\n\nProof. Letp∈Ω be arbitrarily close to a strongly pseudoconvex boundary point and letφ(p) =q∈∆∂Ωq . Also denotepjj(p) andV = ∆∂Ωq .\n\nLetξ1, . . . , ξm be mlinearly independent complex tangent vectors to V and use the intrinsic measure defined with respect to the complex unit m-ball, i.e., U is the complex unitm-ball in Definition 2. We may assumeV lies in thez2. . . zm+1plane, where Rez1 is the outward normal direction. Letπbe the projection mapping ofCn onto thez1. . . zm+1 plane and ˜pj =π(pj). Forj large enough, one can findV such that q∈V ⊂⊂V and that one can move V into Ω using the translation mapping that mapsq to ˜pj. LetVj be the image of such translation mapping ofV.\n\nWe may assumeq= 0. Suppose pj = (a1, . . . , an),p˜j = (a1, . . . , am+1,0, . . . ,0).\n\nConsider the holomorphic mappingfj :Cn−→Cn defined as fj(z) = (h1(z), . . . , hn(z)), where\n\nhk=\n\nzk, k= 1, . . . , m+ 1 ak·a1\n\n|a1|2 z1, k=m+ 2, . . . , n.\n\n(10)\n\nThenfj(0) = 0 andfj(˜pj) =pj. We have\n\nCU\n\np; φ−1j ◦fj\n\n(˜pjl\n\nKU(p; (φ−1)(q)ξl) ≤ Cfj(Vj)\n\nU pj; (fj)(˜pjl KU(p; (φ−1)(q)ξl)\n\n≤ CUVj(˜pjl)\n\nKU(p; (φ−1)(q)ξl)≤ CUV(q;ξl) KUV (q;ξl),\n\nwhere ξl stands for the set of m-vectors, ξ1, . . . , ξm. Note that (φ−1)ξj should be interpreted as the pre image vector ofξj, which is well-defined since the rank ofφis m along ∆∂Ωq .\n\nAs j → ∞, one can letV →V. Then the left hand side approaches 1, whereas the right hand side is always less than or equal to 1.\n\nTherefore we have\n\nCUV (q;ξl) KUV (q;ξl) = 1\n\nand henceV is biholomorphic to a complexm-dimensional ball.\n\nIn the following Theorem, we assume that there existsα <∞such that (9) holds for all z and in Theorem 4, we will give a proof without the assumption on α. The proof of Theorem 3 has its own merit, since is uses the invariant measures to compare the domain Ω near a strongly convex boundary point and a flat boundary point.\n\nTheorem 3. LetΩ⊂⊂Cn be a smoothly bounded convex domain. Suppose there exists {φj} ⊂ Aut(Ω) such that φj(z) converges nontangentially to some boundary point for all z ∈Ω. We also assume there exists α <∞ such that (9) holds for all z∈Ω. Ifq∈∂Ωis an orbit accumulation point, then ∆qis trivial and hence there does not exist a complex variety on ∂Ωpassing through q.\n\nProof. Suppose ∆∂Ωq is not trivial and let V = ∆∂Ωq . Let m be the complex dimension ofV. SinceV is convex by Theorem 1, we may assumeV lies on a complex m-dimensional plane.\n\nWe may assume ν = (1,0, . . . ,0) is the constant outward unit normal vector along V and V lies in z2z3· · ·zm+1 plane after a linear change of coordinates. Let π: Ω→ {zm+2=zm+3=· · ·=zn= 0} be the projection mapping.\n\nBy Lemma 6, one can find a strongly convex boundary point p ∈∂Ω such that for any ǫ > 0, there exists {pj} ⊂ B(p, ǫ)∩Ω such that φj(pj) = qj → q ∈ V normally for some q ∈∆∂Ωq . Choose Ωj’s, as a relatively compact exhaustion of Ω, such that ΩjրΩ and thatpj ∈Ωj for allj. LetU = ∆×Bm and choosemlinearly independent vectors ξ1, . . . , ξm ∈ TqCV. Since ∂Ω is geometrically flat alongV, we have ξj ∈ TqC−νǫ(V −νǫ). Hence for j large enough ξj ∈TqCj(V −ν|qj−q|). Let ξj = φ−1j\n\n(qjj andν= φ−1j\n\n(qj)ν We let Γǫ=\n\nz∈C: π2 +ǫ <argz < 2 −ǫ andH ={z∈C: Rez <0}. Then Γǫ →H as ǫ →0. LetVǫ be a subset of ∂Ω such that Vǫ ցV as ǫ→0. Then we\n\n(11)\n\nhave\n\nCUj(pj, ξ1, . . . , ξm)\n\nKU(pj, ξ1, . . . , ξm ) ≥CUφj(Ωj)(qj;ν, ξ1, . . . , ξm) KU(qj;ν, ξ1, . . . , ξm) (10)\n\n≥CUπ(φj(Ωj))(qj;ν, ξ1, . . . , ξm) KU(qj;ν, ξ1, . . . , ξm)\n\n≥ CU(H×Vǫ)∩W(qj;ν, ξ1, . . . , ξm) KUǫ×V)∩W(qj;ν, ξ1, . . . , ξm), (11)\n\nwhereW=W∩Ω,W an open neighborhood ofV. In the last inequality we used the inclusion mapping i :π(φj(Ωj))−→(H×Vǫ)∩W for the numerator and another inclusion mapping ˜i: (Γǫ×Vǫ)∩W −→Ω for the denominator. The left hand side of (10) is strictly less than 1 since we may assumepj is arbitrarily close to a strongly convex boundary point and j is large enough, whereas the right hand side of (11) approaches 1 since one can let ǫ →0 asj → ∞, chooseW small enough, and V is biholomorphic to a ball by Proposition 5, which leads to a contradition.\n\nRemark 3. In the proof of Theorem 3, one can letU =Bm+1instead of ∆×Bm. In this case we should consider the ratioK/C. The left hand side of (10) approaches 1 aspj’s approach a strongly pseudoconvex boundary point, whereas the right hand side of (11) is strictly greater than 1 asqj’s approach a flat boundary point. Hence it gives rise to a contradiction.\n\nAdditionally, we prove a lemma that shows that if a point converges non- tangentially then all the other points must converge non-tangentially in the normal direction.\n\nLemma 7. Let Ω⊂⊂Cn be a smoothly bounded convex domain. Suppose∆∂Ωq is not trivial for someq∈∂Ωand that there exists p∈Ω and{φj} ⊂Aut(Ω)such that φj(p)→q∈∆∂Ωq non-tangentially. Then φj(a)→b∈∆∂Ωq non-tangentially in the normal direction for alla∈Ωfor someb∈∆∂Ωq .\n\nProof. Leta∈Ω. Since Ω is complete hyperbolic, we havedK(p, a) =r <∞for somer >0.\n\nWe may assume q = 0 and the outward normal vector to ∂Ω along ∆∂Ωq is in the direction of Rezn-axis. Let pjj(p) and ajj(a). By Proposition 4,aj’s converge tob∈∆∂Ωq for some b∈∆∂Ωq . Letpj andaj be the projection ofpj andaj\n\nontozn-axis. Thenpj = (0, . . . ,0, sj) andaj = (0, . . . ,0, tj) for somesj, tj ∈C. Let tj =Ajej andsj =Pjej. Since Ω is convex we have Resj, Retj<0.\n\nSincepj →qnon-tangentially, pj∈Γα(q) for someαfor allj large enough. By Lemma 3, we haveπ−θj<arccos (1/α) for allj large enough. Hence\n\n(12) cosθj <−1/α.\n\nWe have\n\n(13) ∞> r=dK(p, a) =dK(pj, aj)≥dK pj, aj\n\n≥dHK(sj, tj),\n\nwhereH ={z∈C: Rez <0}. Using the Poincar´e distance between two pointsz, w∈\n\n∆ given by ln\n\n|1−wz|+|w−z|\n\n|1−wz| − |w−z|\n\nand the biholomorphic mapping f(z) = (z+\n\n(12)\n\n1)/(z−1) that mapsH to ∆, we get\n\ndHK(sj, tj) = ln\n\n|tj+sj|\n\n|sj−1| +|t|sjj−s−1|j|\n\n|tj+sj|\n\n|sj−1||t|sj−sj|\n\nj−1|\n\n. We may assume|sj|,|tj|<1/2. Then we have\n\ndHK(sj, tj)≥ ln 1\n\n3\n\n|tj+sj|+|tj−sj|\n\n|tj+sj| − |tj−sj|\n\n≥ ln1 3 + ln\n\np1 + cos (θjj) +p\n\n1−cos (θj−αj) p1 + cos (θjj)−p\n\n1−cos (θj−αj)\n\n!\n\n→ ∞, (14)\n\nif αj → π/2. From (12), (13), and (14), we conclude that aj → b ∈ ∆∂Ωq non- tangentially for someb∈∆∂Ωq .\n\nRemark 4. From Lemma 7, it is not hard to see counting the dimensions involved that if there exists a pointp∈Ω such that {φj(p)} converges non-tangentially to a boundary point q∈∂Ω, then dim ∆∂Ωq < n−1, wheren= dim Ω.\n\nIn the following theorem we give another proof of Theorem 3 without using the assumption that there existsα <∞such that (9) holds for allz∈Ω.\n\nTheorem 4. LetΩ⊂⊂Cn be a smoothly bounded convex domain. Suppose there exists {φj} ⊂ Aut(Ω) such that φj(z) converges nontangentially to some boundary point for all z∈Ω. Ifq∈∂Ωis an orbit accumulation point, then∆∂Ωq is trivial and hence there does not exist a complex variety on ∂Ωpassing throughq.\n\nProof. As in the proof of Theorem 3, one can assumeV = ∆∂Ωq lies on a complex m-dimensional plane, wheremis the complex dimension ofV.\n\nLet the Re z1-direction be the outward normal direction alongV and V lies on the complex z2z3· · ·zm+1 plane.\n\nLet Γǫ,r be a wedge domain with radius less than r in C defined as Γǫ,r = z∈C: π2 +ǫ <argz < 2 −ǫ, |z|< r . Choose p ∈ Ω close to a strongly pseu- doconvex boundary point. Then φj(p) → q ∈ V non-tangentially for some q. Let V ⊂⊂V andq∈V. Consider the product domain Γǫ,r×V ⊂Ω. LetAǫ,r be the interior of Γǫ,r×V. Let q= 0, pjj(p) and ˜pj be the projection ofpj onto the z1z2· · ·zm+1-plane, i.e. if pj = (a1, . . . , an), then ˜pj = (a1, a2, . . . , am+1,0,· · ·,0).\n\nThen ˜pj →qnontangentially.\n\nConsider the holomorphic mapping fj:Cn −→Cn defined as fj(z) = (h1(z), . . . , hn(z)), where\n\nhk =\n\nzk, k= 1, . . . , m+ 1 ak·a1\n\n|a1|2 z1, k=m+ 2, . . . , n.\n\nNote thatfj is the identity mapping when restricted toV and fj(˜pj) = pj. Since pj →q non-tangentially, one can find ǫ, r >0 such thatfj(Aǫ,r)⊂Ω assuming j is large enough.\n\n(13)\n\nLetU = ∆×Bmjbe the unit vector in thezj-direction and Ωkbe the exhaustion of Ω, i.e, ΩkրΩ. Then we have\n\n(15)\n\nCUk\n\np; φ−1j\n\n(pjl\n\nKU\n\np; φ−1j ◦fj\n\n(˜pjl\n\n≥ CUφj(Ωk)(pjl) KU pj; (fj)(˜pjl\n\n≥ CUAǫ,r(˜pjl) KUAǫ,r(˜pjl),\n\nwhereξlstands for the set ofm+ 1 vectorsξ2, . . . , ξm+1. Note that the first (m+ 1) by (m+ 1) complex Jacobian offj is the identity and hence (fj)ξlis well-defined for l= 1, . . . , m+ 1. The second inequality for the Carath´eodory measure is derived using the projection mapping ofCn onto thez1z2. . . zm+1 plane. Forj andklarge enough we may assume the projection ofφj(Ωk) is inside Aǫ,r for someǫ and r. Note that the Jacobian matrix of the projection is identity along z1. . . zm+1 direction, hence ξ1, . . . , ξm+1remain unchanged.\n\nSince fj is the identity along z1, . . . , zm+1 directions, letting j, k → ∞, we see that the left side of (15) is strictly less than 1, whereas the right hand side converges to 1 as one can letǫ→0 andV →V. Hence a contradiction.\n\nREFERENCES\n\n G. Aladro,The comparability of the Kobayashi approach region and the admissible approach region, Illinois J. Math., 33:1 (1989), pp. 42–63.\n\n S. Fu and B. Wong,On boundary accumulation points of a smoothly bounded pseudoconvex domain inC2, Math. Ann., 310 (1998), pp. 183–196.\n\n I. Graham,Boundary behavior of the Carath´eodory and Kobayashi metrics on strongly pseu- doconvex domains in Cn with smooth boundary, Trans. Amer. Math. Soc., 207 (1975), pp. 219–240.\n\n A. V. Isaev and S. G. Krantz,Domains with non-compact automorphism group: a survey, Adv. Math., 146:1 (1999), pp. 1–38.\n\n K. T. Kim,Domains in Cn with a piecewise Levi flat boundary which possess a noncompact automorphism group, Math. Ann., 292:4 (1992), pp. 575–586.\n\n S. G. Krantz,Function theory of several complex variables. Second edition, The Wadsworth &\n\nBrooks/Cole Mathematics Series. Wadsworth & Brooks/Cole Advanced Books & Software, Pacific Grove, CA, 1992.\n\n H. L. Royden,Remarks on the Kobayashi metric, Several complex variables, II (Proc. Internat.\n\nConf., Univ. Maryland, College Park, Md., 1970), pp. 125–137. Lecture Notes in Math., Vol. 185, Springer, Berlin, 1971.\n\n B. Thomas, Thesis, University of California Riverside, 2012\n\n B. Wong,Characterizations of the ball inCn by its automorhpism group, Invent. Math., 41 (1977), pp. 253–257.\n\n B. Wong,Characterization of the bidisc by its automorphism group, Amer. J. of Math., 117 (1995), pp. 279–288\n\n B. Wong,On complex manifolds with noncompact automorphism groups, Explorations in com- plex and Riemannian geometry, pp. 287–304, Contemp. 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https://www.daniweb.com/programming/software-development/threads/295807/winform-random-info-generation
[ "hi\nim new in vb.net\ni need to make a form like this\nhttp://img138.imageshack.us/img138/8861/randome.png\nhttp://img153.imageshack.us/img153/7952/random2o.png\n\nFor Firstname and Lastname: When you click to random Button it will auto pick random Firstname from Firstname.txt and random Lastname from Lastname.txt\n\nfrom firstname, lastname, string, number like this\n\nPassword = random string + random number\n\nShow us the code you have done so far plz.\nOr you expecting us to write you the whole code?\nshow some effort plz!\n\nhi GeekByChoiCe\nim new in vb.net that why im looking for some similar code like this. can you help :)\n\nok here is a small sample code as Console Application. the names in the two txt files should be one per line, else you have to adjust the code.\n\n``````Sub Main()\nDim fNames() As String = IO.File.ReadAllLines(\"Firstname.txt\")\nDim lNames() As String = IO.File.ReadAllLines(\"Lastname.txt\")\n\nRandomize(DateAndTime.Now.Ticks)\t'shuffle randoms, else the randoms will be the same on each start of the app\nDim fRandom As New Random\n\nDim rFirst As String = fNames(fRandom.Next(0, fNames.Count))\nDim rLast As String = lNames(fRandom.Next(0, lNames.Count))\nDim rNumber As Integer = fRandom.Next(10, 99)\nConsole.WriteLine(\"Random FirstName: {0}\", rFirst)\nConsole.WriteLine(\"Random LastName: {0}\", rLast)\nConsole.WriteLine(\"Created Username: {0}{1}{2}\", rFirst, rLast, rNumber)\n\nDim passArray As ArrayList = fillArray(True)\nDim pass As String = \"\"\nFor x As Integer = 1 To 12 'to lenght of the password\npass = pass & passArray.Item(fRandom.Next(0, passArray.Count)).ToString\nNext\n\nEnd Sub\n\nPrivate Function fillArray(ByVal withSpecialChars As Boolean) As ArrayList\nDim passArray As New ArrayList\n\nFor i As Integer = AscW(\"A\") To AscW(\"Z\") 'add capitals\nNext\n\nFor i As Integer = AscW(\"a\") To AscW(\"z\") 'add lower case\nNext\n\nFor i As Integer = 0 To 9\t\t 'add numbers\nNext\n\nIf withSpecialChars Then\t\t 'create list of special chars\nFor i As Integer = 33 To 126\nIf i < 48 Then\nElseIf i >= 58 AndAlso i <= 64 Then\nElseIf i >= 91 AndAlso i <= 96 Then\nElseIf i >= 123 AndAlso i <= 126 Then" ]
[ null ]
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http://list.seqfan.eu/pipermail/seqfan/2014-October/065556.html
[ "# [seqfan] Re: Extended with a pair of integers\n\nSat Oct 11 20:10:04 CEST 2014\n\n```That should be 8,26, not 8,24.\n\nThe odd bifurcation of this sequence is A005228; and the other half is\nin A030124.\n\n-----Original Message-----\nFrom: Eric Angelini <Eric.Angelini at kntv.be>\nTo: Sequence Discussion list <seqfan at list.seqfan.eu>\nSent: Sat, Oct 11, 2014 11:17 am\nSubject: [seqfan] Extended with a pair of integers\n\nHello SeqFans,\nS starts with 1 and is always extended with a pair of integers:\n- the first integer of the pair is the smallest integer not already in\nS;\n- the second integer is the sum of the two integers that preceed it.\n\nS=1\nS=1,2,3,\nS=1,2,3,4,7\nS=1,2,3,4,7,5,12\nS=1,2,3,4,7,5,12,6,18\nS=1,2,3,4,7,5,12,6,18,8,24\nS=1,2,3,4,7,5,12,6,18,8,24,9,33\nS=1,2,3,4,7,5,12,6,18,8,24,9,33,10,43\netc.\nS is a permutation of the intergers > 0\nEvery integer having an odd position in S\nis the sum of the two integers before it\nand the absolute difference of the two\nintegers after it (except for 1, the first term of S):\n3, for instance, is 1+2 and also |4-7|\nBest,\nÉ.\n\n```" ]
[ null ]
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https://documen.tv/question/urgent-help-pls-pls-pls-a-solid-sphere-of-radius-3-cm-is-of-mass-9-kg-find-the-mass-of-a-sphere-20707587-43/
[ "## URGENT HELP PLS PLS PLS A solid sphere of radius 3 cm is of mass 9 kg. Find the mass of a sphere of the same material of radius 5 cm\n\nQuestion\n\nURGENT HELP PLS PLS PLS\nA solid sphere of radius 3 cm is of mass 9 kg. Find the mass of a sphere of the same material of radius 5 cm\n\nin progress 0\n5 months 2021-08-18T16:27:56+00:00 1 Answers 0 views 0\n\n15 kg\n\nStep-by-step explanation:\n\nRadius of a solid sphere : mass of a solid sphere\n\n3 cm : 9 kg\n\nFind the mass of a sphere of the same material of radius 5 cm\n\nRadius of a solid sphere : mass of a solid sphere\n\n5 cm : x kg\n\nEquate both ratios to find x\n\n3 cm : 9 kg = 5 cm : x kg\n\n3/9 = 5/x\n\n3 * x = 9 * 5\n\n3x = 45\n\nx = 45/3\n\nx = 15 kg\n\nThe mass of a sphere of the same material of radius 5 cm is 15 kg" ]
[ null ]
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https://builtin.com/machine-learning/agglomerative-clustering
[ "Clustering is an unsupervised machine learning technique that groups data points based on the similarity between them. The data points are grouped by finding similar patterns/features such as shape, color, behavior, etc. of the data points.\n\n## 4 Uses For Clustering\n\n• Market Segmentation\n• Anomaly Detection\n• Image segmentation\n• Geo-Spatial Analysis\n\nFor a day-to-day life example of clustering, consider a store such as Walmart, where similar items are grouped together.\n\nThere are different types of clustering algorithms, including. centroid-based clustering algorithms, connectivity-based clustering algorithms (hierarchical clustering), distribution-based clustering algorithms and density-based clustering algorithms.\n\nIn this article, we will discuss connectivity-based clustering algorithms, also called hierarchical clustering. It is based on the core idea that similar objects lie nearby to each other in a data space while others lie far away. It uses distance functions to find nearby data points and group the data points together as clusters.\n\nThere are two major types of approaches in hierarchical clustering:\n\n• Agglomerative clusteringDivide the data points into different clusters and then aggregate them as the distance decreases.\n• Divisive clustering: Combine all the data points as a single cluster and divide them as the distance between them increases.\n\n## Agglomerative Clustering\n\nAgglomerative clustering is a bottom-up approach. It starts clustering by treating the individual data points as a single cluster then it is merged continuously based on similarity until it forms one big cluster containing all objects. It is good at identifying small clusters.\n\nThe steps for agglomerative clustering are as follows:\n\n1. Compute the proximity matrix using a distance metric.\n2. Use a linkage function to group objects into a hierarchical cluster tree based on the computed distance matrix from the above step.\n3. Data points with close proximity are merged together to form a cluster.\n4. Repeat steps 2 and 3 until a single cluster remains.\n\nThe pictorial representation of the above steps would be:\n\nIn the above figure,\n\n• The data points 1,2,...6 are assigned to each individual cluster.\n• After calculating the proximity matrix, based on the similarity the points 2,3 and 4,5 are merged together to form clusters.\n• Again, the proximity matrix is computed and clusters with points 4,5 and 6 are merged together.\n• And again, the proximity matrix is computed, then the clusters with points 4,5,6 and 2,3 are merged together to form a cluster.\n• As a final step, the remaining clusters are merged together to form a single cluster.\n\nThe proximity matrix is a matrix consisting of the distance between each pair of data points. The distance is computed by a distance function. Euclidean distance is one of the most commonly used distance functions.\n\nThe above proximity matrix consists of n points named x, and the d(xi,xj) represents the distance between the points.\n\nIn order to group the data points in a cluster, a linkage function is used where the values in the proximity matrix are taken and the data points are grouped based on similarity. The newly formed clusters are linked to each other until it forms a single cluster containing all the data points.\n\nThe most common linkage methods are as follows:\n\n• Complete linkage: The maximum of all pairwise distance between elements in each pair of clusters is used to measure the distance between two clusters.\n• Single linkage: The minimum of all pairwise distance between elements in each pair of clusters is used to measure the distance between two clusters.\n• Average linkage: The average of all pairwise distances between elements in each pair of clusters is used to measure the distance between two clusters.\n• Centroid linkage: Before merging, the distance between the two clusters’ centroids are considered.\n• Ward’s Method: It uses squared error to compute the similarity of the two clusters for merging.\n\n### Agglomerative Clustering Implementation\n\nAgglomerative clustering can be implemented in Python using sklearn and scipy. Let’s implement Agglomerative clustering on the Iris dataset. The dataset can be found here. As a first step, import the necessary libraries and read the dataset.\n\n``````import pandas as pd\nimport numpy as np\n\ndata_path = \"/content/Iris.csv\"\n#select only feature columns\nX = df[['SepalLengthCm', 'SepalWidthCm', 'PetalLengthCm', 'PetalWidthCm']]``````\n\nNow we can use agglomerative clustering class from sklearn to cluster the data points.\n\n``````#import the class\nfrom sklearn.cluster import AgglomerativeClustering\n#instantiate the model\nmodel = AgglomerativeClustering(n_clusters = 3, affinity = 'euclidean', linkage = 'ward')\n#fit the model and predict the clusters\ny_pred = model.fit_predict(X)``````\n\nIn the above code, at first we import the agglomerative clustering class and instantiate the model with the required parameters. We use the clusters of three since there are three classes in the Iris dataset and we use the ward linkage function with the euclidean function as a distance metric which is specified in affinity parameter.\n\nSince we know the number of species in the Iris dataset, we were able to specify the number of clusters in the parameter but in most cases of unsupervised learning, we don’t know the number of clusters beforehand. So how can we find the optimal number of clusters for hierarchical clustering?\n\n### Dendrogram Charts\n\nThe way to find the optimal number of clusters in hierarchical clustering is to use a dendrogram chart. Let’s see how we can identify an optimal number of clusters for the Iris dataset. For this, we can use the scipy library in Python.\n\n``````#import the necessary libraries\nimport matplotlib.pyplot as plt\n\nplt.figure(figsize=(15,6))\nplt.title('Dendrogram')\nplt.xlabel('Flowers')\nplt.ylabel('Euclidean distances')\nplt.savefig(\"iris.png\")\nplt.show()``````\n\nIn the above code, the dendrogram chart is created as follows:\n\n• Import the necessary libraries.\n• Create a linkage matrix with the linkage function from scipy. Here we use the ward method.\n• Pass the linkage matrix to the dendrogram function to plot the chart.\n\nThe chart is shown here:\n\nFrom the above chart we can visualize the hierarchical technique, so how to find the optimal number of clusters from the above chart?\n\nTo find it, draw a horizontal line where there is no overlap in the vertical lines of the bars. The number of bars without the overlap below the line is the optical number of the clusters. Refer to the figure below for a clear illustration.\n\nFrom the above figure, we have three bars below the horizontal line, so the optimal number of clusters is three. Also, if you recall, the Iris dataset has three classes and we got the same number from the above chart.\n\n## Divisive Clustering\n\nDivisive clustering works just the opposite of agglomerative clustering. It starts by considering all the data points into a big single cluster and later on splitting them into smaller heterogeneous clusters continuously until all data points are in their own cluster. Thus, they are good at identifying large clusters. It follows a top-down approach and is more efficient than agglomerative clustering. But, due to its complexity in implementation, it doesn’t have any predefined implementation in any of the major machine learning frameworks.\n\n### Steps in Divisive Clustering\n\nConsider all the data points as a single cluster.\n\n1. Split into clusters using any flat-clustering method, say K-Means.\n2. Choose the best cluster among the clusters to split further, choose the one that has the largest Sum of Squared Error (SSE).\n3. Repeat steps 2 and 3 until a single cluster is formed.", null, "In the above figure,\n\n• The data points 1,2,...6 are assigned to large cluster.\n• After calculating the proximity matrix, based on the dissimilarity the points are split up into separate clusters.\n• The proximity matrix is again computed until each point is assigned to an individual cluster.\n\nThe proximity matrix and linkage function follow the same procedure as agglomerative clustering, As the divisive clustering is not used in many places, there is no predefined class/function in any Python library.\n\nFurther Reading on Machine LearningHow Is Python Used in Machine Learning?\n\n## Limits of Hierarchical Clustering\n\nHierarchical clustering isn’t a fix-all; it does have some limits. Among them:\n\n• It has high time and space computational complexity. For computing proximity matrix, the time complexity is O(N2), since it takes N steps to search, the total time complexity is O(N3)\n• There is no objective function for hierarchical clustering.\n• Due to high time complexity, it cannot be used for large datasets.\n• It is sensitive to noise and outliers since we use distance metrics.\n• It has difficulty handling large clusters.\n\nClustering helps with the analysis of an unlabelled dataset to group the data points based on their similarity. In terms of business needs, clustering helps quickly segment customers and get insightful decisions.\n\nIn this article, we discussed hierarchical clustering, which is a type of unsupervised machine learning algorithm that works by grouping clusters based on the distance measures and similarity. We also learned about the types of hierarchical clustering, how it works and implementing the same using Python.\n\nExpert Contributors\n\nBuilt In’s expert contributor network publishes thoughtful, solutions-oriented stories written by innovative tech professionals. It is the tech industry’s definitive destination for sharing compelling, first-person accounts of problem-solving on the road to innovation." ]
[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]
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https://beyerhunde.de/2020/12/06/mod-01-lec-01-introduction/
[ "# Mod-01 Lec-01 Introduction", null, "", null, "problems It is going to be somewhat different and more challenging than some of the algorithms you have implemented in other courses, especially you may have implemented search algorithms, standard computer algorithms like sorting, some graph algorithms, m s t The main difference between implementing a geometric algorithm this cemetery algorithms is that you are dealing with points in Euclidian space when therefore; you have dealing with real numbers. So, once you deal with real numbers which are actually very unreal because the computers do not have real numbers, then you have to struggle with issues like over flow, under flow, stability; all those things, the things that you normally learn in numerical analysis course, something that will encounter when you try to actually implement the algorithms that we will design in the developing the class And to tackle those problems, you know of to some handle those problems, you will need to think beyond just the aspects of the algorithms, but also see how the points will be represented in, what kind of precision we need etcetera So, those also will become issues and these are serious issues when people try to implement geometrical algorithms In this course, we are not we are going to kind of avoid those issues for most parts, may be once I will maybe elsewhere a lecture to just trying to give you an over view of what are the kind of problems that arise when you implement geometric algorithms, but the theory of actually trying to implement algorithms in a way where you minimize errors, I said is very similar to numerical computing and that requires completely separate discussions In fact, I would go as far as to say although this area is about a that a more than thirty years old now, but I think people are still trying to come up with a clean model of implementing geometric algorithms So, it is very much a research topic you know how to do the clean implementation given that you have to handle real numbers. Anyway so, outline of what I expect from you in terms of assignment. So, there will be two or three written assignments and I would say one or two programming assignments depending you know I have not decided yet how big, but they will take some time. I will probably limit myself to one midterm exam rather two minus and we will have a major exam. So, I haven’t quite given much thought to how the marks we will distributed, but let us say, I am I am just saying, let us say. So, have minors sorry let say midterm actually not a minor, in the midterm you have a major, you have assignments. So, there are three components and I am saying assignments, there will be some basic theoretical assignments in problems and some programming assignments. So, let see. So, maybe I will do something like perhaps something like this looks let us say, twenty-five, forty and may be twenty and fifteen; that decides up to two hundred. So, let say, this is about the rough distribution that that we have in the course. I have described the syllabus in details in the in the course page I do not want to go through it because most of the terms would not make sense today. Rather I would give you start with an example of a geometric problem. Also let me also clarify that I said computation geometry is about designing and implementing algorithms for problems that are geometric in nature and that is quite distinct from trying to prove geometric theorems automatically So, that as that is completely different area That is done by also there is a community of people doing that who do research in theorem proving and actually they have done impressive work, but like any other theorem proving exercise, this actually trying to prove a new theorem is extremely hard and certainly not feasible algorithmically for most of the algorithms So, people are struggling with that, but just to even get that frame work going you know it is a very impressive achievements, but that is done again by I am saying that people", null, "in the theorem proving community So, we are not going to try to prove theorems using computers. We will try to prove the theorems by hand. Whatever we prove about the algorithms are the running time the analysis we will do by hand Feel free to ask me any questions if I miss something that is very relevant. So, let me proceed with an example to illustrate in what kind of what geometric problems mean to us in what kind of techniques people try to use So, let me take up a very natural example like which is essentially a navigating. So, navigating in a geometric environment which is basically that you have given. So, you can think about this room as a geometric environment where there are all of us can be describe with some geometry primitives So, the furniture, the people, the cameras; whatever. So, this entire that geometric description of this room is what a mean the geometric environment and then we may want to move some object from one part of the room to the other part. That is what a mean by navigation And of course, this navigation has to be done in a way such that it is not allowed to pass through solid objects. So, you can pass through the air, but you cannot pass through a solid object. So, you cannot. So, if you collide, you can only set of at best you have to circumvent that object So, to make things easy and for also you know from the point of view of we have to draw So, let me draw the 2 dimensional log of this So, suppose I take the projection of this room on the plane. So, what does it look like? So, it could look like. So, it could look like, we have his people sitting on this like this here, some rows of this here, again other row and here is where I am seated and we want to move let us say, let us say to begin with, it is just a point, which is easier because it has zero width. So, suppose I want to move a point and it may indicate that by cross, from this point to let us say to this corner of the row. This is the starting point start and this is the end. So, this is the game well. So, there are various ways of doing this. Let me choose different color. So, start walking, start walking you I cannot I have to way around this. So, I go, I go here may be I have to go around this. Alright I found a path So, this is how I have navigated from the start to end. Someone else we decide to do in some other way. So, here, there we will go, that is also a legal path. Someone more may want to do this way and this is not a legal path. It just crashes into furnaces, we cannot do that, one has to go around So, what is observed about these paths is that you can have several ways of navigating from this starting point to the end point without colliding into the obstacles, but which is the way that you are going to choose? Suppose you know I have write a program. So, I can certainly describe these objects in some way, these are let us say for simplicity all these are rectangles. So, how do I move this point and how do I write a program that moves this point from the starting to the end point that goes around obstacles. So, on what on it is to choose a random path, even to choose a random path I have to ensure in the minimum that it does not collide with the with the obstacles So, somehow we need to capture that fact and the one difficulty or let us say one initial", null, "difficulty is that the possible number of paths seems to be just too large. In fact, the possible numbers of paths that will infinite because I can take any legal path here and may just decide to just perturb this little bit just move it like this and that is it So, that is another legal path. So, with every small perturbation of one path, I can get another legal path. So, I am dealing with an infinite number of paths and this is what sets it to path from a path problem in graphs So, if I have a graph. So, suppose I have a graph like… So, I could have lots of edges in this graph does not matter I can have lots of edges this. This starting point and this is the ending point, but still the number of different ways I can move from the starting to the end point is finite. So, I could do like this, I could go like this, but again it just a infinite number and how a large it is it is till finite It may be exponential that it suppose I have n vertices there could be you know it could have you know this is a very common example to show that the number of paths could be exponential So, you could have these choices at every point, this is the starting point, this is the end point. Every point you have two possibilities to pursue and in if there are n such rhombus, you basically have 2 to the power n possibilities to provides, but still it is a finite number, it’s exponential, but still finite Here if a parameter is my problem by seeing that have n obstacles, I have n obstacles, I were knew obstacles, I cannot say that I have less than 2 d power n paths. It is not true is not true in the case of geometry So, the first difficulty that we will face in trying to model any geometric problem is to somehow cut down the number of possibilities something that we can deal with. You cannot deal with infinite numbers So, somehow we need to restrict that. So, we need to base on observations. So, let us try to make certain observations about this particular problem navigating a point from the start to the end through these obstacles and when you do that we must also may be just define some kind of a more clean objective function. It is not simply about let us say going from a start to the end point, but let us say we would like to economies something You know may be we would like to economies space, sorry economies time or the cost of travel somehow that. So, let us say. So, in the shortest path probably may try to we have actually a weights on the edges and we try to seek a path that will minimize the sum of the weights on the edges of the path So, similarly here we can think about the solve trying to solve the shortest path problem between start to end and the national notion of the distance here is equidistance. So, whatever is the path, I will just consider that well I mean strictly speaking, you have to prior define measures because it may be it can be any kind of path, but you understand what I mean. Basically the total length of the path is what we would like to minimize with. That you could go to do of the object function. The other objective function could be that I really do not care as long as it is some path, but here if you can let us try to solve the some more interesting problem, we want to actually minimize or find the shortest path find shortest path .So now, it becomes a better defined problem And when you talk about shortest paths. So, the shortest path between two points in a Euclidean space is what. A straight line segment right, well great. So, at least we are getting somewhere, but the straight line segment between this probably passes through or collides in to some obstacles. So, that does not seem like the solution to this problem, but it is a good starting point. So, this red line is not possible, but then we would like to be as close to this is possible. That that is basically what we give as a shortest path in tutorial shortest path and if that is so, then one way to think about it is that, suppose this red segment is like a rubber band that we can bend around the obstacles", null, "So, if it is allowed to bend, then this you know it would not pass through the objects, but my first attempt could be I bend it something like this. So, there are these two bends and that seems to set of be fairly close to the straight line, but do you think this is a shortest path why. Because that shortest path between the starting point and the first bend point will be a straight line Great. So, is this an intrusion you use basically because you live in the geometric world? You know exactly it is not like if I live in a graph world, you would not be able to figure out, but you were in the geometric world So, to quickly figure out that something like something like you know. So, this bend, let me blow this up. So, if I blow that up, it is like you have a rectangle and we are talking about a path. So, we do a path like this Now what you are saying is that, this bend that you took, why did you take this bend because if I take something like this, this the doted should be a short cut of these two bends and why is that? What is the property that you are using; the triangle inequality exactly. So, you are using triangular inequality. So, you already have a kind of a proof that this dotted line, it should be shorter than this plus this and therefore, the one that we had before, the solid line cannot be the shortest path So, this is another observation that then we will clean up this shortest path that we are trying to find between the start and the end point should have this property that one is that can it consists of curves, can that path have some curves like this No same thing, if there is a curve we know that the shortest distance between two points is straight lines I am going to short cut the curve. So, there is no, so, we do not have to deal with any curves. So, we can leave it ourselves to only straight lines and therefore, the shortest path between the starting and the end point is going to be a piece wise linear curve essentially piecewise linear, now, even that piecewise linear, we have further properties that we do not have to look for things like this Where are the bends, where can the bends occur the bends can only occur, where, see here there was a bend basically when we collide with an object,, but then that bend is not a bend that should appear in the shortest path the bend can be moved or pushed till it meets a corner point Right in other word what I am saying is this red line; let me draw a dotted line again So, this red line instead of this solid line we should be probably looking for something like this of course, that may crash in to other thing. So, the places that it can bend are only the corners. So, in other words we blew this up you know if I am trying to get from let us say here to here around just suppose we had only one obstacle. So, this straight line when you actually distort and think about it like a rubber band it is going to be basically bend here So, the bends only going to be appear at the corner points and we are actually argued in kind of proved it. So, with these two observations in mind now we can business why. So, now, I can actually model this problem, the shortest path problem as a graph problem where the edges or the vertices are defined by the corner points of the obstacles the four. So, these are rectangular obstacles. So, I can consider the corner points, the rectangles as vertices and the edges must be basically are defined the piecewise linear path. So, the edges must be between two such vertices because there", null, "cannot be any other bends, but then there is this thing that not all of them of a legal edges because some. So, the straight line segment between this and this these two points is not legal between it actually crashes in to an obstacle But that again we can verify that you know we will try. So, suppose as I said there are n obstacles, which means that there are 4 n corner points or which I am basically calling the vertices. So, 4 n corner points and then out of the 4 n choose two segments there are some we will define some legal edges that do not crash into obstacles. So, clearly the number of… So, number of vertices. So, if you go back to the graph terminology the number of vertices is 4 n corner points number of edges is bounded by 4 n choose to. So, about let us say about some order n square So, if this is or let me use v square etcetera So, now, you are in the in the graph world and once we are in the graph world some of you find it easier to deal with So, we have a graph with number of vertices I am saying 4 n vertices where n is the number of obstacles. So, this is a special case of course, we have the obstacles all rectangles essentially the corner points. So, all they have obstacles will be the set of vertices, and the number of edges is less than or let us say I will use big o notation, is big O of n square and you still need another parameter which is basically the weights on the edges So, the weights on edges should be what, Euclidean distance. So, the Euclidean distance between the end points let us say v 1, v 2 and every such end point or corner point, v 1 is defined by a coordinate pair. So, there is some x coordinate of v 1 and there is some y coordinate of v 1 So, when I am saying refer to the Euclidean distance between the end points v 1 and v 2, essentially we are finding the distance between and the corresponding v 2 and y v 2.So the Euclidian distance between these two points should be the weights on the edges So, there is a formula for the Euclidean distance; the square root of the sum of the difference of the… So, x v 1 v 2 minus v 1 square plus y v 2 minus y v 1 square the square root of that And if you do not like square roots you may not even want to do the square roots you compute the shortest paths in this squares because after it is just a relative computation you want to know the shortest path you may not want to know the exact distance. So, we are happy to find a shortest path. So that is it. So now, you are you done your favorite shortest path algorithm on this graph. So, from that space of infinite possible paths you got it down to a very finite description some something like that we are familiar with it in terms of graphs and then you run your shortest path algorithm algorithm can be done because actually all the distance are non-negative So, any questions till now I mean. So, what did we do, we I mean if you want to write a formal proof whatever I have said can be", null, "", null, "", null, "", null, "" ]
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https://in.mathworks.com/matlabcentral/answers/362554-find-index-of-first-zero-searching-from-left-first-column-first-row-then-find-index-of-first-zero-s
[ "# Find index of first zero searching from left first column first row, then find index of first zero searching from last column last row\n\n40 views (last 30 days)\nmonkey_matlab on 22 Oct 2017\nEdited: Cedric Wannaz on 22 Oct 2017\nHello,\nIn the code below, I am attempting to find the index of first zero searching from the left first column first row, going down the column, then find the index of first zero from last column last row searching up (as shown in the arrows below)?", null, "In the picture above, rstart = 1, cstart = 1, rstop = 10, cstop = 20.\nThis is what I have so far:\nn = 10;\nm = 20;\nM = randi([0 1], n,m);\nfor i = 1:1:n\nfor j = 1:1:m\nif M(i,j) == 0\nrstart = i;\ncstart = j;\nbreak\nend\nend\nend\nfor ii = n:-1:1\nfor jj = m:-1:1\nif M(ii,jj) == 0\nrstop = ii;\ncstop = jj;\nbreak\nend\nend\nend\n\nCedric Wannaz on 22 Oct 2017\nEdited: Cedric Wannaz on 22 Oct 2017\nMATLAB stores data column first in memory. Accessing your 2D array linearly follows this structure. Evaluate\n>> M(:)\nand you will see a column vector that represents the content of the array read column per column. This means that you can FIND the linear index of these 0s using the 'first' (default) or 'last' options of FIND, by operating on M(:):\n>> M = randi( 10, 4, 5 ) > 5\nM =\n4×5 logical array\n1 1 1 1 0\n1 0 1 0 1\n0 0 0 1 1\n1 1 1 0 1\n>> linId = find( M(:) == 0, 1 )\nlinId =\n3\n>> linId = find( M(:) == 0, 1, 'last' )\nlinId =\n17\nWhat remains is to convert linear indices back to subscripts, and you can do this using IND2SUB. For example:\n>> [r,c] = ind2sub( size( M ), find( M(:) == 0, 1, 'last' ))\nr =\n1\nc =\n5" ]
[ null, "https://www.mathworks.com/matlabcentral/answers/uploaded_files/168580/image.png", null ]
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https://stacks.math.columbia.edu/tag/0AWN
[ "Lemma 48.22.4. Let $X/A$ with $\\omega _ X^\\bullet$ and $\\omega _ X$ be as in Example 48.22.1. Then\n\n1. $H^ i(\\omega _ X^\\bullet ) \\not= 0 \\Rightarrow i \\in \\{ -\\dim (X), \\ldots , 0\\}$,\n\n2. the dimension of the support of $H^ i(\\omega _ X^\\bullet )$ is at most $-i$,\n\n3. $\\text{Supp}(\\omega _ X)$ is the union of the components of dimension $\\dim (X)$, and\n\n4. $\\omega _ X$ has property $(S_2)$.\n\nProof. Let $\\delta _ X$ and $\\delta _ S$ be the dimension functions associated to $\\omega _ X^\\bullet$ and $\\omega _ S^\\bullet$ as in Lemma 48.21.2. As $X$ is proper over $A$, every closed subscheme of $X$ contains a closed point $x$ which maps to the closed point $s \\in S$ and $\\delta _ X(x) = \\delta _ S(s) = 0$. Hence $\\delta _ X(\\xi ) = \\dim (\\overline{\\{ \\xi \\} })$ for any point $\\xi \\in X$. Hence we can check each of the statements of the lemma by looking at what happens over $\\mathop{\\mathrm{Spec}}(\\mathcal{O}_{X, x})$ in which case the result follows from Dualizing Complexes, Lemmas 47.16.5 and 47.17.5. Some details omitted. The last two statements can also be deduced from Lemma 48.22.3. $\\square$\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)." ]
[ null ]
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https://www.topcoder.com/challenges/16958
[ "# Marathon Match 94 - Marathon Match 94\n\n## Key Information\n\nRegister\nSubmit\nThe challenge is finished.\n\n## Challenge Overview\n\n### Problem Statement\n\nYou are given an S x S matrix M. Each element of the matrix is an integer between -9 and 9, inclusive.\n\nA permutation of numbers 0..S-1 p0, p1, ... pS-1 defines a new matrix MP as a permutation of rows and columns of the matrix M: the element of the new matrix in row i and column j MPi, j = Mp_i, p_j.\n\nConsider all 4-connected regions of non-zero elements of the matrix MP which are bordered either by the boundaries of the matrix or by zero elements. Your task is to find a permutation which maximizes the sum of the elements multiplied by the square root of the number of elements in one of these connected regions. Note that you can't exclude negative elements at will; all non-zero elements of a region count towards the sum of its elements.\n\nHere is an example for test 0 with p = (4 3 8 7 2 0 1 9 6 5).", null, "### Implementation\n\nYour code should implement a single method permute(int[] m). m[i*S+j] = Mi, j (you can deduce S as sqrt(length(m))). You should return an array of S integers specifying the permutation p.\n\n### Scoring\n\nYour score for an individual test case will be the greatest of (the sum of the elements multiplied by square root of the number of elements) of the connected regions in the permuted matrix MP. If your return has invalid format or no connected regions in the matrix have positive sum of the elements, your score for the test case will be 0.\n\nYour overall score will be calculated in the following way: for each test case where your score is not 0, you get 1 point for each competitor you beat on this test case (i.e., your score on a test case is larger than this competitor's score) and 0.5 points for each competitor you tie with (a tie with yourself is not counted); finally, the sum of points is divided by (the number of competitors - 1), then multiplied by 1000000 and divided by the number of test cases.\n\n#### Tools\n\nAn offline tester is available here. You can use it to test/debug your solution locally. You can also check its source code for exact implementation of test case generation and score calculation. That page also contains links to useful information and sample solutions in several languages.\n\n### Definition\n\n Class: ConnectedComponent Method: permute Parameters: int[] Returns: int[] Method signature: int[] permute(int[] m) (be sure your method is public)\n\n### Notes\n\n-The time limit is 10 seconds per test case (this includes only the time spent in your code). The memory limit is 1024 megabytes.\n-There is no explicit code size limit. The implicit source code size limit is around 1 MB (it is not advisable to submit codes of size close to that or larger). Once your code is compiled, the binary size should not exceed 1 MB.\n-The compilation time limit is 30 seconds. You can find information about compilers that we use and compilation options here.\n-There are 10 example test cases and 100 full submission (provisional) test cases. There will be 2000 test cases in the final testing.\n-The match is rated.\n\n### Constraints\n\n-S will be between 50 and 500, inclusive.\n\n### Examples\n\n0)\n\n `\"1\"`\n```Returns:\n\"seed = 1\n10\n6 -6 4 0 -3 0 0 -1 8 0\n-1 -9 0 1 5 -1 0 0 1 4\n0 9 0 0 3 2 0 0 0 0\n0 0 0 0 -3 0 5 0 1 -1\n0 0 1 -7 0 0 6 0 -7 0\n-7 -5 3 0 0 0 -5 0 0 0\n0 0 -3 3 0 0 0 0 -6 0\n1 5 0 0 6 0 -1 -2 -9 3\n0 0 0 0 0 0 0 0 0 2\n0 -4 0 0 2 -7 -8 0 0 0\n\"```\n1)\n\n `\"2\"`\n```Returns: \"seed = 2\nS = 50\n\"```\n2)\n\n `\"3\"`\n```Returns: \"seed = 3\nS = 500\n\"```\n3)\n\n `\"4\"`\n```Returns: \"seed = 4\nS = 102\n\"```\n4)\n\n `\"5\"`\n```Returns: \"seed = 5\nS = 120\n\"```\n5)\n\n `\"6\"`\n```Returns: \"seed = 6\nS = 411\n\"```\n6)\n\n `\"7\"`\n```Returns: \"seed = 7\nS = 218\n\"```\n7)\n\n `\"8\"`\n```Returns: \"seed = 8\nS = 354\n\"```\n8)\n\n `\"9\"`\n```Returns: \"seed = 9\nS = 357\n\"```\n9)\n\n `\"10\"`\n```Returns: \"seed = 10\nS = 270\n\"```\n\nThis problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2020, TopCoder, Inc. All rights reserved.\n\n## Payments\n\nTopcoder will compensate members in accordance with our standard payment policies, unless otherwise specified in this challenge. For information on payment policies, setting up your profile to receive payments, and general payment questions, please refer to ‌Payment Policies and Instructions.\n\n## LEARN:\n\nTopcoder Challenges Explained" ]
[ null, "http://www.topcoder.com/contest/problem/ConnectedComponent/1.png", null ]
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https://export.arxiv.org/abs/1209.5992?context=hep-ph
[ "Full-text links:\n\nhep-ph\n\n# Title: Predicting the existence of a 2.9 GeV $Df_0(980)$ molecular state\n\nAbstract: A $D$-like meson resonance with mass around 2.9 GeV has been found in the $DK\\bar K$ system using two independent and different model calculations based on: (1) QCD sum rules and (2) solution of Fadeev equations with input interactions obtained from effective field theories built by considering both chiral and heavy quark symmetries. The QCD sum rules have been used to study the $D_{s^*0}(2317) \\bar{K}$ and $D f_0(980)$ molecular currents. A resonance of mass 2.926 GeV is found with the $D f_0(980)$ current. Although a state in the $D_{s^*0}(2317) \\bar{K}$ current is also obtained, with mass around 2.9 GeV, the coupling of this state is found to be two times weaker than the one formed in $D f_0(980)$. On the other hand, few-body equations are solved for the $D K \\bar{K}$ system and its coupled channels with the input $t$-matrices obtained by solving Bethe-Salpeter equations for the $D K$, $D\\bar{K}$ and $K \\bar{K}$ subsystems. In this study a $D$-like meson with mass 2.890 GeV and full width $\\sim$ 55 MeV is found to get dynamically generated when $D K \\bar{K}$ gets reorganized as $D f_0(980)$. However, no clear signal appears for the $D_{s^*0}(2317) \\bar{K}$ configuration. The striking similarity between the results obtained in the two different models indicates strongly towards the existence of a $D f_0 (980)$ molecule with mass nearly 2.9 GeV.\n Comments: Improved error estimation made, which lead to better conclusions. Version accepted for publication in PRD Subjects: High Energy Physics - Phenomenology (hep-ph); Nuclear Theory (nucl-th) DOI: 10.1103/PhysRevD.87.034025 Cite as: arXiv:1209.5992 [hep-ph] (or arXiv:1209.5992v2 [hep-ph] for this version)\n\n## Submission history\n\nFrom: Kanchan Khemchandani [view email]\n[v1] Wed, 26 Sep 2012 16:32:25 GMT (346kb)\n[v2] Mon, 18 Feb 2013 20:23:05 GMT (463kb)\n\nLink back to: arXiv, form interface, contact." ]
[ null ]
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https://perl6advent.wordpress.com/2016/12/10/
[ "# Day 10 — Let’s learn and try double-ended priority queue with Perl 6\n\nHello! My name is Itsuki Toyota. I’m a web developer in Japan.\n\nIn this post, let me introduce Algorithm::MinMaxHeap.\n\n# Introduction\n\nAlgorithm::MinMaxHeap is a Perl 6 implementation of double-ended priority queue (i.e. min-max heap).1\n\nFig.1 shows an example of this data structure:", null, "Fig.1\n\nThis data structure has an interesting property, that is, it has both max-heap and min-heap in a single tree.\n\nMore precisely, the min-max ordering is maintained in a single tree as the below description:\n\n• values stored at nodes on even levels (i.e. min levels) are smaller than or equal to the values stored at their descendants (if any)\n• values stored at nodes on odd levels (i.e. max levels) are greater than or equal to the values stored at their descendants (if any)\n\n# Algorithm::MinMaxHeap Internals\n\nI’ll shortly explain what this double-ended priority queue does internally when some frequently used methods (e.g. insert, pop) are called on.2\nThey maintain the min-max ordering by the following operations:\n\n## insert(\\$item)\n\n1. Pushes the given item at the first available leaf position.\n2. Checks whether it maintains the min-max ordering by traversing the nodes from the leaf to the root. If it finds a violating node, it swaps this node with a proper node and continues the checking operation.\n\n## pop-max\n\n1. Extracts the maximum value node.\n2. Fills the vacant position with the last node of the queue.\n3. Checks whether it maintains the min-max ordering by traversing the nodes from the filled position to the leaf. If it finds a violating node, it swaps this node with a proper node and continues the checking operation.\n\n## pop-min\n\n1. Extracts the minimum value node.\n2. Fills the vacant position with the last node of the queue.\n3. Checks whether it maintains the min-max ordering by traversing the nodes from the filled position to the leaf. If it finds a violating node, it swaps this node with a proper node and continues the checking operation.\n\n# Let’s try !\n\nThen let me illustrate how to use Algorithm::MinMaxHeap.\n\n## Example 1\n\nThe first example is the simplest one.\n\nSource:\n\n```use Algorithm::MinMaxHeap;\n\nmy \\$queue = Algorithm::MinMaxHeap[Int].new; # (#1)\nmy Int @items = (^10).pick(*);\[email protected]; # [1 2 6 4 5 9 0 3 7 8]\n\\$queue.insert(\\$_) for @items; # (#2)\n\\$queue.pop-max.say; # 9 (#3)```\n\nIn this example, `Algorithm::MinMaxHeap[Int].new` creates a queue object, where all of its nodes are type `Int`. (#1)\n\n`\\$queue.insert(\\$_) for @items` inserts the numbers in the list. (#2)\n\n`\\$queue.pop-max` pops a node which stores the maximum value. (#3)\n\n## Example 2\n\nThe second example defines user-defined type constraints using subset:\n\nSource (chimera.p6):\n\n```use Algorithm::MinMaxHeap;\n\nmy subset ChimeraRat of Cool where Num|Rat; # (#1)\nmy \\$queue = Algorithm::MinMaxHeap[ChimeraRat].new;\n\n\\$queue.insert(10e0); # ok\n\\$queue.insert(1/10); # ok\n\\$queue.insert(10); # die # (#2)```\n\nOutput:\n\n```\\$ perl6 chimera.p6\nType check failed in assignment to @!nodes; expected ChimeraRat but got Int (10)\nin method insert at /home/itoyota/.rakudobrew/moar-nom/install/share/perl6/site/sources/240192C19BBAACD01AB9686EE53F67BC530F8545 (Algorithm::MinMaxHeap) line 12\nin block at chimera.p6 line 8```\n\nIn this example, subset `ChimeraRat` is a `Cool`, but only allows `Num` or `Rat`. (#1)\n\nTherefore, when you insert `10` to the queue, it returns the error message and die.\n\nIt’s because `10` is a `Int` object. (#2)\n\n## Example 3\n\nThe third example inserts user-defined classes to the queue.\n\nSource (class.p6):\n\n```use Algorithm::MinMaxHeap;\n\nmy class State {\nalso does Algorithm::MinMaxHeap::Comparable[State]; # (#1)\nhas DateTime \\$.time;\nmethod compare-to(State \\$s) { # (#2)\nif \\$!time == \\$s.time {\nreturn Order::Same;\n}\nif \\$!time > \\$s.time {\nreturn Order::More;\n}\nif \\$!time < \\$s.time {\nreturn Order::Less;\n}\n}\n}\n\nmy @items;\n\[email protected](State.new(time => DateTime.new(:year(1900), :month(6)),\[email protected](State.new(time => DateTime.new(:year(-300), :month(3)),\[email protected](State.new(time => DateTime.new(:year(1963), :month(12)),\[email protected](State.new(time => DateTime.new(:year(2020), :month(6)),\n\nmy Algorithm::MinMaxHeap[Algorithm::MinMaxHeap::Comparable] \\$queue .= new;\n\\$queue.insert(\\$_) for @items;\n\\$queue.pop-max.say until \\$queue.is-empty;```\n\nOutput:\n\n``` `\\$ perl6 class.p6`\n` State.new(time => DateTime.new(2020,6,1,0,0,0), payload => \"Jack\")`\n` State.new(time => DateTime.new(1963,12,1,0,0,0), payload => \"Hanako\")`\n` State.new(time => DateTime.new(1900,6,1,0,0,0), payload => \"Rola\")`\n` State.new(time => DateTime.new(-300,3,1,0,0,0), payload => \"Taro\")````\n\nIn this example, the class `State` does the role `Algorithm::MinMaxHeap::Comparable[State]`, where `Algorithm::MinMaxHeap::Comparable[State]` defines the stub method `compare-to` as:\n`method compare-to(State) returns Order:D { ... };`\n(#1)\n\nTherefore, in this case, the `State` class must define a method `compare-to` so that it accepts a `State` object and returns an `Order:D` object. (#2)\n\n# Footnote\n\n• 1 Atkinson, Michael D., et al. “Min-max heaps and generalized priority queues.” Communications of the ACM 29.10 (1986): 996-1000." ]
[ null, "https://perl6advent.files.wordpress.com/2016/12/minmaxheap.jpg", null ]
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https://hal.archives-ouvertes.fr/hal-01478317
[ "Skip to Main content Skip to Navigation\n\n# Optimal algorithms for smooth and strongly convex distributed optimization in networks\n\n2 SIERRA - Statistical Machine Learning and Parsimony\nDI-ENS - Département d'informatique de l'École normale supérieure, CNRS - Centre National de la Recherche Scientifique, Inria de Paris\n4 DYOGENE - Dynamics of Geometric Networks\nDI-ENS - Département d'informatique de l'École normale supérieure, CNRS - Centre National de la Recherche Scientifique : UMR 8548, Inria de Paris\nAbstract : In this paper, we determine the optimal convergence rates for strongly convex and smooth distributed optimization in two settings: centralized and decentralized communications over a network. For centralized (i.e. master/slave) algorithms, we show that distributing Nesterov's accelerated gradient descent is optimal and achieves a precision $\\varepsilon > 0$ in time $O(\\sqrt{\\kappa_g}(1+\\Delta\\tau)\\ln(1/\\varepsilon))$, where $\\kappa_g$ is the condition number of the (global) function to optimize, $\\Delta$ is the diameter of the network, and $\\tau$ (resp. $1$) is the time needed to communicate values between two neighbors (resp. perform local computations). For decentralized algorithms based on gossip, we provide the first optimal algorithm, called the multi-step dual accelerated (MSDA) method, that achieves a precision $\\varepsilon > 0$ in time $O(\\sqrt{\\kappa_l}(1+\\frac{\\tau}{\\sqrt{\\gamma}})\\ln(1/\\varepsilon))$, where $\\kappa_l$ is the condition number of the local functions and $\\gamma$ is the (normalized) eigengap of the gossip matrix used for communication between nodes. We then verify the efficiency of MSDA against state-of-the-art methods for two problems: least-squares regression and classification by logistic regression.\nKeywords :\nDocument type :\nPreprints, Working Papers, ...\nDomain :\nComplete list of metadata\n\nhttps://hal.archives-ouvertes.fr/hal-01478317\nContributor : Kevin Scaman <>\nSubmitted on : Tuesday, February 28, 2017 - 9:59:55 AM\nLast modification on : Tuesday, May 4, 2021 - 2:06:02 PM\nLong-term archiving on: : Monday, May 29, 2017 - 1:06:46 PM\n\n### Files\n\ndistributed_dual_axiv.pdf\nFiles produced by the author(s)\n\n### Identifiers\n\n• HAL Id : hal-01478317, version 1\n• ARXIV : 1702.08704\n\n### Citation\n\nKevin Scaman, Francis Bach, Sébastien Bubeck, Yin Tat Lee, Laurent Massoulié. Optimal algorithms for smooth and strongly convex distributed optimization in networks. 2017. ⟨hal-01478317⟩\n\nRecord views\n\nFiles downloads" ]
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http://imzifeng.cn/2018/07/notebook-r-3/
[ "# R语言学习笔记（三）- 运算符与函数\n\n2018-07-28 492 次浏览 Comments Off on R语言学习笔记（三）- 运算符与函数\n\n3.1 运算符\n\nR语言中拥有如下几种运算符类型：\n• 算术运算符\n• 关系运算符\n• 逻辑运算符\n• 赋值运算符\n• 其他运算符\n\n3.1.1 算数运算符\n+        两个向量相加\n-         两个向量相减\n两个向量相乘\n/         将第一个向量与第二个向量相除\n％％    两个向量求余\n％/％   两个向量相除求商\n^        将第二向量作为第一向量的指数\n\n3.1.2 关系运算符\n\n> 大于                     <小于\n\n== 等于                  != 不等于\n\n= 大于等于           <= 小于等于\n\n3.1.3 逻辑运算符\n& 逻辑 与 运算符，只有当两个元素均为TRUE时，输出为TRUE\n| 逻辑 或 运算符，两个元素中有一个为TRUE时，输出TRUE\n! 逻辑 非 运算符，给出每个元素相反的逻辑值\n\n3.1.4 赋值运算符\n\n<−\nor\n=\nor\n<<− 右分配： ->\nor\n->>\n\n3.1.5 其他运算符\n\n: 冒号运算符。 它为向量按顺序创建一系列数字。\n\nv <- 2:8\nprint(v)\n\n2 3 4 5 6 7 8\n\n%in% 此运算符用于标识元素是否属于向量。\n\nv1 <- 8\nv2 <- 12\nt <- 1:10\nprint(v1 %in% t)\nprint(v2 %in% t)\n\nTRUE\nFALSE\n\n%*% 此运算符用于将矩阵与其转置相乘。\n\nM = matrix( c(2,6,5,1,10,4), nrow = 2,ncol = 3,byrow = TRUE)\nt = M %*% t(M)\nprint(t)\n\n[,1] [,2]\n[1,] 65 82\n[2,] 82 117\n\n3.2 函数\n\n3.2.1 函数定义\n\nfunction_name <- function(arg_1, arg_2, ...) {\nFunction body\n}\n\n3.2.2函数组件\n\n 函数名称 -这是函数的实际名称。 它作为具有此名称的对象存储在R环境中。\n 参数 -参数是一个占位符。 当函数被调用时，你传递一个值到参数。 参数是可选的; 也就是说，一个函数可能不包含参数。 参数也可以有默认值。\n 函数体 -函数体包含定义函数的功能的语句集合。\n 返回值 -函数的返回值是要评估的函数体中的最后一个表达式。\nR语言有许多内置函数，可以在程序中直接调用而无需先定义它们。我们还可以创建和使\n\n3.2.3 内置功能\n\n# Create a sequence of numbers from 32 to 44.\nprint(seq(32,44))\n\n# Find mean of numbers from 25 to 82.\nprint(mean(25:82))\n\n# Find sum of numbers frm 41 to 68.\nprint(sum(41:68))\n\n 32 33 34 35 36 37 38 39 40 41 42 43 44\n 53.5\n 1526\n\n3.2.4 函数的其他常用功能\n\n# Create a function with arguments.\nnew.function <- function(a = 3, b = 6) {\nresult <- a * b\nprint(result)\n}\n\n# Create a function with arguments.\nnew.function <- function(a,b,c) {\nresult <- a * b + c\nprint(result)\n}\n\n# Call the function by position of arguments.\nnew.function(5,3,11)\n\n# Call the function by names of the arguments.\nnew.function(a = 11, b = 5, c = 3)\n\n3.2.5 函数的延迟评估\n\n# Create a function with arguments.\nnew.function <- function(a, b) {\nprint(a^2)\nprint(a)\nprint(b)\n}\n\n# Evaluate the function without supplying one of the arguments.\nnew.function(6)\n\n 36\n 6\nError in print(b) : argument \"b\" is missing, with no default", null, "", null, "" ]
[ null, "https://bshare.optimix.asia/barCode", null, "http://photo.imzifeng.cn/wp-content/uploads/2018/06/2018060817381628.jpg", null ]
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https://www.naukrijobsindeed.com/maths-syllabus-class-12-term-2-pdf-2021-22/
[ "Maths Syllabus Class 12 term 2 | CBSE Class 12 Maths term 2 Reduced Syllabus PDF Free Download\n\nMaths Syllabus Class 12 term 2. Complete Details of CBSE Class 12 Maths term 2 Reduced Syllabus 2021-22. Full List of deleted questions and exercises from each chapter.\n\nIn this post, we have shared the complete details of the Maths Syllabus Class 12 term 2. We have also attached the official syllabus PDF by CBSE. And a list of exercises and questions deleted from the Maths Syllabus Class 12 term 2 PDF Free Download.\n\nMaths Class 12 term 2 Syllabus\n\nThere are a total of 6 chapters in the CBSE Maths Syllabus Class 12 term 2. 3 chapters from Calculus, 2 chapters from Vectors and 3-D Geometry, and 1 chapter from the unit Probability.\n\nCBSE Class 12 Maths term 2 Reduced Syllabus\n\nHere is the list of deleted topics, exercises, and questions from the Maths Syllabus Class 12 term 2.\n\nChapter 7 – Integrals\n\n∫ √ax2 + bx + c dx\n∫(ax + b)√ax2 + bx + c dx\n\n• Definite integrals as a limit of a sum\n• Exercise 7.7 – Q3-Q7\n• Exercise 7.8\n\nChapter 8 – Application of Integrals\n\n• Area between the curves listed in Integrals\n\nChapter 9 – Differential Equations\n\n• Formation of differential equation whose general solution is given\n• Solutions of linear differential equation of the type – dx/dy +px = q, where p and q are functions of y.\n• Exercise 9.3\n• Exercise 9.6 – Q11, Q12, Q19\n• Miscellaneous Exercise – Q3, Q5, Q17\n\nChapter 10 – Vector Algebra\n\n• Scalar Triple Product of vectors (Do it, because it comes in 3D Geometry Coplanarity of 2 Lines)\n• Supplementary\n\nChapter 11 – Three Dimensional Geometry\n\n• Angle between (i) two lines, (ii) two planes, (iii) a line and a plane\n• Exercise 11.2 – Q10, Q11\n• Exercise 11.3 – Q12, Q13\n• Miscellaneous Exercise – Q3, Q5\n\nChapter 13 – Probability\n\n• Mean and Variance of random variable. Binomial probability distribution.\n• Exercise 13.4\n• Exercise 13.5\n• Miscellaneous Exercise – Q2, Q3" ]
[ null ]
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https://defelement.com/elements/bogner-fox-schmitt.html
[ "an encyclopedia of finite element definitions\n\n# Bogner–Fox–Schmitt\n\n Orders $$k=3$$ Reference elements quadrilateral Polynomial set $$\\mathcal{Q}_{k}$$↓ Show polynomial set definitions ↓ DOFs On each vertex: point evaluations,point evaluations of derivatives in coordinate directions, and point evaluation of mixed second derivative Number of DOFs quadrilateral: $$16$$ Mapping identity continuity Function values and derivatives are continuous. Categories Scalar-valued elements\n\n## Implementations\n\n Symfem \"BFS\"↓ Show Symfem examples ↓\n\n## Examples\n\n quadrilateralorder 3", null, "(click to view basis functions)\n\n## References\n\n• Bogner, F. K., Fox, R. L., and Schmit, L. A. The generation of interelement compatible stiffness and mass matrices by the use of interpolation formulae, Proceedings of the Conference on Matrix Methods in Structural Mechanics, 397–444, 1965. [BibTeX]\n\n## DefElement stats\n\n Element added 06 March 2021 Element last updated 02 August 2022" ]
[ null, "https://defelement.com/img/element-Bogner-Fox-Schmit-quadrilateral-3-dofs.png", null ]
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https://www.padmad.org/2014/07/time-work-basic-question-solving-11.html
[ "## Thursday, July 10, 2014\n\n### Time & Work Basic Question Solving - 11 July 2014\n\n1. X is twice good a workman as Y and together they finish a piece of work in 18 days. In how many days will X alone finish the work ?\n\nSolution - According to the condition given if X is twice good a workman as Y then the time taken by X to finish a particular work would be half of the time taken by Y to finish the same work.\n\nSo if we suppose that Y takes 2k days to finish a work then X would take k days to finish the same work.\n\nSolving,  1/(2k) + 1/(k) = 1/18\n\nk = 27 days\n\nAnswer - X would finish the work in 27 days.\n\n2. If 10 men complete a work in 12 days. How many days would be taken by 18 men to complete the same work ?\n\nSolution - Suppose the time taken by 18 men to complete the same work is t days.\n\nUsing the concept of Mandays -\n\n10*12 = 18*t\n\nt = 20/3 days\n\nAnswer - 18 men would take 20/3 days to complete the same work.\n\n3. If 100 men can do 100 jobs in 100 days, 1 men can do one job in how many days ?\n\nSolution - Using the concept M*D = W\n\n(100*100)/(1*D) = 100/1\n\nSolving we get D = 100 days\n\nAnswer - 1 men can do one job in 100 days.\n\n4. 1 man or 2 boys or 3 girls can do a piece of work in 121 days. In how many days can one man, one girl and one boy do the same work ?\n\nSolution - Let one man does m units of work in 1 day.\n\nLet one boy does b units of work in 1 day.\n\nLet one girl does g units of work in 1 day.\n\nSo 1m*121 = 2b*121 = 3g*121 = W, where W is the total units of work.\n\nm = W/121 , b= W/242 , g = W/ 363\n\nSo 1 man, 1 girl and 1 boy can do the same work in   (W) / ((W/121) + (W/242) + (W/363)) = 66\n\nAnswer - So 1 man, 1 girl and 1 boy can do the same work in 66 days.\n\n5. If the ratio of work efficiency of P & Q is 6:5 and that of Q & R is 6:5. If P can complete the piece of work in 2 days, In how many days the same work can be completed by Q & R separately ?\n\nSolution - If the work efficiency of P & Q is 6:5 then time taken would be in the ratio 5:6\n\nIf the work efficiency of Q & R is 6:5 then time taken would be in the ratio 5:6\n\nLet the time taken by P = 5K and by Q = 6K, where K is some constant.\n\n5K = 2, So 6K = 12/5 Days\n\nSo the time taken by Q to complete the work is 12/5 days\n\nNow let the time taken by Q = 5X and by R = 6X, where X is some constant.\n\nQ = 12/5 days, So R = 72/25 days\n\nAnswer - So the time taken by R to complete the work is 72/25 days.\n\n1.", null, "Can u plz explain. Soln of question 3\n\n2.", null, "Its (M1*H1*D1)/W1 = (M2*H2*D2)/W2 just substitute value M=men H =hours D=days W=work accordingly\n\n1.", null, "This comment has been removed by a blog administrator.\n\n2.", null, "This comment has been removed by a blog administrator.\n\n3.", null, "nice solution." ]
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https://www.rfwireless-world.com/calculators/parabolic-dish-antenna-calculator.html
[ "# RF Wireless World\n\n## Home of RF and Wireless Vendors and Resources\n\nOne Stop For Your RF and Wireless Need\n\n## Parabolic Antenna calculator | Parabolic dish antenna calculator\n\nThis page of converters and calculators section covers Parabolic Antenna calculator. The gain,half power beamwidth and effective aperture are calculated by parabolic antenna calculator. Tt takes operating frequency(GHz) and diameter as inputs.\n\nFrequency of operation in GHz (input1) :\n\nAntenna diameter in meter (input2) :\n\nAntenna Gain in dB (Output1):\n\nHalf Power Beamwidth in degrees (Output2):\n\nEffective Aperture in meter2 (Output3):\n\nParabolic antenna calculator example:\nINPUTS: Frequency = 14.5 GHz, Diameter = 3.5 meter\nOUTPUTS: Gain = 52.3 dB, Beamwidth= 0.35 degrees, Aperture= 5.77 meter2\n\n### Parabolic Antenna calculator equation\n\nParabolic dish is widely used as satellite TV channel receiver across the world. It uses cassegrain type of design where in horn antenna is placed at the focal point of the dish. It usually will have two reflectors, a sub reflector and a main reflector. The rays to be trasmitted first emits from the horn antenna and reflects first at the sub reflector then it reflects at the parabolic dish main reflector. Following equation or formula is used for Parabolic Antenna calculator.", null, "### Useful converters and calculators\n\nFollowing is the list of useful converters and calculators.\n\ndBm to Watt converter\nStripline Impedance calculator\nMicrostrip line impedance\nAntenna G/T\nNoise temp. to NF" ]
[ null, "https://www.rfwireless-world.com/images/parabolic-dish-antenna-equation.jpg", null ]
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https://ch.mathworks.com/matlabcentral/cody/problems/42597-uicbioe240-problem-1-3/solutions/1067680
[ "Cody\n\n# Problem 42597. UICBioE240 problem 1.3\n\nSolution 1067680\n\nSubmitted on 1 Dec 2016 by jang hyen kim\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nx = [1 1 1 1 1]; y_correct = 5; assert(isequal(your_fcn_name(x),y_correct))%% x = [1 1 1 1 1 1 1]; y_correct = 7; assert(isequal(your_fcn_name(x),y_correct))\n\n2   Pass\nx = ; y_correct = 1; assert(isequal(your_fcn_name(x),y_correct))" ]
[ null ]
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https://www.kseebsolutions.com/kseeb-solutions-class-8-maths-chapter-9-ex-9-4/
[ "# KSEEB Solutions for Class 8 Maths Chapter 9 Commercial Arithmetic Ex 9.4\n\nStudents can Download Maths Chapter 9 Commercial Arithmetic Ex 9.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.\n\n## Karnataka Board Class 8 Maths Chapter 9 Commercial Arithmetic Ex 9.4\n\nQuestion 1.\nSindu sells her scooty for Rs. 28,000 through a broker. The rate of brokerage is 2%. Find the commission that the agent gets and the net amount Sindhu gets.\nPrice of the scooty Rs. 28,000", null, "Commission = Rs. 700\nThe agent gets Rs. 700\nNet amount sindhu gets = 28000 – 700 = Rs. 27,300\n\nQuestion 2.\nA share agent sells 2000 shares at Rs. 45 each and gets the commission at the rate of 1.5%. Find the amount the agent gets.\nThe S.R of 1 share = Rs. 45\nThe S.P. of 2000 shares = 2000 × 45 = Rs. 90,000\nThe rate of commission = 1.5%\n∴ The commission =", null, "$\\frac{1.5 \\times 90000}{100}$ = Rs. 1350.0\nThe agent gets Rs. 1350.\n\nQuestion 3.\nA person insures Rs. 26,000 through an insurance agent. If the agent gets Rs. 650 as the commission. find the rate of commission.\nFor Rs. 26,000 the commission is Rs. 650\nFor Rs. 100 the commission", null, "Rate of commission = 2.5%\n\nQuestion 4.\nA selling agent gets Rs. 10,200 in a month. This includes his monthly salary of Rs. 6000 and 6% commission for the sales. Find the value of the goods he sold.", null, "$\\frac{100 \\times 4200}{6}$" ]
[ null, "https://i0.wp.com/www.kseebsolutions.com/wp-content/uploads/2019/06/KSEEB-Solutions-for-Class-8-Maths-Chapter-9-Commercial-Arithmetic-Ex.-9.4-1.png", null, "https://s0.wp.com/latex.php", null, "https://i2.wp.com/www.kseebsolutions.com/wp-content/uploads/2019/06/KSEEB-Solutions-for-Class-8-Maths-Chapter-9-Commercial-Arithmetic-Ex.-9.4-2.png", null, "https://s0.wp.com/latex.php", null ]
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http://paste.tclers.tk/5544?v=raw
[ "### Posted to tcl by oldlaptop at Tue Sep 14 13:00:22 GMT 2021view pretty\n\n```\\$ valgrind ./sqltclsh\n==5160== Memcheck, a memory error detector\n==5160== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.\n==5160== Using Valgrind-3.16.1 and LibVEX; rerun with -h for copyright info\n==5160== Command: ./sqltclsh\n==5160==\n% sqlite3 db :memory:\n% db transaction { db close }\n==5160== Invalid read of size 1\n==5160== at 0x144BB0: sqlite3SafetyCheckOk (sqltclsh.c:33318)\n==5160== by 0x1A3411: sqlite3_exec (sqltclsh.c:126586)\n==5160== by 0x1C6071: DbTransPostCmd (sqltclsh.c:240427)\n==5160== by 0x48B0E31: TclNRRunCallbacks (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B2CF3: ??? (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3462: Tcl_EvalEx (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3485: Tcl_Eval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3B52: Tcl_GlobalEval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x113AAF: main (sqltclsh.c:243200)\n==5160== Address 0x4de8f9d is 109 bytes inside a block of size 712 free'd\n==5160== at 0x48399AB: free (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)\n==5160== by 0x1763C3: sqlite3Close (sqltclsh.c:166915)\n==5160== by 0x1763C3: sqlite3Close (sqltclsh.c:166869)\n==5160== by 0x1972DC: sqlite3_close (sqltclsh.c:166958)\n==5160== by 0x1972DC: DbDeleteCmd (sqltclsh.c:239725)\n==5160== by 0x48AEDD5: Tcl_DeleteCommandFromToken (in /usr/lib/libtcl8.6.so)\n==5160== by 0x1CA502: DbObjCmd (sqltclsh.c:241416)\n==5160== by 0x48B0E31: TclNRRunCallbacks (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B2CF3: ??? (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3462: Tcl_EvalEx (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3485: Tcl_Eval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3B52: Tcl_GlobalEval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x113AAF: main (sqltclsh.c:243200)\n==5160== Block was alloc'd at\n==5160== at 0x483877F: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)\n==5160== by 0x143AE2: sqlite3MemMalloc (sqltclsh.c:24557)\n==5160== by 0x125767: sqlite3Malloc (sqltclsh.c:28283)\n==5160== by 0x125767: sqlite3Malloc (sqltclsh.c:28269)\n==5160== by 0x125767: sqlite3MallocZero (sqltclsh.c:28522)\n==5160== by 0x1919BE: openDatabase (sqltclsh.c:168845)\n==5160== by 0x1922FB: sqlite3_open_v2 (sqltclsh.c:169130)\n==5160== by 0x1922FB: DbMain (sqltclsh.c:243013)\n==5160== by 0x48B0E31: TclNRRunCallbacks (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B2CF3: ??? (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3462: Tcl_EvalEx (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3485: Tcl_Eval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3B52: Tcl_GlobalEval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x113AAF: main (sqltclsh.c:243200)\n==5160==\n==5160== Invalid read of size 1\n==5160== at 0x144C21: sqlite3SafetyCheckSickOrOk (sqltclsh.c:33331)\n==5160== by 0x144C21: sqlite3_errmsg (sqltclsh.c:168219)\n==5160== by 0x1C608F: DbTransPostCmd (sqltclsh.c:240439)\n==5160== by 0x48B0E31: TclNRRunCallbacks (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B2CF3: ??? (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3462: Tcl_EvalEx (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3485: Tcl_Eval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3B52: Tcl_GlobalEval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x113AAF: main (sqltclsh.c:243200)\n==5160== Address 0x4de8f9d is 109 bytes inside a block of size 712 free'd\n==5160== at 0x48399AB: free (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)\n==5160== by 0x1763C3: sqlite3Close (sqltclsh.c:166915)\n==5160== by 0x1763C3: sqlite3Close (sqltclsh.c:166869)\n==5160== by 0x1972DC: sqlite3_close (sqltclsh.c:166958)\n==5160== by 0x1972DC: DbDeleteCmd (sqltclsh.c:239725)\n==5160== by 0x48AEDD5: Tcl_DeleteCommandFromToken (in /usr/lib/libtcl8.6.so)\n==5160== by 0x1CA502: DbObjCmd (sqltclsh.c:241416)\n==5160== by 0x48B0E31: TclNRRunCallbacks (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B2CF3: ??? (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3462: Tcl_EvalEx (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3485: Tcl_Eval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3B52: Tcl_GlobalEval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x113AAF: main (sqltclsh.c:243200)\n==5160== Block was alloc'd at\n==5160== at 0x483877F: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)\n==5160== by 0x143AE2: sqlite3MemMalloc (sqltclsh.c:24557)\n==5160== by 0x125767: sqlite3Malloc (sqltclsh.c:28283)\n==5160== by 0x125767: sqlite3Malloc (sqltclsh.c:28269)\n==5160== by 0x125767: sqlite3MallocZero (sqltclsh.c:28522)\n==5160== by 0x1919BE: openDatabase (sqltclsh.c:168845)\n==5160== by 0x1922FB: sqlite3_open_v2 (sqltclsh.c:169130)\n==5160== by 0x1922FB: DbMain (sqltclsh.c:243013)\n==5160== by 0x48B0E31: TclNRRunCallbacks (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B2CF3: ??? (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3462: Tcl_EvalEx (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3485: Tcl_Eval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3B52: Tcl_GlobalEval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x113AAF: main (sqltclsh.c:243200)\n==5160==\n==5160== Invalid read of size 1\n==5160== at 0x144BB0: sqlite3SafetyCheckOk (sqltclsh.c:33318)\n==5160== by 0x1A3411: sqlite3_exec (sqltclsh.c:126586)\n==5160== by 0x1C60BA: DbTransPostCmd (sqltclsh.c:240442)\n==5160== by 0x48B0E31: TclNRRunCallbacks (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B2CF3: ??? (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3462: Tcl_EvalEx (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3485: Tcl_Eval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3B52: Tcl_GlobalEval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x113AAF: main (sqltclsh.c:243200)\n==5160== Address 0x4de8f9d is 109 bytes inside a block of size 712 free'd\n==5160== at 0x48399AB: free (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)\n==5160== by 0x1763C3: sqlite3Close (sqltclsh.c:166915)\n==5160== by 0x1763C3: sqlite3Close (sqltclsh.c:166869)\n==5160== by 0x1972DC: sqlite3_close (sqltclsh.c:166958)\n==5160== by 0x1972DC: DbDeleteCmd (sqltclsh.c:239725)\n==5160== by 0x48AEDD5: Tcl_DeleteCommandFromToken (in /usr/lib/libtcl8.6.so)\n==5160== by 0x1CA502: DbObjCmd (sqltclsh.c:241416)\n==5160== by 0x48B0E31: TclNRRunCallbacks (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B2CF3: ??? (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3462: Tcl_EvalEx (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3485: Tcl_Eval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3B52: Tcl_GlobalEval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x113AAF: main (sqltclsh.c:243200)\n==5160== Block was alloc'd at\n==5160== at 0x483877F: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)\n==5160== by 0x143AE2: sqlite3MemMalloc (sqltclsh.c:24557)\n==5160== by 0x125767: sqlite3Malloc (sqltclsh.c:28283)\n==5160== by 0x125767: sqlite3Malloc (sqltclsh.c:28269)\n==5160== by 0x125767: sqlite3MallocZero (sqltclsh.c:28522)\n==5160== by 0x1919BE: openDatabase (sqltclsh.c:168845)\n==5160== by 0x1922FB: sqlite3_open_v2 (sqltclsh.c:169130)\n==5160== by 0x1922FB: DbMain (sqltclsh.c:243013)\n==5160== by 0x48B0E31: TclNRRunCallbacks (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B2CF3: ??? (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3462: Tcl_EvalEx (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3485: Tcl_Eval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x48B3B52: Tcl_GlobalEval (in /usr/lib/libtcl8.6.so)\n==5160== by 0x113AAF: main (sqltclsh.c:243200)\n==5160==\nError: bad parameter or other API misuse\n% ==5160==\n==5160== HEAP SUMMARY:\n==5160== in use at exit: 438,676 bytes in 41 blocks\n==5160== total heap usage: 148 allocs, 107 frees, 516,661 bytes allocated\n==5160==\n==5160== LEAK SUMMARY:\n==5160== definitely lost: 0 bytes in 0 blocks\n==5160== indirectly lost: 0 bytes in 0 blocks\n==5160== possibly lost: 404,480 bytes in 22 blocks\n==5160== still reachable: 34,196 bytes in 19 blocks\n==5160== suppressed: 0 bytes in 0 blocks\n==5160== Rerun with --leak-check=full to see details of leaked memory\n==5160==\n==5160== For lists of detected and suppressed errors, rerun with: -s\n==5160== ERROR SUMMARY: 3 errors from 3 contexts (suppressed: 0 from 0)```" ]
[ null ]
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https://physics.stackexchange.com/questions/404912/is-a-dipole-fusion-reactor-feasible
[ "# Is a dipole fusion reactor feasible?\n\nIt is well known that the earth's dipole-like magnetic field is able to confine charged particles to paths that spiral along the field lines. That's why we have auroras and the Van Allen radiation belts. Could we use the same idea to confine a fusion plasma?", null, "This isn't a new idea; several research groups have investigated similar proposals over the last few decades. Most notably, a team at MIT made a fairly serious attempt to build a dipole fusion reactor, the LDX (Levitated Dipole Experiment), but abandoned the project in 2011. There has also been some research on storing positron-electron plasmas in dipole fields. Overall, the consensus in the literature seems to be that plasma confinement in a magnetic dipole field is MHD stable. (1 2 3) I won't try to reproduce all of the results in those papers, but here's a quick summary of the main argument:\n\nSuppose that the dipole field is created by a magnetically levitated current-carrying ring, and that the plasma in the field is in MHD equilibrium with zero pressure at the surface of the ring. (Numerical simulations confirm that an equilibrium solution exists. The pressure gradient from the interior of the plasma to the edge of the ring is opposed by the Lorentz force from the diamagnetic current.) To find out if the equilibrium is stable, we can apply the extended energy principle. If the plasma is perturbed by a small amount $\\tilde{\\xi}(\\mathbf{r}, t)$, the total energy of the plasma changes by\n\n\\begin{align} \\delta W_F = \\frac{1}{2} \\int_P d\\mathbf{r} \\left[ \\frac{|\\mathbf{Q_{\\perp}}|^2}{\\mu_0} + \\frac{B^2}{2\\mu_0}|\\nabla \\cdot \\xi_{\\perp} + 2\\xi_{\\perp} \\cdot \\kappa|^2 + \\gamma p |\\nabla \\cdot \\xi_{\\perp}|^2 \\\\ - 2(\\xi_{\\perp} \\cdot \\nabla p)(\\kappa \\cdot \\xi_{\\perp}^*) - J_{\\parallel}(\\xi_{\\perp}^* \\times \\mathbf{b}) \\cdot \\mathbf{Q}_{\\perp} \\right] \\end{align}\n\nwhere $\\mathbf{Q}$ is the first-order perturbation in the magnetic field $\\mathbf{B}$; $\\mathbf{Q}_{\\perp}$ is the component perpendicular to the $\\mathbf{B}$-field; $\\kappa = \\mathbf{b} \\cdot \\nabla \\mathbf{b}$ is the magnetic field curvature; $\\xi ^*$ is the complex conjugate of $\\xi$, which is just $\\xi$ here since the perturbation is real. If the change in energy is positive, the perturbation shrinks; if it is negative, the perturbation grows.\n\nApplying this equation to a magnetic dipole-confined plasma, we notice that the equilibrium poloidal current $J_{\\parallel}$ is 0, so the last term vanishes. That leaves us with three stabilizing terms representing the energy required to bend the magnetic field lines, compress the magnetic field, and compress the plasma itself; and one destabilizing term representing the curvature drive. Thus, to ensure that our plasma is MHD stable, all we need to do is make sure that the stabilizing terms are larger than the destabilizing term. We can do this by increasing $B$; by making the ring larger (thus decreasing $\\kappa$); or by reducing the pressure gradient $\\nabla p$ in the most instability-prone regions of the initial equilibrium.\n\nSince we have several degrees of freedom to play with, it's not particularly surprising that stable solutions exist with reasonable parameter values. Indeed, Garnier et al. found a specific example using numerical methods and reported it in this paper. A paper by Krasheninnikov justifies this conclusion on a theoretical basis, showing that the plasma can be stabilized against interchange and ballooning modes if certain conditions are met.\n\nIt's worth noting that many of the most troublesome instabilities that plague tokamak design do not exist in a dipole reactor. In particular, the kink and tearing modes are automatically ruled out because there is no current parallel to the B-field.\n\nNevertheless, despite the obvious technical advantages described above, physicists seem to have lost interest in the dipole fusion reactor. In 2011, all funding was cut to the LDX — Wikipedia says the money was redirected to tokamak research — and it appears that few or no papers have been published on the topic since that date. Assuming this decision was not made for purely political reasons, what problems with the dipole design led MIT and the rest of the fusion community to give up on it? Are there additional instabilities that I'm overlooking? Did the engineering issues (e.g. heat management, keeping the superconducting ring levitated away from the vacuum chamber walls) prove overwhelming?\n\n• the Polywell concept is based also on this idea, but if there are companies developing it, they are exceedingly secretive about it – lurscher May 9 '18 at 17:15\n• @lurscher The polywell is a very different design that relies on a combination of electrostatic and magnetic confinement. – Thorondor May 10 '18 at 0:19\n• Tokamak (and stellarator) simply yielded better confinement times, i.e. they performed better. In an ideal world, with lot's of money available for research, sure, investigating levitating dipole experiments sounds like fun. Note, a series engineering issue is the levitating coil being emerged into the plasma (it needs to be properly shielded). – Alf May 10 '18 at 11:54\n• @Alf I see. Is there a compelling theoretical reason we should expect worse confinement times, or is it more of an experimental result? Also, do you have a source? – Thorondor May 10 '18 at 12:00\n• @Thorondor mostly, it is because they are in an early stage of evolution (the first tokamaks and stellarators were bad as well). I don't have a source at the moment (out of office today) - I'll try to find some resources when I'm back at my desk (and have access to publications). – Alf May 10 '18 at 12:06\n\n2. Getting rid of neutrons can be ''easily'' done by considering a different fusion reaction, an aneutronic reaction, for example D+$^3$He. But those require (a) much higher temperatures, (b) a different method to extract the energy, and (c) in the case of $^3$He the fuel is very rare on earth (this is why science fiction literature tells us to harvest the moon for $^3$He)." ]
[ null, "https://i.stack.imgur.com/x1RUW.png", null ]
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http://www.szielenski.com/2018/05/algorithm-approaches.html
[ "## 5 Methods for Problem Solving\n\nWhen developing new software algorithms, a systematic approach helps. When confronted with a software problem use the five methods below to come up with an algorithm to solve the problem.\n\n#### Examplify\n\nWrite out specific examples of the problem and see if you can derive a general rule from there.\n\n#### Pattern Matching\n\nConsider what problems the algorithm is similar to and try to modify the solution to the related problem to develop an algorithm for this problem.\n\n#### Simplify and Generalize\n\nThis has a multi-step approach. First, we change a constraint such as the data type or amount of data. Doing this helps simplify the problem. Then we solve this new simplified version.\n\n#### Base Case and Build\n\nWe solve the problem first for the base case: n=1. This usually just means recording the correct result. Then we try to solve the problem for n=2, assuming you have the answer for n=1. Next try for n=3 and then for n=4. Eventually we can build a solution that can solve the result for N if we know the correct result for N-1. This often leads to a naturally recursive algorithm.\n\n#### Data Structure and Brainstorm\n\nThis is a hacky approach but often works. We can simply run through a list of data structures and try to apply each one. This approach is useful because solving a problem may be trivial once it occurs to us to use, say, a binary search tree. 2\n\n#### Algorithm Implementation Notes\n\nImplement primitive data structures first such as Node or Record, then the data structures that use them, such as List or HashTable. Implement the insert function first, then search, remove, size, print, etc. For recursive functions, implement a shell insert function that then calls a private recursive helper function: “def insert” (public) then “def __insert” (private). The basic template for recursive functions is to do a basic “null” check on input data and then call a recursive helper function. When implementing if/elif/else statements, handle the simplest case first: node is None, then edge cases. Be consistent in how you develop a solution. Write unit tests as you implement each function, that way you catch bugs as you are developing and not all at the end." ]
[ null ]
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https://www.colorhexa.com/0394ed
[ "# #0394ed Color Information\n\nIn a RGB color space, hex #0394ed is composed of 1.2% red, 58% green and 92.9% blue. Whereas in a CMYK color space, it is composed of 98.7% cyan, 37.6% magenta, 0% yellow and 7.1% black. It has a hue angle of 202.8 degrees, a saturation of 97.5% and a lightness of 47.1%. #0394ed color hex could be obtained by blending #06ffff with #0029db. Closest websafe color is: #0099ff.\n\n• R 1\n• G 58\n• B 93\nRGB color chart\n• C 99\n• M 38\n• Y 0\n• K 7\nCMYK color chart\n\n#0394ed color description : Vivid blue.\n\n# #0394ed Color Conversion\n\nThe hexadecimal color #0394ed has RGB values of R:3, G:148, B:237 and CMYK values of C:0.99, M:0.38, Y:0, K:0.07. Its decimal value is 234733.\n\nHex triplet RGB Decimal 0394ed `#0394ed` 3, 148, 237 `rgb(3,148,237)` 1.2, 58, 92.9 `rgb(1.2%,58%,92.9%)` 99, 38, 0, 7 202.8°, 97.5, 47.1 `hsl(202.8,97.5%,47.1%)` 202.8°, 98.7, 92.9 0099ff `#0099ff`\nCIE-LAB 59.261, -0.202, -53.685 25.91, 27.311, 84.022 0.189, 0.199, 27.311 59.261, 53.685, 269.784 59.261, -36.303, -85.424 52.26, -2.957, -58.743 00000011, 10010100, 11101101\n\n# Color Schemes with #0394ed\n\n• #0394ed\n``#0394ed` `rgb(3,148,237)``\n• #ed5c03\n``#ed5c03` `rgb(237,92,3)``\nComplementary Color\n• #03edd1\n``#03edd1` `rgb(3,237,209)``\n• #0394ed\n``#0394ed` `rgb(3,148,237)``\n• #031fed\n``#031fed` `rgb(3,31,237)``\nAnalogous Color\n• #edd103\n``#edd103` `rgb(237,209,3)``\n• #0394ed\n``#0394ed` `rgb(3,148,237)``\n• #ed031f\n``#ed031f` `rgb(237,3,31)``\nSplit Complementary Color\n• #94ed03\n``#94ed03` `rgb(148,237,3)``\n• #0394ed\n``#0394ed` `rgb(3,148,237)``\n• #ed0394\n``#ed0394` `rgb(237,3,148)``\n• #03ed5c\n``#03ed5c` `rgb(3,237,92)``\n• #0394ed\n``#0394ed` `rgb(3,148,237)``\n• #ed0394\n``#ed0394` `rgb(237,3,148)``\n• #ed5c03\n``#ed5c03` `rgb(237,92,3)``\n• #0265a1\n``#0265a1` `rgb(2,101,161)``\n• #0275bb\n``#0275bb` `rgb(2,117,187)``\n• #0384d4\n``#0384d4` `rgb(3,132,212)``\n• #0394ed\n``#0394ed` `rgb(3,148,237)``\n• #0ea1fc\n``#0ea1fc` `rgb(14,161,252)``\n• #27abfc\n``#27abfc` `rgb(39,171,252)``\n• #40b5fd\n``#40b5fd` `rgb(64,181,253)``\nMonochromatic Color\n\n# Alternatives to #0394ed\n\nBelow, you can see some colors close to #0394ed. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #03cfed\n``#03cfed` `rgb(3,207,237)``\n• #03bbed\n``#03bbed` `rgb(3,187,237)``\n• #03a8ed\n``#03a8ed` `rgb(3,168,237)``\n• #0394ed\n``#0394ed` `rgb(3,148,237)``\n• #0381ed\n``#0381ed` `rgb(3,129,237)``\n• #036ded\n``#036ded` `rgb(3,109,237)``\n• #0359ed\n``#0359ed` `rgb(3,89,237)``\nSimilar Colors\n\n# #0394ed Preview\n\nThis text has a font color of #0394ed.\n\n``<span style=\"color:#0394ed;\">Text here</span>``\n#0394ed background color\n\nThis paragraph has a background color of #0394ed.\n\n``<p style=\"background-color:#0394ed;\">Content here</p>``\n#0394ed border color\n\nThis element has a border color of #0394ed.\n\n``<div style=\"border:1px solid #0394ed;\">Content here</div>``\nCSS codes\n``.text {color:#0394ed;}``\n``.background {background-color:#0394ed;}``\n``.border {border:1px solid #0394ed;}``\n\n# Shades and Tints of #0394ed\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000305 is the darkest color, while #f0f9ff is the lightest one.\n\n• #000305\n``#000305` `rgb(0,3,5)``\n• #000f18\n``#000f18` `rgb(0,15,24)``\n• #011b2b\n``#011b2b` `rgb(1,27,43)``\n• #01273f\n``#01273f` `rgb(1,39,63)``\n• #013352\n``#013352` `rgb(1,51,82)``\n• #013f65\n``#013f65` `rgb(1,63,101)``\n• #024b79\n``#024b79` `rgb(2,75,121)``\n• #02588c\n``#02588c` `rgb(2,88,140)``\n• #0264a0\n``#0264a0` `rgb(2,100,160)``\n• #0270b3\n``#0270b3` `rgb(2,112,179)``\n• #037cc6\n``#037cc6` `rgb(3,124,198)``\n• #0388da\n``#0388da` `rgb(3,136,218)``\n• #0394ed\n``#0394ed` `rgb(3,148,237)``\n• #089ffc\n``#089ffc` `rgb(8,159,252)``\n• #1ba7fc\n``#1ba7fc` `rgb(27,167,252)``\n• #2eaefc\n``#2eaefc` `rgb(46,174,252)``\n• #42b6fd\n``#42b6fd` `rgb(66,182,253)``\n• #55bdfd\n``#55bdfd` `rgb(85,189,253)``\n• #69c5fd\n``#69c5fd` `rgb(105,197,253)``\n• #7cccfd\n``#7cccfd` `rgb(124,204,253)``\n• #8fd4fe\n``#8fd4fe` `rgb(143,212,254)``\n• #a3dbfe\n``#a3dbfe` `rgb(163,219,254)``\n• #b6e3fe\n``#b6e3fe` `rgb(182,227,254)``\n• #c9eafe\n``#c9eafe` `rgb(201,234,254)``\n• #ddf2ff\n``#ddf2ff` `rgb(221,242,255)``\n• #f0f9ff\n``#f0f9ff` `rgb(240,249,255)``\nTint Color Variation\n\n# Tones of #0394ed\n\nA tone is produced by adding gray to any pure hue. In this case, #72797e is the less saturated color, while #0394ed is the most saturated one.\n\n• #72797e\n``#72797e` `rgb(114,121,126)``\n• #697c87\n``#697c87` `rgb(105,124,135)``\n• #5f7e91\n``#5f7e91` `rgb(95,126,145)``\n• #56809a\n``#56809a` `rgb(86,128,154)``\n• #4d82a3\n``#4d82a3` `rgb(77,130,163)``\n• #4485ac\n``#4485ac` `rgb(68,133,172)``\n• #3a87b6\n``#3a87b6` `rgb(58,135,182)``\n• #3189bf\n``#3189bf` `rgb(49,137,191)``\n• #288bc8\n``#288bc8` `rgb(40,139,200)``\n• #1f8dd1\n``#1f8dd1` `rgb(31,141,209)``\n• #1590db\n``#1590db` `rgb(21,144,219)``\n• #0c92e4\n``#0c92e4` `rgb(12,146,228)``\n• #0394ed\n``#0394ed` `rgb(3,148,237)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0394ed is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]
[ null ]
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http://keio-ocw.sfc.keio.ac.jp/j/economics/02C-005_j/index.html
[ "", null, "", null, "", null, "#### 慶應義塾OCW >>コース一覧 >> 経済学部>> ADVANCED FINANCE (PCP)（2007春学期）", null, "", null, "【担当教員】\n\n【コース・ガイド】\n\nThe course, which is the sequel to Introduction to finance, deals with an option pricing theory and its exercises. First of all, two-period binomial models are discussed. Moreover, we extend them to multi-period binomial models. Next, in order to learn the Black-Sholes model, we prepare several topics of a probability theory, which include normal distribution, random walk, the central limit theorem, Brownian motion and a basic guide of stochastic differential equations (SDE). Finally, the Black-Sholes model and formula are introduced.\n\n［講義］春季（4～7月）13回\n\n［レベル］学部学生\n\n［フィードバック］\nご意見はこちらまでお寄せください。\n\nコーストップへ＞＞", null, "" ]
[ null, "http://keio-ocw.sfc.keio.ac.jp/j/economics/images/eco_header2_j.jpg", null, "http://keio-ocw.sfc.keio.ac.jp/j/images/course_navi4.gif", null, "http://keio-ocw.sfc.keio.ac.jp/j/images/clear.gif", null, "http://keio-ocw.sfc.keio.ac.jp/j/images/clear.gif", null, "http://keio-ocw.sfc.keio.ac.jp/economics/images/02C-005.gif", null, "http://i.creativecommons.org/l/by-nc-sa/2.1/jp/88x31.png", null ]
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https://momath.org/home/varsity-math/varsity-math-week-134/
[ "", null, "## ________________\n\nTwo puzzles with numbers are on the coach’s workout for today.\n\n## ________________", null, "### Maximization\n\nUse the digits 1, 2, 3, and 4 once and only once to make a mathematical expression. You may use +, -, ×, ÷, exponents, decimal points, and parentheses. No roots, factorials, repeating decimals, or other mathematical functions are permitted.\n\nWhat is the largest mathematical expression possible using these rules?\n\n### Multiplication Table\n\nEach digit in the multiplication table below represents a digit other than itself.\n\nCreate a new table with correct digits.", null, "## Solutions to week 133\n\nIn Maximizing Links, the maximum number of links is 9. For Seven Points, see the diagram below. All lengths shown are 1 foot.", null, "", null, "" ]
[ null, "https://momath.org/wp-content/uploads/2018/04/Varsity-Math-134.png", null, "https://momath.org/wp-content/uploads/2017/10/Math-Symbol-Balls.jpg", null, "https://momath.org/wp-content/uploads/2018/04/MultiplicationTableStraight2.png", null, "https://momath.org/wp-content/uploads/2018/01/Seven-Points.png", null, "https://momath.org/wp-content/uploads/2018/01/Maximizing-Links-Solution.png", null ]
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https://www.statsmodels.org/stable/generated/statsmodels.sandbox.distributions.transformed.Transf_gen.html
[ "# statsmodels.sandbox.distributions.transformed.Transf_gen¶\n\nclass statsmodels.sandbox.distributions.transformed.Transf_gen(kls, func, funcinv, *args, **kwargs)[source]\n\na class for non-linear monotonic transformation of a continuous random variable\n\nMethods\n\n cdf(x, *args, **kwds) Cumulative distribution function of the given RV. entropy(*args, **kwds) Differential entropy of the RV. expect([func, args, loc, scale, lb, ub, …]) Calculate expected value of a function with respect to the distribution by numerical integration. fit(data, *args, **kwds) Return MLEs for shape (if applicable), location, and scale parameters from data. fit_loc_scale(data, *args) Estimate loc and scale parameters from data using 1st and 2nd moments. freeze(*args, **kwds) Freeze the distribution for the given arguments. interval(alpha, *args, **kwds) Confidence interval with equal areas around the median. isf(q, *args, **kwds) Inverse survival function (inverse of sf) at q of the given RV. logcdf(x, *args, **kwds) Log of the cumulative distribution function at x of the given RV. logpdf(x, *args, **kwds) Log of the probability density function at x of the given RV. logsf(x, *args, **kwds) Log of the survival function of the given RV. mean(*args, **kwds) Mean of the distribution. median(*args, **kwds) Median of the distribution. moment(n, *args, **kwds) n-th order non-central moment of distribution. nnlf(theta, x) Return negative loglikelihood function. pdf(x, *args, **kwds) Probability density function at x of the given RV. ppf(q, *args, **kwds) Percent point function (inverse of cdf) at q of the given RV. rvs(*args, **kwds) Random variates of given type. sf(x, *args, **kwds) Survival function (1 - cdf) at x of the given RV. stats(*args, **kwds) Some statistics of the given RV. std(*args, **kwds) Standard deviation of the distribution. support(*args, **kwargs) Return the support of the distribution. var(*args, **kwds) Variance of the distribution.\n\nProperties\n\n random_state Get or set the RandomState object for generating random variates." ]
[ null ]
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https://artofproblemsolving.com/wiki/index.php/2005_AIME_I_Problems/Problem_7
[ "# 2005 AIME I Problems/Problem 7\n\n## Problem\n\nIn quadrilateral", null, "$ABCD,\\ BC=8,\\ CD=12,\\ AD=10,$ and", null, "$m\\angle A= m\\angle B = 60^\\circ.$ Given that", null, "$AB = p + \\sqrt{q},$ where", null, "$p$ and", null, "$q$ are positive integers, find", null, "$p+q.$\n\n## Solution\n\n### Solution 1", null, "$[asy]draw((0,0)--(20.87,0)--(15.87,8.66)--(5,8.66)--cycle); draw((5,8.66)--(5,0)); draw((15.87,8.66)--(15.87,0)); draw((5,8.66)--(16.87,6.928)); label(\"A\",(0,0),SW); label(\"B\",(20.87,0),SE); label(\"E\",(15.87,8.66),NE); label(\"D\",(5,8.66),NW); label(\"P\",(5,0),S); label(\"Q\",(15.87,0),S); label(\"C\",(16.87,7),E); label(\"12\",(10.935,7.794),S); label(\"10\",(2.5,4.5),W); label(\"10\",(18.37,4.5),E); [/asy]$\n\nDraw line segment", null, "$DE$ such that line", null, "$DE$ is concurrent with line", null, "$BC$. Then,", null, "$ABED$ is an isosceles trapezoid so", null, "$AD=BE=10$, and", null, "$BC=8$ and", null, "$EC=2$. We are given that", null, "$DC=12$. Since", null, "$\\angle CED = 120^{\\circ}$, using Law of Cosines on", null, "$\\bigtriangleup CED$ gives", null, "$$12^2=DE^2+4-2(2)(DE)(\\cos 120^{\\circ})$$ which gives", null, "$$144-4=DE^2+2DE$$. Adding", null, "$1$ to both sides gives", null, "$141=(DE+1)^2$, so", null, "$DE=\\sqrt{141}-1$.", null, "$\\bigtriangleup DAP$ and", null, "$\\bigtriangleup EBQ$ are both", null, "$30-60-90$, so", null, "$AP=5$ and", null, "$BQ=5$.", null, "$PQ=DE$, and therefore", null, "$AB=AP+PQ+BQ=5+\\sqrt{141}-1+5=9+\\sqrt{141} \\rightarrow (p,q)=(9,141) \\rightarrow \\boxed{150}$.\n\n### Solution 2", null, "Draw the perpendiculars from", null, "$C$ and", null, "$D$ to", null, "$AB$, labeling the intersection points as", null, "$E$ and", null, "$F$. This forms 2", null, "$30-60-90$ right triangles, so", null, "$AE = 5$ and", null, "$BF = 4$. Also, if we draw the horizontal line extending from", null, "$C$ to a point", null, "$G$ on the line", null, "$DE$, we find another right triangle", null, "$\\triangle DGC$.", null, "$DG = DE - CF = 5\\sqrt{3} - 4\\sqrt{3} = \\sqrt{3}$. The Pythagorean Theorem yields that", null, "$GC^2 = 12^2 - \\sqrt{3}^2 = 141$, so", null, "$EF = GC = \\sqrt{141}$. Therefore,", null, "$AB = 5 + 4 + \\sqrt{141} = 9 + \\sqrt{141}$, and", null, "$p + q = \\boxed{150}$.\n\n### Solution 3", null, "Extend", null, "$AD$ and", null, "$BC$ to an intersection at point", null, "$E$. We get an equilateral triangle", null, "$ABE$. We denote the length of a side of", null, "$\\triangle ABE$ as", null, "$s$ and solve for it using the Law of Cosines:", null, "$$12^2 = (s - 10)^2 + (s - 8)^2 - 2(s - 10)(s - 8)\\cos{60}$$", null, "$$144 = 2s^2 - 36s + 164 - (s^2 - 18s + 80)$$ This simplifies to", null, "$s^2 - 18s - 60=0$; the quadratic formula yields the (discard the negative result) same result of", null, "$9 + \\sqrt{141}$.\n\n### Solution 4\n\nExtend", null, "$BC$ and", null, "$AD$ to meet at point", null, "$E$, forming an equilateral triangle", null, "$\\triangle ABE$. Draw a line from", null, "$C$ parallel to", null, "$AB$ so that it intersects", null, "$AD$ at point", null, "$F$. Then, apply Stewart's Theorem on", null, "$\\triangle CFE$. Let", null, "$CE=x$.", null, "$$2x(x-2) + 12^2x = 2x^2 + x^2(x-2)$$", null, "$$x^3 - 2x^2 - 140x = 0$$ By the quadratic formula (discarding the negative result),", null, "$x = 1 + \\sqrt{141}$, giving", null, "$AB = 9 + \\sqrt{141}$ for a final answer of", null, "$p+q=150$.\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", null, "" ]
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https://www.qsstudy.com/chemistry/self-ionization-or-autoionization-of-water
[ "", null, "# Self-Ionization (or autoionization) of Water\n\nSelf-Ionization (or autoionization) of Water\n\nWater is known to be a non-electrolyte (non-conductor of electricity), although precise measurements indicate that water is a weak conductor of electricity. The conductivity of water arises due to a phenomenon is known as self-ionization. In water two molecules of water may react to give ions as shown below:\n\nH2O (l) + H2O (l) ↔ H3O+ (aq) + OH (aq)\n\nThe equilibrium constant for self-ionization of eater is given by,\n\nK = {[H3O+] [OH]} / [H2O]2 …. …. (1)\n\nSince the concentration of the ions formed by self-ionization is very small, the concentration of H2O remains practically constant and equation (1) can be rewritten as,\n\n[H2O]2 x K = [H3O+] x [OH] …. …. (2)\n\nThe quantities on the left-hand side of equation (2) are constants and maybe, replaced by a new constant term KW, known as the ionic product of water.\n\nKW = [H3O+] x [OH] …. …. (3)", null, "KW is a very important quantity of water. In all ionic equilibrium in water, KW is constant as long as the temperature is constant. In pure water\n\n[H3O+] (or simply [H+]) = [OH]\n\nso that, KW = [H+]2 = [OH]2 … …. …. (4)\n\nThe value of KW has been carefully determined at different temperatures by various methods. The value of KW at 298 K is about 1.0 x10-14 mol2 L-6. Whether the solution is acidic or alkaline the value of [H+] x [OH] will always be 1.0 x 10-14 at 298 K. The ionic product KW of water may be calculated from conductance data as follows:\n\nThe specific conductance of pure water at 298 K is 5.50 x 10-8 ohm-1 cm-1. The ion conductances λ0H+ and λ0OH- at 298 K and at infinite dilutions are 349.8 cm2 ohm-1mol-1 and 198.0 cm2 ohm-1mol-1 respectively. So the molar conductance is-\n\nΛ0 = λ0H+ + λ0OH-\n\n= [349.8 cm2 ohm-1mol-1] + [198.0 cm2 ohm-1mol-1]\n\nUsing equation,\n\nC = 1000k / Λ0 = (1000 x 5.5 x 10-8) / 547.8 = 1.00 x 10-7\n\nwhere c is the concentration of ionized water, As each molecule of H2O gives one H+ ion and one OH ion we can write;\n\nc = [H+] = [OH]\n\nThat is to say that in pure water the concentrations of H+ and OH are equal.\n\nHence,\n\nKW = [H+]2 = (1.00 x 10-7)2\n\n= 1.0 x 10-14\n\nThe ionic product of water at various temperatures is given in Table.\n\nTable: Values of KW at different temperatures", null, "" ]
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https://answers.everydaycalculation.com/simplify-fraction/105-196
[ "Solutions by everydaycalculation.com\n\n## Reduce 105/196 to lowest terms\n\nThe simplest form of 105/196 is 15/28.\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 105 and 196 is 7\n2. Divide both the numerator and denominator by the GCD\n105 ÷ 7/196 ÷ 7\n3. Reduced fraction: 15/28\nTherefore, 105/196 simplified to lowest terms is 15/28.\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\nEquivalent fractions:\n\nMore fractions:" ]
[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]
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https://www.hpmuseum.org/forum/thread-8729-post-76590.html
[ "12c Solving for n\n07-24-2017, 11:27 AM (This post was last modified: 07-28-2017 12:21 AM by Zac Bruce.)\nPost: #1\n Zac Bruce", null, "Junior Member Posts: 37 Joined: Jul 2017\n12c Solving for n\nHi all,\n\nI'm hoping someone can lend their wisdom here. As most of you will well know, when using the 12c to solve for n, it will only solve as an integer, always rounding up. I tried to search but did not come across a solution.\n\nThe user manual offers the following solution for when a PMT amount is involved;\n10.5 g i\n35000 PV\n325 CHS PMT\nn (=328)\n\nTo find the partial last payment;\nFV (181.89)\nRCL PMT (-325)\n+ (-143.11) which is the final fractional payment\n\nTo then make the answer correlate with other financial calculators I came up with the following;\n(with the -143.11 still on screen)\nRCL PMT X><Y %T (44.03) e.g. .4403 of a total payment.\nSo the answer is 327.44, which is what other calculators report.\nThis would be really simple to program, however it's simple enough to just remember or re-logic my way through\n\nHowever, when a PMT value is not involved, I can't find a simple way of performing the same calculation. My math just isn't strong enough.\ne.g. PV 1250, FV 2500, i=9% compounded annually, solve for n?\n12c gives 9, and when you re-solve for FV it gives 2714.87\nOther calculators will give n=8.04\n\nIs there a simple way to calculate what the extra 214.87 represents in terms of time?\nI have read that there was a program solution for this, but could not find it in part 3 of the manual or in the solutions handbook. If someone could point it out for me that would be greatly appreciated.\n\nMany thanks,\n\nZac\n\nEDIT to fix values.\n07-24-2017, 01:10 PM (This post was last modified: 07-24-2017 03:31 PM by Dieter.)\nPost: #2\n Dieter", null, "Senior Member Posts: 2,397 Joined: Dec 2013\nRE: 12c Solving for n\nNote: edited to address some more questions.\n\n(07-24-2017 11:27 AM)Zac Bruce Wrote:  To then make the answer correlate with other financial calculators I came up with the following;\n(with the -143.11 still on screen)\nRCL PMT X><Y %T (44.03) e.g. .4403 of a total payment.\nSo the answer is 33.44, which is what other calculators report.\n\n?!? Where do you get this 33,44 from? I'd say this is 327 full periods plus 0,4403 = 327,44 periods.\n\nBut this is only an approximate answer, it is a linear interpolation of a nonlinear function. The true result here is not 327,4403 but 327,4393. Rounded to two decimals the difference in this particular case does not show up. But don't try this with smaller n. Anyway, at least you get an approximate result.\n\n(07-24-2017 11:27 AM)Zac Bruce Wrote:  However, when a PMT value is not involved, I can't find a simple way of performing the same calculation. My maths just isn't strong enough.\ne.g. PV 1250, FV 2500, i=9% compounded annually, solve for n?\n12c gives 9, and when you re-solve for FV it gives 2714.87\nOther calculators will give n=8.04\n\nWithout PMT the solution is trivial. If you really need the fractional answer for n this can be calculated directly, either manually or with a small program:\n\nRCL FV\nRCL PV\n/\nCHS\ng LN\n1\nRCL i\n%\n+\ng LN\n/\n\nFor the example this yields n = 8,0432.\n\nOf course you can always use your own little program for calculating n without rounding up to the next higher integer. The formulas are known, so just \"code your own\". ;-) As shown, the PMT=0 case is trivial, but even with PMT the formula is quite simple. For your example (no FV, \"end\" mode assumed) and i decimal (i.e. 0,00875 here) n = – ln(1 + i·PV/PMT) / ln(1+i).\n\n(07-24-2017 11:27 AM)Zac Bruce Wrote:  Is there a simple way to calculate what the 714.87 represents in terms of time?\n\nWhat 714,87? I just see 2714,87 which is the FV after 9 periods.\n\nDieter\n07-25-2017, 02:38 AM (This post was last modified: 07-25-2017 08:56 AM by Zac Bruce.)\nPost: #3\n Zac Bruce", null, "Junior Member Posts: 37 Joined: Jul 2017\nRE: 12c Solving for n\nIt was late last night when I posted, so yes the figure you gave for the first question is correct and I have edited the post. The second figure I meant 214.87 I.e the amount over what we expected or wanted.\n\nI'm guessing that there is not a simple solution similar to the first problem.\n\nCan you explain a little more about it being only an approximate answer? Is this the case will other financial calculators? And what causes it to be so?\n\nI notice using your program method it give the same answer as my HP Prime,(edit) and using the program solving for n that Paul suggested below (which is just a more complex version of yours to allow for PMT values) it gives the same values as my Prime. These answers also correspond with the correct value you suggested in the first example.\n\nZac\n07-25-2017, 04:52 AM\nPost: #4", null, "paul0207", null, "Junior Member Posts: 7 Joined: May 2014\nRE: 12c Solving for n\nHi Zac,\nPlease check \"HP 12c Calculator - Actuarial Calculations\"\nat https://support.hp.com/us-en/document/bpia5043 for a program to calculate n as a real number.\n\nRegards,\n\nPaul\n07-25-2017, 05:51 AM (This post was last modified: 07-25-2017 08:58 AM by Zac Bruce.)\nPost: #5\n Zac Bruce", null, "Junior Member Posts: 37 Joined: Jul 2017\nRE: 12c Solving for n\n(07-25-2017 04:52 AM)paul0207 Wrote:  Hi Zac,\nPlease check \"HP 12c Calculator - Actuarial Calculations\"\nat https://support.hp.com/us-en/document/bpia5043 for a program to calculate n as a real number.\n\nRegards,\n\nPaul\n\nThank-you for this Paul,\nSomehow I had not come across those articles before in my searches, I have a dozen of them that were presented as training modules that I believe I downloaded from some HP fan site, but there is a great deal more contained there.\n\nRegards,\nZac\n07-25-2017, 09:44 AM (This post was last modified: 07-25-2017 01:02 PM by Dieter.)\nPost: #6\n Dieter", null, "Senior Member Posts: 2,397 Joined: Dec 2013\nRE: 12c Solving for n\n(07-25-2017 02:38 AM)Zac Bruce Wrote:  It was late last night when I posted, so yes the figure you gave for the first question is correct and I have edited the post. The second figure I meant 214.87 I.e the amount over what we expected or wanted.\n\nOK, then this is the additional amount you get after full 9 periods. You can also enter n=8 and get the (lower) amount after 8 periods.\n\n(07-25-2017 02:38 AM)Zac Bruce Wrote:  I'm guessing that there is not a simple solution similar to the first problem.\n\nYou mean a solution for the PV=1250 and FV=2500 problem? There is a very simple solution. I even posted the steps to calculate n as a fractional result. ;-)\n\n(07-25-2017 02:38 AM)Zac Bruce Wrote:  Can you explain a little more about it being only an approximate answer?\n\nNote: in the following formulas all values are unsigned, e.g. PMT=325 and not –325 due to the HP sign convention.\n\nI assume you mean the first problem here, the one with a given PMT.\nThe basic annuity formula for this case is\n$PV=PMT \\cdot \\frac{1-(1+i)^{-n}}{i}$\nYour approximate approach in effect calculates PV for n=327 (=34991,78) and n=328 (=35010,44) and then interpolates the \"correct\" n for PV=35000 between these two (your %T method is mathematically equivalent). But PV is not a linear function of n. For instance, the PV for n=11 is not the mean of the PVs for 10 resp. 12 periods. The exact value can be calculated with the given formula:\n$n=- \\frac{\\ln(1 - \\frac{i \\cdot PV}{PMT}} {\\ln(1+i)}$\n(07-25-2017 02:38 AM)Zac Bruce Wrote:  Is this still the case using the formula you presented as a program as a solution to the second problem?\n\nNo, this calculation is exact.\nThe basic formula is  $FV=PV \\cdot (1+i)^{n}$\nSolve for n and get  $n=\\frac{\\ln \\frac{FV}{PV}}{\\ln(1+i)}$\nThat's what these steps calculate.\nHere is a short program that does these two calculations, depending on whether PMT=0 or not.\n\nCode:\n01 RCL PMT 02 x=0? 03 GTO 15 04 1 05 RCL i 06 RCL PV 07 x 08 RCL PMT 09 / 10 % 11 + 12 LN 13 CHS 14 GTO 20 15 RCL FV 16 RCL PV 17 / 18 CHS 19 LN 20 1 21 RCL i 22 % 23 + 24 LN 25 / 26 GTO 00\n\n(07-25-2017 02:38 AM)Zac Bruce Wrote:  Is it just beyond the capabilities of the calculator to give an exact answer? Or is further information needed? I know it works with certain assumptions e.g. periods of equal length, payments always made on either exactly the end or start of a period, all of which would effect real world accuracy.\n\nThe point is: there are two \"exact\" answers. One of them is mathematically exact. That's the n that solves the TVM equation, resp. the FV resp. PV equations given above. Here the answers are n=8,0432 and n=327,4393.\n\nBut these answers are not valid in the real world. They assume than n is a continuous variable, which is not true. There are no 0,0432 or 0,4393 periods. You can have either 8 or 9 periods at the end of which interest is accumulated. You can make either 327 or 328 payments to get a little less or a little more than $35000. And this is the approach of the 12C. It's a tool for real world applications, and less so for students that want a textbook answer for a mathematical formula. The 12C tells you that, if you deposit$1250 at 9% it will take 9 years until you have $2500. After 8 years it's only 2490,70 and after 9 years it's 2714,87. After 8 years your bank will not pay the missing$9,30 for another 0,0432 years, so you have to wait one more year before the next interest payment gets you beyond $2500. So the 12C handles fractional periods different from the theoretical textbook solution. That's not less exact, it's simply a real-life approach for real-life problems. Dieter 07-25-2017, 10:52 AM Post: #7 Zac Bruce", null, "Junior Member Posts: 37 Joined: Jul 2017 RE: 12c Solving for n Deiter, Thankyou very much for that very thorough and detailed reply. I feel I have a much better understanding of the topic now. I can see my flaw in logic with regards to the problem of e.g. n=8 or n=9, thinking that there would indeed be some period in between where the e.g. bank/investment would pay an amount of interest closer to the desired value, but of course this is not the case. So I can now see that in my original calculation it is more applicable to the real world to then recalculate FV for n=9, and see what the amount will be, or recalculate for n=8 and make an informed decision whether you are happy to be$9.30 short of your goal of $2500, or wait the extra period and have an extra$214.87\n\nWith the first question, where PMT is involved, it is still important to be able to calculate the final fractional payment. As in when you have loaned an amount of money, it may take 328 payments (n) to fully pay the loan, but it is important to know that the final payment will indeed be a fractional payment (not a fraction of time). Is there anything wrong with the solution given by HP for calculating that final, fractional payment? (i.e. recalculate FV, RCL PMT +) I now understand that taking that final fractional payment and trying to relate it to a fractional period of time is the wrong (or not practical) way to be thinking about it, but I want to be sure that the calculation of the fractional payment is correct. In place of my original approximate solution (n=327.44), it would instead be interpreted as \"327 full payments of $325, plus a final payment of$143.11\" and not in terms of time.\n\nThe site that Paul suggested included a program to solve for a mathematically correct value of n, which is slightly different to your suggestion (allows for END or BEGIN by storing 1 in either STO 1 or STO 2) but comes up with the same results.\n\nQuote: ...There is a simple solution\nI guess if I do the work to internalize and memorize the equation, then yes, very simple! I'm perhaps too blessed to have constant access to electronics and the internet to do the \"hard\" work for me!\n\nI bought a copy of Gene Wright's book, I think I might go and get it printed and bound tomorrow, start reading and try actually understand the maths, rather than just understanding which buttons to press!\n\nRegards,\n\nZac\n07-25-2017, 11:24 AM (This post was last modified: 07-25-2017 12:59 PM by Dieter.)\nPost: #8\n Dieter", null, "Senior Member Posts: 2,397 Joined: Dec 2013\nRE: 12c Solving for n\n(07-25-2017 10:52 AM)Zac Bruce Wrote:  So I can now see that in my original calculation it is more applicable to the real world to then recalculate FV for n=9, and see what the amount will be, or recalculate for n=8 and make an informed decision whether you are happy to be $9.30 short of your goal of$2500, or wait the extra period and have an extra $214.87 Exactly. (07-25-2017 10:52 AM)Zac Bruce Wrote: With the first question, where PMT is involved, it is still important to be able to calculate the final fractional payment. As in when you have loaned an amount of money, it may take 328 payments (n) to fully pay the loan, but it is important to know that the final payment will indeed be a fractional payment (not a fraction of time). Is there anything wrong with the solution given by HP for calculating that final, fractional payment? (i.e. recalculate FV, RCL PMT +) I'd say this is fine. Especially if this is the way suggested by HP. ;-) You essentially calculate FV as the amount you have overpaid the loan with full 328 payments of$325. So the final payment can be lowered by this.\n\nHere is another approach: Enter n=327 and get PV=$34991,7839. This is the amount you have paid off after 327 periods. So$8,2161 are still missing. For the 328th payment this means you have to pay (1+i)328 times this, which again is $143,11. Clear Fin 35000 PV –325 PMT 10,5 g i n => 328 327 n PV => 34991,78 35000 – PV 0 PMT 328 n FV => 143,11 OK, that's a lot more complicated. #-) (07-25-2017 10:52 AM)Zac Bruce Wrote: In place of my original approximate solution (n=327.44), it would instead be interpreted as \"327 full payments of$325, plus a final payment of $143.11\" and not in terms of time. I'd say that your approximate result of n=327,4403 may be interpreted as 327,4403 payments, i.e. 327 full and one final payment of 0,4403... times$325 = \\$143,11.\n\n(07-25-2017 10:52 AM)Zac Bruce Wrote:  The site that Paul suggested included a program to solve for a mathematically correct value of n, which is slightly different to your suggestion (allows for END or BEGIN by storing 1 in either STO 1 or STO 2) but comes up with the same results.\n\nYes, my little program does only a very basic calculation for this particular case. Real TVM programs do a lot more stuff. ;-)\nOn the other hand the HP-41 standard pac's TVM program (cf. line 06...20) shows how various scenarios (OK, END mode only) can be handled with one simple formula for n. Here is a translation for the 12C:\n\nCode:\n01 RCL FV 02 CHS 03 RCL PMT 04 RCL i 05 / 06 EEX 07 2 08 x 09 + 10 RCL PV 11 LstX 12 + 13 / 14 LN 15 1 16 RCL i 17 % 18 + 19 LN 20 / 21 GTO 00\n\n(07-25-2017 10:52 AM)Zac Bruce Wrote:  I guess if I do the work to internalize and memorize the equation, then yes, very simple! I'm perhaps too blessed to have constant access to electronics and the internet to do the \"hard\" work for me!\n\nC'mon, this compound interest formula is as basic at it gets. ;-)\n\n(07-25-2017 10:52 AM)Zac Bruce Wrote:  I bought a copy of Gene Wright's book, I think I might go and get it printed and bound tomorrow, start reading and try actually understand the maths, rather than just understanding which buttons to press!\n\nI think there are three levels involved here:\n(0) Knowing which buttons to press\n(1) Knowing the math behind this\n(2) Knowing the meaning of the math. ;-)\n\nRegarding the latter I'm sometimes a bit lost myself.\n\nDieter\n07-27-2017, 12:19 PM\nPost: #9\n Zac Bruce", null, "Junior Member Posts: 37 Joined: Jul 2017\nRE: 12c Solving for n\nDieter,\n\nI've made some progress through Gene's book now, and you're right about it being pretty simple. My math is not strong so I didn't know that the log of a number to a power is equal to the power multiplied by the log of that number. What a mouthful.\n\nGene does offer a quick and dirty way to approximate, but it's really no simpler than just working through the formula.\n\nI still don't understand was a logarithm really is, but at least compound interest is starting to make sense. So I guess I'm at stage two, at least!\n\nRegards,\n\nZac\n07-27-2017, 07:00 PM (This post was last modified: 07-27-2017 07:02 PM by Dieter.)\nPost: #10\n Dieter", null, "Senior Member Posts: 2,397 Joined: Dec 2013\nRE: 12c Solving for n\n(07-27-2017 12:19 PM)Zac Bruce Wrote:  My math is not strong so I didn't know that the log of a number to a power is equal to the power multiplied by the log of that number. What a mouthful.\n(...)\nI still don't understand was a logarithm really is, but at least compound interest is starting to make sense.\n\nPower and exponential functions as well as their inverses (roots and logs) are basic math that is not too hard to understand. If it can be done at school in grade 8 or 9 you will be able to get it as well. All this stuff is required to understand the concept of the time value of money, both in simple compoud interest problems, in annuities and in other basic concepts like NPV or IRR. So every minute you spend on this for a better understanding of these basics will pay off later. Financial math simply is not possible without this. There is a reason why the 12C has only a few scientific functions while it does have y^x, e^x and ln x. ;-)\n\nDieter\n07-27-2017, 09:40 PM (This post was last modified: 07-28-2017 12:19 AM by Zac Bruce.)\nPost: #11\n Zac Bruce", null, "Junior Member Posts: 37 Joined: Jul 2017\nRE: 12c Solving for n\nDieter,\n\nWhen I tell people that I'm studying accounting/finance, usually the first question I get asked is, \"Are you good at maths? (sic)\". Usually I just laugh and say, \"Yeah.\"\n\nThe truth being that I'm very systematic and I enjoy processes and logic. So, I would have been good at math if only I'd been paying more attention! I took advanced math in year 10, but I don't remember ever bringing my notebook or doing exercises. I did pass, but I remember very little.\n\nLast trimester I did my first statistics for business course and really enjoyed it, and topped my class. The math involved in (basic) probabilities and statistics is not so difficult to understand.\n\nBut I realize that I lack any strong foundation, so I'm currently doing a self-paced bridging course offered by my university. Then it's on to quantitative skills with applications, which covers logarithmic, exponential and inverse functions in greater detail, among other things.\n\nRegards,\n\nZac\n « Next Oldest | Next Newest »\n\nUser(s) browsing this thread: 1 Guest(s)" ]
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https://codereview.stackexchange.com/questions/119837/normalized-random-position-generator/119920
[ "# Normalized random position generator\n\nI place various divs on screen by adding random x and y positions successively normalized with this function. My question is, how well is it written? Can I improve it? If so, then how?\n\nI am worried not only about the coding style but also performance since I'm using multiple Math.random()s.\n\n function normalize(n) {\nvar x = 0, i = 0;\nwhile (i < n) {\nx += Math.random();\ni++;\n}\nreturn (x / n);\n}\n\n• What values of n do you intend to use? 5? 100? – 200_success Feb 13 '16 at 3:31\n• @200_success 0 to 20 – Asperger Feb 13 '16 at 16:14\n\nWith totally zero context of what's going on and I'm not a math person, why \"normalize\"? What does \"normalize\" mean? As a front-end developer, normalize.css comes into mind, but it's not related to CSS. So I'm clueless. If you can give a better name, that would be great.\n\nAlso, a more compact way to write this is using a for loop\n\nfunction normalize(n) {\nfor(var x = 0, i = 0; i < n; i++){\nx += Math.random();\n}\nreturn (x / n);\n}\n\n• it normalizes the range of possible output values into the interval of [0, 1) (the same interval of the return values from Math.random(), see my answer for the math in all its g(l)ory. – I'll add comments tomorrow Feb 14 '16 at 2:02\n\nI place various divs on screen by adding random x and y positions successively normalized with this function.\n\nThis is a rather vague requirement, yet the resulting code is quite specific.\n\nI am worried not only about the coding style but also performance since I'm using multiple Math.random()s.\n\nI am worried about if the generated output is actually the desired one. (Are you aware of what this function does?)\n\n## What's going on\n\nLet's start by recaping what Math.random() does by quoting MDN:\n\nThe Math.random() function returns a floating-point, pseudo-random number in the range [0, 1)\n\nBasically speaking, that means it's uniformly distributed between 0 and 1 with every value in between equally likely.\n\nNow what happens if you add multiple such uniformly distributed variables together as you do with variable x?\n\nFigure 7.6 of this pdf from Section 7.2 (page 15 in the pdf, page 299 in the document) shows the density function of the sum of n such random variables:", null, "The X-axis shows what values are possible and the Y-axis shows how the probability is to get this value.1\n\nFor n=2, most values that you get for x will be around 1. For n=4, the peak is at 2. In general (as you can see from the picture, the peak is at n/2. Again, for n=1, you get a uniform distribution with everything in [0, 1) equally likely. I added the density for n=1 to the plot in blue color:", null, "On top of that, you do the normalisation of (x / n), which squishes all the curves in the plot into the interval [0, 1) on the X-axis and does indeed normalise the range of possible values. I allowed myself to retrace two of the plots:", null, "and to normalize them by scaling each one into the interval [0, 1):", null, "ignoring my mad image editing skills, it looks like both plots are pretty similar now. They both have the peak in the middle. Sure, the first one has a higher peak, but that doesn't make them much different: both probabilities will produce values near their peak with a higher likelyhood, which is in the middle for both of them.\n\nSo far it looks like the several calls to Math.random() create very similarly distributed random variables.\n\n## A practical demonstration\n\nI created the above plots for all n up to 10 with this piece of octave code:\n\nmax = 10;\nplots = [];\n\nfor n = 1:max\n\nr = zeros(1000000,1);\n\nfor i = 1:n\nr += rand(1000000, 1) * 1000;\nendfor\n\nr /= n;\n\ny = zeros(1000,1);\nfor random = r'\ny(ceil(random)) += 1;\nendfor\n\nplots = [plots y/n];\n\nendfor\n\n\nI work in the interval up to 1000, but qualitatively, the plots that plot(plots) yields are the same as what you see in the plot above:", null, "What values of n do you intend to use? 5? 100? – 200_success♦\n\n@200_success 0 to 20 – Asperger\n\nI wouldn't go as far to say it doesn't make a difference what value n has, but considering how little the distribution changes for the cost of another call to Math.random(), I don't think it's worth it.\n\nThe parameter n is also kind of meaningless. I say it's impossible to predict a certain outcome given a certain value for it. It's more like a knob that has to be adjusted by trial and error until desirable values are obtained.\n\n## summary, tl, dr;\n\n• The function does indeed normalize the range of values into [0, 1)\n• The parameter has not a significant influence on the shape probability density function that's used to create the values (this could be desired behaviour)\n• There's no meaningful way to influence the probability density function. A normal distribution for example is defined by how thick (or thin) the bell shape is.\n\n1 Strictly mathematically speaking, this is horribly wrong. But for the intuitive understanding, it's perfect. Sorry math nerds." ]
[ null, "https://i.stack.imgur.com/AoyPu.png", null, "https://i.stack.imgur.com/thWR6.png", null, "https://i.stack.imgur.com/2TroK.png", null, "https://i.stack.imgur.com/DbfJD.png", null, "https://i.stack.imgur.com/Mpydy.png", null ]
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http://publications.cirad.fr/une_notice.php?dk=580059
[ "Publications des agents du Cirad\n\nModeling lactation curve of Saudi camels using a linear and non-linear forms of the incomplete Gamma function\n\nAziz M.A., Faye B., Al-eknah M., Musaad A.. 2016. Small Ruminant Research, 137 : p. 40-46.\n\nThe objectives of this study were to describe lactation curve of Saudi camels using the linear and non-linear forms of the incomplete Gamma function, to estimate lactation curve characteristics and to determine the appropriate form describing lactation curve. Records of weekly milk yield representing 9 parities were collected from 56 she-camels. Data were collected over the period 2007¿2013. The results revealed that averages of total milk yield, lactation length and daily milk yield of the nine parities, ranged between 967.3 and 3107.21 kg, between 273 and 416 days and between 2.96 and 7.40 kg/day, respectively. Initial milk yield (a) estimated by the linear and non-linear forms of the nine parities ranged between 0.194 and 1.775 and between 0.155 and 1.818, respectively. The increasing rate (b) ranged between 0.754 and 1.414 and between 0.750 and 1.533, while the decreasing rate (c) ranged between ?0.054 and ?0.020 and between ?0.059 and ?0.036, respectively. Days to attain peak production, peak production and persistency, as well as persistency percentage were also estimated. It is concluded that the linear form of the incomplete Gamma function is more appropriate for describing lactation curve of camel milk yield than its non-linear counterpart. (Résumé d'auteur)\n\nMots-clés : modèle de simulation; modèle mathématique; lait de chamelle; production laitière; rendement laitier; lactation; dromadaire; arabie saoudite; camelus dromedarius\n\nThématique : Physiologie animale : croissance et développement; Physiologie animale : reproduction; Méthodes mathématiques et statistiques\n\nArticle de revue" ]
[ null ]
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https://research.nii.ac.jp/graphgolf/2019/problem.html
[ "# Graph Golf\n\nThe Order/degree Problem Competition\n\n## Problem statement\n\nUpdate 2019-07-29\n\n### Definition\n\nThe order/degree problem with parameters n and d: Find a graph with minimum diameter over all undirected graphs with the number of vertices = n and degree ≤ d. If two or more graphs take the minimum diameter, a graph with minimum average shortest path length (ASPL) over all the graphs with the minimum diameter must be found.\n\nThe order/degree problem on a grid graph with a limited edge length r: Do the same as above, but on a w × h grid in a two-dimensional Euclidean space, keeping the lengths of the edges ≤ r in Manhattan distance. Here a \"grid\" implies that (1) the vertices are located at integer coordinates and (2) the edges run along the grid lines. Note that (3) a vertex is not necessarily connected to its nearest neighbors and (4) an edge may change its direction at a grid point but must not run diagonally.\n\n### Example\n\nLet us show examples for the order/degree problem with parameters n = 16 and d = 4. The figure illustrates three examples of 4-regular graphs with 16 nodes. The diameter of the first and the second graphs is 3, while that of the third one is 4. Thus, we should select the first or the second one. Next, we compare the average shortest path length (ASPL) of those two graphs. Since the ASPL of the first one is smaller, this is the best solution of the problem among the three.\n\n### Why order/degree problem?\n\nGraph design has a rich variety of application fileds of computer systems. In particular, it is just meeting a network design of future supercomputers and future high-end datacenters in terms of hop counts , since their networks are topologically modeled as undirected regular graphs. Low latency is preconditioned on small hop counts, but existing network topologies have hop counts much larger than theoretical lower bounds. Therefore, computer network designers desire to find a graph that has a small number of hops between any pair of vertices. The same solution is beneficial for desining such a micro network, though the layout of a graph on two-dimensional surface (on a chip) must be considered .\n\n### Why not degree/diameter problem?\n\nThe degree/diameter problem (DDP) has been studied for decades and consists in generating the largest possible graph given degree and diameter constraints, striving to approach theoretical upper bounds . However, DDP solutions may not be directly usable for network topologies in supercomputer and high-end datacenter systems because they are for particular number of compute nodes, whereas the number of nodes in a real system is determined based on practical considerations, e.g., budget.\n\n### Lower bound of diameter & ASPL\n\nFor a general graph with order n and degree d, the lower bound Kn,d on diameter k is computed as", null, "Similarly, the lower bound Ln,d on ASPL l is computed as", null, "For a grid graph, a tighter lower bound can be calculated by the algorithm shown in . We employ this tighter lower bound for the grid graph category of the competition.\n\nPrograms for calculating those lower bounds are provided below.\n\n### Programs\n\nA Python script that generates a regular random graph and calculates its diameter and ASPL is provided here. This may be a good starting point if you are new to creating graphs.\n\nA C program that generates a grid graph and a Perl script that calculates the tighter lower bound of diameter and ASPL of a grid graph are contributed by Koji Nakano and Daisuke Takafuji.\n\nFor a faster calculation of ASPL, a C++ program was developed by Ryuhei Mori, and an MPI-based parallel program was developed by Masahiro Nakao." ]
[ null, "https://research.nii.ac.jp/graphgolf/2019/img/eq_low-diam.png", null, "https://research.nii.ac.jp/graphgolf/2019/img/eq_low-aspl.png", null ]
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https://se.mathworks.com/help/signal/ref/zp2ss.html
[ "# zp2ss\n\nConvert zero-pole-gain filter parameters to state-space form\n\n## Syntax\n\n``[A,B,C,D] = zp2ss(z,p,k)``\n\n## Description\n\nexample\n\n````[A,B,C,D] = zp2ss(z,p,k)` finds a state-space representation$\\begin{array}{l}\\stackrel{˙}{x}=Ax+Bu\\\\ y=Cx+Du\\end{array}$such that it is equivalent to a system in factored transfer function form$H\\left(s\\right)=\\frac{Z\\left(s\\right)}{P\\left(s\\right)}=k\\frac{\\left(s-{z}_{1}\\right)\\left(s-{z}_{2}\\right)\\cdots \\left(s-{z}_{n}\\right)}{\\left(s-{p}_{1}\\right)\\left(s-{p}_{2}\\right)\\cdots \\left(s-{p}_{n}\\right)}$Column vector `p` specifies the pole locations, and matrix `z` the zero locations with as many columns as there are outputs. The gains for each numerator transfer function are in vector `k`. The `A`, `B`, `C`, and `D` matrices are returned in controller canonical form.```\n\n## Examples\n\ncollapse all\n\nGenerate the state-space representation of a damped mass-spring system that obeys the differential equation\n\n`$\\underset{}{\\overset{¨}{w}}+0.01\\underset{}{\\overset{˙}{w}}+w=u\\left(t\\right).$`\n\nThe measurable quantity is the acceleration, $y=\\underset{}{\\overset{¨}{w}}$, and $u\\left(t\\right)$ is the driving force. In Laplace space, the system is represented by\n\n`$Y\\left(s\\right)=\\frac{{s}^{2}\\phantom{\\rule{0.2777777777777778em}{0ex}}U\\left(s\\right)}{{s}^{2}+0.01s+1}.$`\n\nThe system has unit gain, a double zero at $s=0$, and two complex-conjugate poles.\n\n```z = [0 0]; p = roots([1 0.01 1])```\n```p = 2×1 complex -0.0050 + 1.0000i -0.0050 - 1.0000i ```\n`k = 1;`\n\nUse `zp2ss` to find the state-space matrices.\n\n`[A,B,C,D] = zp2ss(z,p,k)`\n```A = 2×2 -0.0100 -1.0000 1.0000 0 ```\n```B = 2×1 1 0 ```\n```C = 1×2 -0.0100 -1.0000 ```\n```D = 1 ```\n\n## Input Arguments\n\ncollapse all\n\nZeros of the system, specified as a vector. The zeros must be real or come in complex conjugate pairs.\n\n`Inf` values may be used as place holders in `z` if some columns have fewer zeros than others.\n\nExample: `[1 (1+1j)/2 (1-1j)/2]'`\n\nData Types: `double`\nComplex Number Support: Yes\n\nPoles of the system, specified as a vector. The poles must be real or come in complex conjugate pairs.\n\nExample: `[1 (1+1j)/2 (1-1j)/2]'`\n\nData Types: `double`\nComplex Number Support: Yes\n\nScalar gain of the system, specified as a scalar.\n\nData Types: `double`\n\n## Output Arguments\n\ncollapse all\n\nState matrix, returned as a matrix. If the system is described by n state variables, then `A` is n-by-n.\n\nData Types: `single` | `double`\n\nInput-to-state matrix, returned as a matrix. If the system is described by n state variables, then `B` is n-by-1.\n\nData Types: `single` | `double`\n\nState-to-output matrix, returned as a matrix. If the system has q outputs and is described by n state variables, then `C` is q-by-n.\n\nData Types: `single` | `double`\n\nFeedthrough matrix, returned as a matrix. If the system has q outputs, then `D` is q-by-1.\n\nData Types: `single` | `double`\n\n## Algorithms\n\n`zp2ss`, for single-input systems, groups complex pairs together into two-by-two blocks down the diagonal of the `A` matrix. This requires the zeros and poles to be real or complex conjugate pairs." ]
[ null ]
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https://www.journaldev.com/44690/python-convert-time-hours-minutes-seconds
[ "# Convert time into hours minutes and seconds in Python\n\nFiled Under: Python", null, "In this tutorial, we will be talking about time. Don’t worry, this isn’t a boring history tutorial, rather we will be looking at different ways of converting time in seconds to time in hours, minutes, and seconds.\n\nMoving forwards we will be referring to time in hours, minutes and seconds as time in the preferred format.\n\nIt will look like :\n\n```2:46:40\n```\n\nLet’s take some ‘time’ and think about the problem at hand. No doubt python has amazing modules to do the conversion for us. But let’s try and write our own program first before we move on to the in-built modules.\n\n## Building a Custom function to convert time into hours minutes and seconds\n\nTo write our own conversion function we first need to think about the problem mathematically.\n\nHow do you convert seconds into the preferred format?\n\nYou need to get the value of hours, minutes and seconds.\n\nWe are going to assume that the time in seconds doesn’t exceed the total number of seconds in a day. If it does we will divide it with the total number of seconds in a day and take the remainder.\n\nThis is mathematically represented as :\n\n```seconds = seconds % (24 * 3600)\n```\n\n% operator gives the remainder.\n\n24*3600 since one hour has 3600 seconds (60*60) and one day has 24 hours.\n\nAfter this we can proceed and calculate the hour value from seconds.\n\n### 1. Get the hour value\n\nTo get the hour value from seconds we will be using the floor division operator (//).\n\nIt returns the integral part of the quotient.\n\nSince we need the number of hours, we will divide the total number of seconds (n) by the total number of seconds in an hour (3600).\n\nMathematically this is represented as :\n\n```hour = seconds // 3600\n```\n\nAfter this we need to calculate the minutes.\n\n### 2. Get the minute value\n\nTo calculate the value of minutes we need to first divide the total number of seconds by 3600 and take the remainder.\n\nMathematically this is represented as :\n\n``` seconds = seconds % 3600\n```\n\nNow to calculate the value of minutes from the above result we will again use the floor operator.\n\n```minutes = seconds // 60\n```\n\nA minute has sixty seconds hence we floor the seconds value with 60.\n\nAfter calculating minute value we can move forward to calculate the seconds’ value for our preferred format.\n\n### 3. Get the seconds value\n\nTo get seconds value we again need to divide the total number of seconds by the number of seconds in one minute (60) and take the remainder.\n\nMathematically this is done as follows :\n\n``` seconds = seconds % 60\n```\n\nThis will give the second value that we need for our preferred format.\n\n### 4. Complete code\n\nLet’s compile the above knowledge in a python function.\n\n```def convert_to_preferred_format(sec):\nsec = sec % (24 * 3600)\nhour = sec // 3600\nsec %= 3600\nmin = sec // 60\nsec %= 60\nprint(\"seconds value in hours:\",hour)\nprint(\"seconds value in minutes:\",min)\nreturn \"%02d:%02d:%02d\" % (hour, min, sec)\n\nn = 10000\nprint(\"Time in preferred format :-\",convert(n))\n```\n\nOutput :\n\n```seconds value in hours: 2\nseconds value in minutes: 46\nTime in preferred format :- 02:46:40\n```\n\n## Using the Time module\n\nNow let’s look at an inbuilt module that lets us convert seconds into our preferred format in one line of code.\n\nThe time module defines the epoch as January 1, 1970, 00:00:00 (UTC) in Unix systems (system dependent). Epoch is basically the start of time for a computer. Think of it as day 0. Whenever we convert seconds using the time module, this epoch is used as the reference point.\n\nTo output the epoch in your system, use the following line of code :\n\n```time.gmtime(0)\n```\n\nTo convert seconds into preferred format use the following line of code:\n\n```time.strftime(\"%H:%M:%S\", time.gmtime(n))\n```\n\nThis line takes the time in seconds as ‘n’ and then lets you output hour, minute, and second value separately.\n\nThe complete python code is as follows:\n\n```import time\nn=10000\ntime_format = time.strftime(\"%H:%M:%S\", time.gmtime(n))\nprint(\"Time in preferred format :-\",time_format)\n\n```\n\nOutput :\n\n```Time in preferred format :- 02:46:40\n```\n\nThe time module also gives you the option to display some extra information such as day, month, and year.\n\nLet’s try using %a and %b.\n\n```import time\nn=100000000000\ntime_format = time.strftime(\"Day: %a, Time: %H:%M:%S, Month: %b\", time.gmtime(n))\nprint(\"Time in preferred format :-\",time_format)\n```\n\nOutput :\n\n```Time in preferred format :- Day: Wed, Time: 09:46:40, Month: Nov\n```\n\n## Using Datetime module\n\nYou can also use the timedelta method under the DateTime module to convert seconds into the preferred format.\n\nIt displays the time as days, hours, minutes, and seconds elapsed since the epoch.\n\nThe python code to convert seconds into the preferred format using Datetime module is as follows:\n\n```import datetime\nn= 10000000\ntime_format = str(datetime.timedelta(seconds = n))\nprint(\"Time in preferred format :-\",time_format)\n```\n\nOutput :\n\n```Time in preferred format :- 115 days, 17:46:40\n```\n\n## Conclusion\n\nThis tutorial looked at three different ways you can use to convert seconds into hours, minutes, and seconds. Broadly there are two different ways to go about the problem.\n\nEither you write your own function or use an inbuilt module. We started by writing our own function and then looked at the time and DateTime module.\n\nclose\nGeneric selectors\nExact matches only\nSearch in title\nSearch in content" ]
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https://www.dreamcancel.com/wiki/index.php?title=Verse_(XIV)&diff=prev&oldid=24180
[ "# Movelist\n\n(*) = EX OK\n\n(!) = MAX OK\n\nThrows\n\nCommand Normals\n\nSpecial Moves\n\n(Forward) -", null, "(Backward) -", null, "Super Special Moves\n\nClimax Super Special Moves\n\n# Quick Combo Reference\n\n## Quick Combo Reference\n\nNotation Key (place notation here)\n\n0 Meter\n\nAnywhere\nLow, Anywhere\nLow, Anywhere\n\n(Combo 1) = (damage number here) dmg\n(Combo 2) = (damage number here) dmg\n(Combo 3) = (damage number here) = 231 dmg\n\n1 Meter\n\nCorner\nAnywhere\nLow, Anywhere\n\n(Combo 1) = (damage number here) dmg\n(Combo 2) = (damage number here) dmg\n(Combo 3) = (damage number here) dmg\n\n2 Meters Anywhere (Combo 1) = (damage number here) dmg\n\nStanding\n\n• st. A\n• st. B\n• st. C\n• st. D\n\nClose\n\n• cl. A\n• cl. B\n• cl. C\n• cl. D\n\nCrouching\n\n• cr. A\n• cr. B\n• cr. C\n• cr. D\n\nJumping\n\n• j. A\n• j. B\n• j. C\n• j. D\n\nBlowback\n\n• st. CD\n• j. CD\n\n# Throws\n\nGiga Power Bomb - (b/f+C) close\n\n• (description)\n• Can be broken\n• (soft or hard knockdown)\n\nBrain Stem Buster - (b/f+D) close\n\n• (description)\n• Can be broken\n• (soft or hard knockdown)\n\n# Command Moves\n\nStar Crash - - (f+B)\n\n• Verse flips forward then stomps on the ground.\n• Special cancels when Verse lands, and used for some combos.\n• Landing has a low hitbox, and can sometimes cross-up the opponent.\n\n# Special Moves\n\nname of special move - (place motion here example qcf + A/C)\n\n• (description here)\n• (description here)\n• (description here)\n\nEX: (description here)\n\nname of special move - (place motion here)\n\n• (description here)\n• (description here)\n• (description here)\n\nEX: (description here)\n\nname of move - (motion here)\n\n• (description here)\n• (description here)\n• (description here)\n\nEX: (description here)\n\nname of follow up move - (motion here)\n• (description here)\nname of follow up move - (motion here)\n• (description here)\n• EX: (description here)\n\nname of move here - (motion here)\n\n• (description here)\n• (description here)\n• (description here)\n\nEX: (description here)\n\nname of move - (motion here)\n\n• (description here)\n• (description here)\n• (description here)\n\nEX: (description here)\n\n# Super Special Moves\n\nname of move - (motion here)\n\n• (description here)\n• (description here)\n• (description here)\n\nMax:\n\nfollow up move - (motion if possible qcfx4 + B/D)\n• (description here)\n• (description here)\n• (description here)\n\n# Climax Super Special Moves\n\nname of move - (motion here qcf, qcf + B+D)\n\n• (description here)\n• (description here)\n• (description here)\n\n# Combos\n\n• (any special information about the character combos if needed)\n\n### Rush Auto Combo\n\nMeterless: description here\n\n1 Meter: description here\n\nEX: description here\n\n## 0 meter\n\n• (place combo here) = (place damage amount here)\n\n(place combo description here)\n\n## 1 meter\n\n• (place combo here) = (place damage amount here)\n\n(place combo description here)\n\n## 2 meters\n\n• (place combo here) = (place damage amount here)\n\n(place combo description here)\n\n## 3 meters\n\n• (place combo here) = (place damage amount here)\n\n(place combo description here)\n\n## 4 meters\n\n• (place combo here) = (place damage amount here)\n\n(place combo description here)\n\n## 5 meters\n\n• (place combo here) = (place damage amount here)\n\n(place combo description here)" ]
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https://factorization.info/factors/0/factors-of-349.html
[ "Factors of 349", null, "We have all the information you will ever need about the Factors of 349. We will provide you with the definition of Factors of 349, show you how to find the Factors of 349, give you all the Factors of 349, tell you how many Factors 349 has, and supply you with all the Factor Pairs of 349 to prove that our answer is solved correctly.\n\nFactors of 349 definition\nThe Factors of 349 are all the integers (positive and negative whole numbers) that you can evenly divide into 349. 349 divided by a Factor of 349 will equal another Factor of 349.\n\nHow to find the Factors of 349\nSince the Factors of 349 are all the numbers that you can evenly divide into 349, we simply need to divide 349 by all numbers up to 349 to see which ones result in an even quotient. When we did that, we found that these calculations resulted in an even quotient:\n\n349 ÷ 1 = 349\n349 ÷ 349 = 1\n\nThe Postive Factors of 349 are therefore all the numbers we used to divide (divisors) above to get an even number. Here is the list of all Postive Factors of 349 in numerical order:\n\n1 and 349.\n\nFactors of 349 include negative numbers. Therefore, all the Positive Factors of 349 can be converted to negative numbers. The list of Negative Factors of 349 are:\n\n-1 and -349.\n\nHow many Factors of 349?\nWhen we counted the Factors of 349 that we listed above, we found that 349 has 2 Positive Factors and 2 Negative Factors. Thus, the total number of Factors of 349 is 4.\n\nFactor Pairs of 349\nFactor Pairs of 349 are combinations of two factors that when multiplied together equal 349. Here are all the Positive Factor Pairs of 349\n\n1 × 349 = 349\n349 × 1 = 349\n\nLike we said above, Factors of 349 include negative numbers. Minus times minus equals plus, thus you can convert the Positive Factor Pair list above by simply putting a minus in front of every factor to get all the Negative Factor Pairs of 349:\n\n-1 × -349 = 349\n-349 × -1 = 349\n\nFactor Calculator\nDo you need the factors for a particular number? You can submit a number below to find the factors for that number with detailed explanations like we did with Factors of 349 above.\n\nFactors of 350\nWe hope this step-by-step tutorial to teach you about Factors of 349 was helpful. Do you want to see if you learned something? If so, give the next number on our list a try and then check your answer here." ]
[ null, "https://factorization.info/images/factors-of.png", null ]
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https://www.jpost.com/israel/welfare-officials-probe-allegations-of-sexual-relations-among-10-yr-olds
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https://dokumen.pub/abstract-algebra-an-introductory-course-9783319776491-3319776495.html
[ "• Commentary\n##### Citation preview\n\nGregory T. Lee\n\nAbstract Algebra An Introductory Course\n\nSpringer Undergraduate Mathematics Series Advisory Board M.A.J. Chaplain, University of St. Andrews A. MacIntyre, Queen Mary University of London S. Scott, King’s College London N. Snashall, University of Leicester E. Süli, University of Oxford M.R. Tehranchi, University of Cambridge J.F. Toland, University of Bath\n\nGregory T. Lee\n\nAbstract Algebra An Introductory Course\n\n123\n\nGregory T. Lee Department of Mathematical Sciences Lakehead University Thunder Bay, ON Canada\n\nIn memory of my father\n\nPreface\n\nThis book is intended for students encountering the beautiful subject of abstract algebra for the first time. My goal here is to provide a text that is suitable for you, whether you plan to take only a single course in abstract algebra, or to carry on to more advanced courses at the senior undergraduate and graduate levels. Naturally, I wish to encourage you to study the subject further and to ensure that you are prepared if you do so. At many universities, including my own, abstract algebra is the first serious proof-based course taken by mathematics majors. While it is quite possible to get through, let us say, a course in calculus simply by memorizing a list of rules and applying them correctly, without really understanding why anything works, such an approach would be disastrous here. To be sure, you must carefully learn the definitions and the statements of theorems, but that is nowhere near sufficient. In order to master the material, you need to understand the proofs and then be able to prove things yourself. This book contains hundreds of problems, and I cannot stress strongly enough the need to solve as many of them as you can. Do not be discouraged if you cannot get all of them! Some are very difficult. But try to figure out as many as you can. You will only learn by getting your hands dirty. As different universities have different sequences of courses, I am not assuming any prerequisites beyond the high school level. Most of the material in Part I would be covered in a typical course on discrete mathematics. Even if you have had such a course, I urge you to read through it. In particular, you absolutely must understand equivalence relations and equivalence classes thoroughly. (In my experience, many students have trouble with these concepts.) From time to time, throughout Parts II and III, some examples involving matrices or complex numbers appear. These can be bypassed if you have not studied linear algebra or complex numbers, but in any case, the material you need to know is not difficult and is discussed in the appendices. In Part IV, it is necessary to know some linear algebra, but all of the theorems used are proved in the text.\n\nvii\n\nviii\n\nPreface\n\nThe fundamental results about groups are covered in Chaps. 3 and 4, those about rings are in Chaps. 8 and 9, and the introductory theorems concerning fields and polynomials are found in Chap. 11. I think that these chapters are essential in any course. Beyond that, there is a fair amount of flexibility in the choice of topics. I confess my first encounter with abstract algebra was a joyous experience. I found (and still find!) the subject fascinating, and I will consider the time I put into this book well spent if you emerge with an appreciation for the field. I would like to thank Lynn Brandon and Anne-Kathrin Birchley-Brun at Springer for their help in making this book a reality. Also, thanks to the reviewers for their many useful suggestions. I thank my wife and family for their ongoing support. Finally, thanks to my teacher, Prof. Sudarshan Sehgal, both for his advice concerning this book and for all of his help over the years. Thunder Bay, ON, Canada\n\nGregory T. Lee\n\nContents\n\nPart I\n\nPreliminaries\n\n1\n\nRelations and Functions . . . . 1.1 Sets and Set Operations . 1.2 Relations . . . . . . . . . . . 1.3 Equivalence Relations . . 1.4 Functions . . . . . . . . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n3 3 5 6 10\n\n2\n\nThe Integers and Modular Arithmetic 2.1 Induction and Well Ordering . . . . 2.2 Divisibility . . . . . . . . . . . . . . . . . 2.3 Prime Factorization . . . . . . . . . . . 2.4 Properties of the Integers . . . . . . . 2.5 Modular Arithmetic . . . . . . . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n15 15 19 24 26 27\n\nPart II\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\nGroups\n\n3\n\nIntroduction to Groups . . . . . . . . . . . 3.1 An Important Example . . . . . . . 3.2 Groups . . . . . . . . . . . . . . . . . . . 3.3 A Few Basic Properties . . . . . . . 3.4 Powers and Orders . . . . . . . . . . 3.5 Subgroups . . . . . . . . . . . . . . . . 3.6 Cyclic Groups . . . . . . . . . . . . . 3.7 Cosets and Lagrange’s Theorem\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n. . . . . . . .\n\n35 35 38 42 44 48 54 57\n\n4\n\nFactor Groups and Homomorphisms 4.1 Normal Subgroups . . . . . . . . . . 4.2 Factor Groups . . . . . . . . . . . . . . 4.3 Homomorphisms . . . . . . . . . . . . 4.4 Isomorphisms . . . . . . . . . . . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n61 61 65 69 72\n\nix\n\nx\n\nContents\n\n4.5 4.6 5\n\nThe Isomorphism Theorems for Groups . . . . . . . . . . . . . . . . . . Automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n78 81\n\nDirect Products and the Classification of Finite Abelian Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 The Fundamental Theorem of Finite Abelian Groups 5.3 Elementary Divisors and Invariant Factors . . . . . . . . 5.4 A Word About Infinite Abelian Groups . . . . . . . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n. . . . .\n\n85 85 88 93 97\n\n6\n\nSymmetric and Alternating Groups . . . . . . . . . 6.1 The Symmetric Group and Cycle Notation . 6.2 Transpositions and the Alternating Group . . 6.3 The Simplicity of the Alternating Group . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n101 101 105 108\n\n7\n\nThe Sylow Theorems . . . . . . . . . . . . . . . . . . . . . 7.1 Normalizers and Centralizers . . . . . . . . . . . 7.2 Conjugacy and the Class Equation . . . . . . . 7.3 The Three Sylow Theorems . . . . . . . . . . . . 7.4 Applying the Sylow Theorems . . . . . . . . . . 7.5 Classification of the Groups of Small Order\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n115 115 119 122 125 128\n\nPart III\n\nRings\n\n8\n\nIntroduction to Rings . . . . . . . . . 8.1 Rings . . . . . . . . . . . . . . . . . 8.2 Basic Properties of Rings . . 8.3 Subrings . . . . . . . . . . . . . . . 8.4 Integral Domains and Fields 8.5 The Characteristic of a Ring\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n135 135 138 140 142 146\n\n9\n\nIdeals, Factor Rings and Homomorphisms 9.1 Ideals . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Factor Rings . . . . . . . . . . . . . . . . . . . 9.3 Ring Homomorphisms . . . . . . . . . . . 9.4 Isomorphisms and Automorphisms . . . 9.5 Isomorphism Theorems for Rings . . . 9.6 Prime and Maximal Ideals . . . . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n149 149 152 155 159 165 167\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n171 171 176 182 185 188\n\n. . . . . .\n\n. . . . . .\n\n10 Special Types of Domains . . . . . . . . 10.1 Polynomial Rings . . . . . . . . . . 10.2 Euclidean Domains . . . . . . . . . 10.3 Principal Ideal Domains . . . . . 10.4 Unique Factorization Domains . Reference . . . . . . . . . . . . . . . . . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\nContents\n\nPart IV\n\nxi\n\nFields and Polynomials\n\n11 Irreducible Polynomials . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Irreducibility and Roots . . . . . . . . . . . . . . . . . . . . . 11.2 Irreducibility over the Rationals . . . . . . . . . . . . . . . 11.3 Irreducibility over the Real and Complex Numbers . 11.4 Irreducibility over Finite Fields . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n. . . . . .\n\n191 191 195 200 202 205\n\n12 Vector Spaces and Field Extensions 12.1 Vector Spaces . . . . . . . . . . . . . 12.2 Basis and Dimension . . . . . . . 12.3 Field Extensions . . . . . . . . . . . 12.4 Splitting Fields . . . . . . . . . . . . 12.5 Applications to Finite Fields . . Reference . . . . . . . . . . . . . . . . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n207 207 210 215 221 225 229\n\nPart V\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\n. . . . . . .\n\nApplications\n\n13 Public Key Cryptography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 13.1 Private Key Cryptography . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 13.2 The RSA Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 14 Straightedge and Compass Constructions . . . . 14.1 Three Ancient Problems . . . . . . . . . . . . . 14.2 The Connection to Field Extensions . . . . . 14.3 Proof of the Impossibility of the Problems\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n. . . .\n\n241 241 244 250\n\nAppendix A: The Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253 Appendix B: Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297\n\nPart I\n\nPreliminaries\n\nChapter 1\n\nRelations and Functions\n\nWe begin by introducing some basic notation and terminology. Then we discuss relations and, in particular, equivalence relations, which we shall see several times throughout the book. In the final section, we talk about various sorts of functions.\n\n1.1 Sets and Set Operations A set is a collection of objects. We will see many sorts of sets throughout this course. Perhaps the most common will be sets of numbers. For instance, we have the set of natural numbers, N = {1, 2, 3, . . .}, the set of integers, Z = {. . . , −2, −1, 0, 1, 2, . . .} and the set of rational numbers \u0003 \u0002a : a, b ∈ Z, b \u0003= 0 . Q= b We also write R for the set of real numbers and C for the set of complex numbers. But sets do not necessarily consist of numbers. Indeed, we can consider the set of all letters of the alphabet, the set of all polynomials with even integers as coefficients or the set of all lines in the plane with positive slope. The objects in a set are called its elements. We write a ∈ S if a is an element of a set S. Thus, −3 ∈ Z but −3 ∈ / N. The set with no elements is called the empty set, and denoted ∅. Any other set is said to be nonempty.\n\n© Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_1\n\n3\n\n4\n\n1 Relations and Functions\n\nIf S and T are sets, then we say that S is a subset of T , and write S ⊆ T , if every element of S is also an element of T . Of course, S ⊆ S. We say that S is a proper subset of T , and write S \u0003 T , if S ⊆ T but S \u0003= T . Thus, it is certainly true that N ⊆ Z, but we can be more precise and write N \u0003 Z. For any two sets S and T , their intersection, S ∩ T , is the set of all elements that lie in S and T simultaneously. Example 1.1. Let S = {1, 2, 3, 4, 5} and T = {2, 4, 6, 8, 10}. Then S ∩ T = {2, 4}. We can extend this notion to the intersection of an arbitrary collection\u0004of sets. If I is a nonempty set and, for each i ∈ I , we have a set Ti , then we write i∈I Ti for the set of elements that lie in all of the Ti simultaneously. \u0004 Example 1.2. For each q ∈ Q, let Tq = {r ∈ R : r < 2q }. Then q∈Q Tq = {r ∈ R : r ≤ 0}. Also, for any sets S and T , their union, S ∪ T , is the set of all elements that lie in S or T (or both). Example 1.3. Using the same S and T as in Example 1.1, we have S ∪ T = {1, 2, 3, 4, 5, 6, 8, 10}. Furthermore, if I is a nonempty set and we have a set Ti for each i ∈ I , then we \u0005 write i∈I Ti for the union of all of the Ti ; that is, the set of all elements that lie in at least one of the Ti . \u0005 Example 1.4. If we use the same sets Tq as in Example 1.2, we have q∈Q Tq = R. In addition, for any two sets S and T , the set difference (or relative complement) is the set S\\T = {a ∈ S : a ∈ / T }. Example 1.5. Once again using S and T as in Example 1.1, we have S\\T = {1, 3, 5}. We will need one more definition. The following construction is named after René Descartes. Definition 1.1. Let S and T be any sets. Then the Cartesian product S × T is the set of all ordered pairs (s, t), with s ∈ S and t ∈ T . Example 1.6. Let S = {1, 2, 3} and T = {2, 3}. Then S × T = {(1, 2), (1, 3), (2, 2), (2, 3), (3, 2), (3, 3)}. There is also a Cartesian product of finitely many sets. For any sets T1 , T2 , . . . , Tn , we let T1 × T2 × · · · × Tn be the set of all ordered n-tuples (t1 , t2 , . . . , tn ), with ti ∈ Ti for all i.\n\n1.1 Sets and Set Operations\n\n5\n\nExample 1.7. Let T1 = {1, 2}, T2 = {a, b} and T3 = {2, 3}. Then T1 × T2 × T3 is the set {(1, a, 2), (1, a, 3), (1, b, 2), (1, b, 3), (2, a, 2), (2, a, 3), (2, b, 2), (2, b, 3)}. Exercises 1.1. Let S = {1, 2, 3} and T = {3, 4}. Find S ∩ T , S ∪ T , S\\T , T \\S and S × T . 1.2. Let R = {a, b, c}, S = {a, c, d} and T = {c, e, f }. Find R ∩ S, R ∩ (S\\T ), S ∪ T , S ∩ (R ∪ T ) and R × S. 1.3. Let R, S and T be sets with R ⊆ S. Show that R ∪ T ⊆ S ∪ T . 1.4. Let S = {1, 2, . . . , n}, for some positive integer n. Show that S has 2n subsets. 1.5. Let R, S and T be any sets. Show that R ∪ (S ∩ T ) = (R ∪ S) ∩ (R ∪ T ). 1.6. For each positive integer n, let Tn = { an : a ∈ Z}. \u0005 1. What is \u0004∞ n=1 Tn ? 2. What is ∞ n=1 Tn ?\n\n1.2 Relations We are going to use relations (in particular, the equivalence relations and functions that we will see in the next two sections) quite a few times in this course. Definition 1.2. Let S and T be sets. Then a relation from S to T is a subset ρ of S × T . If s ∈ S and t ∈ T , then we write sρt if (s, t) ∈ ρ; otherwise, we write s \u0003 ρ t. In particular, a relation on S is a relation from S to S. Example 1.8. Let S = {1, 2, 3} and T = {1, 2, 3, 4}. Define a relation ρ from S to T via sρt if and only if st 2 ≤ 4. Then ρ = {(1, 1), (1, 2), (2, 1), (3, 1)}. In particular, 3ρ1 but 1 \u0003 ρ 3. We will focus on relations on a set. Let us discuss a few properties enjoyed by some relations. Definition 1.3. Let ρ be a relation on S. We say that ρ is reflexive if aρa for all a ∈ S. Example 1.9. On Z, the relation ≤ is reflexive, but < is not. Indeed, a ≤ a for all integers a, but 1 is not less than 1. Definition 1.4. A relation ρ on a set S is symmetric if aρb implies bρa.\n\n6\n\n1 Relations and Functions\n\nExample 1.10. On Z, neither ≤ nor < is symmetric, as 1 < 2 but 2 is not less than 1 (and similarly for ≤). Define ρ via aρb if and only if |a − b| ≤ 10. Then ρ is symmetric. Indeed, if aρb, then |a − b| ≤ 10, and so |b − a| = |a − b| ≤ 10; thus, bρa. Definition 1.5. Let ρ be a relation on a set S. We say that ρ is transitive if, whenever aρb and bρc, we also have aρc. Example 1.11. On Z, the relations ≤ and < are both transitive. (If a ≤ b and b ≤ c, then a ≤ c.) However, the relation ρ from Example 1.10 is not, since 1ρ8 and 8ρ13, but 1 \u0003 ρ 13. These three properties lead us directly to the next section. Exercises 1.7. Let S = {1, 2, 3} and T = {3, 4, 5, 6, 7, 8}. Define a relation ρ from S to T via aρb if and only if |a 2 − b| ≤ 1. Find all pairs (a, b) ∈ S × T such that aρb. 1.8. Define a relation ρ on Z via aρb if and only if ab is even. Is ρ reflexive? Symmetric? Transitive? 1.9. Define a relation ρ on R via aρb if and only if a − b ∈ Q. Is ρ reflexive? Symmetric? Transitive? 1.10. Define a relation ρ on R via aρb if and only if a − b ∈ N. Is ρ reflexive? Symmetric? Transitive? 1.11. 1. How many relations are there on {1, 2, 3}? 2. How many of these relations are symmetric? 1.12. For each of the eight subsets of {reflexive, symmetric, transitive}, find a relation on {1, 2, 3} that has the properties in that subset, but not the properties that are not in the subset.\n\n1.3 Equivalence Relations Definition 1.6. An equivalence relation on a set S is a relation that is reflexive, symmetric and transitive. We will use the symbol ∼ to denote an equivalence relation. Example 1.12. On Z, let us say that a ∼ b if and only if a + b is even. We claim that ∼ is an equivalence relation. If a ∈ Z, then a + a is certainly even, so a ∼ a, and ∼ is reflexive. If a ∼ b, then a + b is even. But this also means that b + a is even, and hence b ∼ a. Thus, ∼ is symmetric. Finally, suppose that a ∼ b and b ∼ c. Then a + b and b + c are both even. This means that their sum, a + 2b + c, is even. As 2b is even, we see that a + c is even, and hence a ∼ c. That is, ∼ is transitive.\n\n1.3 Equivalence Relations\n\n7\n\nExample 1.13. On the set S = {a ∈ Z : 1 ≤ a ≤ 20}, let a ∼ b if and only if a = 2m b for some m ∈ Z. Let us verify that this is an equivalence relation. Reflexivity: Note that a = 20 a, and hence a ∼ a. Symmetry: If a ∼ b, say a = 2m b, then b = 2−m a, and hence b ∼ a. Transitivity: If a ∼ b and b ∼ c, say a = 2m b and b = 2n c, then a = 2m+n c, and therefore a ∼ c. Example 1.14. On R, let us say that a ∼ b if and only if a − b ∈ Z. Let us check that it is an equivalence relation. Reflexivity: If a ∈ R, then a − a = 0 ∈ Z, and hence a ∼ a. Symmetry: Let a ∼ b. Then a − b ∈ Z, and hence b − a = −(a − b) ∈ Z. Thus, b ∼ a. Transitivity: Suppose that a ∼ b and b ∼ c. Then a − b, b − c ∈ Z, and hence a − c = (a − b) + (b − c) ∈ Z. That is, a ∼ c. Let us try something slightly more complicated. Example 1.15. Let S = Z × (Z\\{0}). Define ∼ on S via (a, b) ∼ (c, d) if and only if ad = bc. We must verify that ∼ is an equivalence relation. Reflexivity: As ab = ba, we have (a, b) ∼ (a, b) for all integers a and nonzero integers b. Symmetry: Suppose that (a, b) ∼ (c, d). Then ad = bc, and this also tells us that (c, d) ∼ (a, b). Transitivity: Let (a, b) ∼ (c, d) and (c, d) ∼ (e, f ). Then ad = bc and c f = de. Thus, ad f = bc f = bde. Since we are assuming that d \u0003= 0, this means that a f = be. Therefore, (a, b) ∼ (e, f ). Equivalence relations are very special. Definition 1.7. Let ∼ be an equivalence relation on a set S. If a ∈ S, then the equivalence class of a, denoted [a], is the set {b ∈ S : a ∼ b}. Why are equivalence classes so interesting? We need another definition. Definition 1.8. Let S be a set, and let T be a set of nonempty subsets of S. We say that T is a partition of S if every a ∈ S lies in exactly one set in T . Example 1.16. Let S = {1, 2, 3, 4, 5, 6, 7} and T = {{1, 3, 4, 6}, {2, 7}, {5}}. Then T is a partition of S. What is the connection between these concepts? Theorem 1.1. Let S be a set, and ∼ an equivalence relation on S. Then the equivalence classes with respect to ∼ form a partition of S. In particular, if a ∈ S, then a ∈ [a] and, furthermore, a ∈ [b] if and only if [a] = [b]. Proof. As ∼ is reflexive, a ∼ a, and hence a ∈ [a] for every a ∈ S. In particular, the equivalence classes are not empty, and every element of S is in at least one of them. Suppose that d ∈ [a] ∩ [c]. We must show that [a] = [c]. If e ∈ [a], then a ∼ e. Also, d ∈ [a] means that a ∼ d, and hence d ∼ a by symmetry. Also, c ∼ d. By transitivity, c ∼ a, and then c ∼ e. Thus, e ∈ [c], and therefore [a] ⊆ [c]. By the same argument, [c] ⊆ [a], and hence [a] = [c]. Thus, the equivalence classes\n\n8\n\n1 Relations and Functions\n\ndo indeed form a partition. To prove the final statement of the theorem, note that if a ∈ [a] ∩ [b], then [a] = [b] and, conversely, if [a] = [b], then a ∈ [a] = [b]. \u0002 So, the equivalence classes break the set down into subsets having no elements in common. It is important to note that, unless there is only one element in an equivalence class, the representative chosen for that class is not unique. That is, if b ∈ [a], then we could just as easily write [b] instead of [a]. They are the same class. This complicates matters a bit when we define operations on equivalence classes, as we will find ourselves doing throughout the course. We must make sure that our operations are well-defined; that is, that they do not depend upon the particular representative of the class that we use. Let us discuss the equivalence classes determined by the relations in our earlier examples. The plan is always the same. We know that each element of the set is in exactly one class. Thus, we will keep looking for elements of the set that are not in any classes we have constructed, and obtain new classes in this way. Example 1.17. In Example 1.12, let us start with 0. We know that a ∼ 0 if and only if a is even. Thus, = {. . . , −6, −4, −2, 0, 2, 4, 6, . . .}. (Note that we would have obtained the same class had we started, for instance, with 14. Since 14 ∈ , we have = .) We have not yet found 1, so we note that a ∼ 1 if and only if a + 1 is even; that is, if and only if a is odd. Therefore, = {. . . , −5, −3, −1, 1, 3, 5, . . .}. (Again, we could just as easily have used [−3].) We have now found all elements of Z. Thus, there are only two equivalence classes, and . Example 1.18. In Example 1.13, we may as well start with 1. We have = {1, 2, 4, 8, 16}. As we have not yet found 3, = {3, 6, 12}. We still do not have 5, and thus we take = {5, 10, 20}. Similarly, we obtain = {7, 14}, = {9, 18}, = {11}, = {13}, = {15}, = {17}, and = {19}. Once again, we could have used in place of , for instance.\n\n1.3 Equivalence Relations\n\n9\n\nThe other two examples are a bit trickier, since there are infinitely many equivalence classes. But we can attempt to describe them. Example 1.19. In Example 1.14, we see that b ∈ [a] if and only if the difference between a and b is an integer. Thus, for instance, [23.86] = {. . . , −2.14, −1.14, −0.14, 0.86, 1.86, 2.86, . . .}. Listing the classes is an impossible task. How, then, to describe them? We note that for any real number a, there is certainly an integer k such that 0 ≤ a − k < 1. Now, a ∼ (a − k), and hence every element of R is in a class [b], for some 0 ≤ b < 1. Furthermore, if 0 ≤ b, c < 1, then 0 ≤ |b − c| < 1 and therefore b − c can only be an integer if b = c. That is, if 0 ≤ b, c < 1 and b \u0003= c, then [b] \u0003= [c]. Thus, the equivalence classes are precisely {[b] : b ∈ R, 0 ≤ b < 1}. Example 1.20. What about Example 1.15? We note that (c, d) ∈ [(a, b)] if and only if ad = bc. Another way to say this is that ab = dc . Thus, [(a, b)] consists of all ordered pairs (c, d), with c, d ∈ Z and d \u0003= 0, such that ab = dc . This is, in fact, exactly how the rational numbers are constructed! We need to ensure that 23 and 46 are treated as the same fraction, and these equivalence classes make that happen. We obtain one equivalence class for each fraction ab . For instance, [(2, 3)] = {. . . , (−6, −9), (−4, −6), (−2, −3), (2, 3), (4, 6), (6, 9), . . .}. Exercises 1.13. Define a relation ∼ on N via a ∼ b if and only if a − b = 3k, for some k ∈ Z. Is ∼ an equivalence relation? If so, what are the equivalence classes? 1.14. Define a relation ∼ on {1, 2, 3, 4, 5, 6, 7} via a ∼ b if and only if a and b are both even or both odd. Is ∼ an equivalence relation? If so, what are the equivalence classes? 1.15. Define a relation ∼ on Z via a ∼ b if and only if |a| = |b|. Is ∼ an equivalence relation? If so, what are the equivalence classes? 1.16. Define a relation ∼ on Z via a ∼ b if and only if ab > 0. Is ∼ an equivalence relation? If so, what are the equivalence classes? 1.17. Let S be the set of all subsets of Z. Define a relation ∼ on S via T ∼ U if and only if T ⊆ U . Is ∼ an equivalence relation? If so, what are the equivalence classes? 1.18. Let S be the set of all subsets of Z. Define a relation ∼ on S via T ∼ U if and only if T \\U and U \\T are both finite. Show that ∼ is an equivalence relation and describe [{1, 2, 3}] and [{. . . , −4, −2, 0, 2, 4, . . .}].\n\n10\n\n1 Relations and Functions\n\n1.19. On the plane R2 , define a relation ∼ via (a, b) ∼ (c, d) if and only if 3a − b = 3c − d. Show that ∼ is an equivalence relation, and describe [(4, 2)]. 1.20. Let S be a nonempty set. Show that for any partition of S, there is an equivalence relation on S having the sets in the partition as its equivalence classes. 1.21. Find an equivalence relation on N having exactly two equivalence classes, one of which contains exactly three elements. 1.22. Suppose there is a relation ρ on a set S, such that ρ is both reflexive and transitive. Define ∼ on S via a ∼ b if and only if aρb and bρa. Show that ∼ is an equivalence relation.\n\n1.4 Functions Let us give two equivalent definitions of a function. Formally, if S and T are sets, then a function from S to T is a relation ρ from S to T such that, for each s ∈ S, there is exactly one t ∈ T such that sρt. In practice, nobody really thinks of functions in this way. The working definition follows. Definition 1.9. Let S and T be any sets. Then a function α : S → T is a rule assigning, to each s ∈ S, an element α(s) of T . Readers who have studied calculus will no doubt be familiar with functions from R to R. Example 1.21. We can define a function α : R → R via α(a) = 5a 3 − 4a 2 + 7a + 3 for all a ∈ R. But we do not need to go from R to R. Example 1.22. We can define a function α : Z → Q via α(a) = (−2)a for all a ∈ Z. In fact, the sets involved do not have to be sets of numbers. Example 1.23. Let S be the set of all English words and T the set of letters in the alphabet. We can define α : S → T by letting α(w) be the first letter of the word w, for every w ∈ S. A few properties enjoyed by certain functions are important. Definition 1.10. A function α : S → T is one-to-one (or injective) if α(s1 ) = α(s2 ) implies s1 = s2 , for all s1 , s2 ∈ S. Putting this another way, a one-to-one function sends different elements to different places.\n\n1.4 Functions\n\n11\n\nExample 1.24. Define functions α and β from R to R via α(a) = a 2 and β(a) = a 3 , for all a ∈ R. Then α is not one-to-one, since α(1) = α(−1), but β is one-to-one, since if a 3 = b3 , then taking the cube root of both sides, we have a = b, for any a, b ∈ R. Definition 1.11. A function α : S → T is onto (or surjective) if, for every t ∈ T , there exists at least one s ∈ S such that α(s) = t. Example 1.25. Define α and β as in Example 1.24. Then α√ is not onto, since there is no a ∈ R such that α(a) = −1. However, if b ∈ R, then β( 3 b) = b; thus, β is onto. We should not get the idea that one-to-one and onto always occur together. Example 1.26. Define α : R → R via α(a) = 2a . Then α is one-to-one, for if 2a = 2b , then taking the base 2 logarithm of both sides, we see that a = b. On the other hand, there is no a ∈ R such that 2a = −1, so α is not onto. However, it is nice when we can combine the two properties. Definition 1.12. A function α : S → T is bijective if it is one-to-one and onto. An equivalent way of expressing this property is that for each t ∈ T , there is exactly one s ∈ S such that α(s) = t. There must be such an s, since α is onto, but if α(s1 ) = α(s2 ) = t, for some s1 , s2 ∈ S, then since α is one-to-one, s1 = s2 . For this reason, a bijective function is also known as a one-to-one correspondence. Example 1.27. Combining Examples 1.24 and 1.25, we see that α : R → R given by α(a) = a 3 is bijective. Let us discuss how to combine functions. Definition 1.13. Let R, S and T be sets, and let α : R → S and β : S → T be functions. Then the composition, β ◦ α, or simply βα, is the function from R to T given by (βα)(r ) = β(α(r )) for all r ∈ R. Note that when we write βα, we are applying α first, then β. The order is important! Indeed, depending upon the sets involved, it is possible that applying β first, then α, would not make sense. But even if it did make sense, the result would not necessarily be the same. Example 1.28. Define functions α and β from R to R via α(a) = a 3 + 1 and β(a) = a 2 , for all a ∈ R. Then (βα)(a) = β(a 3 + 1) = a 6 + 2a 3 + 1, whereas (αβ)(a) = α(a 2 ) = a 6 + 1, for all a ∈ R. That is, βα and αβ are different functions. We can list a few important properties of the composition of functions. Theorem 1.2. Let α : R → S, β : S → T and γ : T → U be functions. Then 1. (γβ)α = γ (βα); 2. if α and β are one-to-one, then so is βα;\n\n12\n\n1 Relations and Functions\n\n3. if α and β are onto, then so is βα; and 4. if α and β are bijective, then so is βα. Proof. (1) Take any r ∈ R. Then ((γβ)α)(r ) = (γβ)(α(r )) = γ (β(α(r ))). Similarly, (γ (βα))(r ) = γ ((βα)(r )) = γ (β(α(r ))). (2) Suppose that (βα)(r1 ) = (βα)(r2 ) for some r1 , r2 ∈ R. Then β(α(r1 )) = β(α(r2 )). Since β is one-to-one, α(r1 ) = α(r2 ). Since α is one-to-one, r1 = r2 . (3) Take any t ∈ T . Since β is onto, there exists an s ∈ S such that β(s) = t. But α is also onto, so there exists an r ∈ R such that α(r ) = s. Thus, (βα)(r ) = β(α(r )) = β(s) = t. (4) Combine (2) and (3). \u0002 The following additional property of bijective functions can be useful. Theorem 1.3. Let α : S → T be a bijective function. Then there exists a bijective function β : T → S such that (βα)(s) = s for all s ∈ S and (αβ)(t) = t for all t ∈ T. Proof. Since α is bijective, for any t ∈ T , there is a unique s ∈ S such that α(s) = t. Define β : T → S via β(t) = s. By definition, we have (βα)(s) = β(α(s)) = s, for all s ∈ S. Also, if t ∈ T , then choosing s such that α(s) = t, we have β(t) = s, and therefore (αβ)(t) = α(β(t)) = α(s) = t, as required. It remains to show that β is bijective. But if β(t1 ) = β(t2 ), then t1 = (αβ)(t1 ) = α(β(t1 )) = α(β(t2 )) = (αβ)(t2 ) = t2 , so β is one-to-one. Furthermore, if s ∈ S, then β(α(s)) = (βα)(s) = s, and hence β is onto. \u0002 Example 1.29. Let α : R → R be given by α(a) = a 3 for all a. Example 1.27 showed us that √ α is bijective. It is easily checked that if we let β : R → R be given by β(a) = 3 a for all a, then (αβ)(a) = (βα)(a) = a for all a. We close the chapter by defining two special types of functions. Definition 1.14. A permutation of a set S is a bijective function from S to S. Example 1.30. By Example 1.27, the function α : R → R given by α(a) = a 3 is a permutation of R. Example 1.31. Let S = {1, 2, 3, 4}. Define α : S → S via α(1) = 3, α(2) = 2, α(3) = 4 and α(4) = 1. Then α is a permutation of S. As this last example illustrates, a permutation is simply a rearrangement of the elements of S. Definition 1.15. Let S be a set. Then a binary operation on S is a function from S × S to S.\n\n1.4 Functions\n\n13\n\nExample 1.32. We can define a binary operation ∗ on R via a ∗ b = 2a 2 b − 3b4 + 5, for all a, b ∈ R. (Putting this in terms of functions, we could write α((a, b)) = 2a 2 b − 3b4 + 5 for all a, b ∈ R.) Note that in order to obtain a binary operation, we must stay within our original set. For instance, we would not get a binary operation on N if we tried to let a ∗ b = ab , / N. for the simple reason that 1 ∗ 2 = 21 ∈ Exercises 1.23. Define α : {1, 2, 3, 4} → {1, 2, 3, 4, 5, 6, 7} via α(a) = 2a − 1. Is this function one-to-one? Is it onto? √ 1.24. Define α : R → R via α(a) = 3 a + 1 − 2. Is this function one-to-one? Is it onto? 1.25. Let S be the set of real numbers and T the set of positive real numbers. Define α : S → T via α(a) = 23a−5 . Show that α is a bijection and find β : T → S such that (βα)(a) = a for all a ∈ S. 1.26. Define α : R → R via \u0006 α(a) =\n\n4a − 3, a ≤ 1 a2, a > 1.\n\nShow that α is bijective and find β : R → R such that (βα)(a) = a for all a ∈ R. 1.27. Which of the following are binary operations on N? 1. a ∗ b = ab 2. a ∗ b = a − b 3. a ∗ b = 3 for all a and b 1.28. Let S be a finite set, and suppose that α : S → S is a one-to-one function. Show that α is a permutation of S. Construct an explicit counterexample to show that this need not be true if S is infinite. 1.29. Let α : R → S and β : S → T be functions, and suppose that βα is onto. Must α be onto? Must β? 1.30. Let α : R → S and β : S → T be functions, and suppose that βα is one-toone. Must α be one-to-one? Must β? 1.31. Let S be a set with m elements and T a set with n elements, for some positive integers m and n. 1. How many functions are there from S to T ? 2. How many of these functions are one-to-one? 1.32. Let S and T be sets and α : S → T a function. Show that there exist a set R and functions β : S → R and γ : R → T such that β is onto, γ is one-to-one and α = γβ.\n\nChapter 2\n\nThe Integers and Modular Arithmetic\n\nIn this chapter, we begin with a discussion of mathematical induction. Next, we examine a number of properties of the integers, with an emphasis on divisibility and prime factorization. We conclude by introducing modular arithmetic.\n\n2.1 Induction and Well Ordering We begin with an important property of the set of natural numbers. Property 2.1 (Well Ordering Axiom). If S is a nonempty set of positive integers, then S has a smallest element. This seems so obvious, but it is actually a rather special property of N. Indeed, Z has no smallest element; neither, for that matter, does the set of positive real numbers. There is an equivalent form of the Well Ordering Axiom that is especially useful. To state it, we need a definition. A proposition is a statement that is either true or false. For instance, “Ottawa is the capital of Canada” is a true proposition, and “There are only finitely many even integers” is a false one. We avoid statements having no truth value, such as “This statement is false” as well as statements that are a matter of opinion, such as “Xena: Warrior Princess was a great television program”1 . What we would like to do is define a sequence of propositions, P(1), P(2), P(3) and so on, and prove that all of them are true at once. This is where induction comes in. Theorem 2.1 (Principle of Mathematical Induction). Suppose that, for each positive integer n, we have a proposition P(n). Further suppose that\n\n1 Of\n\ncourse, any reasonable person would agree with this statement, but in principle, it is a matter of opinion. © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_2\n\n15\n\n16\n\n2 The Integers and Modular Arithmetic\n\n1. P(1) is true; and 2. for each n ∈ N, if P(n) is true, then so is P(n + 1). Then P(n) is true for every positive integer n. Proof. Suppose the theorem is false, and let S be the set of all positive integers n such that P(n) is false. Then S is a nonempty subset of N. By the Well Ordering Axiom, S has a smallest element k. Now, we are assuming that P(1) is true, so k > 1. Then k − 1 ∈ / S, and hence P(k − 1) is true. By our assumption, P(k) is true as well, giving us a contradiction and completing the proof. \u0002 Induction is a powerful tool! We can prove infinitely many propositions in just two steps. Here is a simple example. Example 2.1. We claim that for every positive integer n, we have 12 + 22 + · · · + n 2 =\n\nn(n + 1)(2n + 1) . 6\n\nWe proceed by induction. For each n ∈ N, the proposition P(n) is the statement 12 + 22 + · · · + n 2 =\n\nn(n + 1)(2n + 1) . 6\n\nFirst, we must prove P(1). But it states that 12 =\n\n1(1 + 1)(2 · 1 + 1) , 6\n\nwhich is obvious. Now, we assume P(n) and prove P(n + 1). But 12 + 22 + · · · + n 2 + (n + 1)2 =\n\nn(n + 1)(2n + 1) + (n + 1)2 , 6\n\nby our inductive hypothesis, P(n). Simplifying, we have (n + 1)(n(2n + 1) + 6(n + 1)) 6 (n + 1)(2n 2 + 7n + 6) = 6 (n + 1)(n + 2)(2n + 3) = 6 (n + 1)((n + 1) + 1)(2(n + 1) + 1) . = 6\n\n12 + 22 + · · · + (n + 1)2 =\n\nBut this is precisely P(n + 1). Thus, the proof is complete.\n\n2.1 Induction and Well Ordering\n\n17\n\nThere is another result that we can prove by induction, and which we will need later. A bit of notation is required. For any positive integer n, we define n! (read “n factorial”) via n! = n(n − 1)(n −\u0002 2) \u0003 · · · (2)(1). Also, 0! = 1.\u0002 If\u0003 n andn!k are . integers, with n ≥ k ≥ 0, then we define nk (read “n choose k”) via nk = (n−k)!k! Example 2.2. We have 5! = 5 · 4 · 3 · 2 · 1 = 120 and\n\n\u00026\u0003 2\n\n=\n\n6! 4!2!\n\n=\n\n720 24·2\n\n= 15.\n\nTheorem 2.2 (Binomial Theorem). Let a and b be real numbers and n a positive integer. Then\n\n\u0004 \u0005 \u0004 \u0005 \u0004 \u0005 n n−1 n n−2 2 n (a + b) = a + a b+ a b + ··· + abn−1 + bn . 1 2 n−1 n\n\nn\n\nProof. Let us proceed by induction on n. When n = 1, both sides of the equation are a + b, so there is nothing to do. Assume the result for n, and prove it for n + 1. But (a + b)n+1 = (a + b)n (a + b) \u0005 \u0004 \u0005 \u0004 \u0005 \u0004 n n−1 n a b + ··· + abn−1 + bn (a + b), = an + 1 n−1 by our inductive hypothesis. When we expand this product, we obtain a sum of terms consisting of a coefficient n+1 multiplied by a n+1−k bk , where 0 ≤ k ≤ n + 1. The coefficients of a n+1 and \u0003 \u0002n \u0003 b \u0002 nare n+1−k k b is k + k−1 , clearly 1, whereas if 0 < k < n + 1, then the coefficient of a \u0002 n \u0003 n−(k−1) k−1 \u0002\u0003 b )b. However, a since these terms arise from ( nk a n−k bk )a and ( k−1 \u0004 \u0005 \u0004 \u0005 n! n n n! + + = (n − k)!k! (n − k + 1)!(k − 1)! k k−1 n!(n − k + 1) + n!k = (n − k + 1)!k! \u0004 \u0005 n+1 = . k That is, (a + b)\n\nn+1\n\n=a\n\nn+1\n\n\u0004 \u0005 \u0004 \u0005 n+1 n n+1 + a b + ··· + abn + bn+1 , 1 n\n\nand the proof is complete.\n\n\u0002\n\nSometimes, a slightly different form of induction is required. Theorem 2.3 (Strong Induction). Suppose that, for each positive integer n, we have a proposition P(n). Further suppose that\n\n18\n\n2 The Integers and Modular Arithmetic\n\n1. P(1) is true; and 2. for each integer n > 1, if P(k) is true for every k < n, then P(n) is true. Then P(n) is true for every positive integer n. Proof. Suppose that the theorem is false, and let S be the set of positive integers n such that P(n) is false. Then S is a nonempty subset of N. By the Well Ordering Axiom, it has a smallest element j. As P(1) is true, j > 1. But then by the minimality of j, we see that P(k) is true whenever k < j. Thus, P( j) is true, giving us a contradiction. \u0002 As before, we must prove the first proposition. But after that, instead of just assuming that the previous case is true, we assume that all prior cases are true. This can give us more to work with. Example 2.3. Define a sequence via a1 = 1, a2 = 3, a3 = 7 and, for each n ≥ 4, an = an−1 + an−2 + an−3 . We claim that an < 2n for all n ∈ N. We need strong induction here, because when we consider an , we require information not just about an−1 , but about the terms before it as well. When n = 1, there is nothing to do. Assume that n > 1 and that the claim is true for smaller values of n. If n = 2 or 3, again, the result is obvious, so assume that n ≥ 4. Then an = an−1 + an−2 + an−3 < 2n−1 + 2n−2 + 2n−3 , by our inductive hypothesis. However, 2n−1 + 2n−2 + 2n−3 = 7 · 2n−3 < 2n . We are done. Exercises 2.1. Show that for every positive integer n, 1 + 2 + ··· + n =\n\nn(n + 1) . 2\n\n2.2. Show that for every positive integer n, 1 · 2 + 2 · 3 + · · · + n(n + 1) =\n\nn(n + 1)(n + 2) . 3\n\n2.3. Show that for every positive integer n, the following two identities hold. 1.\n\n\u0004 \u0005 \u0004 \u0005 \u0004 \u0005 \u0004 \u0005 n n n n + + + ··· + = 2n 0 1 2 n\n\n2.\n\n\u0004 \u0005 \u0004 \u0005 \u0004 \u0005 \u0004 \u0005 \u0004 \u0005 n n n n n − + − + · · · + (−1)n =0 0 1 2 3 n\n\n2.4. In the plane R2 , let us draw n lines, no two of which are parallel and no three of which meet at a point. Into how many regions do they divide the plane?\n\n2.1 Induction and Well Ordering\n\n19\n\n2.5. Show that for all integers n ≥ 2, we have 1. √ (1 + a)n > 1 + na, for all positive real numbers a; and 2. n n < 2 − n1 . \u0002 \u0003 is less than 4n−1 for all positive integers n ≥ 5. 2.6. Show that 2n n 2.7. We define the Fibonacci sequence via f 1 = f 2 = 1, and if n > 2, then f n = f n−1 + f n−2 . Show that, for every positive integer n, f n ≤ (7/4)n−1 . 2.8. With f n as in the preceding exercise, show that for every positive integer n, \u0006 fn =\n\n√ \u0007n 1+ 5 2\n\n\u0006 √ \u0007n − 1−2 5 . √ 5\n\n2.9. A bar of chocolate is a rectangular array consisting of r rows and c columns of unit square chocolate pieces, with thin lines separating the rows and columns. A single action consists of taking one bar, and breaking it along a line separating two rows or two columns, producing two smaller bars. Show that it will take precisely r c − 1 such actions to turn the bar into r c square pieces. (This can be done using strong induction, or with no induction at all.) 2.10. Show that for every positive integer n, there exist a positive integer k, and integers ai ∈ {0, 1}, such that n = a0 + 2a1 + 22 a2 + 23 a3 + · · · + 2k ak .\n\n2.2 Divisibility The following theorem simply formalizes the usual division process in the integers. Theorem 2.4 (Division Algorithm). Let a, b ∈ Z with b > 0. Then there exist unique integers q and r such that a = bq + r , with 0 ≤ r < b. Proof. We will prove the existence of q and r first, and then worry about their uniqueness. Let S = {a − bt : t ∈ Z, a − bt ≥ 0}. If 0 ∈ S, then a − bq = 0 for some q ∈ Z, and hence a = bq + 0, as desired. Therefore, we may assume that S ⊆ N. We claim that S is nonempty. Let t = −|a|. Then a − bt = a + |a|b. If a ≥ 0, then a + |a|b ≥ 0, since b > 0. If a < 0, then a + |a|b = a(1 − b). But a < 0 and since b ≥ 1, we have 1 − b ≤ 0. Thus, a(1 − b) ≥ 0. Either way, the claim is proved. In view of the Well Ordering Axiom, S has a least element, say r = a − bq. By definition, r ≥ 0. If r ≥ b, then 0 ≤ r − b < r , but also r − b = a − bq − b = a − b(q + 1), and therefore a − b(q + 1) is a smaller element of S than r , contradicting the choice of r . Thus, a = bq + r , with 0 ≤ r < b. As to uniqueness, suppose that a = bq1 + r1 = bq2 + r2 , with qi , ri ∈ Z and 0 ≤ ri < b. Then b(q1 − q2 ) = r2 − r1 . In particular, b|q1 − q2 | = |r2 − r1 |. If q1 \u0007= q2 , then b|q1 − q2 | ≥ b. But 0 ≤ r1 , r2 < b, so |r2 − r1 | < b, which is impossible. Therefore, q1 = q2 . But then r1 = r2 as well. \u0002\n\n20\n\n2 The Integers and Modular Arithmetic\n\nWe call q and r in the preceding theorem the quotient and remainder respectively. Example 2.4. Using b = 5, we have 68 = 5(13) + 3 and −21 = 5(−5) + 4. The case in which the remainder is 0 is of particular interest. Definition 2.1. Let a and b be integers. We say that a divides b (or b is a multiple of a) if there exists an integer c such that b = ac. In this case, we write a|b. Example 2.5. As 84 = 6(14) and 84 = −3(−28), we write 6|84 and −3|84. On the other hand, 10 \u0002 84. Here are a few basic properties of divisibility, the proofs of which are left as Exercise 2.14. Lemma 2.1. Let a, b, c ∈ Z. Then 1. if a|b and b|c, then a|c; 2. if a|b and b \u0007= 0, then a ≤ |b|; and 3. if a|b and a|c, then a|(bu + cv) for any u, v ∈ Z. Definition 2.2. Let a and b be integers, not both 0. Then the greatest common divisor (or gcd) of a and b, written (a, b), is the largest positive integer g such that g|a and g|b. Example 2.6. We have (60, 170) = 10 and (42, −55) = 1. Note that the gcd must always exist. As 1 divides everything, a and b must have a common divisor. Also, by Lemma 2.1, if a \u0007= 0, then (a, b) ≤ |a|. Thus, only the numbers from 1 to |a| need to be considered. We specifically exclude the case a = b = 0, since everything divides 0. Let us mention a couple of easy facts about gcds. Lemma 2.2. Take any integers a and b with a \u0007= 0. Then 1. (a, b) = (−a, b); and 2. (a, 0) = |a|. Proof. (1) Any divisor of a also divides −a, and vice versa. (2) Clearly |a| divides both a and 0, and Lemma 2.1 shows that no larger integer can do so. \u0002 One particular case is important. Definition 2.3. Let a, b ∈ Z, not both 0. Then we say that a and b are relatively prime if (a, b) = 1. Example 2.7. By Example 2.6, 60 and 170 are not relatively prime, but 42 and −55 are.\n\n2.2 Divisibility\n\n21\n\nWhy is the gcd so significant? The following theorem gives us an idea. Theorem 2.5. Let a and b be integers, not both 0. Then there exist u, v ∈ Z such that (a, b) = au + bv. Furthermore, (a, b) is the smallest positive integer that can be written in this way. Proof. Let S = {ax + by : x, y ∈ Z, ax + by > 0}. Clearly S ⊆ N. Without loss of generality, we may assume that a \u0007= 0. Then a 2 + b(0) = a 2 ∈ S, and hence S is not empty. By the Well Ordering Axiom, S has a least element, say g = au + bv. We claim that g = (a, b). This will complete the proof. Suppose that c|a and c|b. By Lemma 2.1, c|g and, hence, c ≤ g. It remains only to show that g divides both a and b. Using the division algorithm, write a = gq + r , where q and r are integers and 0 ≤ r < g. Then r = a − gq = a − (au + bv)q = a(1 − uq) + b(−vq). Thus, if r > 0, then r ∈ S. But r < g, contradicting the minimality of g. Therefore, r = 0 and g|a. By the same argument, g|b. \u0002 The following is an immediate consequence. Corollary 2.1. Let a, b ∈ Z, not both 0. Then a and b are relatively prime if and only if there exist integers u and v such that au + bv = 1. We can now prove a couple of useful results for relatively prime numbers. Corollary 2.2. Let a, b, c ∈ Z with a and b not both 0. If (a, b) = 1 and a|bc, then a|c. Proof. By the preceding corollary, we may write au + bv = 1, for some u, v ∈ Z. Then acu + bcv = c. But a|a and a|bc hence, by Lemma 2.1, a|(acu + bcv) = c. \u0002 Corollary 2.3. Let a, b ∈ Z, not both 0. If a and b are relatively prime, and for some integer n, we have a|n and b|n, then ab|n. Proof. See Exercise 2.18.\n\n\u0002\n\nBe careful not to apply the last two corollaries if a and b are not relatively prime! For instance, 6|4 · 3, but 6 \u0002 4 and 6 \u0002 3. Also, 4|12 and 6|12 but 24 \u0002 12. What we have not yet discussed is how to find (a, b) and the numbers u and v from Theorem 2.5. We could certainly list the common divisors of a and b and see which one is largest, but if the numbers are large, this would be rather time-consuming. It would also give us no insight into finding u and v. Happily, there is a better way. The following technique is attributed to the ancient Greek mathematician Euclid.\n\n22\n\n2 The Integers and Modular Arithmetic\n\nTheorem 2.6 (Euclidean Algorithm). Let a and b be integers, with b positive. If b|a, then (a, b) = b. Otherwise, apply the division algorithm repeatedly. Let a = bq1 + r1 b = r 1 q2 + r 2 r 1 = r 2 q3 + r 3 .. . rk−2 = rk−1 qk + rk rk−1 = rk qk+1 + 0, where qi , ri ∈ Z for all i and 0 < rk < rk−1 < · · · < r1 < b. Then (a, b) = rk . Proof. If b|a, then b is a common divisor of a and b. In view of Lemma 2.1, it is the largest possible common divisor. Assume that b \u0002 a. Note that we will only apply the division algorithm finitely many times, as each ri+1 < ri , and all are positive. Suppose that c|a and c|b. By Lemma 2.1, c|(a − bq1 ) = r1 . Thus, every common divisor of a and b is also a common divisor of b and r1 . But if d|b and d|r1 , then d|(bq1 + r1 ) = a. That is, the common divisors of a and b are precisely the same as those of b and r1 . In particular, (a, b) = (b, r1 ). But by exactly the same argument, (a, b) = (b, r1 ) = (r1 , r2 ) = (r2 , r3 ) = · · · = (rk , 0) = rk , \u0002\n\nby Lemma 2.2.\n\nWe do require b to be positive in the Euclidean algorithm, but we can use the fact that (a, b) = (−a, b) if neither a nor b is positive. In fact, the Euclidean algorithm is doubly useful, because if we start with the penultimate equation and work our way backwards, we can find integers u and v such that (a, b) = au + bv. Indeed, we have (a, b) = rk = rk−2 (1) + rk−1 (−qk ), and so (a, b) is a multiple of rk−2 plus a multiple of rk−1 . But then rk−1 = rk−3 − rk−2 qk−1 and substitution yields (a, b) = rk−2 (1) + (rk−3 − rk−2 qk−1 )(−qk ) = rk−2 (1 + qk−1 qk ) + rk−3 (−qk ). That is, (a, b) is a multiple of rk−3 plus a multiple of rk−2 . Eventually, we will write it in the desired form.\n\n2.2 Divisibility\n\n23\n\nExample 2.8. Let a = 45 and b = 33. Applying the Euclidean algorithm, we have 45 = 33(1) + 12 33 = 12(2) + 9 12 = 9(1) + 3 9 = 3(3) + 0. Thus, (a, b) = 3. Let us find u and v such that au + bv = 3. We have 3 = 12(1) + 9(−1) = 12(1) + (33(1) + 12(−2))(−1) = 12(3) + 33(−1) = (45(1) + 33(−1))(3) + 33(−1) = 45(3) + 33(−4). That is, (a, b) = 3a − 4b. Exercises 2.11. In each case, use the Euclidean algorithm to find (a, b). 1. a = 57, b = 20 2. a = 117, b = 51 2.12. For each of the two parts of the preceding problem, find integers u and v such that (a, b) = au + bv. 2.13. Let a and b be integers such that a|b and b|a. Show that a ∈ {b, −b}. 2.14. Prove Lemma 2.1. 2.15. Show that if a, b and c are positive integers, with (a, b) = 1, and c|a, then (c, b) = 1. 2.16. Show that n 5 − n is divisible by 5 for every positive integer n. 2.17. Let a and n be positive integers. Show that there exists an integer u such that n|(au − 1) if and only if a and n are relatively prime. 2.18. Let a, b ∈ Z, not both 0. If a and b are relatively prime, and for some integer n, we have a|n and b|n, show that ab|n. 2.19. Take f n as in Exercise 2.7. Show that 3| f n if and only if 4|n. 2.20. Take f n as in Exercise 2.7. Show that 4| f n if and only if 6|n.\n\n24\n\n2 The Integers and Modular Arithmetic\n\n2.3 Prime Factorization Prime numbers will have a special importance throughout the course. Definition 2.4. A natural number p > 1 is said to be prime if its only positive divisors are 1 and p. Otherwise, it is composite. Note that 1 is neither prime nor composite. Example 2.9. The first few primes are 2, 3, 5, 7, 11, 13, 17, . . .. An equivalent way of defining a prime number is given in the following result due to Euclid. Theorem 2.7 (Euclid’s Lemma). Let p > 1 be a positive integer. Then the following are equivalent: 1. p is prime; and 2. if a and b are integers such that p|ab, then p|a or p|b. Proof. Suppose that p is prime and p|ab. Now, ( p, a)| p, so ( p, a) = 1 or p. If ( p, a) = p, then since ( p, a)|a, we have p|a. Otherwise, by Corollary 2.1, there exist integers u and v such that pu + av = 1. But then pbu + abv = b. Now, p| p and p|ab, so by Lemma 2.1, p|b. On the other hand, if p is composite, then let p = cd, where 1 < c, d < p. In this case, p|cd, but by Lemma 2.1, p \u0002 c and p \u0002 d. \u0002 Corollary 2.4. Let p be a prime number and a1 , . . . , an ∈ Z. If p|a1 a2 · · · an , then p|ai , for some i. Proof. Exercise 2.24.\n\n\u0002\n\nIn fact, every positive integer larger than 1 can be written as a product of primes, called its prime factorization. Theorem 2.8 (Fundamental Theorem of Arithmetic). If a ∈ N and a > 1, then there exist primes p1 , . . . , pn (not necessarily distinct) such that a = p1 p2 · · · pn . Furthermore, this product is unique up to order. That is, if a = q1 q2 · · · qm , for some primes qi , then m = n and, after rearranging the primes, pi = qi for all i. Proof. Let us prove the existence of the prime factorization and then handle the uniqueness. We will prove the result by strong induction on a. We are excluding the case a = 1, so start with a = 2. There is nothing to do here, since 2 is prime. Thus, let a > 2 and assume that the theorem is true for smaller numbers. If a is prime, there is nothing to do. Otherwise, we can write a = bc, with 1 < b, c < a. But then by our inductive hypothesis, b and c are both products of primes, and hence a is a product of primes.\n\n2.3 Prime Factorization\n\n25\n\nNow let us prove the uniqueness. Suppose that a = p1 · · · pn = q 1 · · · q m , for some primes pi and qi . Without loss of generality, say n ≤ m. Now, p1 |a. Thus, by Corollary 2.4, p1 |qi , for some i. Rearranging the primes as needed, we may assume that p1 |q1 . But q1 is prime, so p1 = 1 or q1 . As 1 is not prime, p1 = q1 . Cancelling p1 and q1 from the two sides of our equation, we have p2 · · · pn = q 2 · · · q m . Now do the same for p2 and repeat. We find that, after rearranging, pi = qi , 1 ≤ i ≤ n. If m = n, we are done. Otherwise, we are left with 1 = qn+1 · · · qm . But then \u0002 qm |1, which is impossible, as qm > 1. Example 2.10. We can write 1400 = 2 · 2 · 2 · 5 · 5 · 7, and there is no other way to write 1400 as a product of primes, except by rearranging (for instance, 2 · 5 · 7 · 2 · 2 · 5). Note that this gives us one good reason not to consider 1 as a prime: we would have to abandon uniqueness, as we could multiply by 1 as many times as we wanted. We can use the existence of prime factors to prove a handy fact. Corollary 2.5. Let a, b and n be integers, with n \u0007= 0. If (a, n) = (b, n) = 1, then (ab, n) = 1. Proof. If (ab, n) > 1, then by Theorem 2.8, there exists a prime p dividing (ab, n). Since p|ab, Theorem 2.7 tells us that p|a or p|b. But p|n as well; thus, (a, n) ≥ p or (b, n) ≥ p. Either way, we have a contradiction. \u0002 Exercises 2.21. Factor each of the following numbers into a product of primes: 3528, 30030 and 220000. 2.22. Show that for every prime p > 3, there exists a positive integer k such that p = 6k + 1 or p = 6k − 1. 2.23. Let p be a prime and n an integer. Show that either p|n or ( p, n) = 1. 2.24. Use induction to prove Corollary 2.4. 2.25. Let p1 , . . . , pk be any primes. Show that for each i, pi \u0002 ( p1 p2 · · · pk + 1). 2.26. Use the preceding exercise to show that there are infinitely many primes. 2.27. Let p be a prime and a, n ∈ N. Suppose that p|a n . Show that p n |a n . 2.28. Let p1 , . . . , pk be distinct primes, and let m i , n i be nonnegative integers. Find the gcd of p1m 1 · · · pkm k and p1n 1 · · · pkn k .\n\n26\n\n2 The Integers and Modular Arithmetic\n\n2.4 Properties of the Integers This section may seem a tad underwhelming. Indeed, there are no proofs at all and we will not really learn any new facts about the integers. The whole point is to establish some terminology that we will see many times in different settings. While our discussion will take place in Z, it is worth noting that we could just as easily use Q, R or C. First, we observe that Z is closed under addition and multiplication. That is, a + b, ab ∈ Z for all a, b ∈ Z. Next, addition and multiplication on Z are both associative. This means that (a + b) + c = a + (b + c) and (ab)c = a(bc) for all a, b, c ∈ Z. In particular, we can write a + b + c and abc without fear of ambiguity. Furthermore, addition and multiplication are both commutative on Z. In other words, a + b = b + a and ab = ba for all a, b ∈ Z. We also have the distributive law. Specifically, a(b + c) = ab + ac for all a, b, c ∈ Z. The numbers 0 and 1 are rather special. We call 0 the additive identity for Z and 1 the multiplicative identity. This is because a + 0 = a and a · 1 = a for all a ∈ Z. Finally, if a ∈ Z, then −a is its additive inverse. This means that a + (−a) = 0. It is important to note that we do not have multiplicative inverses for all integers; that is, if a ∈ Z, it does not follow that there exists a b ∈ Z such that ab = 1. In fact, this only happens if a is 1 or −1. (The sets Q, R and C are a bit different on this last point. Every element other than 0 has a multiplicative inverse in these sets. For instance, in Q, the multiplicative inverse of 29 is 29 .)\n\n2.4 Properties of the Integers\n\n27\n\nExercises 2.29. For each of the following binary operations on Z, decide if it is commutative; that is, do we have a ∗ b = b ∗ a for all a, b ∈ Z? 1. a ∗ b = ab + 1 2. a ∗ b = a + b + ab 3. a ∗ b = a 2.30. For each part of the preceding exercise, are the operations associative? That is, do we have (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ Z? 2.31. For parts (1) and (2) from Exercise 2.29, decide if ∗ has an identity; that is, does there exist an e ∈ Z such that a ∗ e = e ∗ a = a for all a ∈ Z? 2.32. Define a binary operation ∗ on Q via a ∗ b = a + b − ab. Find an identity e; that is, find e ∈ Q such that a ∗ e = e ∗ a = a for all a ∈ Q. Then decide which elements of Q have inverses. That is, determine for which b ∈ Q there exists a c ∈ Q such that b ∗ c = c ∗ b = e.\n\n2.5 Modular Arithmetic When we perform modular arithmetic, we choose an integer n ≥ 2 and then for any integer a, we concern ourselves only with the remainder when a is divided by n. As the only possible remainders are 0, 1, 2, . . . , n − 1, these are the only numbers to worry about. Definition 2.5. Let n ≥ 2 be an integer. If a, b ∈ Z, then we say that a is congruent to b modulo n, and write a ≡ b (mod n), if n|(a − b); that is, if a and b have the same remainder when divided by n. Example 2.11. As 8|(53 − 21), we have 53 ≡ 21 (mod 8). Putting this another way, 53 and 21 both have remainder 5 when divided by 8. We reduce to the remainder and write 53 ≡ 5 (mod 8) and 21 ≡ 5 (mod 8). We add and multiply modulo n in the usual way, simply reducing to the remainder. Example 2.12. We observe that 5 + 8 ≡ 1 (mod 12) and 5 · 8 ≡ 4\n\n(mod 12).\n\nOf course, we should be a bit careful here. For instance, since 5 ≡ 17 (mod 12), we had better make sure that 5 + 8 ≡ 17 + 8 (mod 12). This is certainly the case, but it will help if we express things in terms of equivalence classes.\n\n28\n\n2 The Integers and Modular Arithmetic\n\nTheorem 2.9. Let n ≥ 2 be an integer. Then a ≡ b (mod n) is an equivalence relation on Z. The equivalence class of a consists of all integers having the same remainder as a when divided by n. Proof. Reflexivity: We have n|0 = a − a, so a ≡ a (mod n). Symmetry: Suppose that a ≡ b (mod n). Then n|(a − b), and hence n| − (a − b) = b − a. Thus, b ≡ a (mod n). Transitivity: Suppose that a ≡ b (mod n) and b ≡ c (mod n). Then n|(a − b) and n|(b − c). Hence, n|((a − b) + (b − c)) = a − c. That is, a ≡ c (mod n). The statement about the equivalence classes follows from the definition. \u0002 Definition 2.6. Let n ≥ 2 be an integer. The set of integers modulo n, denoted Zn , is the set of all equivalence classes of Z with respect to the equivalence relation a ≡ b (mod n). We call these the congruence classes modulo n. Specifically, Zn = {, , , . . . , [n − 1]}. Example 2.13. The elements of Z4 are , , and , where = {. . . , −, 8, −4, 0, 4, 8, . . .} = {. . . , −7, −3, 1, 5, 9, . . .} = {. . . , −6, −2, 2, 6, 10, . . .} = {. . . , −5, −1, 3, 7, 11, . . .}. As usual, in dealing with equivalence classes, the choice of the representative of the class is not unique. For instance, in the above example, we could just as easily have written [−5] or instead of . It is, however, customary to reduce final answers in Zn to the form [a], where 0 ≤ a < n. We can now define addition and multiplication on Zn . These work in the obvious way. Specifically, [a] + [b] = [a + b] [a][b] = [ab]. Example 2.14. In Z7 , we have + = = and = = . Theorem 2.10. For any integer n ≥ 2, addition and multiplication on Zn are welldefined. Proof. Suppose that a1 ≡ a2 (mod n) and b1 ≡ b2 (mod n). Then (a1 + b1 ) − (a2 + b2 ) = (a1 − a2 ) + (b1 − b2 ). Since n|(a1 − a2 ) and n|(b1 − b2 ), we see that n|((a1 + b1 ) − (a2 + b2 )). That is, [a1 + b1 ] = [a2 + b2 ], so addition is well-defined. Also, a1 b1 − a2 b2 = (a1 b1 − a1 b2 ) + (a1 b2 − a2 b2 ) = a1 (b1 − b2 ) + (a1 − a2 )b2 . Since n|(a1 − a2 ) and n|(b1 − b2 ), we find that n|(a1 b1 − a2 b2 ). That is, [a1 b1 ] = [a2 b2 ], and multiplication is well-defined. \u0002\n\n2.5 Modular Arithmetic\n\n29\n\nLet us discuss a few properties of addition and multiplication in Zn . These should be compared with the properties of Z mentioned in Section 2.4. Theorem 2.11. Let n ≥ 2 be an integer, and take any [a], [b], [c] ∈ Zn . Then 1. 2. 3. 4. 5.\n\n[a] + [b] ∈ Zn (closure under addition); [a] + ([b] + [c]) = ([a] + [b]) + [c] (associativity); [a] + [b] = [b] + [a] (commutativity); [a] + = [a] (additive identity); and [a] + [−a] = (additive inverse).\n\nProof. (1) is clear from the definition. The other parts all work because they work in Z. For instance, [a] + [b] = [a + b] = [b + a] = [b] + [a], proving (3). The remaining parts are left as Exercise 2.35. \u0002 And now, some properties of multiplication. Theorem 2.12. Let n ≥ 2 be an integer, and [a], [b], [c] ∈ Zn . Then 1. 2. 3. 4. 5.\n\n[a][b] ∈ Zn (closure under multiplication); [a]([b][c]) = ([a][b])[c] (associativity); [a][b] = [b][a] (commutativity); [a]([b] + [c]) = [a][b] + [a][c] (distributive law); and [a] = [a] (multiplicative identity).\n\nProof. (1) follows from the definition, and the other parts are true because they are true in Z. For instance, [a]([b] + [c]) = [a][b + c] = [a(b + c)] = [ab + ac] = [ab] + [ac] = [a][b] + [a][c], proving (4). The rest is left as Exercise 2.36.\n\n\u0002\n\nAs in Z, we do not necessarily have multiplicative inverses. For instance, in Z14 , we find that = , but there is no integer a such that [a] = . However, Z5 behaves more like Q; indeed, if [a] \u0007= , then there exists a [b] ∈ Z5 such that [a][b] = . More on this later! It is worth mentioning that Zn does not behave exactly like Z. For instance, in Z, we are used to the fact that if ab = 0, then a = 0 or b = 0. But in Z12 , we have = . We are also accustomed to cancellation in Z; that is, if ab = ac, and a \u0007= 0, then b = c. Not necessarily true in Zn ! For example, in Z12 , we have = , but \u0007= and \u0007= . So we must be careful with our assumptions. And now, having acquainted ourselves with Zn , we are going to make a change in notation. It is rather cumbersome to have to write [a] or [a] + [b] all the time. Therefore, when working in Zn , we will normally simply write a or a + b, as long as the context is clear. We will include the equivalence class brackets if they are needed for greater clarity. Example 2.15. When working in Z10 , we simply write 3 + 8 = 1 and 3 · 8 = 4.\n\n30\n\n2 The Integers and Modular Arithmetic\n\nWe need to prove one last property of modular arithmetic, which dates back to ancient China. Theorem 2.13 (Chinese Remainder Theorem). Let n 1 , . . . , n k be positive integers, all larger than 1, such that (n i , n j ) = 1 whenever i \u0007= j. If a1 , . . . , ak ∈ Z, then there exists an integer b such that b ≡ ai (mod n i ) for all i. Furthermore, if c ≡ ai (mod n i ) for all i, then b ≡ c (mod n 1 n 2 · · · n k ). Proof. For each i, let di be the product of all of the n j except for n i ; that is, di = n 1 n 2 ···n k . Since (n i , n j ) = 1 when i \u0007= j, Corollary 2.5 shows us that (n i , di ) = 1. By ni Corollary 2.1, there exist integers u i and vi such that n i u i + di vi = 1. Thus, di vi ≡ 1 (mod n i ). Let b = d1 v1 a1 + d2 v2 a2 + · · · + dk vk ak . Then since n i |d j if i \u0007= j, we have b ≡ di vi ai ≡ ai\n\n(mod n i ),\n\nfor all i, as required. Finally, if c ≡ ai (mod n i ) for all i as well, then b ≡ c (mod n i ) for all i; that \u0002 is, n i |(b − c) for all i. By Corollary 2.3, n 1 n 2 · · · n k |(b − c). Example 2.16. Let us solve the congruences b ≡ 3 (mod 5), b ≡ 4 (mod 11) and b ≡ 6 (mod 14). We have d1 = 154, d2 = 70 and d3 = 55. Solving 5u 1 + 154v1 = 1 using the Euclidean algorithm, we get u 1 = 31, v1 = −1. When we solve 11u 2 + 70v2 = 1, we get u 2 = −19, v2 = 3. Finally, a solution to 14u 3 + 55v3 = 1 is u 3 = 4, v3 = −1. Therefore, b = 154(−1)(3) + 70(3)(4) + 55(−1)(6) = 48. Thus, the solution is b ≡ 48 (mod 770). Exercises 2.33. Perform each calculation in Z7 . The final answer should be a nonnegative integer no larger than 6. 1. 2 − 3 · 4 2. (4 · 5)25 2.34. Perform each calculation in Z15 . The final answer should be a nonnegative integer no larger than 14. 1. 5 · 11 − 3 · 4 2. 282 2.35. Complete the proof of Theorem 2.11. 2.36. Complete the proof of Theorem 2.12. 2.37. For each nonzero element a ∈ Z20 , decide if there is a nonzero b ∈ Z20 such that ab = 0 in Z20 . If so, provide such an element b.\n\n2.5 Modular Arithmetic\n\n31\n\n2.38. For each element a ∈ Z20 , decide if there exists a b ∈ Z20 such that ab = 1 in Z20 . If so, provide such an element b. 2.39. Show that if p is prime, then there are at most two elements a ∈ Z p such that a 2 = 1 in Z p . Find an example of a composite p where there are more than two solutions. 2.40. Let a and b be integers. Show that if a ≡ b (mod p) for every prime p, then a = b. 2.41. Find a ∈ Z such that a ≡ 2 (mod 3), a ≡ 4 (mod 7) and a ≡ 3 (mod 10) simultaneously. 2.42. Find a ∈ Z such that a ≡ 3 (mod 8), a ≡ 4 (mod 11) and a ≡ 7 (mod 15) simultaneously.\n\nPart II\n\nGroups\n\nChapter 3\n\nIntroduction to Groups\n\nWe now begin our study of abstract algebra in earnest! A group is one of the simplest algebraic structures; we take a set, assign an operation to it, impose four basic rules, and see what we can deduce. And yet, the possibilities are endless. Groups show up everywhere, and not just in mathematics. Indeed, it would be difficult to study physics or chemistry without an understanding of group theory. The solution to the famous Rubik’s cube is also a problem in groups. In this chapter, we will define the notion of a group, and give a number of examples. We will also prove several basic properties of groups and subgroups.\n\n3.1 An Important Example In the next section, we will give the definition of a group. For now, we will look at a motivating example. Let A be the set {1, 2, 3}. We would like to consider all of the permutations of A; that is, all the ways of rearranging the numbers 1, 2 and 3. For example, we have the permutation σ , where σ (1) = 2, σ (2) = 1 and σ (3) = 3. We can easily see that there are going to be exactly 6 such permutations, as there are 3 choices for σ (1), then 2 remaining choices for σ (2), and once those are known, σ (3) is determined. A bit of notation would be helpful. Let us denote a permutation σ by writing two rows. The elements of A go in the first row, and the numbers to which each of them is sent in the second; that is, we take \u0002 \u0003 123 σ = abc to mean that σ (1) = a, σ (2) = b and σ (3) = c. Then the permutation we mentioned above would be denoted © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_3\n\n35\n\n36\n\n3 Introduction to Groups\n\n\u0002\n\n\u0003 123 . 213\n\nIn fact, the complete list of permutations is \u0002 \u0003 \u0002 \u0003 \u0002 \u0003 \u0002 \u0003 \u0002 \u0003 \u0002 \u0003 123 123 123 123 123 123 , , , , and . 123 132 213 231 312 321 Let us now discuss the composition of two permutations. For instance, if \u0002\n\n123 σ = 213\n\n\u0003\n\n\u0002 \u0003 123 and τ = , 312\n\nthen we see that (σ ◦ τ )(1) = σ (τ (1)) = σ (3) = 3, (σ ◦ τ )(2) = σ (τ (2)) = σ (1) = 2 and (σ ◦ τ )(3) = σ (τ (3)) = σ (2) = 1. Thus, σ ◦τ =\n\n\u0002 \u0003 123 . 321\n\n(It is worth noting here that we apply τ first, then σ .) We can now consider some properties that these permutations enjoy with respect to this composition operation. As we discuss them, please compare with the properties of Z or Zn , under addition, with which we are already familiar. First of all, we have closure. That is, if we take two permutations of A and compose them, we obtain another permutation of A. In fact, we proved this in Theorem 1.2, where we saw that the composition of two bijections is a bijection. Next, we have associativity; that is, for any permutations ρ, σ and τ , we have ρ ◦ (σ ◦ τ ) = (ρ ◦ σ ) ◦ τ . We have seen this before as well; by Theorem 1.2, the composition of functions is always associative. Also, we have an identity. In particular, if σ is any permutation of A, then \u0002 \u0003 \u0002 \u0003 123 123 σ◦ = ◦ σ = σ. 123 123 Composing with the permutation that moves nothing cannot change a function. Finally, we have inverses; that is, for each permutation σ , there is another permutation τ such that \u0002 \u0003 123 σ ◦τ =τ ◦σ = , 123 the identity. This is easy enough to calculate directly; for instance, \u0003 \u0003 \u0002 \u0003 \u0002 \u0003 \u0002 \u0003 \u0002 123 123 123 123 123 . ◦ = ◦ = 123 231 312 312 231\n\n\u0002\n\n3.1 An Important Example\n\n37\n\nHowever, the existence of such an inverse is guaranteed by Theorem 1.3. Given our discussion in Sections 2.4 and 2.5, we can agree that all of these properties are shared by Z and Zn under addition. However, we also noted that the addition operation is commutative. Not so here! For instance, \u0002\n\n\u0003 \u0002 \u0003 \u0002 \u0003 123 123 123 ◦ = , 231 132 213\n\n\u0002 \u0003 \u0002 \u0003 \u0002 \u0003 123 123 123 ◦ = . 132 231 321\n\nwhereas\n\nThus, in general, σ ◦ τ \u0003= τ ◦ σ . These permutations under the composition operation give us a nice example of a group, as we shall see momentarily. There is, of course, nothing very magical about the set A = {1, 2, 3} here. Indeed, we could just as easily have used {1, 2, 3, . . . , n}, for any positive integer n. The set of all permutations of this set, under the composition operation, is called the symmetric group on n letters, and is denoted Sn . Exercises\n\n\u0002\n\n\u0003 \u0002 \u0003 1234 1234 3.1. In S4 , let σ = and τ = . Calculate the following. 3142 3412\n\n1. σ τ 2. τ σ 3. the inverse of σ\n\n\u0002 \u0003 \u0002 \u0003 12345 12345 and τ = . Calculate the following. 3.2. In S5 , let σ = 53214 24135\n\n1. σ τ σ 2. σ σ τ 3. the inverse of σ 3.3. How many permutations are there in Sn ? In S5 , how many permutations α satisfy α(2) = 2? 3.4. Let H be the set of all permutations α ∈ S5 satisfying α(2) = 2. Which of the properties we have discussed (closure, associativity, identity, inverses) does H enjoy under composition of functions? 3.5. Consider the set of all functions from {1, 2, 3, 4, 5} to {1, 2, 3, 4, 5}. Which of the properties (closure, associativity, identity, inverses) does this set enjoy under composition of functions? 3.6. Let G be the set of all permutations of N. Which of the properties (closure, associativity, identity, inverses) does G enjoy under composition of functions?\n\n38\n\n3 Introduction to Groups\n\n3.2 Groups We can now give the general definition of a group. Definition 3.1. A group is a set G, together with a binary operation ∗, satisfying the following conditions: 1. a ∗ b ∈ G for all a, b ∈ G (closure); 2. (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ G (associativity); 3. there exists an e ∈ G such that a ∗ e = e ∗ a = a for all a ∈ G (existence of identity); and 4. for each a ∈ G, there exists a b ∈ G such that a ∗ b = b ∗ a = e (existence of inverses). We will refer to G as a group under ∗. The element e is called the identity of the group. If a ∈ G, and a ∗ b = b ∗ a = e, then b is called the inverse of a, and we write b = a −1 . As we discussed in the previous section, the group operation does not have to be commutative. We have a special term for groups that do have this property, named after mathematician Niels H. Abel. Definition 3.2. A group G is said to be abelian if a ∗ b = b ∗ a for all a, b ∈ G. We devote the remainder of this section to examples of groups. Example 3.1. As we saw in Sections 2.4 and 2.5, Z and Zn (for any integer n ≥ 2) are abelian groups under addition. Indeed, 0 is the identity, and the inverse of a is −a. In fact, the same can be said for Q, R and C under addition. When a group G has only finitely many elements, we can represent it with a group table. We write the elements of G down the first column and along the first row of the table. Then the entry in the row headed by a and the column headed by b is a ∗ b. For instance, the group table for Z5 is given in Table 3.1. Table 3.1 Group table for the additive group Z5 0 1 2 3 4\n\n0 0 1 2 3 4\n\n1 1 2 3 4 0\n\n2 2 3 4 0 1\n\n3 3 4 0 1 2\n\n4 4 0 1 2 3\n\nExample 3.2. Let G be the set of nonzero rational numbers. Then G is an abelian group under multiplication. Indeed, we see that the product of two nonzero rationals is a nonzero rational, hence closure is satisfied. Also, multiplication of rationals is both\n\n3.2 Groups\n\n39\n\nassociative and commutative. Clearly, a · 1 = a, for all a ∈ G, so 1 is the identity. Finally, if a = m/n ∈ G, with m and n nonzero integers, then a −1 = n/m ∈ G, since (m/n)(n/m) = 1. This last example merits a second look. In particular, it is worth noting that we cannot do the same thing with the set of nonzero integers. To be sure, the product of two nonzero integers is a nonzero integer, and the multiplication is associative. Also, 1 is the identity. But 2 has no inverse; that is, there is no integer a such that 2a = 1. In fact, the only integers that would have inverses in this set are 1 and −1. The set {1, −1} is easily seen to be a group under multiplication. Let us see how the integers modulo n compare. Example 3.3. Let n ≥ 2 be a positive integer. Let U (n) denote the set of all elements a ∈ Zn such that (a, n) = 1. (For instance, U (10) = {1, 3, 7, 9}.) Let us ensure that this makes sense. That is, if a ≡ b (mod n), and (a, n) = 1, then it had also better be the case that (b, n) = 1. But a = b + nk, for some integer k. Then, if c divides both b and n, then c divides a as well. We claim that U (n) is an abelian group under the multiplication operation in Zn . First, closure. By Corollary 2.5, if (a, n) = 1 and (b, n) = 1, then (ab, n) = 1. We also know that multiplication in Zn is associative and commutative, and 1 (which obviously lies in U (n)) is the identity. What about inverses? If a ∈ U (n), then since (a, n) = 1, there exist integers u and v such that au + nv = 1. That is, in Zn , au = ua = 1. The group table of U (10) is given in Table 3.2.\n\nTable 3.2 Group table for the multiplicative group U (10) 1 3 7 9\n\n1 1 3 7 9\n\n3 3 9 1 7\n\n7 7 1 9 3\n\n9 9 7 3 1\n\nNote that we will use the notation U (n) from the above example throughout the book. Example 3.4. Let n be a positive integer, and let G be the set of all complex numbers w satisfying wn = 1. Then we claim that G is an abelian group under multiplication. Of course, we know that the multiplication is both associative and commutative, and 1 ∈ G will serve as a multiplicative identity. If v, w ∈ G, then (vw)n = vn wn = 1, so vw ∈ G, and we have closure. Also, if w ∈ G, then we know that 1/w ∈ C. But (1/w)n = 1/(wn ) = 1, and therefore 1/w ∈ G. In particular, if n = 4 in the above example, then we get the group {1, −1, i, −i}. Also, if n = 1, then we just get the group consisting of the identity. This is known as the trivial group. Of course, not all groups are abelian. Two useful examples follow.\n\n40\n\n3 Introduction to Groups\n\nExample 3.5. As we illustrated in the previous section, Sn is a group under composition. If n ≥ 3, then the group is nonabelian. Example 3.6. The set of all invertible 2 × 2 matrices with entries in R is called the general linear group over R, and denoted G L 2 (R). It is a group under matrix multiplication. The identity matrix I2 is the identity of G L 2 (R). Also, if A, B ∈ G L 2 (R), then AB(B −1 A−1 ) = A(B B −1 )A−1 = AI2 A−1 = A A−1 = I2 and, similarly, (B −1 A−1 )AB = I2 . Thus, AB is invertible as well, so G L 2 (R) is closed under multiplication. Also, matrix multiplication is associative. By definition of G L 2 (R), every element A has an inverse, and since (A−1 )−1 = A, we know that A−1 ∈ G L 2 (R). Thus, G L 2 (R) is indeed a group. However, the group is nonabelian. For instance, \u0002 \u0003\u0002 \u0003 \u0002 \u0003 11 10 21 = , 01 11 11 whereas\n\n\u0002\n\n10 11\n\n\u0003\u0002\n\n\u0003 \u0002 \u0003 11 11 = . 01 12\n\nBy changing the entries in the matrices, we can obtain other general linear groups, such as G L 2 (Q). We can also use invertible n × n matrices and obtain G L n (R). Let us also present a useful way of obtaining new groups from old ones. Definition 3.3. Let G be a group with operation ∗ and H a group with operation •. On the Cartesian product G × H , define an operation \b via (g1 , h 1 ) \b (g2 , h 2 ) = (g1 ∗ g2 , h 1 • h 2 ), for all gi ∈ G, h i ∈ H . Under this operation, we call G × H the direct product of G and H . Theorem 3.1. The direct product of two groups is a group. Proof. Let us adopt the same notation as in the definition. First, we must check that the direct product is closed. But if g1 , g2 ∈ G, h 1 , h 2 ∈ H , then (g1 , h 1 ) \b (g2 , h 2 ) = (g1 ∗ g2 , h 1 • h 2 ) ∈ G × H , since g1 ∗ g2 ∈ G, h 1 ∗ h 2 ∈ H . The associativity of \b follows from the associativity of ∗ and •. Indeed, if g1 , g2 , g3 ∈ G, h 1 , h 2 , h 3 ∈ H , then ((g1 , h 1 ) \b (g2 , h 2 )) \b (g3 , h 3 ) = (g1 ∗ g2 , h 1 • h 2 ) \b (g3 , h 3 ) = ((g1 ∗ g2 ) ∗ g3 , (h 1 • h 2 ) • h 3 ) = (g1 ∗ (g2 ∗ g3 ), h 1 • (h 2 • h 3 )) = (g1 , h 1 ) \b ((g2 , h 2 ) \b (g3 , h 3 )).\n\n3.2 Groups\n\n41\n\nLet eG and e H be the identities of G and H respectively. Then for any g ∈ G, h ∈ H , we have (g, h) \b (eG , e H ) = (g ∗ eG , h • e H ) = (g, h) and, similarly, (eG , e H ) \b (g, h) = (g, h). Thus, (eG , e H ) is the identity for G × H . Furthermore, (g, h) \b (g −1 , h −1 ) = (g ∗ g −1 , h • h −1 ) = (eG , e H ) and, similarly, (g −1 , h −1 ) \b (g, h) = (eG , e H ). Thus, (g, h)−1 = (g −1 , h −1 ). The proof is complete. \u0002 Example 3.7. Suppose that G = Z5 and H = S3 . Then in G × H , \u0002 \u0002 \u0003\u0003 \u0002 \u0002 \u0003\u0003 \u0002 \u0002 \u0003 \u0002 \u0003\u0003 \u0002 \u0002 \u0003\u0003 123 123 123 123 123 4, \b 3, = 4 + 3, ◦ = 2, . 231 321 231 321 132 Before concluding this section, it seems to be worth mentioning that part of the definition of a group is redundant. We specify that ∗ is a binary operation on G, and then require closure. But closure is part of the definition of a binary operation! Nevertheless, it is a good idea to emphasize this point, as closure must be checked whenever a new group is defined, and it is easy to forget about it if it is buried inside another definition. Exercises 3.7. Give group tables for the following additive groups. 1. Z3 2. Z3 × Z2 3.8. Give group tables for the following groups. 1. U (12) 2. S3 3.9. Show that G × H is abelian if and only if G and H are abelian. 3.10. Let G be a group containing at most three elements. Show that G is abelian. 3.11. Explain why neither of the following is a group. 1. the set of positive rational numbers under division 2. the set of rational numbers q ≥ 1 under multiplication 3.12. Is either of the following a group under addition? 1. the set of even integers 2. the set of odd integers 3.13. Let G = {a + bi ∈ C : a 2 + b2 = 1}. Is G a group under multiplication?\n\n42\n\n3 Introduction to Groups\n\n3.14. Let G be the following subset of Z15 , namely {3, 6, 9, 12}. Show that G is a group under multiplication in Z15 . Find the identity, and the inverse of each group element. 3.15. Let p be a prime and G = {a/ p n : a ∈ Z, n ∈ N}. Is G a group under addition? ⎛ ⎞ 1ab 3.16. Let G be the set of all matrices of the form ⎝0 1 c ⎠, with a, b, c ∈ Z. Show 001 that G is a group under matrix multiplication. Is it abelian?\n\n3.3 A Few Basic Properties Let us begin with a small notational change. Usually, when we are working inside a group, we suppress the symbol for the group operation. That is, we write ab instead of a ∗ b. The major exception is where the operation is addition, in which case this multiplicative notation would be confusing. In that case, we will use additive notation and continue to write a + b instead of ab, 0 instead of e and −a instead of a −1 . In the preceding section, we glossed over the uniqueness of the group identity and inverses of group elements. These are important points, if we are to speak of “the” identity, or write a −1 and have it mean something. Let us take care of this problem. Theorem 3.2. Let G be a group. Then 1. the identity of G is unique; and 2. if a ∈ G, then a −1 is unique. Proof. (1) Suppose that e and f are both identities in G. Then as f is an identity, e f = e. But as e is an identity, we also have e f = f . Therefore, e = f . (2) Suppose that b and c are both inverses of a. Then as b is an inverse of a, (ba)c = ec = c. However, as c is an inverse of a, we have b(ac) = be = b. Given that our group operation is associative, b = b(ac) = (ba)c = c. \u0002 We know that in any group, (ab)c = a(bc). Thus, we can write abc without worrying about ambiguity. But we would like to be able to write abcd, for instance. To that end, we have the following result. Theorem 3.3. Let G be any group, and a1 , a2 , . . . , an ∈ G. Then regardless of how the product a1 a2 · · · an is bracketed, the result equals (· · · (((a1 a2 )a3 )a4 ) · · · an−1 )an . Proof. Our proof is by strong induction upon n. If n is 1 or 2, no bracketing is needed, so there is nothing to do. When n = 3, this is the associativity from the definition of a group. Therefore, let n ≥ 4, and suppose that the theorem is true for any product of fewer than n group elements. Take any bracketing of w = a1 · · · an , and look at the\n\n3.3 A Few Basic Properties\n\n43\n\nlast operation to be performed. Then w = x y, where x is the product a1 · · · am and y is the product am+1 · · · an , each with some bracketing. By our inductive hypothesis, x = (· · · ((a1 a2 )a3 ) · · · am−1 )am and y = (· · · ((am+1 am+2 )am+3 ) · · · an−1 )an . If m = n − 1, then writing x y in this way, we have our desired conclusion. If not, then by associativity, x y = ((· · · ((a1 a2 )a3 ) · · · am )(· · · ((am+1 am+2 )am+3 ) · · · an−1 ))an . Now applying our inductive hypothesis to the product of the first n − 1 terms, we obtain the desired bracketing. \u0002 Therefore, we do not have to use brackets when we write a product of group elements. However, we must always remember that unless our group is abelian, we cannot rearrange terms at will. For instance, (ab)(cd) = (a(bc))d, and we can write both as abcd, but we cannot write abcd = cdba. Let us also prove a couple of useful facts about inverses. Theorem 3.4. Let G be a group, with a, b ∈ G. Then 1. (a −1 )−1 = a; and 2. (ab)−1 = b−1 a −1 . Proof. (1) Since aa −1 = a −1 a = e, we see from the definition of inverses that the inverse of a −1 is a. (2) Notice that (ab)(b−1 a −1 ) = a(bb−1 )a −1 = aea −1 = aa −1 = e and, simi\u0002 larly, (b−1 a −1 )(ab) = e. Therefore, b−1 a −1 is the inverse of ab. Do not make the mistake of thinking that the inverse of ab is a −1 b−1 ! In ordinary arithmetic with real numbers, we know that if ab = ac, and a \u0003= 0, then b = c. We have something similar for groups. Theorem 3.5. (Cancellation Law). Let G be a group and a, b, c ∈ G. If either ab = ac or ba = ca, then b = c. Proof. If ab = ac, then a −1 ab = a −1 ac. As a −1 a = e, we have eb = ec, and therefore b = c. The proof is similar if ba = ca. \u0002 When our group has finitely many elements, the cancellation law has important implications for the group table. Suppose that, in the row headed by a, the group element b occurs twice. Then there exist group elements c and d such that ac = b = ad. But then we know that c = d. Therefore, a group element can occur only once in each row. In the same way, there will be no repetitions in any column. Corollary 3.1. Let G be a group and a, b ∈ G. Then there exists exactly one c ∈ G such that ac = b, and exactly one d ∈ G such that da = b. Proof. We showed the uniqueness of c and d above. To show the existence of c and d, let c = a −1 b and d = ba −1 . Then ac = aa −1 b = eb = b, and da = ba −1 a = be = b. \u0002\n\n44\n\n3 Introduction to Groups\n\nExample 3.8. Suppose G is a group with four elements, a, b, c and d. If we are given the partial group table shown in Table 3.3, we can fill in the missing elements. Indeed, examining the first row, we see that ad cannot be b or d. The last column tells us that it also cannot be a, so ad = c. As there must be an a in the first row, ab = a. Filling in the rest of the table is left as an exercise.\n\nTable 3.3 Incomplete group table abcd ad b b b a c d a\n\nExercises 3.17. Simplify each of the following expressions as far as possible in an arbitrary group G, leaving no brackets. 1. (acb)(cbab)−1 2. (a −1 bca)−1 3.18. Repeat the preceding exercise, assuming that G is abelian. 3.19. Fill in the rest of Table 3.3. 3.20. Let G = {v, w, x, y, z} be a group with five elements. Further suppose that vw = y, vy = v, wx = z, xv = w and zw = v. Fill in the group table for G. 3.21. Show that the following are equivalent for a group G: 1. for every a, b, c ∈ G satisfying ab = ca, we have b = c; and 2. G is abelian. 3.22. Suppose that in the definition of a group, we replace the third part with the following weaker condition: (3’) There exists an e ∈ G such that for every a ∈ G, ae = a. (That is, we do not insist that ea = a.) Show that we still get a group.\n\n3.4 Powers and Orders In group theory, the word order is used in two different, but related, ways. One is easy.\n\n3.4 Powers and Orders\n\n45\n\nDefinition 3.4. If G is a group, then its order, |G|, is the number of elements in the set G. We say that G is a finite group if its order is finite; otherwise, it is an infinite group. Example 3.9. If G = Z5 , then |G| = 5, and therefore G is a finite group. On the other hand, Z is an infinite group. To understand the other use of the word, we need to know about powers of group elements. Let G be any group, and a ∈ G. Then for any positive integer n, we let a n = aa · · · a . \b\n\nn times\n\n(Alternatively, we could define the powers recursively. That is, let a 1 = a, and for each positive integer n, let a n+1 = a n a.) Also, let a 0 = e and, for each positive integer n, let a −n = (a n )−1 . Example 3.10. In U (20), we calculate 73 = 7 · 7 · 7 = 9 · 7 = 3. If we wanted to know 7−3 , then we would calculate (73 )−1 = 3−1 = 7, since 3 · 7 = 1. Powers behave in a rather nice manner, as the following theorem tells us. Theorem 3.6. Let G be a group, with a ∈ G, and let m and n be any integers. Then 1. a m a n = a m+n ; 2. (a m )n = a mn ; and, in particular, 3. a −n = (a −1 )n . Proof. Exercise 3.26.\n\n\u0002\n\nWe know that if the group operation is addition, then we will use additive notation, rather than multiplicative. In this case, our exponentiation notation would be confusing, so we will write things in a more familiar manner. Instead of a n , we will write na (that is, add a to itself n times). Example 3.11. In Z12 , since our operation is addition, instead of writing 54 , we would write 4 · 5 = 5 + 5 + 5 + 5 = 8. Sometimes, a group will consist only of powers of a specific group element. Definition 3.5. A group G is said to be cyclic if there exists an element a such that every element of G is a power of a. In particular, we say that G is generated by a, and write G = a . Example 3.12. The additive group Z is cyclic; indeed, Z = 1 . (Remember, in an additive group, the powers are integer multiples, so if a ∈ Z, then a = a · 1.) In fact, Z = −1 as well, so the generator of the cyclic group is not unique. In the same way, Zn is cyclic.\n\n46\n\n3 Introduction to Groups\n\nExample 3.13. Consider the multiplicative group of complex fourth roots of unity discussed in Example 3.4, namely G = {1, −1, i, −i}. Then G is cyclic. Indeed, G = i since the powers of i are i, −1, −i and 1. Not every group is cyclic. For one thing, we have the following fact. Theorem 3.7. Every cyclic group is abelian. Proof. Let G = a . If b, c ∈ G, then b = a m and c = a n , for some m, n ∈ Z. Then \u0002 bc = a m a n = a m+n , but cb = a n a m = a m+n as well. However, abelian groups need not be cyclic. Example 3.14. The group U (10) is cyclic, but U (8) is not. To see this, observe that U (10) = {1, 3, 7, 9}. But the powers of 3 are 3, 9, 7 and 1, so U (10) = 3 . On the other hand, U (8) = {1, 3, 5, 7}. But the powers of 1 are all 1, the powers of 3 are 1 and 3, the powers of 5 are 1 and 5, and the powers of 7 are 1 and 7. Therefore, no element generates U (8). Now, let us discuss the order of a group element. Definition 3.6. Let G be a group and a ∈ G. The order of a, denoted |a|, is the smallest positive integer n such that a n = e, assuming that such an n exists, in which case a has finite order. If no such n exists, then a has infinite order. Example 3.15. The identity of a group is the only element having order 1. \u0002 \u0003 \u0002 \u0003 123 123 2 has order 3; indeed, σ = , Example 3.16. In S3 , the element σ = 231 312 3 whereas σ is the identity. Example 3.17. In Z, every element other than 0 has infinite order. For instance, no matter how many times we add 8 to itself, we will never get 0. Example 3.18. In Z6 , we have |0| = 1, |1| = |5| = 6, |2| = |4| = 3 and |3| = 2. For instance, 1 · 4 = 4 \u0003= 0 and 2 · 4 = 2 \u0003= 0, but 3 · 4 = 0, so |4| = 3. The order of an element tells us a great deal about its powers. Theorem 3.8. Let G be a group and a ∈ G. Suppose i, j ∈ Z. Then 1. if a has infinite order, then a i = a j if and only if i = j; and 2. if |a| = n < ∞, then a i = a j if and only if i ≡ j (mod n). Proof. (1) Suppose that a i = a j , but i \u0003= j. Without loss of generality, say i > j. Then a i (a j )−1 = a j (a j )−1 = e. That is, a i− j = e. But i − j is a positive integer, and this contradicts the assumption that a has infinite order. (2) Suppose that a i = a j . Once again, a i− j = e. Using the division algorithm, write i − j = nq + r , with q, r ∈ Z and 0 ≤ r < n. Then\n\n3.4 Powers and Orders\n\n47\n\ne = a i− j = a nq+r = (a n )q a r . But a n = e. Thus, a r = e. But n is the smallest positive integer having this property, and r < n. Therefore, r = 0. That is, n|(i − j), as required. Conversely, suppose that i ≡ j (mod n). Then let us write i − j = nk, for some k ∈ Z. But we now have a i− j = a nk = (a n )k = ek = e. Thus, a i− j a j = ea j , and hence a i = a j .\n\n\u0002\n\nExample 3.19. In U (10), we see that |3| = 4. Thus, 3i = 3 j if and only if 4|(i − j). That is, 36 = 314 , but 32 \u0003= 311 . Example 3.20. In Z, all the integer multiples of 5 are distinct, because 5 has infinite order. Corollary 3.2. Let G be a group, and let a ∈ G have order n < ∞. Then, for any integer i, 1. a i = e if and only if n|i; and 2. |a i | = n/(i, n). Proof. (1) By the preceding theorem, a i = e = a 0 if and only if i ≡ 0 (mod n). (2) Suppose that, for some positive integer j, we have (a i ) j = e. We see from (1) that since a i j = e, we have n|i j. Write i j = nk, with k ∈ Z. Letting d = (n, i), we have j (i/d) = k(n/d). Now, (n/d, i/d) = 1. Thus, by Corollary 2.2, since n/d| j (i/d), we must have n/d| j. Therefore, |a i | ≥ n/d. But (a i )n/d = a in/d = a n(i/d) . As i/d is \u0002 an integer, this is (a n )i/d = ei/d = e. Thus, |a i | = n/d, as required. Example 3.21. Again considering 3 in U (10), we note that 3i = 1 if and only if i is a multiple of 4. Also, |314 | = 4/(4, 14) = 4/2 = 2. If G is a group, and a, b ∈ G, then we say that b is a conjugate of a if there exists a c ∈ G such that b = c−1 ac. Theorem 3.9. In any group, conjugate elements have the same order. Proof. Suppose that b = c−1 ac and that a n = e, for some positive integer n. Then bn = (c−1 ac)n = c−1 acc−1 acc−1 · · · cc−1 ac = c−1 aeae · · · ac = c−1 a n c = c−1 ec = e. That is, |b| ≤ |a|. But since b = c−1 ac, we have a = (c−1 )−1 bc−1 . Thus, by the same argument, |a| ≤ |b|. Therefore, |a| = |b|. \u0002\n\n48\n\n3 Introduction to Groups\n\nExercises 3.23. Find the order of each group, and the order of every element of each group. 1. Z12 2. Z2 × Z4 3.24. Find the order of every element of each group. Is the group cyclic? If so, list all generators. 1. U (14) 2. S3 3.25. Let G = a be a cyclic group of order 20. Find the orders of a 3 , a 12 and a 15 . 3.26. Prove Theorem 3.6. 3.27. Let a ∈ G and b ∈ H . Suppose that |a| = 12 and |b| = 18. Find the order of (a, b) in G × H . 3.28. Let a and b be elements of odd order in a group. Show that a 2 and b2 commute if and only if a and b commute. Also show that this does not have to hold if a and b have even order. 3.29. Let a and b be elements of a group. Show that the following pairs of elements have the same order: 1. a and a −1 ; and 2. ab and ba. 3.30. Let G = {a1 , . . . , ak } be a finite abelian group. Show that a1 a2 · · · ak has order 1 or 2. 3.31. Show that it is possible for an abelian group to have exactly three elements of order 2, but not exactly four elements of order 2. 3.32. Suppose that G is a group in which every element has order 1 or 2. Show that G must be abelian.\n\n3.5 Subgroups One of the most important ways of obtaining new groups is to consider subgroups of a particular group. Definition 3.7. Let G be a group with operation ∗. Then a subset H of G is called a subgroup of G if H is a group under the same operation ∗. In this case, we write H ≤ G.\n\n3.5 Subgroups\n\n49\n\nExample 3.22. Every group is a subgroup of itself, and {e} is a subgroup of every group. When we refer to a proper subgroup of G, we mean any subgroup other than G itself. Example 3.23. We can see that Z is a subgroup of Q, and both are subgroups of R. We do not have to check the entire definition of a group to see if a subset is a subgroup. For instance, we know that the group operation is associative on the entire group, so it is surely associative on every subset. The following theorem will save us some time. Theorem 3.10. Let G be a group and H a subset of G. Then H is a subgroup of G if and only if 1. e ∈ H (the subset contains the identity); 2. ab ∈ H for all a, b ∈ H (the subset is closed); and 3. a −1 ∈ H for all a ∈ H (the subset contains all inverses). Proof. Let H be a subgroup of G. Then H has an identity, f . Thus, f f = f . But also, e f = f . By cancellation, f = e, giving (1). Then, by definition of a group, (2) and (3) must hold. Conversely, suppose that (1)–(3) hold. We must check that H is a group. But by (2), H is closed. As the group operation is associative on G, it is associative on H . By (1) and (3), we have an identity and inverses as well. Therefore, H is a subgroup of G. \u0002 A remark is in order here. To wit, we could replace condition (1) in the above theorem with the weaker condition (1’) H is not the empty set. Indeed, if a ∈ H , then we see from (3) that a −1 ∈ H , and then (2) tells us that e = aa −1 ∈ H . So why not express it that way? Because sometimes, the subset we are checking is not a subgroup. And we can tell immediately that that is the case if the subset does not contain e. Example 3.24. The set of all even integers, 2Z, is a subgroup of Z. Indeed, we certainly have 0 = 2 · 0 ∈ 2Z. If 2m, 2n ∈ 2Z, then 2m + 2n = 2(m + n) ∈ 2Z, so we have closure. Finally, if 2m ∈ 2Z, then its inverse is −(2m) = 2(−m) ∈ 2Z, and we have inverses. Of course, there is nothing magical about the number 2 here. If a is an integer, than aZ is a subgroup of Z. In fact, this last example is a specific case of a more general phenomenon. We have already encountered cyclic groups.\n\n50\n\n3 Introduction to Groups\n\nDefinition 3.8. Let G be a group and a ∈ G. Then the cyclic subgroup generated by a is the set of all powers of a in G, and we write a = {a n : n ∈ Z}. Of course, the group G is cyclic if and only if there exists an a ∈ G such that G = a . Theorem 3.11. If G is a group and a ∈ G, then a is a subgroup of G. Proof. Certainly e = a 0 ∈ a . Take any a m , a n ∈ a . Then a m a n = a m+n ∈ a . \u0002 Finally, if a m ∈ a , then (a m )−1 = a −m ∈ a . Now apply Theorem 3.10. Example 3.25. In U (10), the powers of 3 are 1, 3, 9 and 7, so 3 = {1, 3, 7, 9} = U (10). Similarly, 7 = U (10). But the only powers of 9 are 1 and 9, so 9 = {1, 9}. Also, 1 = {1}. Example 3.26. In Z12 , the multiples of 8 are 1 · 8 = 8, 2 · 8 = 4 and 3 · 8 = 0. Thus, we have 8 = {0, 4, 8}. Of course, we do not insist upon commutativity in groups, but it can be useful to know which elements commute with everything. Definition 3.9. If G is a group, then the centre of G, denoted Z (G), is the set of elements of G that commute with everything in G. That is, Z (G) = {z ∈ G : az = za for all a ∈ G}. Example 3.27. If G is abelian, then Z (G) = G. Example 3.28. The centre of S3 is the trivial subgroup, {e}. Verifying this is a matter of considering each element of S3 other than the identity, and finding another element that does not commute with it. \u0002 \u0003 a0 , where 0 \u0003= Example 3.29. The centre of G L 2 (R) is the set of all matrices 0a a ∈ R. We leave the proof as Exercise 3.36. Theorem 3.12. If G is a group, then Z (G) is a subgroup of G. Proof. Certainly ea = a = ae for all a ∈ G, so e ∈ Z (G). If y, z ∈ Z (G) and a ∈ G, then yza = yaz = ayz; thus, yz ∈ Z (G). Also, if z ∈ Z (G) and a ∈ G, then a −1 z = za −1 . Inverting both sides, we get z −1 a = az −1 . Thus, z −1 ∈ Z (G). The proof is complete. \u0002 Some shortcuts are possible when it comes to testing whether a subset is a subgroup.\n\n3.5 Subgroups\n\n51\n\nTheorem 3.13. Let G be a group and H a subset of G. Then H is a subgroup if and only if 1. e ∈ H ; and 2. ab−1 ∈ H whenever a, b ∈ H . Proof. Suppose that H is a subgroup. By Theorem 3.10, we know that e ∈ H and if a, b ∈ H , then b−1 ∈ H , and therefore ab−1 ∈ H . Conversely, suppose that H satisfies (1) and (2). Take any a, b ∈ H . Then since e ∈ H , we have ea −1 = a −1 ∈ H and, similarly, b−1 ∈ H . Therefore, a(b−1 )−1 = ab ∈ H . In view of Theorem 3.10, H is a subgroup. \u0002 Once again, instead of checking that e ∈ H , it is enough to verify that H is not empty. We can even make things simpler if H is a finite set. Theorem 3.14. Let G be a group and H a finite subset of G. Then H ≤ G if and only if 1. e ∈ H ; and 2. ab ∈ H whenever a, b ∈ H . Proof. If H is a subgroup of G, then Theorem 3.10 tells us that (1) and (2) hold. Conversely, suppose that (1) and (2) are true. By Theorem 3.10, we only need to show that if a ∈ H then a −1 ∈ H . In view of (2), we have aa = a 2 ∈ H , and hence a 2 a = a 3 ∈ H , and so on; thus, a n ∈ H for all positive integers n. But there are infinitely many such powers, and H is finite. Thus, there exist positive integers m and n, with m > n, such that a m = a n . Then a m−n = e. If m − n = 1, then a = e, in which case a −1 = e ∈ H . So, suppose that m − n > 1. Then aa m−n−1 = a m−n−1 a = a m−n = e. That is, a m−n−1 = a −1 . But m − n − 1 is a positive integer, and therefore \u0002 a m−n−1 ∈ H , as required. We must be careful only to use the above theorem when H is finite. To see why, let G be the additive group of integers, and let H be the set of nonnegative integers. Then H contains 0 and is closed under addition, but H is not a subgroup of G, since 1 has no additive inverse. Example 3.30. Let G = Z8 × Z8 , and let H = {(a, b) ∈ G : 4a = 0}. We claim that H is a subgroup of G. Clearly, (0, 0) ∈ H . Also, if (a, b), (c, d) ∈ H , then (a, b) + (c, d) = (a + c, b + d), where 4(a + c) = 4a + 4c = 0 + 0 = 0. Therefore, H is closed, and hence a subgroup. We conclude the section with an extended, and important, example. Suppose we have a floor consisting of featureless square ceramic tiles. Let us pry up one of the tiles, and then consider all of the ways in which we can move the tile around in three-dimensional space, and then replace it so that it looks exactly as it did when we began. For convenience, let us label the vertices of the square 1, 2, 3 and 4. Then we can see that each vertex moves to the position of some vertex. Also, two vertices will not move to the same place. Once we have positioned the vertices, we\n\n52 2\n\n3 Introduction to Groups 2\n\n1\n\n1\n\n2\n\n1\n\n1\n\n2\n\nF1\n\nR0\n\n3\n\n4\n\n3\n\n4\n\n3\n\n4\n\n4\n\n3\n\n2\n\n1\n\n1\n\n4\n\n2\n\n1\n\n3\n\n4\n\nF2\n\nR90\n\n3\n\n4\n\n2\n\n3\n\n3\n\n4\n\n2\n\n1\n\n2\n\n1\n\n4\n\n3\n\n2\n\n1\n\n2\n\n3\n\nF3\n\nR180\n\n3\n\n4\n\n1\n\n2\n\n3\n\n4\n\n1\n\n4\n\n2\n\n1\n\n3\n\n2\n\n2\n\n1\n\n4\n\n1\n\n3\n\n2\n\nF4\n\nR270\n\n3\n\n4\n\n4\n\n1\n\n3\n\n4\n\nFig. 3.1 The symmetries in the dihedral group D8\n\nare done. Therefore, these symmetries of a square can be regarded as permutations of the set {1, 2, 3, 4}; that is, as elements of S4 . Of course, the identity of S4 is such a symmetry, and if we compose two of these symmetries, then we get another. Thus, by Theorem 3.14, they form a subgroup of S4 , known as the dihedral group of order 8, and denoted D8 . What are the elements of D8 ? They are illustrated in Figure 3.1. There are four rotations, R0 , R90 , R180 and R270 , where Rα is a counterclockwise rotation by α degrees. We also have four flips, F1 through F4 , about the lines shown in the diagram. And that is all! Indeed, vertex 1 can go to any of the 4 vertices, but then vertex 2 must be adjacent to it, not diagonally opposite. Once vertices 1 and 2 are positioned, the others fall into place. Therefore, |D8 | = 8. The group table of D8 is shown in Table 3.4. Remember that when we write R90 F1 = F3 , we mean perform F1 first, then R90 . We note that R90 F1 \u0003= F1 R90 , and therefore D8 is a nonabelian group of order 8. In fact, a quick glance through the table tells us that the centre of D8 is {R0 , R180 }.\n\n3.5 Subgroups\n\n53\n\nTable 3.4 Group table for the dihedral group D8 R0 R90 R180 R270 F1 F2 F3 F4\n\nR0 R0 R90 R180 R270 F1 F2 F3 F4\n\nR90 R90 R180 R270 R0 F4 F3 F1 F2\n\nR180 R180 R270 R0 R90 F2 F1 F4 F3\n\nR270 R270 R0 R90 R180 F3 F4 F2 F1\n\nF1 F1 F3 F2 F4 R0 R180 R90 R270\n\nF2 F2 F4 F1 F3 R180 R0 R270 R90\n\nF3 F3 F2 F4 F1 R270 R90 R0 R180\n\nF4 F4 F1 F3 F2 R90 R270 R180 R0\n\nWe do not have to begin with a square. Indeed, let us consider any regular n-gon, with n ≥ 3. Then the symmetries of this n-gon form a subgroup of Sn . By precisely the same arguments as above, it will consist of n rotations and n flips. (There are n possible locations for a given vertex, and once it is fixed, 2 choices for an adjacent vertex. After fixing those vertices, there are no choices remaining.) We call this group of symmetries the dihedral group of order 2n, and denote it by D2n . In particular, if n = 3, we note that D6 consists of all of S3 , but for larger n, D2n is a proper subgroup of Sn . In any case, we now have an example of a nonabelian group of every even order except 2 and 4. Exercises 3.33. In each case, is H a subgroup of G? 1. G = G L 2 (R), H is the set of matrices with determinant 1 2. G = D10 , H is the set of flips 3. G = Q, H = {a/b : a, b ∈ Z, 2 \u0002 b} 3.34. In each case, is H a subgroup of G? 1. G = D10 , H is the set of rotations 2. G = Q, H is the set of nonnegative rational numbers 3. G is the multiplicative group of nonzero rational numbers, H is the set of positive rational numbers 3.35. For each positive integer n ≥ 3, determine the centre of D2n . \u0002 \u0003 a0 3.36. Show that the centre of G L 2 (R) consists of the matrices , for all 0 \u0003= 0a a ∈ R. 3.37. Show that the intersection of two subgroups of G is also a subgroup. Then extend this to show that if Ni is a subgroup of G for every i in some set T , then N i is a subgroup of G. i∈T 3.38. Let H and K be subgroups of G. Show that H ∪ K is a subgroup of G if and only if either H ⊆ K or K ⊆ H . 3.39. Find every cyclic subgroup of each of the following groups.\n\n54\n\n3 Introduction to Groups\n\n1. Z20 2. U (16) 3.40. Let G be an abelian group and n ∈ N. Let H = {a ∈ G : a n = e} and K = {a n : a ∈ G}. Show that H and K are subgroups of G. 3.41. In any dihedral group, show that a rotation followed by a rotation, or a flip followed by a flip, is a rotation, whereas a rotation followed by a flip or a flip followed by a rotation is a flip. 3.42. Let G be the set of all sequences of integers (a1 , a2 , a3 , . . .). 1. Show that G is a group under (a1 , a2 , . . .) + (b1 , b2 , . . .) = (a1 + b1 , a2 + b2 , . . .). 2. Let H be the set of all elements (a1 , a2 , . . .) of G such that only finitely many ai are different from 0 (and (0, 0, 0, . . .) ∈ H ). Show that H is a subgroup of G.\n\n3.6 Cyclic Groups Cyclic groups have a very straightforward structure. Let us prove a few basic facts. First, we can illustrate the link between the order of an element and the order of a group. Theorem 3.15. Let G = a be cyclic. If a has infinite order, then all powers of a are distinct. If |a| = n < ∞, then the distinct elements of G are e, a, a 2 , . . . , a n−1 . In particular, |a| = | a |. Proof. If a has infinite order, then we can use Theorem 3.8. Suppose |a| = n < ∞. If m ∈ Z, then write m = nq + r , where q, r ∈ Z and 0 ≤ r < n. Then as m ≡ r (mod n), Theorem 3.8 tells us that a m = a r . In particular, every element of G is equal to some a i (0 ≤ i < n). Now, suppose that a i = a j , with 0 ≤ i < j < n. Then by Theorem 3.8, i ≡ j (mod n). But given the range of values for i and j, this is impossible. \u0002 The subgroups of cyclic groups are also easy to determine. Theorem 3.16. Every subgroup of a cyclic group is cyclic. Proof. Let G = a , and let H ≤ G. If H = {e}, then H = e , and we are done, so assume that H is not the trivial subgroup. Then H contains a m , for some nonzero integer m. If m < 0, then H also contains (a m )−1 = a −m , so H contains a positive power of a. Let n be the smallest positive integer such that a n ∈ H . We claim that H = a n . Surely H contains every power of a n , so a n ≤ H . But suppose a k ∈ H . Then write k = nq + r , with q, r ∈ Z and 0 ≤ r < n. Now, H contains a k and (a n )−q , and therefore a k (a n )−q = a k−nq = a r . But n is the smallest positive integer such that a n ∈ H . As r < n, we can only have r = 0. Thus, a k = (a n )q ∈ a n . That \u0002 is, H ≤ a n , proving the claim. We are done.\n\n3.6 Cyclic Groups\n\n55\n\nActually, we can say more. Corollary 3.3. Let G = a , where |a| = n < ∞. Then the order of every subgroup of G is a divisor of n. Furthermore, if m is a positive divisor of n, then G has exactly one subgroup of order m, namely a n/m . Proof. By the preceding theorem, every subgroup is of the form a k , for some k ∈ Z. But Corollary 3.2 tells us that the order of every power of a is a divisor of n. Let m be a positive divisor of n. Again using Corollary 3.2, we see that |a n/m | = n/(n, n/m) = n/(n/m) = m. Thus, a n/m is indeed a subgroup of order m. Let us check that it is unique. Suppose that a k is a subgroup of order m. Then |a k | = m, and so (a k )m = e. That is, a km = e, hence n|km. But then n/m|k. Writing k = (n/m)i, with i ∈ Z, we have a k = (a n/m )i ∈ a n/m . Thus, a k ≤ a n/m . But these two subgroups have the same order. Therefore, they are equal. \u0002 Example 3.31. Let G = a , where |a| = 20. Then G has exactly one subgroup of order 5, namely a 4 = {e, a 4 , a 8 , a 12 , a 16 }. Thus, a cyclic group can only have one subgroup of any given order. This is a special property of cyclic groups; indeed, D8 and U (8) are easily seen to have several different cyclic subgroups of order 2. We can also discuss the number of elements of a particular order in a cyclic group. Some notation will be helpful. The following function is named after Leonhard Euler. Definition 3.10. The Euler phi-function is a function ϕ : N → N, where ϕ(n) is the number of positive integers less than or equal to n that are relatively prime to n. Example 3.32. Of the integers from 1 to 10, only 1, 3, 7 and 9 are relatively prime to 10, so ϕ(10) = 4. The first few values of ϕ are given in Table 3.5. From the definition of the group U (n), we immediately obtain the following. Theorem 3.17. For any positive integer n, |U (n)| = ϕ(n). But we can also use the Euler function to count the elements of a particular order in a finite cyclic group. Theorem 3.18. Let G = a be a cyclic group of order n. Let m be a positive divisor of n. Then the number of elements of order m in G is ϕ(m). Proof. If b is an element of order m in G, then b must be the unique cyclic subgroup of order m. That is, all of the elements of order m in G are in the cyclic subgroup of order m. Thus, we may as well assume that G is cyclic of order m. We must\n\nTable 3.5 Values of the Euler phi-function n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ϕ(n) 1 1 2 2 4 2 6 4 6 4 10 4 12 6 8 8\n\n56\n\n3 Introduction to Groups\n\ntherefore decide which elements of this group have order m. But by Corollary 3.2, the order of a k is m if and only if (k, m) = 1. By definition, the number of such k, with 1 ≤ k ≤ m, is ϕ(m). (By Theorem 3.15, the elements of a are precisely a i , with 0 ≤ i < m, but as a 0 = a m = e, this is the same as considering a i with 1 ≤ i ≤ m.) \u0002 Example 3.33. Let G = a be cyclic of order 50. Then we know that there are ϕ(10) = 4 elements of order 10 in G. They lie in the subgroup of order 10, namely a 50/10 = a 5 . Indeed, the precise elements will be (a 5 )k , where (k, 10) = 1. This means that k ∈ {1, 3, 7, 9}, so the elements of order 10 are a 5 , a 15 , a 35 and a 45 . It is worth noting that the number of elements of order 10 in a cyclic group of order one million is also ϕ(10) = 4. For relatively small numbers, ϕ(n) is easy to determine, but for large n, it would be tedious to go through all the numbers from 1 to n in order to see if they are relatively prime to n. Happily, there is a shortcut. The first part of the following theorem is Exercise 3.45. It will make more sense if we postpone the proof of the second part until Section 4.4. Theorem 3.19. Let p be a prime number, and let m and n be positive integers. Then 1. ϕ( p n ) = p n − p n−1 ; and 2. if (m, n) = 1, then ϕ(mn) = ϕ(m)ϕ(n). Thus, we can determine ϕ(n) by writing n as a product of powers of primes and then using the above theorem. Example 3.34. We have ϕ(81) = 81 − 27 = 54 and ϕ(540) = ϕ(4)ϕ(27)ϕ(5) = (4 − 2)(27 − 9)(5 − 1) = 144. Exercises 3.43. 1. Let G = a be a cyclic group of order 12. List every subgroup of G. 2. List every subgroup of Z12 . 3.44. 1. Let G = a be a cyclic group of order 120. List all of the elements of order 12 in G. 2. How many elements of order 12 are there in a cyclic group of order 1200? 3.45. Let p be a prime and n a positive integer. Show that ϕ( p n ) = p n − p n−1 . 3.46. Find all positive integers n such that |U (n)| = 24. 3.47. Let G be a nonabelian group. If H and K are cyclic subgroups of G, does it follow that H ∩ K is also a cyclic subgroup? Prove that it does, or provide a counterexample. 3.48. Let G = a be infinite cyclic. If m and n are positive integers, find a generator for a m ∩ a n .\n\n3.6 Cyclic Groups\n\n57\n\n3.49. Let n be\n\na positive integer and let T be the set of positive integers that divide n. Show that k∈T ϕ(k) = n. 3.50. For precisely which positive integers n is U (2n ) cyclic? 3.51. Let G be any group and n a positive integer. 1. If H and K are subgroups of order n in G, and H \u0003= K , show that H ∩ K does not contain any elements of order n. 2. Show that the number of elements of order n in G is either a multiple of ϕ(n) or infinite. 3.52. Show that a nontrivial group G has no nontrivial proper subgroups if and only if G is cyclic of prime order. (Do not assume, to begin with, that G is finite.)\n\n3.7 Cosets and Lagrange’s Theorem One important fact we learned in the preceding section is that if G is a finite cyclic group, then the order of every subgroup of G divides the order of G. As it turns out, this is true for all finite groups, but a different proof will be required. To this end, we need some new terminology. Definition 3.11. Let G be a group and H a subgroup. If a, b ∈ G, we say that a is congruent to b modulo H , and we write a ≡ b (mod H ), if a −1 b ∈ H (or, in the case of an additive group, if −a + b ∈ H ). Example 3.35. Let G = Z and H = 5Z. Then as −1 + 16 = 15 ∈ H , we see that 1 ≡ 16 (mod H ). In this particular case, the notion is identical to congruence modulo 5. Example 3.36. Let G = U (20) = {1, 3, 7, 9, 11, 13, 17, 19}, and let H = 3 ; namely, H = {1, 3, 7, 9}. Then we note that 13−1 · 19 = 17 · 19 = 3 ∈ H . Thus, 13 ≡ 19 (mod H ). Lemma 3.1. Let G be a group and H a subgroup. Then congruence modulo H is an equivalence relation on G. Proof. Reflexivity: If a ∈ G, then a −1 a = e ∈ H , and therefore a ≡ a (mod H ). Symmetry: If a, b ∈ G and a ≡ b (mod H ), then a −1 b ∈ H , and therefore (a −1 b)−1 = b−1 a lies in H as well. But this means that b ≡ a (mod H ). Transitivity: Suppose that a, b, c ∈ G, where a ≡ b (mod H ) and b ≡ c (mod H ). Then a −1 b, b−1 c ∈ H . But in this case, H contains their product, a −1 bb−1 c = a −1 c. Thus, a ≡ c (mod H ). We are done. \u0002 What are the equivalence classes? Lemma 3.2. Let G be a group and H a subgroup. If a ∈ G, then its equivalence class with respect to congruence modulo H is the set {ah : h ∈ H }.\n\n58\n\n3 Introduction to Groups\n\nProof. If a ≡ b (mod H ), then a −1 b ∈ H , so a −1 b = h, for some h ∈ H . Thus, b = ah, which is in our set. Conversely, if b = ah, for some h ∈ H , then a −1 b = h ∈ H , and therefore a ≡ b (mod H ). \u0002 We need a name for this set. Definition 3.12. Let G be a group, H ≤ G and a ∈ G. Then the left coset of a with respect to H is the set {ah : h ∈ H }, which is denoted a H . (Note: If the group operation is addition, then we will write a + H = {a + h : h ∈ H }.) Example 3.37. If G = U (20), let H = 9 = {1, 9}. Then 3H = {3 · 1, 3 · 9} = {3, 7}. Also, 7H = {7 · 1, 7 · 9} = {3, 7}, so 3H = 7H . Furthermore, 1H = 9H = H , 11H = 19H = {11, 19} and 13H = 17H = {13, 17}. Note that these left cosets partition G. Example 3.38. Let G = Z and H = 3Z. Then there are three left cosets: 0 + H = H , 1 + H = {· · · , −5, −2, 1, 4, 7, . . .} and 2 + H = {· · · , −4, −1, 2, 5, 8, . . .}. Note that 2 + H = 5 + H = −13 + H , and so on. Again, the left cosets partition G. In general, we know that equivalence classes always partition a set. Therefore, we can record the following result. Theorem 3.20. Let G be a group and H a subgroup. Then the left cosets of H in G partition G. In particular, 1. each a ∈ G is in exactly one left coset, namely a H ; and 2. if a, b ∈ G, then either a H = bH or a H ∩ bH = ∅. Two points should be kept in mind here. First, left cosets are not subgroups! Remember, the left cosets partition G, and therefore the identity can only be in one of them, namely, eH = H . The rest cannot possibly be subgroups. Second, as we have already seen, when we write a H , the element a is not unique. Indeed, since the left cosets are equivalence classes, we have a H = bH if and only if a −1 b ∈ H . We can now prove our first big result on finite groups, due to Joseph-Louis Lagrange. Theorem 3.21. (Lagrange’s Theorem). Let G be a finite group and H a subgroup. Then |H | divides |G|. Proof. We have already seen that G is partitioned into left cosets; in particular, |G| is the sum of the sizes of these left cosets. But for any a ∈ G, a H = {ah : h ∈ H }. Now, if ah 1 = ah 2 , with h 1 , h 2 ∈ H , then by the cancellation law, h 1 = h 2 . Therefore, a H consists of precisely |H | distinct elements. It now follows that the order of G is |H | multiplied by the number of left cosets. In particular, |H | divides |G|. \u0002 Definition 3.13. Let G be a group and H ≤ G. Then the index of H in G, denoted [G : H ], is the number of left cosets of H in G. Corollary 3.4. If G is a finite group and H is a subgroup, then [G : H ] = |G|/|H |.\n\n3.7 Cosets and Lagrange’s Theorem\n\n59\n\nProof. This is immediate from the proof of the above theorem.\n\n\u0002\n\nExample 3.39. Let G = D8 and H = R90 = {R0 , R90 , R180 , R270 }. Then [G : H ] = |G|/|H | = 8/4 = 2. Thus, there are two left cosets. One is R0 H = H . The other must be F1 H = {F1 , F2 , F3 , F4 }. If K = F1 = {R0 , F1 }, then it must have 8/2 = 4 left cosets. One is R0 K = K . To find another, just choose an element of G that we have not yet found, say R90 . Then we get R90 K = {R90 , F3 }. We haven’t yet used F2 , so take F2 K = {F2 , R180 }. Finally, we can take R270 K = {R270 , F4 }. Example 3.40. Note that the subgroups of an infinite group can be of finite or infinite index. For instance, we saw above that 0 + 3Z, 1 + 3Z and 2 + 3Z are the distinct left cosets of 3Z in Z. Thus, [Z : 3Z] = 3. On the other hand, Z has infinite index in Q. To see this, observe that for all positive integers n, the left cosets 1/n + Z are distinct. And there are still more! Lagrange’s theorem has a beautiful consequence. Corollary 3.5. Let G be a finite group, and a ∈ G. Then the order of a divides the order of G. Proof. The order of a is the order of the cyclic subgroup generated by a, and that must divide the order of G. \u0002 Example 3.41. Note that |D8 | = 8, the identity has order 1, R180 and the flips all have order 2 and |R90 | = |R270 | = 4. All of the orders are divisors of 8. Of course, it does not follow that because a number n divides the order of a group, then the group has an element of that order. Indeed, if that were always true, then a group of order n would have to have an element of order n, and therefore every finite group would be cyclic, which is not the case. One important thing that we can do is to try to classify all the groups of some particular order. We can now make a step in that direction. Corollary 3.6. Every group of prime order is cyclic. Proof. Take e \u0003= a ∈ G, where |G| is a prime. As |a| divides |G|, and |a| \u0003= 1, we must have |a| = |G|. But then |G| = | a |, and therefore G = a . \u0002 Not surprisingly, there is also such a thing as a right coset. Indeed, if we had defined a ≡ b (mod H ) to mean that ab−1 ∈ H , then we would have found that this is still an equivalence relation, and the equivalence classes would have been as follows. Definition 3.14. Let G be a group and H ≤ G. Then for any a ∈ G, the right coset of a with respect to H is H a = {ha : h ∈ H }. (If G is an additive group, then we write H + a = {h + a : h ∈ H }.) If G is abelian, then there is no distinction between left and right cosets. In nonabelian groups, right cosets also partition G, but possibly in a different way.\n\n60\n\n3 Introduction to Groups\n\nExample 3.42. Take G, H and K as in Example 3.39. Then we can see that one right coset of H in G is H R0 = H = R0 H and the other must be H F1 = {F1 , F2 , F3 , F4 } = F1 H . Here, the left and right cosets agree. But it is not the same for K . For instance, R90 K = {R90 , F3 }, but K R90 = {R90 , F4 }. Would it have made a difference if we had defined the index of H in G using right cosets instead of left? Fortunately, no. This is clear if G is finite, as Lagrange’s theorem works equally well using right cosets. But what if G is an infinite group having a subgroup H of index n < ∞? Then notice that a H = bH if and only if a −1 b ∈ H , but also H a −1 = H b−1 if and only if a −1 b ∈ H . Thus, if the distinct left cosets of H in G are a1 H, a2 H, . . . , an H , then the distinct right cosets are H a1−1 , H a2−1 , . . . , H an−1 . Exercises 3.53. For each group G and subgroup H , find all the left cosets and right cosets of H in G. 1. G = Z, H = 4Z 2. G = D8 , H = F2\n\n3.54. For each group G and subgroup H , find all the left cosets and right cosets of H in G. 1. G = U (13), H\u000e\u0002 = 8 \u0003\u000f 123 2. G = S3 , H = 213 3.55. Let G be a group whose order is the product of two (not necessarily distinct) primes. Show that every proper subgroup of G is cyclic. 3.56. Let G be a group of order p n , for some prime p and positive integer n. Show that G has an element of order p. 3.57. Let G be a group having a subgroup H of order 28 and a subgroup K of order 65. Show that H ∩ K = {e}. 3.58. Let G be a finite group having an element of order k, for each 1 ≤ k ≤ 10. What is the smallest possible order of G? Show that a group of that order exists having this property. 3.59. Let G = {a1 , . . . , ak } be an abelian group of odd order k. Show that a1 a2 · · · ak = e. 3.60. Show that every group of order 55 contains an element of order 5 and an element of order 11. 3.61. Let G be a group with subgroups H and K . If [G : K ] = n, show that [H : H ∩ K ] ≤ n. 3.62. Let G be a group with subgroups H and K such that K ≤ H . Suppose that [G : H ] = m and [H : K ] = n. Show that [G : K ] = mn. (Do not assume that G is finite.)\n\nChapter 4\n\nFactor Groups and Homomorphisms\n\nIn the previous chapter, we tended to consider just one group at a time. But we need to find ways of relating groups to each other. For instance, we would like to know if two groups are, in every meaningful sense, the same. This would be the case if we took a group and created a new one by simply changing the labels on the group elements, but left the structure otherwise intact. Surely, we would not wish to think of these as different sorts of groups.1 This is where the notion of a group homomorphism and, in particular, an isomorphism, will come into the picture. But first, we will discuss factor groups. These constitute an important way of creating new groups from old ones. As we shall see, there is a natural connection between factor groups and homomorphisms. In order to define a factor group, we require a special sort of subgroup, called a normal subgroup. Let us begin there.\n\n4.1 Normal Subgroups Let H be a subgroup of G. We would like to form a group whose elements are the left cosets aH . Unfortunately, as we shall see in the next section, not just any subgroup will suffice; we need an extra condition. This is where normal subgroups come in. Recall that if H ≤ G, then the left cosets of H do not necessarily coincide with the right cosets. We need to consider subgroups for which they do coincide. Definition 4.1. Let G be a group and N a subgroup. We say that N is a normal subgroup of G if aN = Na for all a ∈ G. Example 4.1. For every group G, G is a normal subgroup of itself, as aG = Ga = G for all a. Also, {e} is normal. Indeed, a{e} = {e}a = {a} for all a. 1 Upon\n\nreading this sentence aloud, the author failed to stop himself from writing “And don’t call me Shirley.” We miss you, Leslie Nielsen! © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_4\n\n61\n\n62\n\n4 Factor Groups and Homomorphisms\n\nExample 4.2. The centre of every group is a normal subgroup. Indeed, writing Z = Z(G), we have aZ = {az : z ∈ Z} = {za : z ∈ Z} = Za. In fact, every subgroup of Z(G) is normal in G, for precisely the same reason. In particular, every subgroup of an abelian group is normal. Be warned: this last example can be a bit misleading. Remember, when we say that aN = Na, we do not necessarily mean that an = na for all n ∈ N . Indeed, we could have an = n1 a, for some different n1 ∈ N . The following example may be helpful. Example 4.3. Refer to Example 3.42. We saw that in D8 , the subgroup \u0004R90 \u0005 is normal. That is, a\u0004R90 \u0005 = \u0004R90 \u0005a for all a ∈ D8 . This does not mean that aR90 = R90 a, however. Indeed, F1 R90 = R270 F1 . But as R270 ∈ \u0004R90 \u0005, this is fine. We also saw in that example that \u0004F1 \u0005 is not a normal subgroup of D8 , as R90 \u0004F1 \u0005 \u0006= \u0004F1 \u0005R90 . There is one special case in which we do not need to worry about normality. Theorem 4.1. If G is a group, then any subgroup of index 2 is normal in G. Proof. Let H be a subgroup of index 2. Then one of the left cosets is H , and the other must consist of everything outside of H . In particular, aH = H if a ∈ H and aH is the other left coset if a ∈ / H . But exactly the same thing can be said for right cosets! So the left and right cosets agree. \u0002 Example 4.4. It is worth noting that if N is a normal subgroup of G and H is a normal subgroup of N , it does not necessarily follow that H is normal in G. For instance, N = {R0 , R180 , F1 , F2 } is a normal subgroup of D8 . (To check that it is a subgroup, use Theorem 3.14. To check that it is normal, use Theorem 4.1.) Also, H = \u0004F1 \u0005 is normal in N . (Again, it has index 2.) But as we saw in Example 4.3, H is not normal in D8 . Let us define a new subgroup. Definition 4.2. Let H be a subgroup of G. Then for any a ∈ G, we write a−1 Ha = {a−1 ha : h ∈ H }. Theorem 4.2. If H is a subgroup of G and a ∈ G, then a−1 Ha is a subgroup of G. Furthermore, |a−1 Ha| = |H |. Proof. We have e ∈ H , and therefore e = a−1 ea ∈ a−1 Ha. If a−1 h1 a, a−1 h2 a ∈ a−1 Ha, then (a−1 h1 a)(a−1 h2 a) = a−1 h1 (aa−1 )h2 a = a−1 h1 eh2 a = a−1 h1 h2 a ∈ a−1 Ha, since h1 h2 ∈ H . Finally, if a−1 ha ∈ a−1 Ha, then (a−1 ha)−1 = a−1 h−1 a ∈ a−1 Ha, since h−1 ∈ H . Thus, a−1 Ha is a subgroup of G. Also, given the definition of a−1 Ha, it is clear that we can only get one element for each element of H . But if a−1 h1 a = a−1 h2 a, then by cancellation, h1 = h2 . Thus, |a−1 Ha| = |H |. \u0002\n\n4.1 Normal Subgroups\n\n63\n\nWe can use this to give several different ways of saying that a subgroup is normal. Theorem 4.3. Let G be a group and H a subgroup. Then the following are equivalent: 1. 2. 3. 4.\n\nH is normal in G; a−1 ha ∈ H for all h ∈ H and all a ∈ G; a−1 Ha ⊆ H for all a ∈ G; and a−1 Ha = H for all a ∈ G.\n\nProof. It is clear that (4) implies (3) and (3) implies (2). Let us show that (2) implies (1). Suppose that (2) holds. Take any a ∈ G. Then for any h ∈ H , we have a−1 ha = h1 , for some h1 ∈ H . Thus, ha = ah1 ∈ aH . That is, Ha ⊆ aH . Also, (a−1 )−1 ha−1 = h2 , for some h2 ∈ H . That is, aha−1 = h2 , and therefore ah = h2 a ∈ Ha. Thus, aH ⊆ Ha, so aH = Ha and (1) is proved. Finally, let us show that (1) implies (4). Let H be a normal subgroup of G. Take any a ∈ G. Then Ha = aH . Thus, for any h ∈ H , we have ha ∈ aH , and therefore ha = ah1 , for some h1 ∈ H . That is, a−1 ha = h1 ∈ H . Therefore, a−1 Ha ⊆ H . But using a−1 in place of a, we also get aHa−1 ⊆ H . Hence, if h ∈ H , then aha−1 = h2 , for some h2 ∈ H . But now h = a−1 h2 a ∈ a−1 Ha. That is, H ⊆ a−1 Ha, and we are done. \u0002 Example 4.5. Let SLn (R) denote the set of all matrices in GLn (R) having determinant 1. We call this the special linear group. In view of Exercise 3.33, we know that SLn (R) is a subgroup of GLn (R). In fact, it is a normal subgroup. Indeed, if A ∈ SLn (R) and B ∈ GLn (R), then det(B−1 AB) = det(B−1 ) det(A) det(B) = det(B−1 ) det(B) det(A), since the determinants are just real numbers. But this is det(B−1 B) det(A) = 1, since B−1 B is the identity matrix and det(A) = 1. Therefore, B−1 AB ∈ SLn (R), and by Theorem 4.3, SLn (R) is indeed normal. Another useful construction is the following. Definition 4.3. If H and K are subgroups of G, then we write HK = {hk : h ∈ H , k ∈ K}. (If the group operation is addition, write H +K = {h+k : h ∈ H , k ∈ K}.) Note that HK is a subset of G, not necessarily a subgroup! It is easy to come up with examples where HK is not a subgroup, but the following theorem will lead us to some that cannot possibly work. Theorem 4.4. If H and K are finite subgroups of a group G, then |HK| =\n\n|H ||K| . |H ∩ K|\n\n64\n\n4 Factor Groups and Homomorphisms\n\nProof. Considering all possible h ∈ H and k ∈ K, it is clear that we can produce at most |H ||K| elements hk, but we must determine how many times each unique group element appears in such a list. Note that if h1 k1 = h2 k2 , with hi ∈ H and −1 ki ∈ K, then h−1 ∈ H ∩ K. Thus, h1 = h2 g and k2 = gk1 , for some 2 h1 = k2 k1 g ∈ H ∩ K. Conversely, if h1 = h2 g and k2 = gk1 , with g ∈ H ∩ K, then h1 k1 = h2 gk1 = h2 gg −1 k2 = h2 k2 . In other words, each hk will occur once for every element of H ∩ K. The result follows. \u0002 Example 4.6. Let G = S3 and let H and K be any two different subgroups of order 2. Then H ∩K can only contain the identity, and therefore |HK| = 4. But by Lagrange’s theorem, a group of order 6 cannot have a subgroup of order 4. Therefore, HK is not a subgroup. But HK will be a subgroup if either H or K is normal. Theorem 4.5. Let H and K be subgroups of G. Then 1. if either H or K is normal in G, then HK is a subgroup of G; and 2. if both H and K are normal in G, then HK is normal as well. Proof. (1) Observe that e = ee ∈ HK. Suppose that H is normal. Let us show closure. If hi ∈ H and ki ∈ K, then (h1 k1 )(h2 k2 ) = h1 (k1 h2 k1−1 )k1 k2 . Since H is normal, k1 h2 k1−1 ∈ H , and therefore h1 k1 h2 k1−1 ∈ H , k1 k2 ∈ K, as required. Also, −1 −1 −1 (h1 k1 )−1 = k1−1 h−1 1 = (k1 h1 k1 )k1 . −1 ∈ HK. If K is normal, the Again, since H is normal, k1−1 h−1 1 k1 ∈ H , so (h1 k1 ) proof is similar and left as an exercise. (2) Take h ∈ H , k ∈ K and a ∈ G. Then\n\na−1 hka = (a−1 ha)(a−1 ka). But a−1 ha ∈ H and a−1 ka ∈ K. Thus, a−1 hka ∈ HK. Exercises 4.1. Is each of the following sets a normal subgroup of GL2 (R)? 1. H = {A ∈ GL2 (R) : det(A) \u0002 ∈ Q}\u0003 a0 2. the set of diagonal matrices in GL2 (R) 0b 4.2. Find every normal subgroup of S3 . 4.3. If N is a normal subgroup of G, and |N | = 2, show that N ≤ Z(G).\n\n\u0002\n\n4.1 Normal Subgroups\n\n65\n\n4.4. Let N be a normal subgroup of G. Let H be the set of all elements h of G such that hn = nh for all n ∈ N . Show that H is a normal subgroup of G. 4.5. Show that the intersection of two normal subgroups of G is also a normal subgroup. Then extend \u0004 this to show that if Ni is a normal subgroup of G for every i in some set T , then i∈T Ni is a normal subgroup of G. \u0005 4.6. Let N1 ≤ N2 ≤ N3 ≤ · · · be normal subgroups of G. Show that ∞ i=1 Ni is a normal subgroup of G. 4.7. Let G be a group having exactly one subgroup H of order n. Show that H is normal in G. 4.8. Let G = H × K. If N is a normal subgroup of H and L is a normal subgroup of K, show that N × L is a normal subgroup of G. Is every normal subgroup of G of this form? 4.9. Suppose that H is a subgroup of G and a−1 b−1 ab ∈ H , for all a, b ∈ G. Show that H is normal. 4.10. Let H and K be subgroups of G. Show that HK is a subgroup if and only if HK = KH .\n\n4.2 Factor Groups We are now in a position to construct a new sort of group. Definition 4.4. Let G be a group and N a normal subgroup. Then the factor group (or quotient group) G/N is the set of all left cosets aN , with a ∈ G, under the operation (aN )(bN ) = abN . The fact that the factor group is indeed a group needs proving. Then we can look at some examples. Theorem 4.6. If G is any group and N is a normal subgroup, then G/N is a group of order [G : N ]. Proof. The main point is to verify that the operation is well-defined. The rest will follow easily from the fact that G is a group. In other words, suppose that a1 N = a2 N and b1 N = b2 N . We must show that a1 b1 N = a2 b2 N . Otherwise, this operation is nonsensical. But as a1 N = a2 N , we have a1−1 a2 = n1 , for some n1 ∈ N . Similarly, b−1 1 b2 = n2 ∈ N . Then −1 −1 −1 −1 −1 (a1 b1 )−1 a2 b2 = b−1 1 a1 a2 b2 = b1 n1 b2 = (b1 n1 b1 )(b1 b2 ) = b1 n1 b1 n2 . −1 Now, as N is normal, b−1 1 n1 b1 ∈ N . Thus, (a1 b1 ) a2 b2 ∈ N , which means that a1 b1 N = a2 b2 N , as required.\n\n66\n\n4 Factor Groups and Homomorphisms\n\nLet us check the group properties. As for closure, if aN and bN are left cosets, then so is abN . Also, for any a, b, c ∈ G, we have aN (bNcN ) = aNbcN = a(bc)N = (ab)cN = (aNbN )cN , so associativity is proved. If a ∈ G, then aNeN = aN = eNaN ; thus, eN is the identity of G/N . Finally, aNa−1 N = eN = a−1 NaN ; that is, a−1 N is the inverse of aN . Therefore, G/N is a group. The group consists of the left cosets, so the order is the number of left cosets, which is [G : N ]. The proof is complete. \u0002 Notice that the proposed group operation would not even be well-defined if N were not a normal subgroup. Example 4.7. Let G = U (15) = {1, 2, 4, 7, 8, 11, 13, 14}, and let N = \u000414\u0005 = {1, 14}. There is no need to worry about normality, since G is abelian. The left cosets are 1N = {1, 14}, 2N = {2, 13}, 4N = {4, 11} and 7N = {7, 8}. Thus, G/N = {1N , 2N , 4N , 7N }. We note that (4N )(7N ) = 13N = 2N and (2N )(4N ) = 8N = 7N . The rest of the group table is given in Table 4.1. We can also use this table to find inverses; for instance, 2N 7N = 1N . Since 1N is the identity, (2N )−1 = 7N .\n\nTable 4.1 Group table for U (15)/\u000414\u0005 1N 2N 4N 7N\n\n1N 1N 2N 4N 7N\n\n2N 2N 4N 7N 1N\n\n4N 4N 7N 1N 2N\n\n7N 7N 1N 2N 4N\n\nExample 4.8. Let G = Z and N = 5Z. Again, N is certainly a normal subgroup of G. Also, G/N = {0 + N , 1 + N , 2 + N , 3 + N , 4 + N }. Addition in the factor group behaves like modular arithmetic; indeed, (1 + N ) + (2 + N ) = 3 + N and (3 + N ) + (4 + N ) = 7 + N = 2 + N . The full group table is given in Table 4.2. Note that Z/5Z has precisely the same group table as Z5 (see Table 3.1).\n\nTable 4.2 Group table for Z/5Z 0+N 1+N 2+N 3+N 4+N\n\n0+N 0+N 1+N 2+N 3+N 4+N\n\n1+N 1+N 2+N 3+N 4+N 0+N\n\n2+N 2+N 3+N 4+N 0+N 1+N\n\n3+N 3+N 4+N 0+N 1+N 2+N\n\n4+N 4+N 0+N 1+N 2+N 3+N\n\n4.2 Factor Groups\n\n67\n\nExample 4.9. Let G = D8 and N = \u0004R90 \u0005. As N has index 2, it is necessarily a normal subgroup, by Theorem 4.1. In fact, there are only two left cosets, R0 N , which consists of all of the rotations, and F1 N , which consists of all of the flips. The group table is given in Table 4.3.\n\nTable 4.3 Group table for D8 /\u0004R90 \u0005 R0 N F1 N R0 N R0 N F1 N F1 N F1 N R0 N\n\nObserve that powers of group elements in a quotient group work as we would expect. Indeed, (aN )m = am N , for any integer m. In particular, (aN )−1 = a−1 N . Let us prove a few other useful facts. Theorem 4.7. Let G be a group, with a ∈ G and N a normal subgroup of G. Then 1. if G is abelian, then so is G/N ; 2. if G is cyclic, then so is G/N ; and 3. if |a| = m < ∞, then |aN | divides m. Proof. (1) If b, c ∈ G, then (bN )(cN ) = bcN = cbN = (cN )(bN ). (2) If G = \u0004b\u0005, then for any cN ∈ G/N , let us say that c = bk . Then cN = bk N = (bN )k . Thus, G/N = \u0004bN \u0005. (3) Note that (aN )m = am N = eN . Thus, by Corollary 3.2, the order of aN divides m. \u0002 A small word of caution is in order. Do not assume that the order of a equals that of aN . All we know is that |aN | divides |a|. Also, if a has infinite order, then we know nothing about |aN |; it could be finite or infinite. The following theorem tells us how to determine the subgroups of a factor group. The proof, however, is left as Exercise 4.18. Theorem 4.8. Let G be a group and N a normal subgroup. Then the subgroups of G/N are precisely of the form H /N , where H is a subgroup of G containing N . Furthermore, H /N is normal in G/N if and only if H is normal in G. Here is one more rather neat fact about factor groups. Theorem 4.9. Let G be any group. If G/Z(G) is cyclic, then G is abelian. Proof. Let Z = Z(G), and suppose that G/Z = \u0004aZ\u0005. Take any b, c ∈ G. Then bZ = am Z, for some integer m, and cZ = an Z, for some integer n. Thus, b = am y and c = an z, for some y, z ∈ Z. But noting that powers of a commute with each other and elements of Z commute with everything, we have bc = am yan z = an zam y = cb. Thus, G is abelian. \u0002\n\n68\n\n4 Factor Groups and Homomorphisms\n\nCorollary 4.1. The centre of a group cannot have prime index in that group. Proof. If G/Z(G) has prime order, then by Corollary 3.6, G/Z(G) is cyclic. But then the preceding theorem tells us that G is abelian; therefore, Z(G) = G has index 1, which is not prime. \u0002 Note that it is entirely possible for G to be nonabelian but G/Z(G) to be abelian. See Exercise 4.13. Exercises 4.11. Let G be a group having a normal subgroup N . Suppose that in G/N , the order of aN is 5. If |N | = 14, what are the possible orders of a? Show that each order you find can actually occur in some group. 4.12. Write the group table for 1. D8 /\u0004R180 \u0005 2. U (40)/\u00043\u0005. 4.13. Find a nonabelian group G such that 1. G/Z(G) is abelian 2. G is infinite, but G/Z(G) is finite. 4.14. Show that an element of the factor group R/Z has finite order if and only if it is in Q/Z. 4.15. Let G be a finite group having a normal subgroup N . If G/N has an element of order 42, show that G has an element of order 42. Does the same hold for infinite groups? 4.16. Let N be a normal subgroup of G. Show that G/N is abelian if and only if a−1 b−1 ab ∈ N for all a, b ∈ G. 4.17. Suppose that G has normal subgroups K and N such that G/K and G/N are abelian. If K ∩ N = {e}, show that G is abelian. 4.18. Let G have a normal subgroup N . Show that the subgroups of G/N are precisely of the form H /N , where H is a subgroup of G with N ⊆ H . Furthermore, show that H is normal in G if and only if H /N is normal in G/N . 4.19. Let G be an abelian group. Show that the elements of finite order in G form a normal subgroup N , and that the only element of finite order in G/N is the identity. 4.20. Let G be a nonabelian group. Show that there exists a subgroup H of G such that Z(G) \u0002 H \u0002 G.\n\n4.3 Homomorphisms\n\n69\n\n4.3 Homomorphisms We would like to talk about functions from one group to another. But an arbitrary function is not necessarily very useful. We need it to respect the group operation. This is the first step towards our goal (realized in the next section) of describing a way of determining if two groups have the same structure. Definition 4.5. Let G and H be groups. Then a group homomorphism (or, simply, homomorphism) from G to H is a function α : G → H such that α(g1 g2 ) = α(g1 )α(g2 ) for all g1 , g2 ∈ G. Note that in the above definition, the product g1 g2 is the product in G, whereas the product α(g1 )α(g2 ) takes place in H . These group operations need not be the same. Definition 4.6. If α : G → H is a homomorphism, then the kernel of α is the set ker(α) = {g ∈ G : α(g) = e}. Example 4.10. If n ≥ 2 is a positive integer, then α : Z → Zn given by α(a) = [a] (where we insert the equivalence class brackets for clarity) is a homomorphism. Indeed, α(a + b) = [a + b] = [a] + [b] = α(a) + α(b), for all a, b ∈ Z. Here, ker(α) = {a ∈ Z : [a] = } = nZ. Example 4.11. Let G be the additive group of integers, and let H be the multiplicative group of nonzero rational numbers. Then the function α : G → H given by α(a) = 2a is a homomorphism. To check this, we had first better verify that α does indeed map G into H . But if a is an integer, then 2a is a nonzero rational number. Also, α(a + b) = 2a+b = 2a 2b = α(a)α(b), as required. We see that ker(α) = {a ∈ Z : 2a = 1} = {0}. Example 4.12. Let G be any group, and consider the map α : G × G → G given by α((g1 , g2 )) = g2 . Then α is a homomorphism. Indeed, if gi ∈ G, then α((g1 , g2 )(g3 , g4 )) = α((g1 g3 , g2 g4 )) = g2 g4 , and this is also equal to α((g1 , g2 ))α((g3 , g4 )). Furthermore, the kernel is {(g, e) : g ∈ G} = G × {e}. Example 4.13. If G and H are any groups, then α : G → H given by α(g) = e for all g ∈ G is a homomorphism. Indeed, α(g1 g2 ) = e, and α(g1 )α(g2 ) = e2 = e. The kernel of α is all of G. We can give a few basic properties of homomorphisms.\n\n70\n\n4 Factor Groups and Homomorphisms\n\nTheorem 4.10. Let α : G → H be a homomorphism, and take any g ∈ G. Then 1. α(e) = e; 2. α(g n ) = (α(g))n , for any integer n; and 3. if |g| = m < ∞, then the order of α(g) divides m. Proof. (1) Note that α(e) = α(ee) = α(e)α(e). Cancelling, we find that α(e) is the identity of H . (2) If n > 0, then note that α(g n ) = α(gg · · · g ) = α(g)α(g) · · · α(g) = (α(g))n . \u0007\b \u0006 \u0007\b \u0006 n times\n\nn times\n\nIf n = 0, then use part (1). If n = −1, then note that α(g)α(g −1 ) = α(gg −1 ) = α(e) = e. Similarly, α(g −1 )α(g) = e. Therefore, α(g −1 ) = (α(g))−1 . Combining what we already know, the case where n < −1 follows immediately. (3) We have (α(g))m = α(g m ) = α(e) = e. Thus, by Corollary 3.2, the order of α(g) divides m. \u0002 The kernel of a homomorphism is rather important, as the following result suggests. Theorem 4.11. Let α : G → H be a homomorphism. Then 1. ker(α) is a normal subgroup of G; and 2. α is one-to-one if and only if ker(α) = {e}. Proof. Let K = ker(α). Let us show that K is a subgroup of G. By Theorem 4.10, α(e) = e, so e ∈ K. Suppose k1 , k2 ∈ K. Then α(k1 k2 ) = α(k1 )α(k2 ) = ee = e; hence, k1 k2 ∈ K. Also, α(k1−1 ) = (α(k1 ))−1 = e−1 = e. Thus, k1−1 ∈ K, and K is a subgroup. If k ∈ K and g ∈ G, then α(g −1 kg) = α(g −1 )α(k)α(g) = α(g −1 )eα(g) = (α(g))−1 α(g) = e. Therefore, g −1 kg ∈ K, and K is normal. Now, suppose that α is one-to-one. Since α(e) = e, we know that if α(g) = e, then g = e. Therefore, the kernel is simply {e}. Conversely, suppose that ker(α) = {e}. If α(g1 ) = α(g2 ), then α(g1 )(α(g2 ))−1 = e. But this means that α(g1 g2−1 ) = e, and \u0002 therefore g1 g2−1 ∈ K = {e}. That is, g1 = g2 , and α is one-to-one. Two other sorts of subgroups are also useful.\n\n4.3 Homomorphisms\n\n71\n\nDefinition 4.7. Let α : G → H be a homomorphism. If L is any subgroup of G, then the image of L is α(L) = {α(l) : l ∈ L}. If M is any subgroup of H , then the preimage (or inverse image) of M is the set α −1 (M ) = {g ∈ G : α(g) ∈ M }. Note that the use of the notation α −1 (M ) does not imply that the function α is invertible. It may or may not be. Example 4.14. Consider Example 4.11. If L = 3Z, then α(L) = {23a : a ∈ Z} = {8a : a ∈ Z}. If M = {±4a : a ∈ Z}, then α −1 (M ) = 2Z. Example 4.15. Let G = Z, and consider α : G × G → G, as in Example 4.12. Let L = 3Z × 5Z. Then α(L) = 5Z. If M = 6Z, then α −1 (M ) = Z × 6Z. We conclude with a few properties of images and preimages. Theorem 4.12. Let α : G → H be a homomorphism. Then 1. 2. 3. 4. 5. 6. 7.\n\nif L is a subgroup of G, then α(L) is a subgroup of H ; if L is normal in G, then α(L) is normal in α(G); if L is cyclic, then α(L) is cyclic; if L is abelian, then α(L) is abelian; α is onto if and only if α(G) = H ; if M ≤ H , then α −1 (M ) ≤ G; and if M is a normal subgroup of H , then α −1 (M ) is normal in G.\n\nProof. (1) We have e ∈ L, so e = α(e) ∈ α(L). If α(l1 ), α(l2 ) ∈ α(L), then α(l1 )α(l2 ) = α(l1 l2 ) ∈ α(L), since l1 l2 ∈ L. Also, (α(l1 ))−1 = α(l1−1 ) ∈ α(L), since l1−1 ∈ L. (2) If l ∈ L, g ∈ G, then (α(g))−1 α(l)α(g) = α(g −1 lg) ∈ α(L), since g −1 lg ∈ L. (3) If L = \u0004k\u0005, then for any α(l) ∈ α(L), we have l = k m , for some integer m. Then α(l) = α(k m ) = (α(k))m . Thus, α(L) = \u0004α(k)\u0005. (4) If l1 , l2 ∈ L, then α(l1 )α(l2 ) = α(l1 l2 ) = α(l2 l1 ) = α(l2 )α(l1 ). (5) This is the definition of “onto”. (6) Notice that α(e) = e ∈ M ; hence, e ∈ α −1 (M ). Also, if g1 , g2 ∈ α −1 (M ), then α(g1 g2 ) = α(g1 )α(g2 ) ∈ M , since α(g1 ), α(g2 ) ∈ M . Thus, g1 g2 ∈ α −1 (M ). Furthermore, α(g1−1 ) = (α(g1 ))−1 ∈ M , since α(g1 ) ∈ M . Thus, g1−1 ∈ α −1 (M ). (7) Take a ∈ α −1 (M ), g ∈ G. Then α(g −1 ag) = (α(g))−1 α(a)α(g) ∈ M , since \u0002 α(a) ∈ M and M is normal. Thus, g −1 ag ∈ α −1 (M ). Exercises 4.21. Are α and β, described below, homomorphisms? If so, are they one-to-one and onto? 1. G is the group of positive real numbers under multiplication, H is R (under addition), α : G → H via α(a) = log10 a 2. β : Z → Z, β(a) = a + 1\n\n72\n\n4 Factor Groups and Homomorphisms\n\n4.22. Let α : Z9 × Z27 → Z27 be given by α((a, b)) = 3b, for all a ∈ Z9 , b ∈ Z27 . Show that α is a homomorphism. Also, find ker(α), and decide if α is onto. 4.23. Define α : U (16) × U (16) → U (16) via α((a, b)) = ab−1 . Show that α is a homomorphism, and find α −1 (\u00047\u0005). 4.24. Describe every homomorphism α : Z10 → Z15 . 4.25. Let G be a finite group and α : G → H an onto homomorphism. 1. If G has an element of order n, must H have one? 2. If H has an element of order n, must G have one? 4.26. Let α : G → H be a homomorphism, and suppose that α(g) = h. For any a ∈ G, show that α(a) = h if and only if a = gk for some k ∈ ker(α). 4.27. Define α : G × G → G via α((g, h)) = gh. If α is a homomorphism, show that G is abelian. 4.28. Show that a group G is cyclic if and only if there exists an onto homomorphism from Z to G. 4.29. Let N be a normal subgroup of G. Show that there exist a group H and a homomorphism α : G → H with kernel N . 4.30. Let G be the multiplicative group of nonzero complex numbers and H the multiplicative group of nonzero real numbers. Does there exist a one-to-one homomorphism from G to H ?\n\n4.4 Isomorphisms One of our goals is to establish if two groups are, in effect, the same. To this end, we need to strengthen the notion of a homomorphism. Definition 4.8. Let G and H be groups. Then a group isomorphism (or, simply, isomorphism) from G to H is a homomorphism from G to H that is bijective. When such an isomorphism exists, we say that G and H are isomorphic groups. Isomorphic groups have precisely the same structure. The isomorphism simply provides new labels for the group elements. Theorem 4.13. On any collection of groups, isomorphism is an equivalence relation.\n\n4.4 Isomorphisms\n\n73\n\nProof. Reflexivity: Use the function α : G → G given by α(g) = g for all g. It is easily seen to be an isomorphism. Symmetry: Suppose that α : G → H is an isomorphism. By Theorem 1.3, there exists a function β : H → G given by β(h) = g, where α(g) = h, and this β is also bijective. We must check that it is a homomorphism. Take any h1 , h2 ∈ H , and suppose that β(hi ) = gi . Then α(g1 g2 ) = α(g1 )α(g2 ) = h1 h2 ; thus, β(h1 h2 ) = g1 g2 = β(h1 )β(h2 ), as required. Transitivity: Suppose that α : G → H and β : H → K are isomorphisms. Let γ = β ◦α. By Theorem 1.2, γ is bijective. We must check that it is a homomorphism. Take any g1 , g2 ∈ G. Then γ (g1 g2 ) = β(α(g1 g2 )) = β(α(g1 )α(g2 )) = β(α(g1 ))β(α(g2 )) = γ (g1 )γ (g2 ). We are done.\n\n\u0002\n\nTherefore, it makes sense to say that G and H are isomorphic; we do not have to specify that G is isomorphic to H . In order to verify that a particular function is an isomorphism, we have to check three things: it must respect the group operation, it must be one-to-one and it must be onto. We can use Theorem 4.11 for the second of these; to show that it is one-to-one, it is enough to show that the kernel is trivial. Example 4.16. Let us show that Z3 × Z5 and Z15 are isomorphic groups. We define α : Z15 → Z3 × Z5 via α(a) = (a, a). First, is this well-defined? If a = b in Z15 , then 15|(a − b), so 3|(a − b) and 5|(a − b), and therefore (a, a) = (b, b) in Z3 × Z5 . Check that it is a homomorphism. If a, b ∈ Z15 , then α(a + b) = (a + b, a + b) = (a, a) + (b, b) = α(a) + α(b). Next, let us show that it is one-to-one. If a ∈ ker(α), then (a, a) = (0, 0). That is, 3|a and 5|a. Therefore, 15|a, so a = 0 in Z15 , and α is one-to-one. In this case, we do not need to check surjectivity, because the 15 elements of Z15 map to 15 different elements of Z3 × Z5 . But Z3 × Z5 only has 15 elements! Hence, the function must be onto. Example 4.17. Lest we get too comfortable, Z24 is not isomorphic to Z4 × Z6 . Why not? Notice that 1 has order 24 in Z24 . If these groups had precisely the same structure, then Z4 × Z6 would have to have an element of order 24 as well. But it is easy to see that 12(a, b) = (0, 0) for every (a, b) ∈ Z4 × Z6 , so every element has order dividing 12. Example 4.18. As we noted following Example 3.4, the set G = {1, −1, i, −i} (where i is the complex number) is a group under multiplication. We claim that it is isomorphic to the additive group Z4 . To see this, we define α : Z4 → G via α(0) = 1, α(1) = i, α(2) = −1 and α(3) = −i. This function is clearly bijective, and we can check that it respects the group operations by comparing the group tables. The tables for Z4 and G are shown in Tables 4.4 and 4.5. Note that if we replace 0 and α(0) with A, 1 and α(1) with B, and so on, we see that both groups have Table 4.6. Thus, α is just a relabelling of the group elements. In fact, we can classify all cyclic groups up to isomorphism.\n\n74\n\n4 Factor Groups and Homomorphisms\n\nTable 4.4 Group table for the additive group Z4 0 1 2 3\n\n0 0 1 2 3\n\n1 1 2 3 0\n\n2 2 3 0 1\n\n3 3 0 1 2\n\nTable 4.5 Group table for the multiplicative group {1, −1, i, −i} 1 i −1 −i\n\n1 1 i −1 −i\n\ni i −1 −i 1\n\n−1 −1 −i 1 i\n\n−i −i 1 i −1\n\nTable 4.6 Group table for both Tables 4.4 and 4.5 after relabelling A B C D\n\nA A B C D\n\nB B C D A\n\nC C D A B\n\nD D A B C\n\nTheorem 4.14. Let G = \u0004a\u0005 be a cyclic group. If a has infinite order, then G is isomorphic to Z. If a has order n < ∞, then G is isomorphic to Zn . Proof. Let G be infinite cyclic. Define α : Z → G via α(i) = ai . We claim that α is an isomorphism. If i, j ∈ Z, then α(i + j) = ai+j = ai aj = α(i)α(j), as required. If i ∈ ker(α), then ai = e = a0 . By Theorem 3.8, i = 0. Thus, α is one-to-one. Furthermore, if i ∈ Z, then ai ∈ α(Z), as α(i) = ai . Thus, α is onto as well, and therefore an isomorphism. Now suppose that |a| = n < ∞. Define α : Zn → G via α(i) = ai . Here, we must check that α is well-defined. But if i = j in Zn , then n|(i − j). Thus, by Theorem 3.8, ai = aj . The fact that α is an onto homomorphism follows as above. If i ∈ ker(α), then ai = e, and by Corollary 3.2, n divides i. Thus, in Zn , i = 0. \u0002 Corollary 4.2. If a group G has prime order p, then G is isomorphic to Zp . Proof. Combine Corollary 3.6 and Theorem 4.14.\n\n\u0002\n\nSo, groups of prime order have as nice a structure as we could ask. With a little more work, we can also classify the groups with order twice a prime. Lemma 4.1. Let G be a group having distinct commuting elements a and b of order 2. Then G has a subgroup isomorphic to Z2 × Z2 .\n\n4.4 Isomorphisms\n\n75\n\nProof. Given the conditions upon a and b, we can see that H = {e, a, b, ab} is a subgroup. (It contains the identity, and closure is easily checked.) Also, H contains four distinct elements. (Clearly, e, a and b are distinct. If ab = e = bb, then a = b. If ab = a = ae, then b = e. If ab = b = eb, then a = e. These are all impossible.) We claim that it is isomorphic to Z2 × Z2 . Let α : H → Z2 × Z2 be given by α(e) = (0, 0), α(a) = (1, 0), α(b) = (0, 1) and α(ab) = (1, 1). This function is clearly bijective, and running through the possible pairs of group elements, we see that it is a homomorphism. \u0002 Corollary 4.3. Every group G of order 4 is isomorphic to either Z4 or Z2 × Z2 . Proof. In a group of order 4, every nonidentity element has order 2 or 4. If there is an element of order 4, G is cyclic and, by Theorem 4.14, isomorphic to Z4 . Otherwise, every nonidentity element has order 2. By Exercise 3.32, G is abelian, and the preceding lemma tells us that G has a subgroup isomorphic to Z2 ×Z2 . Given the order of the group, we are done. \u0002 When p is a prime larger than 2, we are already aware of two possible groups of order 2p; the cyclic group and the dihedral one. In fact, those are all of the options. Theorem 4.15. Let |G| = 2p, where p is an odd prime. Then G is isomorphic to either Z2p or D2p . Proof. The possible orders for nonidentity elements of G are 2, p and 2p. If G has an element of order 2p then it is cyclic and, by Theorem 4.14, isomorphic to Z2p . So, assume that every nonidentity element has order 2 or p. If every nonidentity element has order 2, then once again, G is abelian and, by Lemma 4.1, G has a subgroup of order 4, contradicting Lagrange’s theorem. Therefore, let a ∈ G have order p. Take any b ∈ / \u0004a\u0005. Suppose that |b| = p. Then noting that \u0004a\u0005 ∩ \u0004b\u0005 is a subgroup of both \u0004a\u0005 and \u0004b\u0005 (see Exercise 3.37), Lagrange’s theorem tells us that it can only have order 1 or p. As b ∈ / \u0004a\u0005, it must be 1. Thus, by Theorem 4.4, |\u0004a\u0005\u0004b\u0005| = |\u0004a\u0005||\u0004b\u0005|/|\u0004a\u0005 ∩ \u0004b\u0005| = p2 /1 = p2 . But this exceeds the order of G. Therefore, |b| = 2. Now, as \u0004a\u0005 has index 2, Theorem 4.1 tells us that it is normal. Thus, b−1 ab ∈ \u0004a\u0005, say b−1 ab = ai . But then a = b−2 ab2 = b−1 (b−1 ab)b = b−1 ai b = (b−1 ab)i = (ai )i = ai . 2\n\nAs a has order p, we have i2 ≡ 1 (mod p). That is, p|(i2 − 1) = (i − 1)(i + 1). As p is prime, i ≡ ±1 (mod p). Thus, b−1 ab = a or a−1 . Suppose that b−1 ab = a. Then a and b commute. But consider the order of ab. If (ab)n = e, then an = b−n ∈ \u0004a\u0005 ∩ \u0004b\u0005 = {e}, as a and b have different prime orders. Thus, p|n and 2|n, so 2p|n. That is, ab has order 2p, which we have excluded. Therefore, b−1 ab = a−1 .\n\n76\n\n4 Factor Groups and Homomorphisms\n\nWe now know everything about the group. As \u0004a\u0005 has index 2 and b ∈ / \u0004a\u0005, the elements of G are precisely ai and bai , 0 ≤ i < p. Furthermore, we know how to find the product of any two elements. Indeed, ai aj = ai+j (reducing the exponent modulo p if necessary), bai aj = bai+j , ai baj = b(b−1 ai b)aj = b(b−1 ab)i aj = b(a−1 )i aj = baj−i and bai baj = b(baj−i ) = aj−i . Thus, we can fill in the entire group table for G, and we have precisely the same group structure as in the dihedral group! Indeed, letting F be any flip in D2p , we define α : G → D2p via α(ai ) = R360i/p and α(bai ) = FR360i/p . Then α is certainly bijective and it is a homomorphism as well. \u0002 We can also mop up a proof we postponed. Theorem 4.16. If m and n are relatively prime, then U (mn) is isomorphic to U (m)× U (n). Proof. Define α : U (mn) → U (m) × U (n) via α(a) = (a, a). If (a, mn) = 1, then (a, m) = (a, n) = 1, so we have (a, a) ∈ U (m) ×U (n) whenever a ∈ U (mn). Let us verify that α is well-defined. But if a = b in Zmn , then mn|(a − b), so m|(a − b) and n|(a − b), and therefore (a, a) = (b, b) in U (m) × U (n). It is also a homomorphism; indeed, if a, b ∈ U (mn), then α(ab) = (ab, ab) = (a, a)(b, b) = α(a)α(b). Let us check that α is one-to-one. But if a ∈ ker(α), then (a, a) = (1, 1) in U (m) × U (n); that is, m|(a − 1) and n|(a − 1). As m and n are relatively prime, mn|(a − 1). That is, a = 1 in U (mn); hence, α is one-to-one. Finally, we must show that α is onto. Take any (c, d ) ∈ U (m) × U (n). By the Chinese Remainder Theorem, there exists an a such that a ≡ c (mod m) and a ≡ d (mod n). Furthermore, to show that a is in U (mn), it suffices to show that it is relatively prime to both m and n. Without loss of generality, suppose that (a, m) = k > 1. Then as k|a and k|m, we see that k|c as well. But then (c, m) \u0006= 1, which is impossible. Therefore, a ∈ U (mn) and α(a) = (c, d ). Thus, α is indeed an isomorphism. \u0002 This gives us the second part of Theorem 3.19. Corollary 4.4. If m and n are relatively prime, then ϕ(mn) = ϕ(m)ϕ(n). Proof. The order of U (k) is ϕ(k). As isomorphic groups have the same order, the preceding theorem completes the proof. \u0002 We wish to add one more point to Corollary 4.3 and Theorem 4.15. If we are to classify groups of a particular order up to isomorphism, we had better ensure that the groups we have listed are not isomorphic to each other. Proving that two groups are not isomorphic generally involves finding a property that one has but the other lacks. For instance, Z2p is cyclic for all primes p, but neither Z2 × Z2 nor D2p is cyclic (indeed, D2p is not even abelian). Some properties that can be useful follow.\n\n4.4 Isomorphisms\n\n77\n\nTheorem 4.17. Let G and H be isomorphic groups. Then 1. 2. 3. 4.\n\nG is abelian if and only if H is abelian; G is cyclic if and only if H is cyclic; |G| = |H |; for any positive integer n, G and H have the same number of elements of order n (which could be an infinite number); 5. for any positive integer n, G and H have the same number of subgroups of order n (which could be an infinite number); and 6. for any positive integer n, G and H have the same number of normal subgroups of order n (which could be an infinite number). Proof. Let α : G → H be an isomorphism. (1) As α(G) = H , we see from Theorem 4.12 that if G is abelian, so is H . But the same can be said for α −1 : H → G. (2) Same idea. (3) An isomorphism is a bijection. (4) Take g ∈ G of order n. By Theorem 4.10, |α(g)| divides |g|. But by the same argument, |g| = |α −1 (α(g))| divides |α(g)|. Thus, |g| = |α(g)|. That is, the elements of order n in G are in one-to-one correspondence with the elements of order n in H . (5) Let L be a subgroup of G of order n. Then α(L) is a subgroup of H , and it is isomorphic to L; hence, it has the same order. If M is some other subgroup of G, then since α is one-to-one, α(M ) is a different group. Thus, H has at least as many subgroups of order n as G does. But applying α −1 , we find that G has at least as many subgroups of order n as H does. (6) Let L be a normal subgroup of order n in G. Then by Theorem 4.12, α(L) is a normal subgroup of α(G) = H . Now proceed as in (5). \u0002 Example 4.19. As U (10) = \u00043\u0005 is cyclic of order 4, we know that U (10) is isomorphic to Z4 . Now, U (8) is an abelian group of order 4, but it is not cyclic, so it is not isomorphic to U (10). By Corollary 4.3, U (8) is isomorphic to Z2 × Z2 . Example 4.20. Consider the groups U (20) and U (8) × U (3). Each is an abelian group of order 8 and neither is cyclic; however, they are not isomorphic. To see this, note that U (20) has exactly three elements of order 2; namely, 9, 11 and 19. However, U (8)×U (3) has too many elements of order 2; in fact, all seven nonidentity elements have that order. Exercises 4.31. For each of the following pairs of groups, explain why they are not isomorphic. 1. Z4 × Z4 and Z4 × Z2 × Z2 2. GL2 (R) and R 3. Z and Z × Z 4.32. For each of the following pairs of groups, explain why they are or are not isomorphic.\n\n78\n\n1. 2. 3. 4.\n\n4 Factor Groups and Homomorphisms\n\nZ9 × Z3 and Z3 × Z3 × Z3 Z21 and Z3 × Z7 U (22) and Z10 D20 and Z2 × Z10\n\n\u0002\n\n\u0003 10 4.33. Let G be the set of all matrices of the form , for all integers a. Show a1 that G is a subgroup of GL2 (R). To what familiar group is it isomorphic? 4.34. Show that Z is not isomorphic to Q. 4.35. Let H ≤ G and a ∈ G. Show that H and a−1 Ha are isomorphic. 4.36. Show that G × H is isomorphic to H × G. 4.37. Show that Z is isomorphic to a proper subgroup of itself.\n\n4.38. Let G be any group. Let H consist of the same set of elements as G, but with a new operation given by a ∗ b = ba, for all a and b. Show that H is a group, and that it is isomorphic to G. 4.39. Consider the group G from Exercise 3.42. Show that it is isomorphic to a proper subgroup of itself. 4.40. Consider the group H from Exercise 3.42. Show that it is isomorphic to the multiplicative group of positive rational numbers.\n\n4.5 The Isomorphism Theorems for Groups In this section, we will discuss three theorems that can aid us in showing that certain groups are isomorphic. The first of these theorems is the most important, and is used to prove the other two. Theorem 4.18 (First Isomorphism Theorem for Groups). Let α : G → H be a homomorphism. Then G/ ker(α) is isomorphic to α(G). Proof. Let K = ker(α). We know that K is a normal subgroup of G. Define β : G/K → α(G) via β(aK) = α(a). We claim that β is an isomorphism. First, we must show that β is well-defined. Suppose that aK = bK. Then a−1 b ∈ K, and therefore α(a−1 b) = e. That is, (α(a))−1 α(b) = e, so α(a) = α(b). Thus, β is well-defined. Also, β is a homomorphism. Indeed, β(aKbK) = β(abK) = α(ab) = α(a)α(b) = β(aK)β(bK). Next, let us check that β is one-to-one. Suppose that aK ∈ ker(β). Then α(a) = e, which means that a ∈ K, so aK = eK. That is, ker(β) = {eK}, and β is one-to-one. Finally, we must verify that β is onto. Take α(a) ∈ α(G). Then β(aK) = α(a). We are done. \u0002\n\n4.5 The Isomorphism Theorems for Groups\n\n79\n\nThe First Isomorphism Theorem is a crucial tool in proving that groups are isomorphic. It is also an enormous time-saver! Whenever we are asked to show that something along the lines of G/N is isomorphic to H , all we need to do is find a homomorphism from G onto H with kernel N . We do not need to define a function on cosets and check that it is well-defined. Example 4.21. For any integer n ≥ 2, Z/nZ is isomorphic to Zn . Indeed, define α : Z → Zn via α(a) = [a] (where we insert the equivalence class brackets for clarity). This is a homomorphism, as α(a + b) = [a + b] = [a] + [b] = α(a) + α(b), for all a, b ∈ Z. Also ker(α) = {a ∈ Z : a ≡ 0 (mod n)} = nZ. Finally, if [a] ∈ Zn , then α(a) = [a], so α is onto. The First Isomorphism Theorem completes the proof. Example 4.22. We claim that GL2 (R)/SL2 (R) is isomorphic to the multiplicative group of nonzero real numbers, which we denote by H . Indeed, define α : GL2 (R) → H via α(A) = det(A). As an invertible matrix has a nonzero determinant, the image of GL2 (R) is indeed contained in H . Also, if A, B ∈ GL2 (R), then α(AB) = det(AB) and since determinants respect products, this is det(A) det(B) = α(A)α(B). Thus, α is a homomorphism. By definition, its kernel is SL2 (R). Finally, if a ∈ H , then \u0002\u0002 \u0003\u0003 a0 α = a, 01 and therefore α is onto. Now we apply the First Isomorphism Theorem. Example 4.23. Let us show that if G and H are any groups, then G × H has a factor group isomorphic to G. Define α : G × H → G via α((g, h)) = g, for all g ∈ G, h ∈ H . Check that α is a homomorphism. If gi ∈ G, hi ∈ H , then α((g1 , h1 )(g2 , h2 )) = α((g1 g2 , h1 h2 )) = g1 g2 = α((g1 , h1 ))α((g2 , h2 )). Also, if g ∈ G, then α((g, e)) = g, so α is onto. Therefore, (G × H )/ ker(α) is isomorphic to H . If we wish to specify the group being factored out, note that ker(α) = {(e, h) : h ∈ H } = {e} × H . Theorem 4.19 (Second Isomorphism Theorem for Groups). Let G be a group with H and N subgroups, such that N is normal. Then H /(H ∩ N ) is isomorphic to HN /N . Proof. We will show that H ∩ N is normal in H by demonstrating that it is the kernel of a homomorphism. Also, by Theorem 4.5, HN is a subgroup of G, since N is normal. Define α : H → HN /N via α(h) = hN . As H ⊆ HN , we see that hN ∈ HN /N . Observe that α is a homomorphism. Indeed, if h1 , h2 ∈ H , then α(h1 h2 ) = h1 h2 N = (h1 N )(h2 N ) = α(h1 )α(h2 ). Also, if hn ∈ HN , then α(h) = hN = hnN , since h−1 hn = n ∈ N . Thus, α is onto. Finally, ker(α) = {h ∈ H : hN = eN } = {h ∈ H : h ∈ N } = H ∩ N . The First Isomorphism Theorem finishes the proof. \u0002\n\n80\n\n4 Factor Groups and Homomorphisms\n\nTheorem 4.20 (Third Isomorphism Theorem for Groups). Let G be a group and suppose that N and K are normal subgroups, with K ⊆ N . Then (G/K)/(N /K) is isomorphic to G/N . Proof. Define α : G/K → G/N via α(aK) = aN , for any a ∈ G. First, let us check that this is well-defined. But if aK = bK, then a−1 b ∈ K ⊆ N , so aN = bN . Next, let us show that α is a homomorphism. But α((aK)(bK)) = α(abK) = abN = (aN )(bN ) = α(aK)α(bK). Furthermore, if aN ∈ G/N , then α(aK) = aN , so α is onto. Finally, ker(α) = {aK ∈ G/K : aN = eN } = {aK ∈ G/K : a ∈ N } = N /K. We now apply the First Isomorphism Theorem.\n\n\u0002\n\nExample 4.24. The Third Isomorphism Theorem tells us that (Z/12Z)/(4Z/12Z) is isomorphic to Z/4Z. (Admittedly, we could have worked this out by noting that Z is cyclic, so its factor group is cyclic, and the factor group of the factor group is cyclic, and that every cyclic group of order 4 is isomorphic to Z4 , which in turn is isomorphic to Z/4Z. But isn’t this faster?) Exercises 4.41. Let G = Z × Z and N = {(a, a) : a ∈ Z}. Show that G/N is isomorphic to Z. 4.42. For any groups G and H , show that (G × H )/(G × {e}) is isomorphic to H . 4.43. Show that R/Z is isomorphic to the multiplicative group H = {a + bi ∈ C : a2 + b2 = 1}. 4.44. Let G be an abelian group and n a positive integer. Consider the groups H and K from Exercise 3.40. Show that G/H is isomorphic to K. 4.45. Let G be the group from Exercise 3.16. 1. Find Z(G). 2. Show that G/Z(G) is isomorphic to Z × Z. 4.46. Let G be a group having subgroups N and K of index 2, such that N \u0006= K. 1. Show that [N : N ∩ K] = 2. 2. Show that G/(N ∩ K) is isomorphic to Z2 × Z2 .\n\n4.6 Automorphisms\n\n81\n\n4.6 Automorphisms One particular type of isomorphism deserves special mention. Definition 4.9. Let G be any group. Then an automorphism of G is an isomorphism α : G → G. The set of all automorphisms of G is called the automorphism group of G, and is denoted Aut(G). Example 4.25. Let G be any abelian group. Then the function α : G → G given by α(a) = a−1 for all a ∈ G is an automorphism. To see that α is a homomorphism, note that α(ab) = (ab)−1 = a−1 b−1 = α(a)α(b). (Note that this would not work if G were nonabelian, as (ab)−1 = b−1 a−1 !) If α(a) = e, then a−1 = e, so a = e and α is one-to-one. Also, if a ∈ G, then α(a−1 ) = a. Thus, α is onto as well. Theorem 4.21. For any group G, the automorphism group of G is a group under composition of functions. Proof. As we noted in Theorem 4.13, the composition of two isomorphisms is an isomorphism; therefore, the same follows for automorphisms, so Aut(G) is closed. By Theorem 1.2, the composition of functions is always associative. Certainly, the identity function that fixes every element of G is an automorphism, and serves as an identity for Aut(G). Finally, we saw in Theorem 4.13 that every isomorphism has an inverse isomorphism; thus, each automorphism has an inverse. \u0002 Generally speaking, determining Aut(G) for an arbitrary group G is a difficult problem. But we can, at least, solve it when G is cyclic. Theorem 4.22. Let G = \u0004a\u0005 be a cyclic group. Then 1. if a has infinite order, then Aut(G) is isomorphic to Z2 ; and 2. if |a| = n < ∞, then Aut(G) is isomorphic to U (n). Proof. Let α ∈ Aut(G). If α(a) = ai , then for every j ∈ Z, we have α(aj ) = (α(a))j = (ai )j . In particular, G = α(G) = \u0004ai \u0005. Thus, ai must generate G. Conversely, suppose that G = \u0004ai \u0005, and α(a) = ai . Then we can only have α(aj ) = aij for all integers j. We claim such an α is an automorphism. Indeed, α(aj ak ) = α(aj+k ) = ai(j+k) = aij aik = α(aj )α(ak ), so α is a homomorphism. If α(aj ) = e, then (ai )j = e. If a has infinite order, then ij = 0, and therefore j = 0. If |a| = n < ∞, then n|ij. But as |a| = |ai | = |G|, Corollary 3.2 tells us that (n, i) = 1. This means that n|j, so aj = e. Either way, ker(α) = {e}. As ai is a generator, it follows immediately that α is onto. The claim is proved. Thus, the automorphisms of G are precisely given by α(aj ) = aij , where ai is a fixed generator of G. If a has infinite order, then the only generators of \u0004a\u0005 are a and a−1 . Indeed, if m a were a generator, then we would have to have a = (am )l , for some l ∈ Z. But\n\n82\n\n4 Factor Groups and Homomorphisms\n\nthen ml = 1, which means that m ∈ {1, −1}. It is clear, on the other hand, that both a and a−1 are generators. Thus, Aut(G) has order 2. By Corollary 4.2, Aut(G) is isomorphic to Z2 . Now suppose that |a| = n < ∞. Let us define γ : Aut(G) → U (n) via γ (α) = i, where α(a) = ai . Again, Corollary 3.2 tells us that since |a| = |ai | = |G|, we have (i, n) = 1, so i ∈ U (n). Now, if α, β ∈ Aut(G), with α(a) = ai and β(a) = aj , then (α ◦ β)(a) = α(β(a)) = α(aj ) = (α(a))j = aij . Thus, γ (α ◦ β) = ij (reducing modulo n if necessary). But γ (α)γ (β) = ij as well, so γ is a homomorphism. If γ (α) = 1, then α(a) = a, and hence α is the identity automorphism. Therefore, ker(γ ) is trivial. Finally, if i ∈ U (n), then as we have observed, ai is a generator of G, and we obtain α ∈ Aut(G) such that α(a) = ai . Therefore, γ (α) = i, and γ is onto. Hence, γ is the isomorphism we seek. \u0002 In particular, we see that the automorphism group of a cyclic group is abelian. It would be a mistake to think that the automorphism group of an abelian group is necessarily abelian, as the following example indicates. Example 4.26. Let G = Z2 × Z2 . Then define α : G → G via α((0, 0)) = (0, 0), α((1, 0)) = (0, 1), α((0, 1)) = (1, 0) and α((1, 1)) = (1, 1). Also, let β((0, 0)) = (0, 0), β((1, 0)) = (1, 0), β((0, 1)) = (1, 1) and β((1, 1)) = (0, 1). Clearly, α and β are both bijective. The group is also small enough that one can check all of the possibilities and find that they are homomorphisms. Therefore, α, β ∈ Aut(G). But α(β((1, 0))) = α((1, 0)) = (0, 1), whereas β(α((1, 0))) = β((0, 1)) = (1, 1). Thus, α ◦ β \u0006= β ◦ α, so Aut(G) is nonabelian. In Exercise 4.49, we must show that Aut(G) is isomorphic to D6 . Let us define a particular type of automorphism. Definition 4.10. Let G be a group and a ∈ G. Then the inner automorphism induced by a is θa : G → G given by θa (g) = a−1 ga for all g ∈ G. The inner automorphism group of G is Inn(G) = {θa : a ∈ G}. Inner automorphisms are only interesting when the group G is nonabelian; for abelian groups, every inner automorphism is the identity function, as a−1 ga = g. Let us list a few basic properties of inner automorphisms. Lemma 4.2. Let G be a group and a, b ∈ G. Then 1. θa ∈ Aut(G); 2. θa ◦ θb = θba ; and 3. (θa )−1 = θa−1 . Proof. (1) First, let us show that θa is a homomorphism. If g, h ∈ G, then θa (gh) = a−1 gha = (a−1 ga)(a−1 ha) = θa (g)θa (h).\n\n4.6 Automorphisms\n\n83\n\nIf θa (g) = e, then a−1 ga = e, so g = aea−1 = e; thus, ker(θa ) = {e}, and θa is one-to-one. Finally, if g ∈ G, then θa (aga−1 ) = a−1 aga−1 a = g; thus, θa is onto. (2) If g ∈ G, then θa (θb (g)) = θa (b−1 gb) = a−1 b−1 gba = θba (g). Thus, θa ◦θb = θba . (3) If g ∈ G, then θa (θa−1 (g)) = θa (aga−1 ) = a−1 aga−1 a = g; thus, θa ◦ θa−1 is \u0002 the identity function. Similarly, so is θa−1 ◦ θa . Theorem 4.23. For any group G, Inn(G) is a normal subgroup of Aut(G). Proof. By the preceding lemma, Inn(G) ⊆ Aut(G). Certainly θe ∈ Inn(G) is the identity automorphism. Also, the preceding lemma shows that Inn(G) is closed under composition and the taking of inverses. Therefore, Inn(G) ≤ Aut(G). To show normality, take α ∈ Aut(G) and θa ∈ Inn(G). Then (α −1 ◦ θa ◦ α)(g) = α −1 (θa (α(g))) = α −1 (a−1 α(g)a) = α −1 (a−1 )α −1 (α(g))α −1 (a) = (α −1 (a))−1 gα −1 (a) = θα−1 (a) (g), for all g ∈ G. That is, α −1 ◦ θa ◦ α = θα−1 (a) ∈ Inn(G), and Inn(G) is normal.\n\n\u0002\n\nIt is certainly possible for Aut(G) to be larger than G; indeed, Example 4.26 provides such a group. But there is only one inner automorphism for each group element. However, θa does not have to be different from θb if a \u0006= b. For instance, if a and b are both central, then θa and θb are both equal to the identity automorphism. The following theorem tells the tale. Theorem 4.24. Let G be a group. Then 1. if a, b ∈ G, then θa = θb if and only if ba−1 ∈ Z(G); and 2. G/Z(G) is isomorphic to Inn(G). Proof. (1) Take any a, b ∈ G. Then θa = θb if and only if a−1 ga = b−1 gb for all g ∈ G. But this occurs if and only if ba−1 g = gba−1 for all g ∈ G. In other words, if and only if ba−1 is central. (2) Define α : G → Inn(G) via α(a) = θa−1 . Let us show that α is a homomorphism. If a, b ∈ G, then α(ab) = θ(ab)−1 = θb−1 a−1 = θa−1 ◦ θb−1 = α(a)α(b), making use of Lemma 4.2. Also, if θa ∈ Inn(G), then α(a−1 ) = θa , so α is onto. Furthermore, a ∈ ker(α) if and only if θa = θe . By (1), this happens if and only if \u0002 a = ae−1 ∈ Z(G). Now apply the First Isomorphism Theorem. Example 4.27. As the centre of S3 is trivial, we see that Inn(S3 ) is isomorphic to S3 .\n\n84\n\n4 Factor Groups and Homomorphisms\n\nExample 4.28. As Z(D8 ) = \u0004R180 \u0005, the distinct elements of Inn(D8 ) are of the form θa , where we take one a for each left coset of \u0004R180 \u0005 in D8 . That is, Inn(D8 ) = {θR0 , θR90 , θF1 , θF3 }. In particular, it is a group of order 4, so by Corollary 4.3, it is isomorphic to either Z4 or Z2 × Z2 . But every flip in D8 already has order 2, and the square of every rotation is in Z(D8 ). Therefore, we see that there is no element of order 4 in D8 /Z(D8 ); thus, it must be isomorphic to Z2 × Z2 . Exercises 4.47. Let G be an abelian group of order n, and let m be a positive integer relatively prime to n. Show that α : G → G given by α(a) = am is an automorphism of G. 4.48. Let G be a group with automorphism α and H a group with automorphism β. Show that γ : G × H → G × H given by γ ((g, h)) = (α(g), β(h)) is an automorphism. 4.49. Show that the automorphism group of Z2 × Z2 is isomorphic to D6 . 4.50. Let G and H be isomorphic groups. Show that their automorphism groups are also isomorphic. 4.51. Let α be an automorphism of G. Show that {a ∈ G : α(a) = a} is a subgroup of G. 4.52. Let α and β be any two automorphisms of G. Show that {a ∈ G : α(a) = β(a)} is a subgroup of G. 4.53. For any group G, an automorphism α of G is said to be a power automorphism if α(H ) ⊆ H for every subgroup H of G. If G = \u0004a\u0005 × \u0004b\u0005 is the direct product of two cyclic groups, and α is a power automorphism of G, show that there exists a k ∈ Z such that α(g) = g k for all g ∈ G. 4.54. To what familiar group is the inner automorphism group of D12 isomorphic? 4.55. Let α be an automorphism of Q. Show that for every q ∈ Q, we have α(q) = qα(1). 4.56. Let G be a group such that the automorphism group of G is trivial. 1. Show that G is abelian. 2. Show that a2 = e for every a ∈ G.\n\nChapter 5\n\nDirect Products and the Classification of Finite Abelian Groups\n\nWe can now determine the structure of finite abelian groups. In particular, every such group is isomorphic to a direct product of cyclic groups, each having prime power order. The proof of this result is our main goal in the present chapter.\n\n5.1 Direct Products We defined the direct product of two groups in Definition 3.3. There is no particular reason that we need to restrict ourselves to two. Definition 5.1. Let G 1 , . . . , G k be any groups. Then the (external) direct product G 1 × G 2 × · · · × G k is the Cartesian product of the groups G i under the operation (a1 , . . . , ak )(b1 , . . . , bk ) = (a1 b1 , . . . , ak bk ), for all ai , bi ∈ G i . (We allow k = 1 here, in which case G = G 1 .) Theorem 5.1. If G 1 , . . . , G k are groups, then G 1 × · · · × G k is a group. Proof. The proof is essentially identical to that of Theorem 3.1.\n\n\u0002\n\nThe reason we used the word “external” in the above definition is that the groups G i are not subgroups of the direct product; indeed, they are not even subsets. However, G 1 is, for instance, isomorphic in a natural way to G 1 × {e} × · · · × {e}, which is a subgroup of the direct product. What we would like is a way of showing that a group is isomorphic to the direct product of certain subgroups. To this end, let us consider the following. Definition 5.2. Let G be a group, and let N1 , . . . , Nk be subgroups of G. Then we say that G is the internal direct product of N1 , . . . , Nk if © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_5\n\n85\n\n86\n\n5 Direct Products and the Classification of Finite Abelian Groups\n\n1. each Ni is normal; 2. N1 N2 · · · Nk = G; and 3. for each i, 1 ≤ i < k, we have (N1 N2 · · · Ni ) ∩ Ni+1 = {e}. (Again, we allow k = 1, in which case G = N1 .) In particular, G is the internal direct product of normal subgroups N1 and N2 if and only if N1 N2 = G and N1 ∩ N2 = {e}. Example 5.1. Let G = Z20 , N1 = \u00054\u0006 and N2 = \u00055\u0006. As G is abelian, every subgroup is normal. Also, N1 = {0, 4, 8, 12, 16} and N2 = {0, 5, 10, 15}. Thus, N1 ∩ N2 = {0}. For each a ∈ G, we could find n 1 ∈ N1 and n 2 ∈ N2 such that a = n 1 + n 2 but, in fact, we can avoid this by noting that |N1 + N2 | = |N1 ||N2 |/|N1 ∩ N2 | = 5·4/1 = 20 (see Theorem 4.4). Thus, N1 + N2 = G, and G is the internal direct product of N1 and N2 . Note that if there are more than two groups, then we need to check more than just that each Ni ∩ N j = {e} for the final part of the definition. Example 5.2. Let G = Z30 , N1 = \u000515\u0006, N2 = \u000510\u0006 and N3 = \u00056\u0006. Again, normality is not an issue. It is easy to see that N1 ∩ N2 = {0}. Thus, |N1 + N2 | = |N1 ||N2 | = 2 · 3 = 6. As every element of N1 + N2 is in \u00055\u0006, we see immediately that N1 + N2 = \u00055\u0006. But now we observe that (N1 + N2 ) ∩ N3 = {0}. Then the same argument shows that |N1 + N2 + N3 | = 30, and we know that N1 + N2 + N3 = G. Therefore, G is the internal direct product of N1 , N2 and N3 . Let us see how internal direct products behave. Here are some highly useful facts. Lemma 5.1. Let G be a group with normal subgroups K and N . If K ∩ N = {e}, then kn = nk for all k ∈ K , n ∈ N . Proof. Let h = (nk)−1 (kn) = k −1 n −1 kn. As K is normal, n −1 kn ∈ K , so h ∈ K . As N is normal, k −1 n −1 k ∈ N , so h ∈ N . Since K ∩ N = {e}, we have (nk)−1 (kn) = e, and therefore kn = nk, as required. \u0002 Lemma 5.2. If G is the internal direct product of N1 , . . . , Nk , then every element of G can be written in exactly one way as n 1 n 2 · · · n k , with each n i ∈ Ni . Proof. Since G = N1 · · · Nk , we know that every element of G can be written in such a way. We only need to show uniqueness. Our proof is by induction on k. If k = 1, there is nothing to do, as G = N1 . Assume that k > 1 and the result holds for groups written as an internal direct product of a smaller number of subgroups. Suppose that n 1 · · · n k−1 n k = h 1 · · · h k−1 h k , with n i , h i ∈ Ni . Then −1 h k n −1 k = (h 1 · · · h k−1 ) (n 1 · · · n k−1 ) ∈ Nk ∩ (N1 · · · Nk−1 ) = {e}.\n\nTherefore, n k = h k , and we have n 1 · · · n k−1 = h 1 · · · h k−1 in N1 N2 · · · Nk−1 , which is an internal direct product of k − 1 subgroups. By our inductive hypothesis, n i = h i for all i. \u0002\n\n5.1 Direct Products\n\n87\n\nExample 5.3. As we saw in Example 5.2, Z30 is the internal direct product of \u000515\u0006, \u000510\u0006 and \u00056\u0006. Note, for instance, that 23 = 15 + 20 + 18. By the above lemma, there is no other way to write 23 as a sum of elements in \u000515\u0006, \u000510\u0006 and \u00056\u0006. And now, the big reason why we are interested in these internal direct products. Theorem 5.2. Let G be a group, and suppose that it is the internal direct product of normal subgroups N1 , . . . , Nk . Then G is isomorphic to the external direct product N1 × · · · × Nk . Proof. Define α : N1 × · · · × Nk → G via α((n 1 , . . . , n k )) = n 1 · · · n k . We claim that α is an isomorphism. In view of Lemma 5.2, α is bijective. Thus, it remains to show that it is a homomorphism. Take n i , h i ∈ Ni . Then α((n 1 , . . . , n k )(h 1 , . . . , h k )) = α((n 1 h 1 , . . . , n k h k )) = n 1 h 1 n 2 h 2 n 3 h 3 · · · n k h k . As N1 and N2 are normal subgroups, and N1 ∩ N2 = {e}, Lemma 5.1 says that h 1 n 2 = n 2 h 1 . Thus, n1h1n2 h2n3h3 · · · nk hk = n1n2 h1h2n3h3 · · · nk hk . By Theorem 4.5, N1 N2 is a normal subgroup of G, and we know that (N1 N2 ) ∩ N3 = {e}. Therefore, h 1 h 2 n 3 = n 3 h 1 h 2 . We now have n1h1n2 h2n3h3 · · · nk hk = n1n2n3h1h2 h3n4h4 · · · nk hk . Repeating this procedure, we find that α((n 1 , . . . , n k )(h 1 , . . . , h k )) = n 1 n 2 · · · n k h 1 h 2 · · · h k = α((n 1 , . . . , n k ))α((h 1 , . . . , h k )).\n\nThus, α is a homomorphism, and the proof is complete.\n\n\u0002\n\nAs a result of this theorem, we will engage in a small abuse of notation and write G = N1 × N2 × · · · × Nk when G is the internal direct product of N1 , . . . , Nk , as well as for the external direct product. Example 5.4. By Example 5.2, Z30 = \u000515\u0006 × \u000510\u0006 × \u00056\u0006. Example 5.5. We claim that U (8) = \u00053\u0006 × \u00057\u0006. As the group is abelian, normality is not an issue. Also, |3| = |7| = 2, so the intersection of these cyclic subgroups must be trivial. Furthermore, 1 = 1 · 1, 3 = 3 · 1, 7 = 1 · 7 and 5 = 3 · 7, so U (8) = \u00053\u0006\u00055\u0006 (or just use an order argument). Thus, we have an internal direct product. Exercises 5.1. Write U (32) as the internal direct product of two proper subgroups.\n\n88\n\n5 Direct Products and the Classification of Finite Abelian Groups\n\n5.2. Let G = H × K . If h ∈ H has order m and k ∈ K has order n, what is the order of (h, k)? 5.3. How many elements of order 5 are there in Z5 × Z25 ? How many elements of order 25? 5.4. How many cyclic subgroups of order 5 are there in Z5 × Z25 ? How many cyclic subgroups of order 25? 5.5. Show that D8 is not the internal direct product of two proper subgroups. 5.6. Let |a| = 4 and |b| = 2. Write \u0005a\u0006 × \u0005b\u0006 as the internal direct product of two proper subgroups in every possible way. 5.7. Show that in Definition 5.2, it is not sufficient to replace the third condition with the stipulation that Ni ∩ N j = {e} whenever i \b= j. In particular, find a group G with normal subgroups N1 , N2 and N3 such that N1 N2 N3 = G and Ni ∩ N j = {e} whenever i \b= j, but G \b= N1 × N2 × N3 . 5.8. Let G = \u0005a\u0006 be cyclic of order 84. Show that G = \u0005a 12 \u0006 × \u0005a 21 \u0006 × \u0005a 28 \u0006. 5.9. Suppose that G = N1 × N2 is an internal direct product. If α : G → H is an onto homomorphism, does it follow that H = α(N1 ) × α(N2 )? Prove that it does, or give an explicit counterexample. 5.10. Let G be a group having finite normal subgroups N1 , . . . , Nk , such that the gcd of |Ni | and |N j | is 1 whenever i \b= j. Show that N1 N2 · · · Nk = N1 × N2 × · · · × Nk .\n\n5.2 The Fundamental Theorem of Finite Abelian Groups Let us now classify the finite abelian groups. We will break our proof down into stages. For the first stage, we need a definition. Definition 5.3. Let p be a prime number. Furthermore, let G be a group and a ∈ G. We say that a is a p-element if the order of a is p n for some integer n ≥ 0. If every element of G is a p-element, then G is a p-group. Example 5.6. The dihedral group D8 is a 2-group, as every element has order 1, 2 or 4. On the other hand, Z24 is not a p-group. Indeed, 12 and 18 are both 2-elements and 8 is a 3-element, so it cannot be a p-group. In fact, 1 is not a p-element, for any prime p. Lemma 5.3. Let p be a prime and G an abelian group. Then the p-elements of G form a subgroup.\n\n5.2 The Fundamental Theorem of Finite Abelian Groups\n\n89\n\nProof. Let H be the set of all p-elements of G. As e has order p 0 , we have e ∈ H . Let a, b ∈ H . Then say that |a| = p n and |b| = p m . Let k be the larger of m and k k k n. Then as G is abelian, (ab) p = a p b p = e2 = e, as |a| and |b| both divide p k . k Thus, |ab| divides p , and therefore ab ∈ H . Finally, if a ∈ H , then |a| = |a −1 |, so \u0002 a −1 ∈ H . Thus, H is indeed a subgroup of G. Note that the\u0002 preceding \u0003 lemma \u0002 does \u0003 not work for nonabelian groups. Indeed, \u0002 in S\u00033 , 123 123 123 we can see that and both have order 2, but their product, , 213 132 231 has order 3. The following result is also very handy. Lemma 5.4. Let G be any group and let e \b= a ∈ G be such that a has finite order. Then a = a n 1 a n 2 · · · a n k for some integers n 1 , . . . , n k , where each a ni is a pi -element, for some prime pi dividing |a|. Proof. Our proof is by induction on the number of distinct primes, l, dividing |a|. If l = 1, then a is a p-element, so just let n 1 = 1. Suppose that l > 1 and that the result is true for smaller values of l. Let p be a prime dividing |a|, and say that |a| = p m q, with ( p, q) = 1. By Corollary 2.1, there exist u, v ∈ Z such that p m u + qv = 1. Then m m a = a 1 = a p u+qv = (a p )u (a q )v . m\n\nm\n\nNow, (a q ) p = a p q = e; hence, a q is a p-element and so is (a q )v . So, let p1 = p and m n 1 = qv. Similarly, the order of (a p )u divides q, and q has fewer primes dividing m it than |a|. Thus, by our inductive hypothesis, a up can be written as a product of powers (which are also powers of a) in the manner stated in the theorem. The proof is complete. \u0002 We can now simplify our task by breaking a finite abelian group down into a direct product of p-groups. Lemma 5.5. Let G be a nontrivial finite abelian group, and let p1 , . . . , pk be the distinct primes dividing |G|. Then G = H1 × H2 ×· · ·× Hk , where Hi is the subgroup of G consisting of all of the pi -elements of G. Proof. Lemma 5.3 tells us that the Hi are subgroups and, as G is abelian, we do not have to worry about normality. Let us show that G = H1 H2 · · · Hk . But taking any a ∈ G, we see from Lemma 5.4 that a can be written as a product of elements from various Hi . (If a = e, there is obviously nothing to worry about.) Finally, we must show that for each i, 1 ≤ i < k, we have (H1 · · · Hi ) ∩ Hi+1 = {e}. But suppose that m a ∈ Hi+1 and, simultaneously, a = a1 · · · ai , with a j ∈ H j . Then letting |a j | = p j j , mi m1 m m m and m = p1 · · · pi , we have a = a1 · · · ai , and since each |a j | divides m, we conclude that a m = e. Thus, |a| divides m. But also, a is a pi+1 -element. As (m, pi+1 ) = 1, the only possible conclusion is that a = e, and we have an internal direct product. \u0002\n\n90\n\n5 Direct Products and the Classification of Finite Abelian Groups\n\nWe can now focus our attention on finite abelian p-groups. The following lemma does the biggest part of the work. It is the most difficult proof we have encountered so far, and will take some time to absorb. Lemma 5.6. Let G be a finite abelian p-group, and let a ∈ G be an element of largest possible order. Then G = \u0005a\u0006 × H , for some subgroup H of G. Proof. Our proof is by strong induction on |G|. If |G| = 1, then a = e and using H = \u0005e\u0006 will work. So, assume that |G| > 1 and that the lemma holds for groups of smaller order. Let |a| = p n , with n a positive integer. If \u0005a\u0006 = G, then we can use H = \u0005e\u0006, so assume that \u0005a\u0006 \b= G. Take b ∈ G such that b ∈ / \u0005a\u0006. As b is a p-element, we k know that b p = e ∈ \u0005a\u0006, for some positive integer k. Let m be the smallest positive m m−1 m / \u0005a\u0006, but c p = b p ∈ \u0005a\u0006. In integer such that b p ∈ \u0005a\u0006, and let c = b p . Then c ∈ particular, let us say that c p = a i , with i ∈ Z. Now, as G is a p-group, and the largest element order is p n , we must have pn c = e. Thus, |c p | divides p n−1 . Suppose that ( p, i) = 1. Then by Corollary 3.2, |c p | = |a i | = p n , which is impossible. Thus, p divides i; let us say that i = pj. Then let d = a − j c. Note that a j ∈ \u0005a\u0006; thus, if d ∈ \u0005a\u0006, then c = a j d ∈ \u0005a\u0006, which is a contradiction. Therefore, d ∈ / \u0005a\u0006. However, d p = a − j p c p = (a i )−1 c p = e; thus, |d| = p. Now, let us consider the group M = G/\u0005d\u0006. (As G is abelian, we do not have to worry about \u0005d\u0006 being normal.) We note that M is still abelian (by Theorem 4.7), its order is [G : \u0005d\u0006] = |G|/ p and it is a p-group with the orders of elements dividing orders of elements of G (by Theorem 4.7). Also, we claim that |a\u0005d\u0006| = p n . As n−1 n−1 its order must divide p n , suppose that a p ∈ \u0005d\u0006. Since a p \b= e, we must have n−1 a p = d s , with 0 < s < p. But then (s, p) = 1, so by Corollary 2.1, there exist n−1 u, v ∈ Z such that su + pv = 1. Thus, d = d su+ pv = (d s )u (d p )v = a p u e ∈ \u0005a\u0006, n giving us a contradiction. Therefore, |a\u0005d\u0006| = p , as claimed. It now follows that a\u0005d\u0006 is an element of largest order in M. As M is an abelian p-group of smaller order than G, our inductive hypothesis tells us that there is a subgroup K of M such that M = N × K , where N is the subgroup of M generated by a\u0005d\u0006. By Theorem 4.8, K = H/\u0005d\u0006, where H is a subgroup of G containing \u0005d\u0006. We claim that G = \u0005a\u0006× H . Normality is not an issue. Suppose that a i ∈ \u0005a\u0006∩ H . Then a i \u0005d\u0006 ∈ N ∩K , and as the product N ×K is direct, this means that a i \u0005d\u0006 = e\u0005d\u0006. But we demonstrated above that the order of a\u0005d\u0006 is p n , which means that p n divides i, and therefore a i = e. Thus, \u0005a\u0006 ∩ H = {e}. Now, take any g ∈ G. Then as M = N × K , we have g\u0005d\u0006 = x y, for some x ∈ N , y ∈ K . Let us write x = a t \u0005d\u0006 and y = w\u0005d\u0006, with t ∈ Z and w ∈ H . Then g = a t wd l , for some l ∈ Z. As a t ∈ \u0005a\u0006 and wd l ∈ H , we now see that \u0005a\u0006H = G. Thus, we have the required direct product, and our proof is complete. \u0002 And now, the payoff for our hard work! Theorem 5.3 (Fundamental Theorem of Finite Abelian Groups). Let G be a finite abelian group. Then G is the direct product of subgroups, H1 × · · · × Hk , with\n\n5.2 The Fundamental Theorem of Finite Abelian Groups\n\n91\n\neach Hi cyclic of order pini , where the pi are (not necessarily distinct) primes, and the n i are nonnegative integers. Proof. If G is the trivial group, there is nothing to do. Otherwise, by Lemma 5.5, G is the direct product of p-subgroups. Therefore, we may as well assume that G is a finite abelian p-group. Our proof is by strong induction on |G|. If |G| = 1, again, there is nothing to do, so let G be nontrivial and suppose that our theorem holds for groups of smaller order. Let a be an element of largest possible order in G. Then by Lemma 5.6, G = \u0005a\u0006 × H , for some subgroup H . But then |H | = |G|/|a|, so H has smaller order, and by our inductive hypothesis, H is a direct product of cyclic groups of prime power order. However, \u0005a\u0006 is also a cyclic group of prime power order, and we are done. \u0002 We can express this slightly differently. Corollary 5.1. Let G be a nontrivial finite abelian group. Then G is isomorphic to Z p1n1 × Z p2n2 × · · · × Z pknk , where the pi are some (not necessarily distinct) primes, and the n i are positive integers. Proof. Combine Theorems 5.2 and 5.3 with Theorem 4.14.\n\n\u0002\n\nExample 5.7. Up to isomorphism, the abelian groups of order 16 are Z16 , Z8 × Z2 , Z4 × Z4 , Z4 × Z2 × Z2 and Z2 × Z2 × Z2 × Z2 . Example 5.8. Note that U (32) is an abelian group of order ϕ(32) = 16, so it must be isomorphic to one of the groups in the preceding example. But which one? Examining the orders of the elements, we find that there is no element of order 16, so it is not Z16 . However, |3| = 8. As none of the other groups in the preceding example have an element of order 8, U (32) is isomorphic to Z8 × Z2 . Example 5.9. As 200 = 23 52 , the finite abelian groups of order 200 are all isomorphic to one of the following, namely Z8 × Z25 , Z4 × Z2 × Z25 , Z2 × Z2 × Z2 × Z25 , Z8 × Z5 × Z5 , Z4 × Z2 × Z5 × Z5 and Z2 × Z2 × Z2 × Z5 × Z5 . We might be momentarily concerned about the absence of Z200 in the preceding example. However, it is isomorphic to Z8 × Z25 , as the following theorem shows us. Theorem 5.4. Let G = H1 × · · · × Hk , where each Hi is cyclic of order n i . Then G is cyclic if and only if (n i , n j ) = 1 whenever i \b= j. Proof. Let Hi = \u0005ai \u0006. If the n i are all relatively prime, then we claim that (a1 , . . . , ak ) has order n 1 · · · n k = |G|, and therefore G is cyclic. Suppose that (a1 , . . . , ak )m = (e, . . . , e). Then each aim = e, so n i |m. As the n i are relatively prime, n 1 · · · n k |m, by Corollary 2.3. Since |G| = n 1 · · · n k , the largest possible order of an element is n 1 · · · n k , and the claim is proved. On the other hand, suppose that the n i are not relatively prime. Without loss of generality, say that some prime p divides both n 1 and n 2 . Then for any ri ∈ Z, we have (a1r1 , . . . , akrk )n 1 ···n k / p = (e, . . . , e), since each n i divides n 1 · · · n k / p. (For i = 1, we\n\n92\n\n5 Direct Products and the Classification of Finite Abelian Groups\n\nhave n 1 (n 2 / p)n 3 · · · n k , and for i ≥ 2, we have (n 1 / p)n 2 n 3 · · · n k .) Thus, every element of G has order dividing n 1 n 2 · · · n k / p, and therefore there is no element of order |G|, so G is not cyclic. \u0002 As a result of our classification, we can prove a special case of a famous result due to Augustin-Louis Cauchy. Theorem 5.5 (Cauchy’s Theorem for Abelian Groups). Let G be a finite abelian group, and suppose that p is a prime dividing |G|. Then G has an element of order p. Proof. If |G| is divisible by a prime, then G is not the trivial group. Letting G be as in Corollary 5.1, we see that |G| = p1n 1 p2n 2 · · · pkn k . If p divides |G|, then p = pi , for some i. But then G has a subgroup isomorphic to Z pni , for some n i > 0. However, \u0002 in Z pni , the element p ni −1 has order p. The proof is complete. Corollary 5.2. A finite abelian p-group has order p n , for some n ≥ 0. Proof. Let G be a finite abelian p-group. If the corollary is false, then the order of G is divisible by q, for some prime q \b= p. But then G has an element of order q, which is impossible. \u0002 Exercises 5.11. Give a list of abelian groups of each of the following orders, such that every abelian group of that order is isomorphic to one of the groups in the list. 1. 21 2. 81 3. 9800 5.12. Give a list of abelian groups of each of the following orders, such that every abelian group of that order is isomorphic to one of the groups in the list. 1. 144 2. 243 3. 55125 5.13. Write U (56) as an external direct product of cyclic groups of prime power order, as in Corollary 5.1. 5.14. Write (Z20 × Z6 )/\u0005(10, 2)\u0006 as an external direct product of cyclic groups of prime power order, as in Corollary 5.1. 5.15. Let p be a prime. Suppose that G is a nontrivial finite abelian group in which every element has order 1 or p. Show that G is isomorphic to a group of the form Zp × Zp × · · · × Zp.\n\n5.2 The Fundamental Theorem of Finite Abelian Groups\n\n93\n\n5.16. Suppose that n is an integer that is a product of distinct primes. If G is a finite abelian group, and |G| is divisible by n, show that G has a cyclic subgroup of order n. 5.17. If \u0005a\u0006 is a cyclic group of order 35, write a as the product of a 5-element and a 7-element. 5.18. If \u0005a\u0006 is a cyclic group of order 90, write a as the product of p-elements, for various primes p. 5.19. Prove Theorem 5.5 in a different way, as follows. Let p be a prime dividing |G|. Show that G has an element a of some prime order, say q. If q = p, we are done. Otherwise, what can be said about G/\u0005a\u0006? Complete the proof. 5.20. Let G be a finite abelian group and let n be a positive integer dividing |G|. Show that G has a subgroup of order n.\n\n5.3 Elementary Divisors and Invariant Factors For any positive integer n, we now know all possible abelian groups of order n, up to isomorphism. Indeed, we determine the prime factorization of n, and then proceed as in Examples 5.7 and 5.9. But we have not yet made certain that the groups we found are not isomorphic to each other. Let us work on that. Definition 5.4. Let G be a nontrivial finite abelian group, and say that G = H1 × H2 × · · · × Hk , where each Hi is cyclic of order pini , for some prime pi and positive integer n i . Then the elementary divisors of G are the numbers p1n 1 , p2n 2 , . . . , pkn k , where the order in this list is irrelevant, but each number must be listed as many times as it occurs. The trivial group has no elementary divisors. Example 5.10. The elementary divisors of Z9 × Z9 × Z3 × Z125 are 9, 9, 3, 125. Example 5.11. To find the elementary divisors of Z300 × Z3 , we use Theorem 5.4 to see that the group is isomorphic to Z25 × Z4 × Z3 × Z3 , so the elementary divisors are 4, 3, 3, 25. Definition 5.5. Let G be an abelian group and n a positive integer. Then we write G n = {a n : a ∈ G}. Lemma 5.7. Let G and H be abelian groups and n a positive integer. Then 1. G n is a subgroup of G; and. 2. if α : G → H is an onto homomorphism, then α(G n ) = H n . Proof. (1) See Exercise 3.40. (2) If g n ∈ G n , then α(g n ) = (α(g))n ∈ H n . Also, if h n ∈ H n , then as α is onto, write h = α(g), with g ∈ G. Then h n = (α(g))n = α(g n ) ∈ α(G n ), completing the proof. \u0002\n\n94\n\n5 Direct Products and the Classification of Finite Abelian Groups\n\nThe elementary divisors are very important, as they uniquely determine a finite abelian group, up to isomorphism. Theorem 5.6. Let G and H be finite abelian groups. Then G and H are isomorphic if and only if they have the same elementary divisors. Proof. If G and H have the same elementary divisors, then each is isomorphic to a direct product of cyclic groups, and the groups appearing in the direct product in G have the same order as those appearing in H , so they are isomorphic. (We must be a bit careful, as the cyclic groups may appear in a different order in the direct product, but M × N is always isomorphic to N × M, so this is not a problem. See Exercise 4.36.) Note that if neither G nor H has any elementary divisors, then each is the trivial group, so they are isomorphic. On the other hand, let α : G → H be an isomorphism. Take any prime p. Now, by Lemma 5.3, the p-elements of G form a subgroup, as do those of H . Furthermore, as isomorphisms preserve the orders of group elements, α provides an isomorphism from one of these p-subgroups to the other. As the elementary divisors come from these p-subgroups, we may as well assume to begin with that G and H are both p-groups. We proceed by strong induction on |G|. If |G| = 1, then G and H are both the trivial group, so neither has elementary divisors. Therefore, assume that |G| > 1 and the result holds for groups of smaller order. In particular, say G = G 1 × · · · × G k and H = H1 × · · · × Hl , where G i = \u0005gi \u0006 is cyclic of order p ni , and Hi = \u0005h i \u0006 is cyclic of order p m i . Rearranging the terms if necessary, we may assume that n 1 ≥ n 2 ≥ · · · ≥ n k > 0 and m 1 ≥ m 2 ≥ · · · ≥ m l > p p p p 0. By the above lemma, α(G p ) = H p . Thus, α(G 1 × · · · × G k ) = H1 × · · · × Hl . p p p But G i = \u0005gi \u0006, and since |gi | = p ni , we have |gi | = p ni −1 , by Corollary 3.2. p Similarly, |h i | = p m i −1 . Thus, G p is a p-group of strictly smaller order than G, and by our inductive hypothesis, the elementary divisors of G p and H p are the same. But the elementary divisors of G p are p n 1 −1 , p n 2 −1 , . . . , p nr −1 , where n r > 1 but n u = 1 whenever u > r . (When n u = 1, we have p n u −1 = 1, which does not count as an elementary divisor. If n 1 = 1, then G p has no elementary divisors.) Similarly, the elementary divisors of H p are p m 1 −1 , . . . , p m s −1 , where m s > 1 but m v = 1 whenever v > s. Therefore, r = s and m i − 1 = n i − 1 whenever i ≤ r . But then m i = n i , for all i ≤ r . Also, n i = 1 for all i > r and m i = 1 for all i > s. In order to prove that G and H have the same elementary divisors, it remains only to show that k = l. But |G| = p n 1 · · · p nr p k−r and |H | = p n 1 · · · p nr pl−r . As isomorphic groups have the same order, p k−r = pl−r , and therefore k = l. If G p has no elementary \u0002 divisors, then neither does H p , and we simply get p k = pl , hence k = l. Example 5.12. The five abelian groups of order 16 listed in Example 5.7 are all nonisomorphic, as they have different elementary divisors. Similarly for the six abelian groups of order 200 given in Example 5.9. Example 5.13. Let G = Z200 × Z8 × Z6 , H = Z120 × Z10 × Z4 × Z2 and K = Z25 × Z24 × Z8 × Z2 . These are all abelian groups of order 9600. However, using Theorem 5.4, we see that G is isomorphic to Z8 × Z25 × Z8 × Z3 × Z2 , so its\n\n5.3 Elementary Divisors and Invariant Factors\n\n95\n\nelementary divisors are 8, 8, 2, 3, 25. Similarly, H is isomorphic to Z3 × Z8 × Z5 × Z5 ×Z2 ×Z4 ×Z2 , so its elementary divisors are 8, 4, 2, 2, 3, 5, 5 and K is isomorphic to Z25 × Z3 × Z8 × Z8 × Z2 , so its elementary divisors are 8, 8, 2, 3, 25. Therefore, G and K are isomorphic, but H is not isomorphic to either of them. There is another interesting way to express a finite abelian group as a direct product of cyclic groups. Theorem 5.7 (Invariant Factor Decomposition). Suppose that G is a nontrivial finite abelian group. Then G = H1 × H2 × · · · × Hk , where each Hi is a cyclic subgroup of G of order m i , with m 1 > 1 and m i |m i+1 , for 1 ≤ i < k. Proof. We will explain how to construct the Hi , assuming that G has been expressed as a direct product of cyclic groups of prime power order, as in Corollary 5.1. Let n p1 , . . . , pr be the primes dividing |G|. For each j, find the largest power p j j such that n n 1 2 n Z pn j appears in Corollary 5.1. Letting m k = p1 p2 · · · pr r , Theorem 5.4 says that j Hk = Z p1n1 × · · · × Z prnr is isomorphic to Zm k . Now, delete all of the terms from the direct product in Corollary 5.1 that we have used (deleting only one copy, if multiple s copies of the same group appear). For each j, let p j j be the largest power appearing in the remaining terms (where s j = 0 is entirely possible). Let m k−1 = p1s1 · · · prsr . By construction, each s j ≤ n j , so m k−1 |m k . Again, Hk−1 = Z p1s1 × · · · × Z prsr is isomorphic to Zm k−1 . Delete all of these terms that we have just used, and repeat until we exhaust the entire direct product in Corollary 5.1. \u0002 Definition 5.6. If G is isomorphic to Zm 1 × · · · × Zm k , where m 1 > 1 and m i |m i+1 , for 1 ≤ i < k, then the numbers m 1 , . . . , m k are called the invariant factors of G. Example 5.14. Let us use our work in Example 5.9 to find the invariant factors of the abelian groups of order 200. We apply the method from Theorem 5.7. Considering Z4 × Z2 × Z25 , we see that the highest power of 2 that appears is 4, and the highest power of 5 is 25. Therefore, m k = 4 · 25 = 100. Deleting Z4 and Z25 , we are left with Z2 , so m k−1 = 2, and we are finished. Thus, our group is isomorphic to Z2 × Z100 , so the invariant factors are 2, 100. When we examine Z2 × Z2 × Z2 × Z5 × Z5 , we see that m k = 2 · 5 = 10. Deleting Z2 and Z5 , we are left with Z2 × Z2 × Z5 . Thus, m k−1 = 2 · 5 = 10. Deleting Z2 and Z5 , we are left only with Z2 . Thus, m k−2 = 2, and we are finished. Therefore, our group is isomorphic to Z2 × Z10 × Z10 , which gives invariant factors of 2, 10, 10. Considering Z8 × Z25 , we simply get Z200 , so 200 is the only invariant factor. Looking at Z2 × Z2 × Z2 × Z25 , we have Z2 × Z2 × Z50 , so the invariant factors are 2, 2, 50. When we examine Z8 × Z5 × Z5 , we obtain Z5 × Z40 , so the invariant factors are 5, 40. Finally, if we take Z4 × Z2 × Z5 × Z5 , then we get Z10 × Z20 , so the invariant factors are 10, 20. In the above example, the nonisomorphic groups produced different lists of invariant factors. As it turns out, this always happens. Theorem 5.8. Let G and H be nontrivial finite abelian groups. Then G and H are isomorphic if and only if they have the same invariant factors.\n\n96\n\n5 Direct Products and the Classification of Finite Abelian Groups\n\nProof. Let G be isomorphic to Zm 1 ×· · ·×Zm k with m 1 > 1 and m i |m i+1 , 1 ≤ i < k. Similarly, write H as Zn 1 × · · · × Znl , with n 1 > 1 and n i |n i+1 , 1 ≤ i < l. If G and H have the same invariant factors, then they are both isomorphic to the same direct product, and therefore to each other. On the other hand, suppose that G and H are isomorphic. We will show that they have the same invariant factors. Our proof is by strong induction on |G|. If |G| = 2, then the only possible invariant factor list is 2 for both G and H , so there is nothing to do. Assume that |G| > 2 and that the result is true for groups of smaller order. If we take (g1 , . . . , gk ) ∈ G, then each gi has order dividing m i , and therefore all gi have order dividing m k . On the other hand (0, 0, . . . , 0, 1) has order m k . Thus, m k is the largest possible order of an element of G. Similarly, n l is the largest possible order of any element of H . Therefore, as isomorphisms preserve orders of group elements, m k = n l . Now, expressing each m i as a product of prime powers, we note that the elementary divisors of G are those that come from Zm 1 ×· · ·×Zm k−1 together with those from Zm k . Similarly, the elementary divisors of H are those coming from Zn 1 × · · · × Znl−1 together with those from Znl = Zm k . As G and H are isomorphic, Theorem 5.6 tells us that they have the same elementary divisors. Deleting those from Zm k , the groups Zm 1 × · · · × Zm k−1 and Zn 1 × · · · × Znl−1 have the same elementary divisors. Thus, by Theorem 5.6, these groups are isomorphic. As they have smaller order than G, our inductive hypothesis tells us that k − 1 = l − 1 and each m i = n i . Therefore, the invariant factors are identical. (We have to be a bit careful if either k = 1 or l = 1, as then we have nothing left when we remove the term Zm k or Znl . But in this case, comparing orders, we must \u0002 have k = l = 1, and the only invariant factor is m 1 for both groups.) Exercises 5.21. Find the elementary divisors for each of the following groups. 1. Z42 × Z4200 2. Z6 × Z18 × Z54 5.22. Find the invariant factors for each of the following groups. 1. Z3 × Z3 × Z9 × Z25 × Z11 × Z121 2. Z4 × Z8 × Z8 × Z16 × Z5 × Z25 × Z49 5.23. Let p, q and r be distinct primes. Give the list of elementary divisors for every possible abelian group of order p 3 q 2 r . 5.24. Let p, q and r be distinct primes. Give the list of invariant factors for every possible abelian group of order p 3 q 2 r . 5.25. For which positive integers n are all abelian groups of order n isomorphic? 5.26. Find the smallest positive integer n such that there are exactly four nonisomorphic abelian groups of order n.\n\n5.3 Elementary Divisors and Invariant Factors\n\n97\n\n5.27. Let G 1 , G 2 and G 3 be finite abelian groups, and suppose that G 1 × G 2 is isomorphic to G 1 × G 3 . Show that G 2 and G 3 are isomorphic. 5.28. Let a finite abelian group G have invariant factors n 1 , n 2 , . . . , n k . What are the invariant factors of G × G? 5.29. Let G be a nontrivial finite abelian 2-group. Show that the number of elements of order 2 in G is 2k − 1, for some positive integer k. 5.30. Let G be a finite abelian group. Suppose that, for every n ∈ N, there are at most n elements a ∈ G satisfying a n = e. Show that G is cyclic.\n\n5.4 A Word About Infinite Abelian Groups Unfortunately, that word is “messy”. We have seen that finite abelian groups behave very nicely. To be sure, we cannot possibly expect every infinite abelian group to be a direct product of cyclic groups of prime power order. But even if we allow direct products of infinite cyclic groups such as Z × Z, that does not come close to covering all of the possibilities. While a deep discussion of infinite abelian groups is beyond the scope of an introductory abstract algebra course, we can make a few remarks. Definition 5.7. Let G be a nontrivial group. We say that G is decomposable if it is the direct product of two proper subgroups. If not, then it is indecomposable. We can easily classify the indecomposable finite abelian groups. Theorem 5.9. Let G be a finite abelian group. Then G is indecomposable if and only if G is a cyclic group of order p n , for some prime p and positive integer n. Proof. In view of Theorem 5.3, an indecomposable finite abelian group must indeed be cyclic of prime power order. If G is cyclic of order p n , then suppose that G = H × K , for some subgroups H and K . Then by Lagrange’s theorem, H and K are both p-groups. Furthermore, by Theorem 3.16, they are both cyclic. But since G = H × K is cyclic, it follows from Theorem 5.4 that (|H |, |K |) = 1. As the orders are both powers of p, this means that either H or K is trivial, so either K or H is all of G. Thus, H and K are not both proper and G is indecomposable. \u0002 What about infinite abelian groups? Example 5.15. The additive group Q is indecomposable. Indeed, suppose that Q = H × K , where H and K are proper subgroups. Then neither H nor K is {0}, so take a/b ∈ H , c/d ∈ K , where a, b, c and d are nonzero integers. Note that bc(a/b) = ac ∈ H and ad(c/d) = ac ∈ K . Then H ∩ K is not trivial, so we do not have a direct product. Also, Q is not cyclic. Indeed, if a, b ∈ Z and b > 0, it is clear that 1/(b + 1) ∈ / \u0005a/b\u0006. Thus, Q \b= \u0005a/b\u0006.\n\n98\n\n5 Direct Products and the Classification of Finite Abelian Groups\n\nNow, every element of Q other than the identity has infinite order. What about infinite abelian groups where every element has finite order? Example 5.16. Consider the group Q/Z. Exercise 5.31 asks us to examine some properties of this group. In particular, the distinct elements of the group are precisely of the form q + Z, where q ∈ Q and 0 ≤ q < 1. Also, every element has finite order. But this group is decomposable. Indeed, fix any prime p. Then let H = {a/b + Z : a, b ∈ Z, b = p n , n ≥ 0} and K = {c/d + Z : c, d ∈ Z, (d, p) = 1}. In Exercise 5.32, we also demonstrate that Q/Z = H × K . The group H from the preceding example is named for E.P. Heinz Prüfer. Definition 5.8. Let p be a prime. Then the Prüfer p-group is the subgroup {a/ p n + Z : a, n ∈ Z, n ≥ 0} of the additive group Q/Z. Example 5.17. Let H be the Prüfer p-group. We note that H is an abelian p-group; indeed, p n (a/ p n + Z) = a + Z = 0 + Z; thus, the order of a/ p n + Z divides p n . But H is not cyclic; indeed, 1/ p n + Z has order p n , so H has elements of arbitrarily large order. So if it were cyclic, what order could its generator possibly have? However, Exercise 5.36 asks us to show that every nontrivial subgroup of H contains 1/ p + Z. Thus, H is surely indecomposable. In fact, Q and the Prüfer p-group share another interesting property. Definition 5.9. Let G be an abelian group written additively. We say that G is divisible if, for every element a of G and every positive integer n, there exists a b ∈ G such that nb = a. Note that if G is a nontrivial finite abelian group, then it cannot be divisible. Indeed, if G has order n, then nb = 0 for every b ∈ G. Thus, if 0 \b= a ∈ G, then nb = a has no solution. So, we must look to infinite abelian groups. Example 5.18. The group Q is divisible. Indeed, if a ∈ Q and n is a positive integer, then n(a/n) = a. Example 5.19. For any prime p, the Prüfer p-group is divisible. Indeed, to see this, we note that if G is divisible, so is any factor group of G. (See Exercise 5.35.) Thus, Q/Z is divisible. As in Example 5.16, write Q/Z = H × K , where H is the Prüfer p-group. If a ∈ H , then by the divisibility of Q/Z, for any positive integer n, there exist h ∈ H , k ∈ K such that n(h, k) = (a, 0). But then nh = a. Exercises 5.31. Let G = Q/Z. 1. Show that the elements of G can be uniquely written in the form q + Z, where q ∈ Q and 0 ≤ q < 1. 2. If a, b ∈ Z, b > 0 and (a, b) = 1, what is the order of a/b + Z in G?\n\n5.4 A Word About Infinite Abelian Groups\n\n99\n\n5.32. Show that for any prime p, Q/Z = H × K , where H is the Prüfer p-group and K = {c/d + Z : c, d ∈ Z, (d, p) = 1}. 5.33. Let G be a divisible group, written additively. Show that for every positive integer n, the function α : G → G given by α(a) = na is an onto homomorphism. Is it necessarily an automorphism? 5.34. Let G and H be abelian groups, written additively. Show that G × H is divisible if and only if G and H are both divisible. 5.35. Show that if G is a divisible group, then every factor group of G is divisible, but subgroups need not be. 5.36. Let G be the Prüfer p-group, for some prime p. Show that every nontrivial subgroup of G contains 1/ p + Z. 5.37. Let G be an abelian group having a subgroup N such that G/N is infinite cyclic. Show that G has a subgroup H such that H is infinite cyclic and G = H × N . 5.38. For any prime p, show that every proper subgroup of the Prüfer p-group is finite.\n\nChapter 6\n\nSymmetric and Alternating Groups\n\nWe have seen the definition of the symmetric group Sn , but so far, we do not have too much experience with it. In this chapter, we will introduce the notions of cycles and, in particular, transpositions, which are important elements of the symmetric group. These will help us to understand the group. We will also construct a subgroup of the symmetric group called the alternating group. If n ≥ 5, then the alternating group is very special in that it has no nontrivial proper normal subgroups.\n\n6.1 The Symmetric Group and Cycle Notation Let n be a positive integer. Then we recall that the set of permutations of the set {1, 2, . . . , n} is a group of order n! under composition of functions. It is called the symmetric group and denoted Sn . Why is this group of sufficient interest to merit a chapter on its own? In the earliest years of group theory, the abstract definition of a group had not been written down. Instead, mathematicians worked with groups of permutations. As it turns out, they were not losing much by doing so! If A is any nonempty set, write P(A) for the set of all permutations of A. Then just as we saw that Sn is a group under composition of functions, so is P(A). The following famous result is due to Arthur Cayley. Theorem 6.1 (Cayley’s Theorem). Let G be any group. Then G is isomorphic to a subgroup of P(G). Proof. For each a ∈ G, define ρa : G → G via ρa (g) = ag, for all g ∈ G. We claim that ρa ∈ P(G). Certainly ρa (g) ∈ G. If ρa (g1 ) = ρa (g2 ), for g1 , g2 ∈ G, then ag1 = ag2 , so g1 = g2 . Thus, ρa is one-to-one. If g ∈ G, then ρa (a −1 g) = g, so ρa is also onto. The claim is proved. © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_6\n\n101\n\n102\n\n6 Symmetric and Alternating Groups\n\nNow, define ρ : G → P(G) via ρ(a) = ρa . We claim that ρ is a homomorphism. If a, b ∈ G, then ρ(ab)(g) = ρab (g) = abg and (ρ(a) ◦ ρ(b))(g) = ρa (ρb (g)) = ρa (bg) = abg, for all g ∈ G. Thus, ρ(ab) = ρ(a) ◦ ρ(b), proving the claim. Also, if a ∈ ker(ρ), then ρa is the identity permutation. In particular, ρa (e) = e, and therefore ae = e. Thus, a = e, and ρ is one-to-one. It now follows that G is isomorphic to ρ(G), which is a subgroup of P(G). \u0007 \u0006 Corollary 6.1. Let G be a group of order n < ∞. Then G is isomorphic to a subgroup of Sn . Proof. We know that G is isomorphic to a subgroup of P(G), but replacing G with {1, 2, . . . , n} is just a relabelling. Thus, G is isomorphic to a subgroup of Sn . \u0007 \u0006 The notation we have been using for elements of Sn is rather cumbersome and tends to hide the properties of the permutations. It is time to introduce something better. Definition 6.1. Let k be a positive integer. A permutation σ ∈ Sn is called a k-cycle if there exist distinct elements a1 , a2 , . . . , ak ∈ {1, 2, . . . , n} such that σ (ai ) = ai+1 , / {a1 , . . . , ak }, then σ (a) = a. We use the cycle for 1 ≤ i < k, σ (ak ) = a1 and if a ∈ notation σ = (a1 a2 · · · ak ). A cycle means a k-cycle for some k. \u0002 \u0003 12345 Example 6.1. Let us work in S5 . Then σ = is a 3-cycle; as σ (2) = 5, 15324 σ (5) = 4, σ (4) = 2 and everything else is fixed, we have σ = (2 5 4). Note that it would \u0002 be just as\u0003correct to write σ = (5 4 2) or (4 2 5) (but not (2 4 5)). Similarly, 12345 τ = satisfies τ (1) = 3, τ (3) = 2, τ (2) = 5, τ (5) = 4, τ (4) = 1, and 35214 there are no other values to consider, so τ is the 5-cycle (1 3 2 5 4) (or (3 2 5 4 1), and so on). Note that the only 1-cycle in Sn is the identity permutation, denoted (1). Theorem 6.2. Any k-cycle in Sn has order k. Proof. Simply note that if σ = (a1 · · · ak ), then σ (a1 ) = a2 , σ 2 (a1 ) = σ (a2 ) = a3 , \u0007 \u0006 and so on. It takes k steps to reach a1 again. Similarly for all other ai . Definition 6.2. We say that cycles σ1 , . . . , σr are disjoint if, whenever σi (a) = a, we have σ j (a) = a for all j = i. If σ ∈ Sn and we write σ = σ1 σ2 · · · σr , where the σi are disjoint cycles, then we have a disjoint cycle decomposition for σ . \u0002 \u0003 123456 Example 6.2. Let σ = in S6 . Then noting that σ (1) = 5, σ (5) = 4 563142 and σ (4) = 1, we have a cycle (1 5 4). Also, σ (2) = 6 and σ (6) = 2, so we have another cycle (2 6). The remaining number, 3, is fixed by σ , so a disjoint cycle decomposition for σ is σ = (1 5 4)(2 6).\n\n6.1 The Symmetric Group and Cycle Notation\n\n103\n\n\u0002\n\n\u0003 12345678 . Using the same proce21568437 dure as above, we find that σ = (1 2)(3 5 8 7)(4 6) is a disjoint cycle decomposition. Example 6.3. Similarly, consider σ =\n\nIn fact, we can always apply the procedure from the last two examples. Theorem 6.3. Every element of Sn is a product of disjoint cycles. Proof. Take any σ ∈ Sn . If σ is the identity, then σ = (1) and there is nothing to do. Assume otherwise, and take a1 ∈ {1, . . . , n} such that σ (a1 ) = a2 = a1 . If σ (a2 ) = a1 , then we have a 2-cycle, (a1 a2 ). Otherwise, let σ (a2 ) = a3 . Continue until we find ak such that σ (ak ) ∈ {a1 , . . . , ak }. Now, if σ (ak ) = ai , with 1 < i ≤ k, then σ k (a1 ) = σ i−1 (a1 ). Thus, σ k−i+1 (a1 ) = a1 . In other words, ak−i+2 = a1 . But this is a contradiction. Therefore, σ (ak ) = a1 , and we have a k-cycle, (a1 a2 · · · ak ). If σ = (a1 a2 · · · ak ), then we are done. Otherwise, take b1 which is not in {a1 , . . . , ak } such that σ (b1 ) = b2 = b1 . Now repeat the same procedure, obtaining an l-cycle, (b1 b2 · · · bl ). We must make sure that these cycles are disjoint; that / {a1 , . . . , an }. If is, we cannot have bm ∈ {a1 , . . . , ak }, for any m. By choice, b1 ∈ b2 = σ (b1 ) = at , then since at = σ (as ), for some s, we have σ (b1 ) = σ (as ), and as σ is one-to-one, b1 = as , which is impossible. Proceeding in this way, we see that the cycles are disjoint. If σ = (a1 · · · ak )(b1 · · · bl ), then we are done. Otherwise, take any c1 that does not lie in {a1 , . . . , ak , b1 , . . . , bl } such that σ (c1 ) = c1 and repeat. As there are only n entries in {1, . . . , n}, this procedure must stop eventually. \u0007 \u0006 We were not too concerned about the order in which we wrote the cycles in the last proof. But this is ok. Theorem 6.4. In Sn , disjoint cycles commute. Proof. Let σ = (a1 · · · ak ) and τ = (b1 · · · bm ) be disjoint cycles. We will show that σ τ = τ σ . Take c ∈ {1, . . . , n}. If c ∈ {a1 , . . . , ak }, then as σ and τ are disjoint, τ fixes c. Thus, σ τ (c) = σ (c). But σ (c) ∈ {a1 , . . . , ak } as well. Thus, τ fixes σ (c) too, so τ σ (c) = σ (c). By a similar argument, if c ∈ {b1 , . . . , bm }, then σ τ (c) = τ σ (c) = τ (c). If c is not among the ai or bi , then both σ and τ fix c, so σ τ (c) = τ σ (c) = c. We are done. \u0007 \u0006 Example 6.4. It makes no difference if we write (1 5)(2 6 4) or (2 6 4)(1 5). Both are the same permutation. However, it would be wrong to try to extend this to cycles that are not disjoint! Example 6.5. In S3 , let σ = (1 2) and τ = (1 3). Let us compute σ τ . Now, τ (1) = 3 and σ (3) = 3, so σ τ (1) = 3. Also, τ (3) = 1 and σ (1) = 2, so σ τ (3) = 2. Finally, τ (2) = 2 and σ (2) = 1, so σ τ (2) = 1. There are no other values to consider, so σ τ is the 3-cycle (1 3 2). But proceeding in the same way, we find that τ σ is a different 3-cycle, (1 2 3).\n\n104\n\n6 Symmetric and Alternating Groups\n\nExample 6.6. Let us find a disjoint cycle decomposition for (2 4)(2 5 3 4)(1 3)(1 5). We see (working from right to left) that 1 is mapped by (1 5) to 5, which is fixed by (1 3), which then goes to 3, which is fixed by (2 4). So, 1 goes to 3. Next, 3 is fixed by (1 5), then goes to 1, which is fixed by the other cycles, so we have a 2-cycle (1 3). Next, 2 is fixed by (1 3)(1 5), it then goes to 5, which is fixed by (2 4), so 2 goes to 5. Now, 5 goes to 1 which goes to 3 which goes to 4 and then back to 2. Thus, we have another 2-cycle, (2 5). Finally, 4 goes to 2 then back to 4, so 4 is fixed. Therefore, we have (2 4)(2 5 3 4)(1 3)(1 5) = (1 3)(2 5). We can use the disjoint cycle decomposition to find the order of a permutation. Recall that the least common multiple of positive integers a1 , a2 , . . . , ar is the smallest positive integer m such that ai |m for all i. Theorem 6.5. If σ1 , . . . , σr are disjoint cycles in Sn , then the order of σ1 · · · σr is the least common multiple of the lengths of the σi . Proof. Let k be a positive integer. Then since the σi commute, by Theorem 6.4, we have (σ1 · · · σr )k = σ1k · · · σrk . As the σi move disjoint subsets of {1, . . . , n}, we have σ1k · · · σrk = (1) if and only if each σik = (1). In view of Theorem 6.2, this occurs if \u0007 \u0006 and only if the length of each σi divides k. Exercises 6.1. Write each of the following permutations as a product of disjoint cycles. \u0002 \u0003 1234567 1. 1457326 \u0002 \u0003 12345678 2. 25631487 6.2. Write each of the following permutations as a product of disjoint cycles. 1. (1 3 2)(1 4)(2 5 3) 2. (2 5 3 4)(1 2 6)(3 5 4)(1 2 7) 6.3. Find the inverse of each of the following permutations. Write the answer as a product of disjoint cycles. 1. (1 2 4)(3 5 7 6) 2. (1 2)(2 4 3)(2 3 5) 6.4. Find all possible orders of elements of S7 . 6.5. How many elements of order 3 are there in S9 ? 6.6. Let σ be a k-cycle. If m is a positive integer, show that σ m is a k-cycle if and only if (k, m) = 1. 6.7. Let σ ∈ Sn be a k-cycle. Show that there exists a k-cycle τ ∈ Sn such that τ 2 = σ if and only if k is odd.\n\n6.1 The Symmetric Group and Cycle Notation\n\n105\n\n6.8. If n = 2, show that Z (Sn ) = {(1)}. 6.9. Find the smallest positive integers m and n such that Sm has an element of order 105 and Sn has an element of order 125. 6.10. Find a subgroup of order 120 in S8 .\n\n6.2 Transpositions and the Alternating Group While a disjoint cycle decomposition gives us the clearest picture of the action of a permutation, it is often useful to write the permutation as a different sort of product. Definition 6.3. A transposition is a 2-cycle. Theorem 6.6. If n ≥ 2, then every permutation in Sn is a product of transpositions. Proof. In view of Theorem 6.3, it is sufficient to show that every cycle is a product of transpositions. The identity is (1) = (1 2)(1 2). Let us take a k-cycle σ ; without loss of generality, say σ = (1 2 3 · · · k). We claim that σ = (1 k)(1 (k − 1)) · · · (1 2). Our proof is by induction on k. If k = 2, there is nothing to do. Otherwise, assume that (1 2 · · · k) = (1 k)(1 (k − 1)) · · · (1 2). Then (1 (k + 1))(1 k) · · · (1 2) = (1 (k + 1))(1 2 · · · k), \u0007 and performing the calculation, we see that this is (1 2 · · · (k + 1)), as required. \u0006 Example 6.7. Let us write (1 4 5)(1 3 6 4 5) as a product of transpositions. Using the method described in the above proof, (1 4 5) = (1 5)(1 4) and (1 3 6 4 5) = (1 5)(1 4)(1 6)(1 3), so (1 4 5)(1 3 6 4 5) = (1 5)(1 4)(1 5)(1 4)(1 6)(1 3). It is worth noting that the expression of a permutation as a product of transpositions is by no means unique. For instance, we have seen that (1 2 3 4) = (1 4)(1 3)(1 2). But also, (1 2 3 4) = (1 2)(2 3)(3 4). In fact, the number of transpositions involved does not have to be the same, as both of these are equal to (5 6)(1 2)(2 3)(3 4)(5 6). Nevertheless, we note that all of the products we have just calculated involve an odd number of transpositions. It is a very useful fact that this parity is always preserved; that is, a permutation will be a product of either an even or an odd number of transpositions, not both. The following lemma does most of the work in proving this fact.\n\n106\n\n6 Symmetric and Alternating Groups\n\nLemma 6.1. In Sn , the identity permutation cannot be written as a product of an odd number of transpositions. Proof. Suppose that the lemma is false, and let k be the smallest odd number such that (1) = σ1 σ2 · · · σk , where each σi is a transposition. Now, choose an element of {1, . . . , n} that is not fixed by all of the σi . Without loss of generality, let us say that some σi (1) = 1. Let j be such that σ j (1) = 1 but σr (1) = 1 for all r > j. Among all expressions of (1) as a product of k transpositions such that at least one does not fix 1, we proceed by induction on j. If j = 1, then we note that σ2 · · · σk fixes 1, but σ1 does not, so σ1 · · · σk does not fix 1, which is a contradiction. Therefore, assume that j > 1 and that our result holds for expressions with a smaller j value. Without loss of generality, say that σ j = (1 2). We have four cases to consider for σ j−1 . If σ j−1 = (1 2), then since (1 2)(1 2) is the identity, we can cancel it from our expression. But this contradicts the minimality of k. Suppose that σ j−1 fixes 1 but not 2. Without loss of generality, say σ j−1 = (2 3). Then notice that (2 3)(1 2) = (1 3 2) = (1 3)(2 3). Thus, replacing σ j−1 σ j with (1 3)(2 3), we see that the j value has now decreased to j − 1. By our inductive hypothesis, it is impossible to write the identity as a product in this way. Suppose that σ j−1 fixes 2 but not 1. Without loss of generality, say σ j−1 = (1 3). Then we see that (1 3)(1 2) = (1 2 3) = (1 2)(2 3). Again, replacing σ j−1 σ j with (1 2)(2 3), the j value decreases, and we have a contradiction. Finally, suppose that σ j−1 fixes both 1 and 2. Without loss of generality, say σ j−1 = (3 4). Then by Theorem 6.4, (3 4)(1 2) = (1 2)(3 4), so we can once again decrease the j value. Our proof is complete. \u0007 \u0006 Theorem 6.7. No permutation in Sn can be written as a product of both an even and an odd number of transpositions. Proof. Suppose that σ1 σ2 · · · σk = τ1 τ2 · · · τm , where each σi and τi is a transposition, k is even and m is odd. Then (1) = σk−1 · · · σ1−1 τ1 · · · τm = σk · · · σ1 τ1 · · · τm , since each σi has order 2 (by Theorem 6.2) and is therefore its own inverse. Thus, we have written the identity as a product of k + m transpositions. But k + m is odd, contradicting the preceding lemma. \u0007 \u0006 Definition 6.4. We say that a permutation in Sn is even (respectively, odd) if it is the product of an even (respectively, odd) number of transpositions. Example 6.8. In S5 , we note that (1 2 3)(4 5) is odd, as (1 2 3)(4 5) = (1 3)(1 2)(4 5). Theorem 6.8. A k-cycle is even if and only if k is odd.\n\n6.2 Transpositions and the Alternating Group\n\n107\n\nProof. If k = 1, then we know that the identity is even. If k > 1, then refer to the proof of Theorem 6.6, where we wrote a k-cycle as a product of k − 1 transpositions. \u0007 \u0006 Thus, to determine if a particular permutation is even or odd, we can look at its disjoint cycle decomposition. The preceding theorem tells us whether each cycle is a product of an even or odd number of transpositions, so we can easily determine the answer for the entire permutation. Definition 6.5. The alternating group An is the set of all even permutations in Sn . Example 6.9. We note that S3 consists of the identity (which is even), three transpositions (which are odd) and two 3-cycles (which are even). Thus, A3 = {(1), (1 2 3), (1 3 2)}. Similarly S4 consists of the identity (even), six transpositions (odd), eight 3-cycles (even), six 4-cycles (odd) and three elements that are products of two disjoint transpositions (even). Thus, A4 = {(1), (1 2 3), (1 2 4), (1 3 2), (1 3 4), (1 4 2), (1 4 3), (2 3 4), (2 4 3), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. Theorem 6.9. Let n ≥ 2. Then An is a normal subgroup of Sn , and [Sn : An ] = 2. Proof. Define α : Sn → Z2 as follows. Let α(σ ) = 0 if σ is even and 1 if σ is odd. We claim that α is a homomorphism. Indeed, as the product of two even or two odd permutations is even, and the product of an even and an odd is odd, this follows immediately. By definition, the kernel is An , so An is a normal subgroup. Furthermore, α((1)) = 0 and α((1 2)) = 1, so α is onto. Thus, by the First Isomorphism Theorem, \u0007 \u0006 Sn /An is isomorphic to Z2 . That is, |Sn /An | = 2, so An has index 2. Exercises 6.11. Decide if each of the following permutations is even or odd. 1. (2 3)(1 3 4)(1 4 2 3) 2. (1 4 3 5)(1 2)(1 3 2 4) 6.12. Write each of the following permutations as a product of transpositions. 1. (1 3 2)(1 4)(2 5 3) 2. (2 5 3 4)(1 2 6)(3 5 4) 6.13. Find every possible order of the product of two transpositions. 6.14. Let n ≥ 2 and H ≤ Sn . Show that either every element of H is even, or exactly half of the elements of H are even.\n\n108\n\n6 Symmetric and Alternating Groups\n\n6.15. For which n ≥ 2 does An have a subgroup of order 4? What if we insist that the subgroup be cyclic? 6.16. Find the orders of all the elements of A8 . 6.17. If n ≥ 2, show that every element of odd order in Sn lies in An . 6.18. Show that every permutation other than the identity in Sn is the product of at most n − 1 transpositions. 6.19. For which positive integers n does Sn have 1. more elements of even order than odd order; 2. more elements of odd order than even order; 3. the same number of elements of odd order as even order? 6.20. For which integers n ≥ 2 does there exist a σ ∈ An such that |σ | > n?\n\n6.3 The Simplicity of the Alternating Group Why are we so interested in the group An ? In order to explain this, we must start with a definition. Definition 6.6. A group is simple if it is nontrivial and has no nontrivial proper normal subgroups. If G is abelian, then every subgroup is normal, so we are looking for groups whose only subgroups are G and {e}. But these were determined in Exercise 3.52. Indeed, we saw that these were precisely the cyclic groups of prime order. By Theorem 4.14, we have the following result. Theorem 6.10. Let G be an abelian group. Then G is simple if and only if G is isomorphic to Z p , for some prime p. That was pretty painless! However, the nonabelian case is much much more difficult. Much! The classification of all of the finite simple groups was one of the biggest mathematical projects of the twentieth century. Over one hundred mathematicians contributed to the solution, and the proof consists of many thousands of pages of journal articles. For obvious reasons, we will not be discussing this classification here. We will content ourselves with proving one of the earliest results on the subject; namely, if n ≥ 5 then An is a nonabelian simple group. (Actually, A5 is the smallest nonabelian simple group.) The n = 5 case was established by Évariste Galois in the early nineteenth century. Decades later, M.E. Camille Jordan provided a proof for all n ≥ 5.\n\n6.3 The Simplicity of the Alternating Group\n\n109\n\nWhy are finite simple groups so interesting? Let us look at it this way. Suppose that G is a nontrivial finite group. Let N1 be a proper normal subgroup of largest order in G. (If G is simple, this will be {e}. Otherwise, it will be something larger.) Now, we claim that G/N1 is simple. Indeed, by Theorem 4.8, the normal subgroups of G/N1 are precisely of the form H/N1 , where H is a normal subgroup of G containing N1 . But by definition of N1 , H = N1 or G. Thus, G/N1 has no nontrivial proper normal subgroups, so it is simple. Now, suppose that N1 = {e}. Then in the same way, take a proper normal subgroup N2 of N1 of largest possible order. Then N1 /N2 is simple. We can repeat this procedure and obtain G = N0 ≥ N1 ≥ N2 ≥ N3 ≥ · · · ≥ Nk−1 ≥ Nk = {e}, where each Ni+1 is normal in Ni and Ni /Ni+1 is simple. We know the process must end, as each Ni+1 is properly contained in Ni , and the original group is finite. In a way, then, finite groups can be built up using simple groups. Let us begin the process of proving that An is simple, for n ≥ 5. We start with a general fact about the conjugation of cycles. Lemma 6.2. Let σ = (a1 a2 · · · ak ) be a k-cycle in Sn . If τ ∈ Sn , then τ σ τ −1 = (τ (a1 ) τ (a2 ) · · · τ (ak )). Proof. Suppose that b = τ (ai ). Then τ −1 (b) = ai ; hence, σ (τ −1 (b)) = σ (ai ) = ai+1 (or a1 , if i = k). Therefore, τ σ τ −1 (b) = τ (ai+1 ) (or τ (a1 ), if i = k). That is, τ σ τ −1 permutes the τ (ai ) as described. If b is not among the τ (ai ), then τ −1 (b) is not equal to any ai , which means that it is fixed by σ . Thus, τ σ τ −1 (b) = τ τ −1 (b) = b. \u0007 \u0006 Therefore, τ σ τ −1 is the k-cycle described in the statement of the lemma. Corollary 6.2. Let n and k be positive integers with n ≥ k. Then 1. any two k-cycles are conjugate in Sn ; and 2. if k is odd and n ≥ k + 2, then any two k-cycles are conjugate in An . Proof. (1) Let σ = (a1 · · · ak ) and δ = (b1 · · · bk ) be any two k-cycles. The preceding lemma tells us that in order to show that σ and δ are conjugate, we need only find τ ∈ Sn such that τ (ai ) = bi for all i; in this case, τ σ τ −1 = δ. But Sn contains every possible permutation of {1, . . . , n}. Thus, we can certainly assign / {a1 , . . . , ak }, let the τ ( j) be any distinct values not in τ (ai ) = bi , and for the j ∈ {b1 , . . . , bk }. (2) As k is odd, the k-cycles are even, and therefore lie in An . Let σ and δ be any k-cycles. Without loss of generality, let us say that δ = (1 2 · · · k). Then just as in (1), we can find τ ∈ Sn such that τ σ τ −1 = δ. If τ ∈ An , then we are done. Otherwise, τ is odd, so ((k + 1) (k + 2))τ is even. (Note that this is valid, as n ≥ k + 2.) Thus, letting η = ((k + 1) (k + 2))τ ∈ An , we have ησ η−1 = ((k + 1) (k + 2))τ σ τ −1 ((k + 1) (k + 2)) = ((k + 1) (k + 2))(1 2 · · · k)((k + 1) (k + 2)).\n\n110\n\n6 Symmetric and Alternating Groups\n\nBut disjoint cycles commute, so this is ((k + 1) (k + 2))((k + 1) (k + 2))δ = δ. We are done.\n\n\u0007 \u0006\n\nExample 6.10. The preceding lemma tells us that (1 2 3) and (1 3 2) are conjugate in S3 , and the proof suggests how we might demonstrate it. We need to find τ such that τ (1) = 1, τ (2) = 3 and τ (3) = 2; that is, τ = (2 3). Then (1 3 2) = τ (1 2 3)τ −1 . However, (1 2 3) and (1 3 2) are not conjugate in A3 ; this is obvious, as A3 is abelian, having order 3, so different elements are not conjugate. It is less obvious that they are not conjugate in A4 either; however, it is possible to try conjugating (1 2 3) by all of the elements of A4 . None of these conjugates will equal (1 3 2). However, the preceding lemma tells us that (1 2 3) and (1 3 2) are indeed conjugate in A5 , and the proof tells us that if we take η = (4 5)τ , then η ∈ A5 , and we find that η(1 2 3)η−1 = (1 3 2). We can now simplify our task by showing that if we have a 3-cycle in a normal subgroup of An , then we have all of An . Corollary 6.3. Let n ≥ 3. Then 1. every element of An is a product of 3-cycles; and 2. if a normal subgroup N of An contains any 3-cycle, then N = An . Proof. (1) We know that an element of An is a product of an even number of transpositions. Thus, it is sufficient to show that every product of two transpositions is a product of 3-cycles. (As (1) = (1 2 3)(1 3 2), we need not worry about the identity.) If the two transpositions are equal, then their product is the identity, with which we have just dealt. Suppose they have one number in common. Without loss of generality, say (1 2)(1 3). Then note that (1 2)(1 3) = (1 3 2), which is a 3-cycle. Finally, suppose they have no numbers in common. Without loss of generality, say (1 2)(3 4). Then we observe that (1 2)(3 4) = (1 4 3)(1 2 3), which is a product of 3-cycles. (2) In view of (1), it is sufficient to show that N contains all of the 3-cycles. But it contains one 3-cycle, so as N is normal, it contains all of its conjugates. If n ≥ 5, then Corollary 6.2 tells us that these conjugates are all of the 3-cycles, and we are done. If n = 3, there is little to do, as the only 3-cycles are (1 2 3) and (1 3 2), and they are squares of each other; thus, if N contains one, it contains the other. The n = 4 case requires a little more work, and we leave it as Exercise 6.24. \u0007 \u0006 And now, our main result for this section. Theorem 6.11. If n ≥ 5, then An is a nonabelian simple group. Proof. The fact that (1 2 3)(1 2 4) = (1 2 4)(1 2 3) shows that An is nonabelian, so we can focus on the simplicity. Let N be a nontrivial normal subgroup of An . We\n\n6.3 The Simplicity of the Alternating Group\n\n111\n\nmust prove that N = An . In view of Corollary 6.3, it is sufficient to show that N contains a 3-cycle. Take any (1) = σ ∈ N , and consider the disjoint cycle decomposition of σ . Suppose, first of all, that there are two or more transpositions in this decomposition. Without loss of generality, say σ = (1 2)(3 4)δ, where δ is a product of disjoint cycles which also fix everything in {1, 2, 3, 4} (and δ = (1) is possible). Let τ = (1 2 4) ∈ An . Then as N is normal in An , we have τ σ τ −1 ∈ N . That is, (1 2 4)(1 2)(3 4)δ(1 4 2) ∈ N . (It is easy to check that (1 2 4)−1 = (1 4 2).) As the cycles in δ are disjoint from all the other cycles in the product, we see from Theorem 6.4 that N contains (1 2 4)(1 2)(3 4)(1 4 2)δ = (1 3)(2 4)δ. But N also contains σ −1 , and therefore σ −1 (1 3)(2 4)δ = δ −1 (3 4)(1 2)(1 3)(2 4)δ ∈ N . Again, δ commutes with these other cycles, so we have δ −1 δ(3 4)(1 2)(1 3)(2 4) = (1 4)(2 3) ∈ N . Let η = (1 4 5) ∈ An (since n ≥ 5). Then N must contain η(1 4)(2 3)η−1 = (1 4 5)(1 4)(2 3)(1 5 4) = (2 3)(4 5). Thus, N also contains (1 4)(2 3)(2 3)(4 5) = (1 4 5). But when N contains a 3-cycle, we know that N = An . Thus, from this point on, we may assume that the disjoint cycle decomposition of σ contains at most one transposition. Now, let us consider the length k of the longest cycle appearing in the disjoint cycle decomposition of σ . If k = 2, then σ is a product of an even number of disjoint transpositions, and we have already dealt with this case. Suppose that k = 3. Then σ is a product of some 3-cycles and, possibly, some transpositions. But the product of some 3-cycles and a single transposition is odd, and therefore not in An . Furthermore, multiple transpositions are not allowed. Therefore, we may assume that σ is a product of one or more 3-cycles. If it is just one 3-cycle, then we are done. So assume that it is a product of two or more disjoint 3-cycles. Without loss of generality, say σ = (1 2 3)(4 5 6)δ, where either δ = (1) or δ is a product of disjoint 3-cycles, all of which fix everything in {1, 2, 3, 4, 5, 6}. Let τ = (3 4 5) ∈ An . Then as N is normal, it contains\n\n112\n\n6 Symmetric and Alternating Groups\n\nτ σ τ −1 = (3 4 5)(1 2 3)(4 5 6)δ(3 5 4) = (1 2 4)(3 6 5)δ, since δ commutes with the other cycles. But N also contains σ −1 , so we have σ −1 (1 2 4)(3 6 5)δ = δ −1 (4 6 5)(1 3 2)(1 2 4)(3 6 5)δ = (2 6 4 3 5) ∈ N , again, since δ is disjoint from the other cycles. Replacing σ with (2 6 4 3 5), we can move to our final case. Let us suppose that k ≥ 4. Then without loss of generality, we may write σ = (1 2 3 · · · k)δ, where k ≥ 4 and δ is some product of disjoint cycles, all of which fix everything in {1, 2, . . . , k}. Let τ = (1 2 3) ∈ An . Then by normality, N contains τ σ τ −1 = (1 2 3)(1 2 3 · · · k)δ(1 3 2) = (1 4 5 · · · k 2 3)δ. But N also contains σ −1 , so noting that (1 2 3 · · · k)−1 = (1 k (k − 1) · · · 2), we have σ −1 (1 4 5 · · · k 2 3)δ = δ −1 (1 k (k − 1) · · · 2)(1 4 5 · · · k 2 3)δ = (1 3 k) ∈ N , again using the fact that δ commutes with everything else. Thus, N contains a 3-cycle, and the proof is complete. \u0007 \u0006 We might well ask about An when n < 5. For n = 2, A2 is the trivial group; hence, by definition, not simple. When n = 3, A3 has order 3 and by Corollary 4.2, it is isomorphic to Z3 . By Theorem 6.10, it is an abelian simple group. The big exception is the n = 4 case, as illustrated in the following example. Example 6.11. The alternating group A4 is not simple. To see, this let N = {(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. It simply requires some computation to see that N is a nontrivial proper normal subgroup of A4 . With the exception of S2 , which is abelian of order 2, and hence isomorphic to Z2 , the symmetric groups are not simple. Indeed, An is a nontrivial proper normal subgroup of Sn , whenever n ≥ 3. However, we can state the following result. Corollary 6.4. If n ≥ 5, then the only nontrivial proper normal subgroup of Sn is An . Proof. Let N be a normal subgroup of Sn . Then N ∩ An is a normal subgroup of An . As An is simple, this means that N ∩ An = An or {(1)}. If N ∩ An = An , then An ≤ N . But by Lagrange’s theorem, this implies that |An | divides |N | and |N | divides |Sn |. As |Sn | = 2|An | (because An is of index 2), this can only mean that |N | = |An | or |Sn |. Thus, N = An or Sn , as desired.\n\n6.3 The Simplicity of the Alternating Group\n\n113\n\nOn the other hand, suppose that N ∩ An = {(1)}. Then by Theorem 4.4, |N An | = |N ||An |. As |An | = |Sn |/2 and |N An | ≤ |Sn |, we see that |N | = 1 or 2. If |N | = 1, we are done, so suppose that |N | = 2. But by Exercise 4.3, a normal subgroup of order 2 in a group is central. However, Exercise 6.8 tells us that the centre of Sn is trivial. Thus, we have a contradiction, and the proof is complete. \u0007 \u0006 Exercises 6.21. Show that A5 has no subgroup of order 30. 6.22. In S7 , describe the conjugates of (1 2)(3 4 5). 6.23. Can a nonabelian simple group have a nonabelian simple proper subgroup? Either prove that it cannot, or construct an explicit example. 6.24. Let N be a normal subgroup of A4 containing a 3-cycle. Show that N = A4 . 6.25. Show that the only nontrivial proper normal subgroup of A4 is the one exhibited in Example 6.11. 6.26. Let n ≥ 2. Show that every element of Sn can be written as a product of transpositions of the form (1 i), for various i. 6.27. If n ≥ 2, show that every element of Sn can be written as a product of the transpositions (1 2), (2 3), . . . , ((n − 1) n). 6.28. If n ≥ 2, let σ = (1 2) and τ = (1 2 3 · · · n). Show that every element of Sn can be written in the form σ i1 τ j1 σ i2 τ j2 · · · σ ik τ jk , where the exponents are any integers and k ∈ N.\n\nChapter 7\n\nThe Sylow Theorems\n\nIn this chapter, we will prove the Sylow theorems. These are difficult results, but fundamental to our understanding of the structure of finite groups. In particular, we will show that if p n is the largest power of a prime p dividing the order of a finite group G, then G has at least one subgroup of order p n . Furthermore, we will discover that any two such subgroups are conjugate to each other, and determine a restriction upon the number of such subgroups. We will then explore various applications of these theorems, and conclude the chapter by classifying all groups of order smaller than 16.\n\n7.1 Normalizers and Centralizers We are very familiar with the centre of a group, which consists of all elements that commute with everything. Let us generalize. Definition 7.1. Let G be a group, a ∈ G and H a subgroup of G. Then the centralizer of a is the set of all elements of G that commute with a. We write C(a) = {g ∈ G : ag = ga}. Also, the centralizer of H is C(H ) = {g ∈ G : gh = hg for all h ∈ H }. Example 7.1. If a ∈ Z (G), then C(a) = G. If H ≤ Z (G), then C(H ) = G. In particular, C(e) = G, so we cannot assume that centralizers are necessarily abelian. Example 7.2. Let G = D8 . Then we find that C(R270 ) = \u0004R90 \u0005, C(R180 ) = G and C(F1 ) = {R0 , R180 , F1 , F2 }.\n\n© Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_7\n\n115\n\n116\n\n7 The Sylow Theorems\n\nTheorem 7.1. Let G be a group, a ∈ G and H a subgroup of G. Then \u0002 1. C(H ) = h∈H C(h); 2. C(a) and C(H ) are both subgroups of G; 3. if H is a normal subgroup of G, then so is C(H ); 4. Z (G) is a subgroup of both C(a) and C(H ); and 5. C(a) = C(\u0004a\u0005). Proof. (1) This follows from the definition. (2) Clearly ae = a = ea, so e ∈ C(a). Suppose that b, c ∈ C(a). Then bca = bac = abc, so bc ∈ C(a). Also, ba = ab, so b−1 (ba)b−1 = b−1 (ab)b−1 . Thus, ab−1 = b−1 a, so b−1 ∈ C(a). Hence, C(a) ≤ G. Furthermore, combining this fact with (1) and Exercise 3.37, we see that C(H ) ≤ G. (3) See Exercise 4.4. (4) Central elements commute with everything hence, in particular, they commute with a and elements of H . (5) If b ∈ C(\u0004a\u0005), then since a ∈ \u0004a\u0005, we see that b commutes with a. Thus, b ∈ C(a). Conversely, if b ∈ C(a), then ab = ba. Therefore, a ∈ C(b). As C(b) ≤ G by (2), we see that a i ∈ C(b) for all integers i. That is, a i b = ba i , for all i ∈ Z. In other words, b ∈ C(\u0004a\u0005). \u0002 Suppose we have a subgroup H of G that is not normal. Of course, H is a normal subgroup of H . Furthermore, it is easy to see that H is normal in H Z (G). How big a subgroup of G can we find in which H is a normal subgroup? This is where normalizers come in. Definition 7.2. Let G be a group and H a subgroup. Then the normalizer of H is the set N (H ) = {a ∈ G : a −1 H a = H }. If K is another subgroup of G, then we write N K (H ) = N (H ) ∩ K , and call it the normalizer of H in K . Remember, if a ∈ C(H ), then a −1 ha = h, for all h ∈ H . But if a ∈ N (H ), then a −1 H a = H . In particular, a −1 ha = h 1 , for some (possibly different) h 1 ∈ H . Thus, the normalizer and centralizer are different concepts. Example 7.3. If H is a normal subgroup of G, then N (H ) = G. See Theorem 4.3. Example 7.4. Let G = S4 and H = \u0004(1 2 3 4)\u0005. Then notice that (24) ∈ / C(H ), as (2 4)(1 2 3 4) = (1 4)(2 3), but (1 2 3 4)(2 4) = (1 2)(3 4). However, (2 4)−1 (1 2 3 4)(2 4) = (2 4)(1 2 3 4)(2 4) = (1 4 3 2) = (1 2 3 4)3 ∈ H. Thus, (2 4)−1 (1 2 3 4)i (2 4) = ((2 4)−1 (1 2 3 4)(2 4))i = (1 4 3 2)i ∈ H, for all i ∈ Z. Therefore, (2 4)−1 H (2 4) ≤ H . By Theorem 4.2, |(2 4)−1 H (2 4)| = |H |, so we conclude that (2 4)−1 H (2 4) = H . Thus, (2 4) ∈ N (H ).\n\n7.1 Normalizers and Centralizers\n\n117\n\nTheorem 7.2. Let H be a subgroup of G. Then N (H ) is a subgroup of G containing H . Furthermore, if K is a subgroup of G containing H , then H is normal in K if and only if K is a subgroup of N (H ). Proof. Take any h ∈ H . Then for any c ∈ H , we have h −1 ch ∈ H , so h −1 H h ≤ H . Also, h −1 (hch −1 )h = c, and hch −1 ∈ H . Thus, every element of H is in h −1 H h, so H = h −1 H h, and H ⊆ N (H ). In particular, e ∈ N (H ). Now, take any a, b ∈ N (H ). Then (ab)−1 H ab = b−1 (a −1 H a)b = b−1 H b = H ; thus, ab ∈ N (H ). Also, as a −1 H a = H , we have aa −1 H aa −1 = a H a −1 ; that is, H = (a −1 )−1 H a −1 , hence a −1 ∈ N (H ), and N (H ) ≤ G. Let H ≤ K ≤ G. Then in view of Theorem 4.3, H is a normal subgroup of K if and only if k −1 H k = H for all k ∈ K . By definition of N (H ), this occurs if and only if K ≤ N (H ). \u0002 It is clear that if H is a subgroup of G, then C(H ) ≤ N (H ). They could, of course be equal; indeed, if H ≤ Z (G), then C(H ) = N (H ) = G. But as we saw in Example 7.4, they need not be. An interesting fact about the relationship between these two subgroups is given in the following result. Theorem 7.3 (N /C Theorem). Let G be a group and H a subgroup. Then C(H ) is a normal subgroup of N (H ), and N (H )/C(H ) is isomorphic to a subgroup of Aut(H ). Proof. We will show that C(H ) is a normal subgroup of N (H ) by illustrating that it is the kernel of a homomorphism from N (H ) to Aut(H ). Define α : N (H ) → Aut(H ) via α(a)(h) = aha −1 , for all a ∈ N (H ), h ∈ H . If a is in the normalizer, then so is a −1 , and therefore a H a −1 = H . Thus, we see immediately that α(a) is an onto function from H to H . Also, if ah 1 a −1 = ah 2 a −1 , then h 1 = h 2 by cancellation, so α(a) is one-to-one as well. Furthermore, for any h 1 , h 2 ∈ H , we have α(a)(h 1 h 2 ) = ah 1 h 2 a −1 = ah 1 a −1 ah 2 a −1 = (α(a)(h 1 ))(α(a)(h 2 )). Therefore, α(a) ∈ Aut(H ). We need to show that α is a homomorphism. But if a, b ∈ N (H ), then for any h ∈ H , we have α(ab)(h) = abh(ab)−1 = abhb−1 a −1 = a(α(b)(h))a −1 = α(a)(α(b)(h)). Thus, α(ab) = α(a)◦α(b), as required. Now, the kernel of α is the set of all c ∈ N (H ) such that α(c) acts as the identity on H ; specifically, we must have chc−1 = h, for all h ∈ H . But this is precisely the definition of C(H ). The First Isomorphism Theorem now tells us that N (H )/C(H ) is isomorphic to α(N (H )), which is a subgroup of Aut(H ), as required. \u0002 The following example illustrates a cute application of the N /C Theorem.\n\n118\n\n7 The Sylow Theorems\n\nExample 7.5. Suppose that G is a nonabelian group of order 39. Let us demonstrate that G cannot possibly have a normal subgroup H of order 3. Suppose such a normal subgroup exists. Then N (H ) = G. Also, what can the centralizer of H be? As H has prime order, Corollary 3.6 tells us that it is cyclic, hence abelian. In particular, H centralizes itself. Thus, H ≤ C(H ) ≤ G. By Lagrange’s theorem, we see that 3 divides |C(H )|, which in turn divides 39. The only possibilities are |C(H )| = 3 or 39. Suppose that |C(H )| = 39. Then C(H ) = G, so H is central. In particular, H ≤ Z (G) ≤ G. But again, looking at the orders, we see that Z (G) = H or G. As G is not abelian, Z (G) = H , but that cannot be the case either, as then Z (G) has prime index which, by Corollary 4.1, is impossible. Therefore, |C(H )| = 3, so C(H ) = H . By the preceding theorem, N (H )/C(H ) is isomorphic to a subgroup of Aut(H ). That is, G/H is isomorphic to a subgroup of Aut(H ). But |G/H | = |G|/|H | = 13. As H is cyclic, the structure of Aut(H ) is given by Theorem 4.22. But even if we did not have that resource, H is a set of 3 elements, so there are only 3! = 6 ways to permute them (and not all of those are automorphisms). Thus, we are trying to fit a group of order 13 inside one that is simply too small. Hence, H cannot exist. Exercises\n\n\u0003\n\n7.1. Which matrices lie in the centralizer of\n\n\u0004 11 in G L 2 (R)? 01\n\n7.2. Which permutations lie in the centralizer of (1 2 3) in S5 ? \u0003 \u0004 23 7.3. In G L 2 (R), let H be the subgroup generated by . Show that C(H ) = 56 N (H ). 7.4. Let H1 ≤ G 1 and H2 ≤ G 2 . Show that in G 1 × G 2 , C(H1 × H2 ) = C(H1 ) × C(H2 ). 7.5. If G is a group having a subgroup H of order 2, show that C(H ) = N (H ). 7.6. If G is a nonabelian group, show that G has a subgroup H such that Z (G) \u0002 H \u0002 G. (Yes, this is the same as Exercise 4.20. Solve it using the results in this section.) 7.7. In any group, show that C(a) = C(a −1 ). 7.8. If a ∈ G and a has odd order, show that C(a) = C(a 4 ). 7.9. Let G be a group of order 77 having a normal subgroup H of order 11. 1. If G is not abelian, show that C(H ) = H . 2. Conclude that G must, in fact, be abelian. 7.10. Let G be a group of order 77. 1. Show that G has a subgroup H of order 11. 2. Show that H is unique, and hence normal. 3. Conclude that G is isomorphic to Z77 .\n\n7.2 Conjugacy and the Class Equation\n\n119\n\n7.2 Conjugacy and the Class Equation We are already familiar with the notion of conjugacy in groups. To reiterate, we say that a and b in G are conjugate if there exists a g ∈ G such that g −1 ag = b. Here is a simple fact that we have not mentioned. Theorem 7.4. If G is any group, then conjugacy is an equivalence relation on the elements of G. Proof. Reflexivity: For any a ∈ G, we have e−1 ae = a, so a is conjugate to itself. Symmetry: Suppose that a is conjugate to b, say g −1 ag = b. Then a = (g −1 )−1 bg −1 , and therefore b is conjugate to a. Transitivity: Suppose that a is conjugate to b, and b to c, say g −1 ag = b and h −1 bh = c. Then c = h −1 (g −1 ag)h = (gh)−1 a(gh). Thus, a is conjugate to c. \u0002 We know that equivalence classes partition a set; thus we can break a group down into disjoint sets of elements, all elements in each set being conjugate to each other. Definition 7.3. Let G be a group and a ∈ G. Then the conjugacy class of a is the set Ca = {g −1 ag : g ∈ G}. Conjugacy classes are subsets of G, not subgroups. Indeed, the only one that will contain the identity is Ce . Example 7.6. Note that Ca contains only a if and only if a ∈ Z (G). (This happens if and only if g −1 ag = a for all g ∈ G.) Example 7.7. Let k and n be positive integers with n ≥ k. If G = Sn and σ = (1 2 3 · · · k), then we see from Lemma 6.2 that Cσ is the set of all k-cycles in G. It is important to know the size of a conjugacy class. Lemma 7.1. Let G be a finite group and a ∈ G. Then the number of elements in Ca is the index of the centralizer, [G : C(a)]. Proof. Take g, h ∈ G. Then notice that g −1 ag = h −1 ah if and only if gh −1 a = agh −1 . That is, g −1 ag and h −1 ah produce the same conjugate of a if and only if gh −1 ∈ C(a) or, equivalently, if and only if the right cosets C(a)g and C(a)h are equal. In other words, we get a distinct conjugate of a for each right coset of C(a), so the number of distinct conjugates is the index, [G : C(a)], as required. \u0002 This allows us to establish an important equation, called the class equation. Theorem 7.5 (Class Equation). Let G be a finite group, and let a1 , . . . , ak be representatives of the conjugacy classes in G with more than one element. Then |G| = |Z (G)| + [G : C(a1 )] + · · · + [G : C(ak )].\n\n120\n\n7 The Sylow Theorems\n\nProof. As G is partitioned into conjugacy classes, we know that |G| is the sum of the sizes of these classes. We noted in Example 7.6 that the conjugacy classes of size 1 are precisely those of the central elements. Collecting them together, we obtain |Z (G)| elements. For the remaining classes, we now apply the preceding lemma. \u0002 This has powerful consequences! Corollary 7.1. Let G be a group of order p n , for some prime p and positive integer n. Then the centre of G is not trivial. Proof. In the class equation, each [G : C(ai )] is the size of a conjugacy class with more than one element. But it is also |G|/|C(ai )|, and therefore a divisor of p n . Thus, each [G : C(ai )] is a multiple of p. But |G| is also a multiple of p, and therefore the one remaining term in the equation, |Z (G)|, is a multiple of p. In particular, it is not 1. \u0002 Corollary 7.2. Let G be a group of order p 2 , for some prime p. Then G is isomorphic to either Z p2 or Z p × Z p . Proof. By Lagrange’s theorem, |Z (G)| ∈ {1, p, p 2 }. But by the preceding corollary, it cannot be 1. Suppose it is p. Then [G : Z (G)] = p 2 / p = p. By Corollary 4.1, this is impossible. Therefore, Z (G) = G, and G is abelian. By Corollary 5.1, we are done. \u0002 Theorem 5.6 tells us that Z p2 and Z p × Z p are not isomorphic, so we now have a complete picture for groups of order p 2 . We also need to know about conjugacy of subgroups. Definition 7.4. Let G be a group and H a subgroup. We say that subgroups K and L of G are H -conjugate if there exists an h ∈ H such that h −1 K h = L. When H = G, we simply say that K and L are conjugate. Example 7.8. Let G be S5 and H = \u0004(1 3)(2 5 4)\u0005. Take σ = (1 3)(2 4 5) = ((1 3)(2 5 4))−1 ∈ H . Then we notice that σ −1 (1 2 3 4)σ = (1 3)(2 5 4)(1 2 3 4)(1 3)(2 4 5) = (1 2 3 5). Therefore, for any integer i, σ −1 (1 2 3 4)i σ = (σ −1 (1 2 3 4)σ )i = (1 2 3 5)i . This means that σ −1 \u0004(1 2 3 4)\u0005σ = \u0004(1 2 3 5)\u0005. Thus, \u0004(1 2 3 4)\u0005 and \u0004(1 2 3 5)\u0005 are H -conjugate (and, therefore, conjugate). Theorem 7.6. Let G be a group and H a subgroup of G. Then H -conjugacy is an equivalence relation on the set of all subgroups of G. Proof. Reflexivity: Let K ≤ G. Then e ∈ H and e−1 K e = K . Therefore, K is H -conjugate to itself. Symmetry: Suppose that h −1 K h = L, with h ∈ H . Then K = hh −1 K hh −1 = h Lh −1 = (h −1 )−1 Lh −1 .\n\n7.2 Conjugacy and the Class Equation\n\n121\n\nSince h −1 ∈ H , we see that H -conjugacy is symmetric. Transitivity: Suppose that −1 h −1 1 K h 1 = L and h 2 Lh 2 = M, where h 1 , h 2 ∈ H . Then −1 −1 M = h −1 2 (h 1 K h 1 )h 2 = (h 1 h 2 ) K (h 1 h 2 ).\n\nSince h 1 h 2 ∈ H , we are done.\n\n\u0002\n\nThus, the subgroups of G are partitioned into equivalence classes, in which those in each class are H -conjugate to each other. In a similar fashion to Lemma 7.1, we have a formula for the number of H -conjugates of a subgroup. Theorem 7.7. Let G be a finite group and H a subgroup. Then for any subgroup K of G, the number of H -conjugates of K in G is [H : N H (K )]. −1 −1 −1 Proof. Take h 1 , h 2 ∈ H . Then h −1 1 K h 1 = h 2 K h 2 if and only if h 2 h 1 K h 1 h 2 = −1 −1 −1 K ; that is, if and only if (h 1 h 2 ) K (h 1 h 2 ) = K . But this means precisely that ∈ N H (K ) or, in other words, that N H (K )h 1 = N H (K )h 2 . Thus, we get h 1 h −1 2 one distinct conjugate for each right coset of N H (K ) in H , so the number of such \u0002 conjugates is the index, [H : N H (K )].\n\nExercises 7.11. What are the conjugacy classes of D8 ? 7.12. What are the conjugacy classes of S4 ? 7.13. Let G be a group having subgroups H and K . Suppose that two subgroups of G are both H -conjugate and K -conjugate. Does it follow that they are (H ∩ K )conjugate? Either prove that it does or construct a counterexample. 7.14. Let G be a finite group with normal subgroup N . Show that there are at least as many conjugacy classes in G as in G/N . 7.15. Let G be a group of order p n , where p is a prime and n ≥ 2. Suppose that |Z (G)| = p. Show that there exists an a ∈ G such that |C(a)| = p n−1 . 7.16. If G is a group of order p n , for some prime p and positive integer n, show that G has a subgroup of order p m for each positive integer m ≤ n. 7.17. For each of the following lists, determine if it is the list of sizes of the conjugacy classes of some finite group. If it is, provide such a group. If not, explain why not. 1. 1, 1, 1, 1, 1, 5, 5, 5, 5 2. 1, 2, 3 3. 2, 4, 6 7.18. Let G be a group and H the set of elements of G having only finitely many conjugates. Show that H is a subgroup of G. 7.19. Suppose that G is a finite group and there exists e = a ∈ G such that a −1 ∈ Ca . Show that G has even order. 7.20. Let G be a group of order n > 1. Show that no conjugacy class can have order greater than n/2.\n\n122\n\n7 The Sylow Theorems\n\n7.3 The Three Sylow Theorems We can now present the three major theorems due to P. Ludwig Sylow concerning subgroups of prime power order in a finite group. We will give the statements and proofs in this section, and some applications in the following sections. Definition 7.5. Let G be a finite group, and suppose that |G| = p n r , where p is a prime, n ≥ 0 and r is a positive integer such that ( p, r ) = 1. Then a Sylow p-subgroup of G is any subgroup of order p n . By Lagrange’s theorem, if H is a subgroup of G of order p k , for some k, then the order of H cannot possibly be any larger than that of a Sylow p-subgroup. Example 7.9. If the p-elements of G form a subgroup H , then that is the unique Sylow p-subgroup. By Lemma 5.3, this happens whenever G is a finite abelian group. But it can also occur for certain nonabelian groups. As an obvious example, consider D8 . The entire group is the Sylow 2-subgroup. Example 7.10. As |A4 | = 12, a Sylow 2-subgroup has to have order 4 and a Sylow 3-subgroup has to have order 3. In fact, there is just one Sylow 2-subgroup, namely {(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} (discussed in Example 6.11). However, there are four different Sylow 3-subgroups, namely \u0004(1 2 3)\u0005, \u0004(1 2 4)\u0005, \u0004(1 3 4)\u0005 and \u0004(2 3 4)\u0005. The First Sylow Theorem says that we can always find a Sylow p-subgroup. Theorem 7.8 (First Sylow Theorem). Let G be a finite group. Then for every prime p, G has at least one Sylow p-subgroup. Proof. We will proceed by strong induction on |G|. If |G| = 1, then {e} is the Sylow p-subgroup for any prime p. Therefore, let |G| > 1 and suppose that the theorem holds for smaller groups. Fix a prime p, and let |G| = p n r , where n ≥ 0 and ( p, r ) = 1. If n = 0, then again, the Sylow p-subgroup is {e}, so assume that n ≥ 1. Suppose there exists a noncentral element a ∈ G such that p does not divide [G : C(a)]. Then as |G| = |C(a)|[G : C(a)], we see that p n divides |C(a)| (and certainly no higher power of p can do so, as C(a) ≤ G). But a is not central, so C(a) = G. Therefore, by our inductive hypothesis, C(a) has a subgroup of order p n . But this is also a subgroup of G, completing this case. Therefore, assume that for every noncentral a ∈ G, we have p|[G : C(a)]. Also, p||G|. Therefore, p divides every term in the class equation except for |Z (G)|, which means that p must divide |Z (G)| as well. By Cauchy’s theorem for abelian groups, Z (G) has an element z of order p. Then \u0004z\u0005 is a central, hence normal, subgroup of order p in G. Furthermore, |G/\u0004z\u0005| = |G|/|\u0004z\u0005| = p n r/ p = p n−1r . By our inductive hypothesis, G/\u0004z\u0005 has a subgroup of order p n−1 . But Theorem 4.8 tells us that the subgroups of G/\u0004z\u0005 are of the form H/\u0004z\u0005, where H is a subgroup of G containing \u0004z\u0005. However, |H | = |H/\u0004z\u0005||\u0004z\u0005| = p n−1 p = p n . Therefore, H is a Sylow p-subgroup. \u0002\n\n7.3 The Three Sylow Theorems\n\n123\n\nThe Second Sylow Theorem says that Sylow p-subgroups are always conjugate to each other. Theorem 7.9 (Second Sylow Theorem). Fix a prime p. Let G be a finite group and P a Sylow p-subgroup of G. If H is a subgroup of G of order p k , for some k ≥ 0, then H is conjugate to a subgroup of P. In particular, all Sylow p-subgroups of G are conjugate. Proof. By Theorem 7.7, there are [G : N (P)] different conjugates of P in G. Also, by Theorem 7.2, P ≤ N (P). Therefore, |G|/|P| = (|G|/|N (P)|)(|N (P)|/|P|) = [G : N (P)][N (P) : P]. Hence, [G : N (P)] divides |G|/|P|. But by definition of a Sylow p-subgroup, |G|/|P| is relatively prime to p; thus, the number of conjugates of P is relatively prime to p. Among all of these subgroups conjugate to P (and hence to each other), let us consider those that are H -conjugate. We know that H -conjugacy is an equivalence relation, and the H -conjugacy classes partition the set of all conjugates of P. If all of these H -conjugacy classes contained numbers of elements that are divisible by p, then the total number of conjugates would be divisible by p, which is impossible. Therefore, there is a subgroup K of G, conjugate to P, such that the number of H -conjugates of K is not divisible by p. Now conjugate subgroups have the same order, so K is also a Sylow p-subgroup of G. By Theorem 7.7, the number of H -conjugates of K is [H : N H (K )]. But |H | = p k , and [H : N H (K )] = |H |/|N H (K )| is a divisor of |H |. The only way we can avoid having [H : N H (K )] be a multiple of p is if it is 1. Thus, H = N H (K ). That is, H = H ∩ N (K ), which means that H ≤ N (K ). Theorem 7.2 tells us that N (K ) contains K as a normal subgroup. Thus, by Theorem 4.5, H K is also a subgroup of N (K ). But by Theorem 4.4, |H K | = |H ||K |/|H ∩ K |. However, |K | is the largest power of p dividing |G|, and since |H K | must divide |G|, we conclude that p does not divide |H |/|H ∩ K |. As H has order p k , this means that H ∩ K = H ; thus, H ≤ K . But K is a conjugate of P! That is, H ≤ g −1 Pg, for some g ∈ G. Equivalently, (g −1 )−1 H g −1 ≤ P, as required. The fact that Sylow subgroups are conjugate now follows immediately from the fact that conjugate subgroups have the same order. \u0002 The Third Sylow Theorem imposes restrictions upon the possible numbers of Sylow p-subgroups in a group. Theorem 7.10 (Third Sylow Theorem). Let p be a prime and G a group of order p n r , where n ≥ 0 and ( p, r ) = 1. Then the number of Sylow p-subgroups of G is congruent to 1 modulo p and divides r . Proof. Fix a Sylow p-subgroup P. By the Second Sylow Theorem, every Sylow p-subgroup of G is conjugate to P. Also, as conjugate subgroups have the same\n\n124\n\n7 The Sylow Theorems\n\norder, only Sylow p-subgroups can be conjugate to P. Therefore, the set of Sylow p-subgroups of G is precisely the set of conjugates of P. By Theorem 7.7, there are [G : N (P)] such conjugates. But P ≤ N (P), which means that [G : P] = |G|/|P| = (|G|/|N (P)|)(|N (P)|/|P|). By definition of the Sylow p-subgroup, [G : P] = r , which means that [G : N (P)] divides r , giving us the last part of the theorem. Now, we know that P-conjugacy is an equivalence relation on the set of all Sylow p-subgroups. Thus, the number of Sylow p-subgroups is the sum of the sizes of the P-conjugacy classes. But if H is a Sylow p-subgroup, then by Theorem 7.7, it has precisely [P : N P (H )] P-conjugates. As P has order p k , we see that [P : N P (H )] = |P|/|N P (H )| is also a power of p. Thus, it is in particular a multiple of p, unless it is 1. So, to determine the number of Sylow p-subgroups modulo p, we have only to consider those H such that P = N P (H ). But this happens if and only if P = N (H ) ∩ P; that is, if and only if P ≤ N (H ). Now proceed as in the proof of the Second Sylow Theorem; we see that this happens if and only if P ≤ H . However, P and H have the same order, so this means that P = H . That is, modulo p, the number of Sylow p-subgroups is [P : N P (P)]. But P is normal in itself, so this is [P : P] = 1. The proof is complete. \u0002 Exercises 7.21. Find all Sylow 2-, 5- and 7-subgroups of Z100 × Z14 . 7.22. Find all Sylow 2- and 3-subgroups of A4 . 7.23. Let G be a group of order 294. Show that G has exactly one Sylow 7-subgroup. 7.24. Let G be a finite group. Explain why it is impossible for G to have one Sylow 2-subgroup isomorphic to Z4 and another Sylow 2-subgroup isomorphic to Z2 × Z2 . 7.25. Suppose that p is a prime and p n divides |G|, for some n ∈ N. Show that G has a subgroup of order p n . 7.26. Find a Sylow 2-subgroup of S4 . To what familiar group is it isomorphic? 7.27. Let G be a finite group having a normal subgroup N . If H is a Sylow p-subgroup of G, show that H N /N is a Sylow p-subgroup of G/N . 7.28. If G is a finite group with normal subgroup N , and H is a Sylow p-subgroup of G, show that H ∩ N is a Sylow p-subgroup of N .\n\n7.4 Applying the Sylow Theorems\n\n125\n\n7.4 Applying the Sylow Theorems Let us discuss some interesting consequences of the Sylow theorems. For one thing, we can now complete Cauchy’s theorem, which we previously discussed for abelian groups. Theorem 7.11 (Cauchy’s Theorem). Let G be a finite group, and suppose that a prime p divides its order. Then G has an element of order p. Proof. By the First Sylow Theorem, G has a Sylow p-subgroup P, the order of which is p n , for some positive integer n. Take any e = a ∈ P. By Lagrange’s theorem, k−1 \u0002 |a| = p k , 1 ≤ k ≤ n. Then by Corollary 3.2, |a p | = p. As a consequence, we can now extend Corollary 5.2 to nonabelian groups. Corollary 7.3. Let p be a prime. Then a finite group G is a p-group if and only if |G| = p n , for some n ≥ 0. Proof. If |G| = p n , then by Lagrange’s theorem, every element has order dividing p n , so G is a p-group. On the other hand, if some prime q different from p divides |G|, then by Cauchy’s theorem, G has an element of order q, so it is not a p-group. \u0002 The Third Sylow Theorem tells us about the possible numbers of Sylow p-subgroups. But from the Second Sylow Theorem, we can deduce when there is just one such subgroup. Corollary 7.4. Let p be a prime and G a finite group. Then G has just one Sylow p-subgroup if and only if the Sylow p-subgroup is normal in G. Proof. Let P be a Sylow p-subgroup of G. Then P is normal if and only if a −1 Pa = P for all a ∈ G; in other words, if and only if P has only itself as a conjugate. But Theorem 7.9 tells us that every Sylow p-subgroup is conjugate to P, and since conjugates have the same order, this means that nothing that is not a Sylow p-subgroup can be conjugate to P. That is, P is normal if and only if it is conjugate only to itself, if and only if there is only one Sylow p-subgroup. \u0002 This corollary is highly useful in finding normal subgroups of groups of a particular order. In particular, if we are asked to show that groups of some particular order cannot be simple, then our first step is often to see if some Sylow p-subgroup must be normal. For instance: Theorem 7.12. Let G be a group of order pq, where p and q are primes with p < q. Then the Sylow q-subgroup of G is normal. In particular, G is not simple. Proof. By the Third Sylow Theorem, the number of Sylow q-subgroups is of the form 1 + kq, with k ∈ Z, and divides p. As q > p, the only possibility is k = 0 and 1 + kq = 1. Now apply the preceding corollary. \u0002\n\n126\n\n7 The Sylow Theorems\n\nLet us try something slightly more complicated. Example 7.11. Let us show that there are no simple groups of order 351. As 351 = 33 · 13, we note that the number of Sylow 3-subgroups is 1 + 3k and divides 13 and the number of Sylow 13-subgroups is 1 + 13l and divides 27, with k, l ∈ Z. The only solutions for l are 0 and 2; that is, the number of Sylow 13-subgroups is either 1 or 27. If it is 1, then we know that the Sylow 13-subgroup is normal, and we are done. So let us assume that it is 27. Now, each Sylow 13-subgroup is of order 13. In a group of prime order, everything but the identity has order equal to that of the group; thus, each Sylow 13-subgroup has 12 elements of order 13. Furthermore, if P and Q are different Sylow 13-subgroups, then since |P ∩ Q| must divide |P| = 13, either P = Q (which is impossible) or P ∩ Q = {e}. Thus, each of the 27 Sylow 13-subgroups contributes 12 elements of order 13, and there is no overlap. We have now used up 12 · 27 = 324 elements of the group. This leaves only 351 − 324 = 27 elements. But that is the size of one Sylow 3-subgroup! Thus, there is only room for one such subgroup. In order words, either the Sylow 13-subgroup or the Sylow 3-subgroup must be normal. We do have to be a bit careful in solving problems like the one in the preceding example. It would not have worked well if we had considered the Sylow 3-subgroups first. To be sure, we would have found that the number of such subgroups is 1 or 13, and if it is 1, we are done. But if it is 13, we would have a problem counting the 3-elements we have used, because the Sylow 3-subgroups do not have prime order and, therefore, do not necessarily intersect trivially. Let us consider groups with order the product of three primes. Theorem 7.13. Let G be a group of order pqr , where p, q and r are distinct primes. Then G is not simple. Proof. Without loss of generality, let us say that p < q < r . Then the number of Sylow r -subgroups is of the form 1 + kr , with k ∈ Z, and divides pq. Now, the only positive divisors of pq are 1, p, q and pq. Since r > q > p, we cannot have 1 + kr = p or q. If there is only one Sylow r -subgroup, then by Corollary 7.4, we are done. Thus, let us assume that there are pq of them. Now, these Sylow r -subgroups have prime order, so just as in Example 7.11, they intersect trivially, and provide us with pq(r − 1) elements of order r . Similarly, the number of Sylow q-subgroups is 1 + lq, with l ∈ Z, and divides pr . As q > p, it cannot be p. If it is 1, then once again, we are done. So it is either r or pr . In any case, it is at least r . Therefore, by the same argument, we obtain at least r (q − 1) elements of order q. Finally, the number of Sylow p-subgroups is 1 + mp, with m ∈ Z, and divides qr . If it is 1, then we are done, so we may assume that it is at least q. Thus, we obtain at least q( p − 1) elements of order p. Adding in the identity, we now have at least pq(r − 1) + r (q − 1) + q( p − 1) + 1 = pqr + qr − q − r + 1 > pqr + qr − 2r + 1\n\n7.4 Applying the Sylow Theorems\n\n127\n\nelements (since q < r ). But as q > p, and p is a prime, we have q ≥ 3, so qr > 2r , and we have accounted for more than pqr group elements, which is impossible. \u0002 In the special case where all of the Sylow p-subgroups are normal, we are in an even better position. Theorem 7.14. Let G be a group of order p1n 1 · · · pkn k , where the pi are distinct primes and the n i are positive integers. If, for each i, G has a unique Sylow pi -subgroup Pi , then G = P1 × · · · × Pk . Proof. Let ai be a pi -element of G. Then by the Second Sylow Theorem, there exists a gi ∈ G such that gi−1 \u0004ai \u0005gi ≤ Pi ; say gi−1 ai gi = h i ∈ Pi . Then ai = gi h i gi−1 ∈ Pi since, by Corollary 7.4, Pi is normal. In particular, each Pi is the set of all pi -elements of G. By Lemma 5.4, every element of G can be written as a product of pi -elements, 1 ≤ i ≤ k. Thus, G = P1 P2 · · · Pk . By Exercise 5.10, G is the internal direct product \u0002 of the Pi . Example 7.12. Suppose we wish to classify the groups of order 45. The number of Sylow 3-subgroups is 1 + 3k and divides 5, for some k ∈ Z. Thus, it can only be 1. The number of Sylow 5-subgroups is 1+5l and divides 9, for some l ∈ Z. Therefore, the Sylow 5-subgroup is unique as well. According to the preceding theorem, a group of order 45 must be the direct product of its Sylow subgroups. By Corollary 7.2, a group of order 9 is isomorphic to either Z3 × Z3 or Z9 , and Corollary 4.2 tells us that a group of order 5 is isomorphic to Z5 . Hence, every group of order 45 is isomorphic to either Z3 × Z3 × Z5 or Z9 × Z5 (and these are not isomorphic to each other, by Theorem 5.6). Exercises 7.29. Show that there are no simple groups of order 84. 7.30. Show that there are no simple groups of order 56. 7.31. Let G be a group of order 4352 = 28 ·17. Show that either a Sylow 2-subgroup or a Sylow 17-subgroup of G must be normal. 7.32. Let G be a group of order 870 = 2 · 3 · 5 · 29. Show that at least one of the Sylow p-subgroups of G must be normal, for some prime p dividing |G|. 7.33. Let G be a group of order p 2 q, for some distinct primes p and q. If q \u0003 ( p 2 −1) and p \u0003 (q − 1), show that G is abelian. 7.34. Show that Theorem 7.13 is still true even if the primes p, q and r are not assumed to be distinct. 7.35. Let G be a group of order 57. There are only two possible numbers of elements of order 3 in G. What are they?\n\n128\n\n7 The Sylow Theorems\n\n7.36. Let G be a group of order 935 = 5 · 11 · 17. Show that the Sylow 17-subgroup of G is central. 7.37. Let G be a group of order 595 = 5 · 7 · 17. Show that G has a subgroup of order 119. 7.38. Let G be a nontrivial finite p-group. If H is a proper subgroup of G, show that H is a proper subgroup of N (H ).\n\n7.5 Classification of the Groups of Small Order We conclude our discussion of groups by classifying the groups of order up to 15. Why 15 in particular? Because the classification of the groups of order 16 is a confounded nuisance! There are, in fact, 14 different nonisomorphic groups of that order. We are aware of the five abelian groups (see Example 5.7), the dihedral group D16 , and D8 × Z2 , but constructing the other seven would be a lot of work. Let G be a group of order n. We already know all of the possibilities for most of the values n < 16. If n = 1, there is only the trivial group, {e}. When n ∈ {2, 3, 5, 7, 11, 13}, Corollary 4.2 tells us that G is isomorphic to Zn . If n ∈ {4, 9}, we rely upon Corollary 7.2, which says that if p is a prime and n = p 2 , then G is isomorphic to Z p2 or Z p × Z p . Also, if n ∈ {6, 10, 14}, then we use Theorem 4.15; when n = 2 p, for some odd prime p, we find that G is isomorphic to Z2 p or D2 p . We are left with groups of order 8, 12 and 15. With the aid of the Sylow theorems, the n = 15 case is a piece of cake. Theorem 7.15. Every group of order 15 is isomorphic to Z15 . Proof. By the Third Sylow Theorem, the number of Sylow 3-subgroups is 1 + 3k, for some k ∈ Z, and divides 5. Thus, there is only one Sylow 3-subgroup. By Theorem 7.12, the same is true for the Sylow 5-subgroup. Therefore, by Theorem 7.14, our group is the direct product of these Sylow subgroups. But the Sylow subgroups have prime order and, therefore, are cyclic. By Theorem 5.4, the direct product of cyclic groups of relatively prime order is also cyclic. Thus, by Theorem 4.14, our \u0002 group is isomorphic to Z15 . Unfortunately, when it comes to groups of order 8, the Sylow theorems cannot help us. Indeed, for any finite p-group, the unique Sylow p-subgroup is the whole group. We can, nevertheless, classify the groups of order 8 up to isomorphism. In view of Corollary 5.1, we know that the abelian groups of order 8 are all isomorphic to one of {Z8 , Z4 × Z2 , Z2 × Z2 × Z2 } (and, by Theorem 5.6, these groups are not isomorphic to each other). Let G be a nonabelian group of order 8. By Lagrange’s theorem, every nonidentity element of G has order 2, 4 or 8. If there is an element of order 8 then G is cyclic, hence abelian, which is not the case. Also, if every nonidentity element has order 2,\n\n7.5 Classification of the Groups of Small Order\n\n129\n\nthen G is abelian, by Exercise 3.32. Therefore, we may assume the existence of an element a of order 4. Then \u0004a\u0005 has index 2 and, by Theorem 4.1, is normal in G. Take any element b in G that is not in \u0004a\u0005. Then we observe that the elements of G are precisely a i and ba i , 0 ≤ i ≤ 3. Also, by the normality of \u0004a\u0005, we have b−1 ab = a j , for some j. Now, conjugate elements have the same order, so j = 1 or 3. If j = 1, then a and b commute. But this means that all elements of the form a i and ba i commute as well, so G is abelian, which is not the case. Thus, b−1 ab = a 3 . What is the order of b? We know it is 2 or 4. Suppose that it is 2. We can now follow the final part of the proof of Theorem 4.15, and we see that G is isomorphic to D8 . Therefore, let |b| = 4. Now, G/\u0004a\u0005 has order 2. Thus, (b\u0004a\u0005)2 = e\u0004a\u0005, so b2 ∈ \u0004a\u0005. Furthermore, |b2 | = 2 (by Corollary 3.2), so b2 = a 2 . But now we know everything about the group. We know what the elements are. Furthermore, a k a l = a k+l , ba k a l = ba k+l , a k ba l = b(b−1 a k b)a l = b(b−1 ab)k a l = b(a 3 )k a l = ba 3k+l and (ba k )(ba l ) = b(ba 3k+l ) = a 2 a 3k+l = a 3k+l+2 , for any k, l ∈ Z. What this means is, we can completely fill in the group table so, up to isomorphism, there can be at most one group meeting this description. This does not, however, mean that such a group necessarily exists. As it happens, it does! Example 7.13. The quaternion group is the group Q 8 = {±1, ±i, ± j ± k}, where i 2 = j 2 = k 2 = −1, i j = k = − ji, jk = i = −k j and ki = j = −ik. The element 1 is the identity, and it is easy to see that the group is closed and every element has an inverse (for instance, i −1 = −i). Checking associativity involves verifying a lot of cases, but it does work. Furthermore, i j = ji, so Q 8 is not abelian. Also, we note that the only element of order 2 is −1, whereas D8 has many elements of order 2. Thus, we have a new group, and it must be the one we described above. (In the notation we used, let a = i and b = j.) We now record our classification of the groups of order 8. Theorem 7.16. Every group of order 8 is isomorphic to one of the following, namely Z8 , Z4 × Z2 , Z2 × Z2 × Z2 , D8 or Q 8 . Finally, suppose that G has order 12. The number of Sylow 3-subgroups is 1+3m, m ∈ Z, and divides 4, so it is 1 or 4. As a group of order 3 is cyclic, let H = \u0004d\u0005 be a Sylow 3-subgroup. The number of Sylow 2-subgroups is 1 + 2l, l ∈ Z, and divides 3, so it is 1 or 3. Let K be a Sylow 2-subgroup, which we know is isomorphic either to Z4 or Z2 × Z2 . Let us break our discussion down into cases. CASE I: The Sylow 2- and 3-subgroups are both unique. (Note that this must be the case if G is abelian, as every subgroup is normal.) By Theorem 7.14, G is\n\n130\n\n7 The Sylow Theorems\n\nthe direct product of its Sylow subgroups. Thus, G is isomorphic to Z4 × Z3 or Z2 × Z2 × Z3 . By Theorem 5.6, these groups are not isomorphic. CASE II: There are three Sylow 2-subgroups and four Sylow 3-subgroups. Now, we proceed as in Example 7.11. As the Sylow 3-subgroups have prime order, they intersect trivially, and every element other than the identity has order 3. Thus, we have 4 · 2 = 8 elements of order 3. But this leaves only four elements unaccounted for. Thus, there is only room for one Sylow 2-subgroup. This case cannot occur. CASE III: The Sylow 2-subgroup is unique, but there are four Sylow 3-subgroups. Let us further break this case down. CASE IIIa: K is isomorphic to Z2 × Z2 . Notice that |H K | = |H ||K |/|H ∩ K | = 4 · 3/1 = 12, using Theorem 4.4 and the fact that H ∩ K must have order dividing both |H | and |K |. Thus, H K = G. Also, H and K are abelian, and if d ∈ C(K ), then H ≤ C(K ), and we see that G is abelian which, as we noted above, means we must be in Case I. Thus, there exists e = a ∈ K such that d −1 ad = b = a. As K is normal, b ∈ K . Now, if d −1 ad = d −1 bd, then a = b, which is impossible, so d −1 bd = b. Also, if d −1 bd = a, then d −2 ad 2 = d −1 (d −1 ad)d = d −1 bd = a, so d 2 ∈ C(a), and therefore d = (d 2 )2 ∈ C(a), which is not the case. As K = {e, a, b, ab}, we must have d −1 bd = ab. This means that d −1 abd = d −1 add −1 bd = bab = a, as the Sylow 2-subgroup is isomorphic to Z2 × Z2 . Now, we know that the elements of G are precisely hk, with h ∈ H and k ∈ K . We also know how products work and can construct a group table. For example, (d 2 ab)(d 2 b) = d(d −2 abd 2 )b = d(d −1 (d −1 abd)d)b = d(d −1 ad)b = db2 = d. Thus, there is at most one group in this case, up to isomorphism. The question that remains is, can such a group be constructed? In fact it can, and we have already seen it. If we let G = A4 , a = (1 2)(3 4), b = (1 3)(2 4) and d = (1 2 3), we find that all of our conditions are met. CASE IIIb: K = \u0004c\u0005 is cyclic of order 4. As K is normal, we have d −1 cd ∈ \u0004c\u0005, say d −1 cd = ci . As conjugates have the same order, i = 1 or 3. If i = 1, then d ∈ C(c), so K ≤ C(c). As in CASE IIIa, G = H K , and we see that G is abelian, which is not permitted. Therefore, d −1 cd = c3 . But then d −2 cd 2 = d −1 (d −1 cd)d = d −1 c3 d = (d −1 cd)3 = (c3 )3 = c. Thus, d −3 cd 3 = d −1 cd = c3 . But d 3 = e, so we have a contradiction. Therefore, this case cannot occur. CASE IV: The Sylow 3-subgroup is unique, but G has three Sylow 2-subgroups. Again, let us break this down further. CASE IVa: G has a Sylow 2-subgroup isomorphic to Z2 × Z2 . Now, if K ≤ C(d), then we see that elements of K commute with elements of H and once again, G is abelian, which is not permitted. Therefore, take a ∈ K such that a −1 da = d. Now, \u0004d\u0005 is normal, and given that the only nonidentity elements are d and d 2 , we have a −1 da = d 2 . If b is another nonidentity element of K , we must also have b−1 db = d or d 2 . In the latter case,\n\n7.5 Classification of the Groups of Small Order\n\n131\n\n(ba)−1 d(ba) = a −1 (b−1 db)a = a −1 d 2 a = (a −1 da)2 = (d 2 )2 = d. Thus, one of the nonidentity elements of K centralizes d. Without loss of generality, say bd = db. What is the order of bd? If (bd) j = e, then as b and d commute, we have b j d j = e, so b j = d − j ∈ H ∩ K = {e}, since H and K have relatively prime orders. Thus, j must be divisible by 2 and 3, and hence 6. On the other hand, (bd)6 = b6 d 6 = e, so |bd| = 6. Also, a has order 2 and a −1 bda = (a −1 ba)(a −1 da) = bd 2 = (bd)−1 , / \u0004bd\u0005. since (bd 2 )(bd) = b2 d 3 = e. Since a does not commute with bd, a ∈ Hopefully this situation rings a bell! Refer to the proof of Theorem 4.15. It is at this point that we can conclude that we have constructed D12 . Note that D12 and A4 are not isomorphic, since D12 has an element of order 6 and A4 does not. Finally, we have CASE IVb: K = \u0004c\u0005 is cyclic of order 4. As H is normal, we have c−1 dc = d j , for some j. Since the identity is only conjugate to itself, this means that c−1 dc = d or d 2 . If c−1 dc = d, then we see immediately that all powers of c and d commute. But once again, G = K H , so G is abelian, which is not the case. Thus, c−1 dc = d 2 . But since G = K H , the elements of G are precisely cr d s , 0 ≤ i ≤ 3 and 0 ≤ s ≤ 2. And we now know how to take a product of any two group elements. For instance, (cd)(cd 2 ) = c2 (c−1 dc)d 2 = c2 d 2 d 2 = c2 d. In particular, we can fill in the group table. This means that there is at most one more group of order 12 that is not isomorphic to any of the ones we have constructed so far. In fact, such a group exists. Example 7.14. Let G = S3 × Z4 . This is a group of order 24. Let H be the set of all elements (σ, t) ∈ G such that the permutation σ and the number t are either both even or both odd. In Exercise 7.40, we are asked to show that H is a subgroup of G of order 12 and that it is not isomorphic to any of the other groups of order 12 that we have found. Thus, it must be the group from CASE IVb. In fact, using c = ((1 2), 1) and d = ((1 2 3), 0), we find that it has the desired properties. We have now completed the classification of groups of order 12. Theorem 7.17. Let G be a group of order 12. Then G is isomorphic either to Z4 ×Z3 , Z2 × Z2 × Z3 , A4 , D12 , or the group H from Example 7.14. Exercises 7.39. To which of the groups listed in Theorem 7.17 is D6 × Z2 isomorphic?\n\n132\n\n7 The Sylow Theorems\n\n7.40. Let H be the subset of S3 × Z4 described in Example 7.14. Show that H is a subgroup of order 12 in S3 × Z4 , and that H is not isomorphic to Z4 × Z3 , Z2 × Z2 × Z3 , A4 or D12 . 7.41. Show that every subgroup of Q 8 is normal. 7.42. Let H be a finite abelian group. Show that every subgroup of Q 8 × H is normal if and only if H has no elements of order 4. 7.43. Let p be a prime. If a, b ∈ Z p and a = 0, define αa,b : Z p → Z p via αa,b (x) = ax + b. 1. Show that these αa,b form a group G under composition and, if p > 2, that this group is not abelian. 2. If p = 7, find a nonabelian subgroup H of G such that |H | = 21. 7.44. Show that every group of order 21 is isomorphic either to Z21 or to the group H from the second part of the preceding exercise. 7.45. Generalize Theorem 7.15 as follows. If p and q are primes, with p > q and q \u0003 ( p − 1), show that every group of order pq is isomorphic to Z pq . 7.46. We know from Theorem 6.11 that A5 is a nonabelian simple group of order 60. Show that there are no nonabelian simple groups with order smaller than 60. (The methods we have discussed up to this point are sufficient to deal with every order except for 24, 36 and 48. Here is a hint if |G| = 36: Suppose that G has distinct Sylow 3-subgroups H and K . What is |H ∩ K |? What can you say about |N (H ∩ K )|? Find a nontrivial proper normal subgroup of G.)\n\nPart III\n\nRings\n\nChapter 8\n\nIntroduction to Rings\n\nWe now move on to the second major type of algebraic object that we are considering: the ring. At first blush, rings look a bit more complicated than groups. Indeed, a ring is an abelian group written additively, and we must still impose a multiplication operation along with several new rules. But in another sense, rings are easier to deal with, because they are more familiar. Indeed, when we think of a ring, we tend to think of the integers (although, as we shall see, the integers are actually a special sort of ring). In this chapter, we will define a ring and prove some properties of rings and subrings. We shall also discuss two well-behaved types of rings; namely, integral domains and fields.\n\n8.1 Rings Let us now define a ring. Definition 8.1. A ring is a set R together with two binary operations, written as addition and multiplication, such that 1. 2. 3. 4. 5.\n\nR is an abelian group under addition; if a, b ∈ R, then ab ∈ R (closure under multiplication); if a, b, c ∈ R, then (ab)c = a(bc) (associativity of multiplication); if a, b, c ∈ R, then a(b + c) = ab + ac (distributive law); and if a, b, c ∈ R, then (a + b)c = ac + bc (distributive law).\n\nAs usual when we have an additive group, we will use additive notation. In particular, we write 0 for the additive identity of a ring, and −a for the additive inverse of a. Notice that we do not insist that the multiplication operation be commutative. © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_8\n\n135\n\n136\n\n8 Introduction to Rings\n\nDefinition 8.2. A ring R is said to be a commutative ring if ab = ba for all a, b ∈ R. Also, while there is an identity for the addition operation, there does not have to be one for the multiplication operation. Definition 8.3. A ring R is said to be a ring with identity if R has an element, denoted 1, such that 1a = a1 = a for all a ∈ R. In this case, we call 1 the identity of R. Note that if we refer to the identity in a ring, we mean the multiplicative identity 1 (if it exists), not the additive identity 0. Example 8.1. As we observed in Section 2.4, the sets Z, Q, R and C are all commutative rings with identity, under the usual addition and multiplication operations. Also, we saw in Section 2.5 that the same can be said for Zn , for any positive integer n ≥ 2. Example 8.2. The set of even integers, 2Z, can easily be seen to be a commutative ring without an identity. There is no even integer that can be multiplied by 2 to get 2. Example 8.3. The set of all polynomials with real coefficients is a commutative ring with identity, using the usual polynomial addition and multiplication operations. We denote it by R[x]. The same can be said for the polynomials with integer coefficients, Z[x]. In each case, the identity is the constant polynomial, 1. How about an example of a noncommutative ring? Example 8.4. Let n be a positive integer. Then the n × n matrices with real entries form a ring under matrix addition and multiplication. The identity matrix is the identity of\u0002 the\u0003ring. \u0002 However, \u0003 \u0002 if\u0003 n\u0002 > \u00031, then it is not a commutative ring as, for 11 10 10 11 instance, \u0004= . We denote this ring by Mn (R). In fact, as 01 11 11 01 we observe in Appendix B, we can substitute entries from any ring R in place of the real numbers, and we obtain a new ring, Mn (R). If R is a ring with identity, then we can form the identity matrix, so Mn (R) is also a ring with identity. The conditions under which it is a commutative ring are discussed in Exercise 8.10. We also have a way of constructing new rings from old, simply extending the idea of the direct product of groups. Definition 8.4. Let R and S be rings. Then the direct sum of R and S, denoted R ⊕ S, is the Cartesian product R × S under the operations (r1 , s1 ) + (r2 , s2 ) = (r1 + r2 , s1 + s2 ) and (r1 , s1 )(r2 , s2 ) = (r1r2 , s1 s2 ), for all ri ∈ R, si ∈ S.\n\n8.1 Rings\n\n137\n\nTheorem 8.1. Let R and S be rings. Then R ⊕ S is a ring. Furthermore, if R and S are commutative rings, then so is R ⊕ S. Also, if R and S are rings with identity, then so is R ⊕ S. Proof. The proof is very similar to that of Theorem 3.1. The ring properties all hold in the direct sum because they hold in R and S. We will prove one of the distributive laws, and leave the rest as Exercise 8.6. Take ri ∈ R, si ∈ S. Then (r1 , s1 )((r2 , s2 ) + (r3 , s3 )) = (r1 , s1 )(r2 + r3 , s2 + s3 ) = (r1 (r2 + r3 ), s1 (s2 + s3 )) = (r1r2 + r1r3 , s1 s2 + s1 s3 ) = (r1r2 , s1 s2 ) + (r1r3 , s1 s3 ) = (r1 , s1 )(r2 , s2 ) + (r1 , s1 )(r3 , s3 ). \u0002 Example 8.5. In Z5 ⊕Z6 , we have (3, 5)+(4, 2) = (7, 7) = (2, 1) and (3, 5)(4, 2) = (12, 10) = (2, 4). One additional point is important to keep in mind. A ring is a group under addition, not under multiplication! While the multiplication operation satisfies the closure and associativity properties, a ring does not have to have an identity. And even if it does, elements do not have to have inverses. For instance, Z has an identity, but there is nothing we can multiply by 2 to obtain 1. Exercises 8.1. Write the addition and multiplication tables for the ring Z5 . 8.2. Write the addition and multiplication tables for the ring Z3 ⊕ Z2 . 8.3. Let R = {0, 3, 6, 9, 12} with addition and multiplication in Z15 . Is R a ring? If so, is it commutative, and does it have an identity? 8.4. Let R be the set of all functions from R to R, under addition and multiplication of functions. Is R a ring? If so, is it commutative, and does it have an identity? 8.5. Let R be the set of all functions from R to R. Let the addition operation be the usual addition of functions, but let the multiplication operation be composition. That is, the product of α and β is α ◦ β. Is R a ring? If so, is it commutative, and does it have an identity? 8.6. Complete the proof of Theorem 8.1.\n\n138\n\n8 Introduction to Rings\n\n\u0002 \u0003 ab , for all a, b, c ∈ Z. Is R a ring 0c under matrix addition and multiplication? If so, is it commutative, and does it have an identity? 8.7. Let R be the set of matrices of the form\n\n8.8. Show that every ring with a prime number of elements is commutative. 8.9. Must a ring with a prime number of elements be a ring with identity? 8.10. Let R be a ring and n a positive integer. Under what conditions is Mn (R) commutative?\n\n8.2 Basic Properties of Rings Let us mention a few straightforward properties of rings. Theorem 8.2. Let R be a ring. Then the additive identity, 0, is unique. If R has a multiplicative identity 1, then it too is unique. Proof. As R is a group under addition, we see from Theorem 3.2 that 0 is unique. Suppose that a and b are both multiplicative identities for R. As a is an identity, ab = b. But as b is an identity, ab = a. Thus, a = b. \u0002 Theorem 8.3. Let R be a ring. If a, b ∈ R, then 1. 0a = a0 = 0; 2. (−a)b = a(−b) = −(ab); and 3. (−a)(−b) = ab. Proof. (1) As 0 = 0 + 0, we have 0a = (0 + 0)a = 0a + 0a. Adding −0a to both sides, we get 0 = 0a. The proof that a0 = 0 is similar. (2) Notice that ab + (−a)b = (a + (−a))b = 0b = 0, by (1). As adding (−a)b to ab gives 0, we have (−a)b = −(ab). The proof that a(−b) = −(ab) is similar. (3) By (2), we have (−a)(−b) = −(a(−b)) = −(−(ab)). But remember that R is a group under addition, and hence −(−(ab)) = ab, as required. \u0002 Corollary 8.1. If R is a ring with identity, then (−1)a = −a, for any a ∈ R. Proof. By the preceding theorem, (−1)a = −(1a) = −a.\n\n\u0002\n\nAs a ring is a group under addition, we know from Theorem 3.3 that an expression such as a1 +a2 +· · ·+an is unambiguous, without the need for brackets. Even though the ring is not a group under multiplication, we can apply precisely the same proof as that of Theorem 3.3 to show that the expression a1 a2 · · · an also does not require brackets.\n\n8.2 Basic Properties of Rings\n\n139\n\nTheorem 8.4. Let R be any ring, and a1 , a2 , . . . , an ∈ R. Then regardless of how the product a1 a2 · · · an is bracketed, the result equals (· · · (((a1 a2 )a3 )a4 ) · · · an−1 )an . In order to avoid mistakes, it is also important to recognize which rules cannot be applied in general. For instance, in ordinary arithmetic using the real numbers, we take for granted that if ab = 0, then a = 0 or b = 0. This is simply not the case in an arbitrary ring. Example 8.6. In Z6 , we have 2 · 3 = 0, but 2 \u0004= 0 and 3 \u0004= 0. Example 8.7. In M2 (R), we have \u0002 \u0003\u0002 \u0003 \u0002 \u0003 12 −2 4 00 = , 36 1 −2 00 but\n\n\u0002\n\n\u0003 \u0002 \u0003 \u0002 \u0003 12 00 −2 4 \u0004= \u0004= . 36 00 1 −2\n\nIn dealing with groups, we have the cancellation law. We are used to something similar happening in ordinary arithmetic; that is, if ab = ac and a \u0004= 0, then b = c. Again, this does not have to hold in rings. Example 8.8. In Z12 , we have 3 · 1 = 3 · 5, but 3 \u0004= 0 and 1 \u0004= 5. Finally, in a group G, we note that if there exists a b ∈ G such that ab = b, then a is the identity. (Just multiply on the right by b−1 .) But even if a ring has an identity, the fact that ab = b does not mean that a = 1. Indeed, the previous example points us in the right direction. Example 8.9. In Z12 , we have 5 · 3 = 3, but 5 \u0004= 1. Thus, to check that a ring element a is the identity, we must make sure that ab = b = ba for every b ∈ R, not just for one such b. Exercises 8.11. Let a and b be elements of a ring R. Simplify the following expressions as far as possible. 1. (a + b)(a − b) 2. (a − b)3 8.12. Let R be a ring with identity. Suppose that there exist a, b, c ∈ R such that ab = ba = 1 and ac = 0. Show that c = 0. 8.13. Let R be a ring with identity. Suppose there exist a, b, c ∈ R such that ba = ac = 1. Does it follow that b = c? Show that it does, or find an explicit counterexample.\n\n140\n\n8 Introduction to Rings\n\n8.14. Let R be a ring and n > 2 a positive integer. Show that if there exists 0 \u0004= a ∈ R such that a n = 0, then there exists 0 \u0004= b ∈ R such that b2 = 0. 8.15. Let R be a ring with identity. Suppose that a(a − 1) = 0 for every a ∈ R. Does it follow that a ∈ {0, 1} for every a ∈ R? Either prove that it does, or construct an explicit counterexample. 8.16. Let R be a ring in which a 2 = a for every a ∈ R. 1. Show that a + a = 0 for every a ∈ R. 2. Show that R is commutative.\n\n8.3 Subrings Just as we have the notion of a subgroup, we can discuss subrings. Definition 8.5. Let R be a ring. Then a subset S of R is said to be a subring if S is a ring under the same addition and multiplication operations as in R. Example 8.10. We see that Z is a subring of Q, and both are subrings of R. Example 8.11. The matrix ring M2 (Q) is a subring of M2 (R). Example 8.12. For any ring R, {0} and R are subrings of R. How can we test if a subset is a subring? Theorem 8.5. Let R be a ring and S a subset of R. Then S is a subring of R if and only if 1. 0 ∈ S; 2. if a, b ∈ S, then a − b ∈ S; and 3. if a, b ∈ S, then ab ∈ S. Proof. Suppose that S is a subring of R. Then it is an additive subgroup. By Theorem 3.13, (1) and (2) hold. As a ring is closed under multiplication, (3) holds as well. Conversely, suppose that (1)–(3) hold. Then by Theorem 3.13, S is an additive subgroup of R. By (3), S is closed under multiplication. The remaining ring properties (associativity and the distributive laws) hold in R, hence in any subset of R. Thus, S is indeed a subring. \u0002 Note that for condition (1), it is actually sufficient to check that S is not the empty set. Example 8.13. Let us show that 2Z is a subring of Z. Certainly 0 ∈ 2Z. If 2a, 2b ∈ 2Z, for some a, b ∈ Z, then 2a − 2b = 2(a − b) ∈ 2Z. Also, (2a)(2b) = 2(2ab) ∈ 2Z.\n\n8.3 Subrings\n\n141\n\n\u0004\u0002 \u0003 \u0005 a0 : a ∈ R . Then letting a = 0, we see that S con00 tains the zero matrix. Also, if a, b ∈ R, then \u0002 \u0003 \u0002 \u0003 \u0002 \u0003 a0 b0 a−b 0 − = ∈S 00 00 0 0\n\nExample 8.14. Let S =\n\nand\n\n\u0002 \u0003\u0002 \u0003 \u0002 \u0003 a0 b0 ab 0 = ∈ S. 00 00 0 0\n\nThus, S is a subring of M2 (R). We recall that the centre of every group is a subgroup. A similar thing happens for rings. Definition 8.6. Let R be a ring. Then the centre of R is the set {z ∈ R : az = za for all a ∈ R}; that is, it is the set of elements of R that commute with everything in R. Theorem 8.6. The centre of any ring is a subring. Proof. Let R be a ring and Z its centre. If a ∈ R, then 0a = 0 = a0, so 0 ∈ Z . Take any y, z ∈ Z . Then for any a ∈ R, we have a(y−z) = ay−az = ya−za = (y−z)a, since y and z are central. Thus, y − z ∈ Z . Also, ayz = yaz = yza, and hence yz ∈ Z . By Theorem 8.5, we are done. \u0002 Example 8.15. If R is a commutative ring, then its centre is all of R. Example 8.16. The centre of M2 (R) is the set of all matrices of the form\n\n\u0002 \u0003 r 0 , for 0r\n\nall real numbers r . See Exercise 8.26. One particular type of subring deserves special mention. Definition 8.7. Let R be a ring with identity 1. Then a subring S of R is said to be a unital subring if 1 ∈ S. Example 8.17. We observe that Z is a unital subring of Q, but 2Z is not a unital subring. Note that a subring can fail to be a unital subring because it does not have an identity (as is the case with 2Z above), but it can also have an identity which is not the same as that for R. Example 8.18. Let R = Z6 and S = {0, 3}. Theorem 8.5 shows us that S is a subring of R. It does not contain 1, so it is not a unital subring. However, S is still a ring with identity, as 3 · 0 = 0 and 3 · 3 = 3. That is, 3 is the identity of S.\n\n142\n\n8 Introduction to Rings\n\nExercises 8.17. Let R = {a + bi : a, b ∈ Z}. Show that R is a subring of C. Is it a ring with identity? If so, is it unital? ⎧⎛ ⎫ ⎞ ⎨ 0ab ⎬ 8.18. Let R = ⎝0 0 c ⎠ : a, b, c ∈ R . Show that R a subring of M3 (R). Is it a ⎩ ⎭ 000 ring with identity? If so, is it a unital subring? \u0002 \u0003 00 8.19. Let R be the set of matrices of the form for all real numbers a. Show 0a that R is a subring of M2 (R). Is it a ring with identity? If so, is it a unital subring? 8.20. Let R be a ring with subrings S and T . Show that S ∩ T is a subring. Extend this to show the intersection of any collection of subrings of R is also a subring. 8.21. Let R and S be rings. Show that T = {(r, 0) : r ∈ R} is a subring of R ⊕ S. 8.22. Find a ring R and an additive subgroup S of R such that S is not a subring of R. 8.23. Let R be a ring and a ∈ R. Show that S = {ra : r ∈ R} is a subring of R. 8.24. Let R be a ring and a ∈ R. Let S = {r ∈ R : ra = 0}. Is S necessarily a subring of R? Prove that it is, or find an explicit counterexample. 8.25. Let R be a ring and a ∈ R. Fix a subring S of R, and let T = {r ∈ R : ra ∈ S}. Is T necessarily a subring of R? Prove that it is, or find an explicit counterexample. \u0002 \u0003 r 0 , for 8.26. Show that the centre of M2 (R) is the set of matrices of the form 0r all r ∈ R.\n\n8.4 Integral Domains and Fields Let us discuss a couple of special sorts of rings. Definition 8.8. Let R be a commutative ring. Then a nonzero element a ∈ R is said to be a zero divisor if there exists a nonzero b ∈ R such that ab = 0. Example 8.19. In Z6 , we note that 4 is a zero divisor, as 4 · 3 = 0. On the other hand, 5 is not a zero divisor. Example 8.20. The ring of integers has no zero divisors.\n\n8.4 Integral Domains and Fields\n\n143\n\nAs we mentioned at the beginning of the chapter, while we tend to think of the integers when we work with rings, they are actually rather special, and this is the reason why. Definition 8.9. An integral domain is a commutative ring R with identity 1 \u0004= 0 having no zero divisors. The condition that 1 \u0004= 0 may seem a bit curious. In fact, if 1 = 0, then for any a ∈ R, we have a = 1a = 0a = 0. Thus, R = {0}. So we are only ruling out one ring with that restriction. Example 8.21. The rings Z, Q, R and C are all integral domains. Example 8.22. The polynomial ring R[x] is an integral domain. Indeed, we know that it is a commutative ring with identity. Also, if f (x) = a0 + a1 x + · · · + an x n and g(x) = b0 + b1 x + · · · + bm x m , with ai , bi ∈ R and an \u0004= 0 \u0004= bm , then the unique term of highest degree in f (x)g(x) is an bm x m+n . As R is an integral domain, an bm \u0004= 0. Thus, f (x)g(x) is not the zero polynomial. Example 8.23. The rings 2Z, Z6 and M2 (R) all fail to be integral domains. The first lacks an identity, the second has zero divisors and the third is not commutative. As we discussed in Section 8.2, rings in general do not enjoy a cancellation law. However, integral domains do. Theorem 8.7 (Cancellation Law). Let R be an integral domain. Suppose that a, b, c ∈ R and ab = ac. If a \u0004= 0, then b = c. Proof. If ab = ac, then ab − ac = 0, and hence a(b − c) = 0. Since R is an integral domain, either a = 0 (which is not true), or b − c = 0, as required. \u0002 We also wish to discuss a stronger restriction on the ring. We need a definition first. Definition 8.10. Let R be a ring with identity. Then we say that an element a ∈ R is a unit if there exists an element b ∈ R such that ab = ba = 1. In this case, we call b the inverse of a and write b = a −1 . We write U (R) for the set of all units of R, and call it the unit group of R. Theorem 8.8. Let R be a ring with identity. Then U (R) is a group under multiplication. Proof. Let a, b ∈ U (R). Then abb−1 a −1 = a1a −1 = aa −1 = 1, and b−1 a −1 ab = b−1 1b = b−1 b = 1. Thus, b−1 a −1 = (ab)−1 , and ab ∈ U (R). Multiplication in a ring is associative. Plainly, 1 ∈ U (R), as 1 · 1 = 1. Also, if a ∈ U (R), then aa −1 = a −1 a = 1. That is, a is the inverse of a −1 , hence a −1 ∈ U (R). We are done. \u0002 Example 8.24. By definition, U (Mn (R)) = G L n (R).\n\n144\n\n8 Introduction to Rings\n\nExample 8.25. The unit group of Z is {±1}. Example 8.26. The unit group of Zn is U (n). See Exercise 8.30. Example 8.27. Every element other than 0 in R is a unit. The same can be said for Q and C. This last example leads us to our next definition. Definition 8.11. Let F be a commutative ring with identity 1 \u0004= 0. Then F is said to be a field if U (F) consists of every element of F other than 0. Example 8.28. As we noted above, Q, R and C are fields. Lemma 8.1. Let R be a commutative ring with identity. Then a unit in R cannot be a zero divisor. \u0002\n\nProof. See Exercise 8.12. This immediately yields the following result. Theorem 8.9. Every field is an integral domain.\n\nOf course, the integers are an integral domain, but not a field. However, we can say something for finite integral domains. As we might expect, if a ∈ R, and n is a positive integer, we write a n = aa · · · a\u0013 . \u0010 \u0011\u0012 n times\n\nTheorem 8.10. Let R be a finite integral domain. Then R is a field. Proof. By definition, R is a commutative ring with identity 1 \u0004= 0. It remains only to check that each nonzero element is a unit. Take 0 \u0004= a ∈ R. Consider the set {a i : i > 0}. It consists of infinitely many powers of a. But R is finite. Thus, there cannot be infinitely many distinct powers. Let us say that a i = a j with i > j > 0. Then a j a i− j = a i = a j . More importantly, a j a i− j = a j · 1. Now, a is a nonzero element of an integral domain, and products of nonzero elements in such a domain do not become zero. Thus, a j \u0004= 0. By the cancellation law, a i− j = 1. If i − j = 1, then a = 1, which is surely a unit. Otherwise, aa i− j−1 = 1. Since i − j − 1 is a \u0002 positive integer, a i− j−1 ∈ R, and we have an inverse for a. We can now handle a particular collection of finite rings of interest. Theorem 8.11. Let n ≥ 2 be a positive integer. Then the following are equivalent: 1. Zn is an integral domain; 2. Zn is a field; and 3. n is prime.\n\n8.4 Integral Domains and Fields\n\n145\n\nProof. In view of Theorems 8.9 and 8.10, we know that (1) and (2) are equivalent. We need only show that they are equivalent to (3). If n is composite, then write n = kl, where k and l are positive integers smaller than n. Then k and l are not 0 in Zn , and yet kl = 0 in Zn . Thus, Zn is not an integral domain. On the other hand, suppose that n is prime. Surely Zn is a commutative ring with identity 1 \u0004= 0. Suppose we have integers i and j such that i j = 0 in Zn . Then n|i j. By Theorem 2.7, n|i or n| j. That \u0002 is, i = 0 or j = 0 in Zn . Thus, Zn is an integral domain. Just as we have subrings, it will also be necessary to know about subfields. Definition 8.12. Let F be a field. Then a subring K of F is said to be a subfield if it is a field using the same addition and multiplication operations. Example 8.29. Q is a subfield of R, which in turn is a subfield of C. But how do we test if a subset is a subfield? Theorem 8.12. Let F be a field. Then a subset S of F is a subfield of F if and only if 1. 1 ∈ S; 2. if a, b ∈ S, then a − b ∈ S; and 3. if a, b ∈ S, and b \u0004= 0, then ab−1 ∈ S. Proof. Suppose that S is a subfield of F. Then S contains an identity f \u0004= 0. We must check that f is 1, the identity of F. But as f is the identity for S, we have f f = f . Now, f is a unit in F, so multiplying by f −1 , we get f = 1. Thus, (1) is proved. Since S is a subring of F, (2) follows from Theorem 8.5. As S is a field, every element except 0 has an inverse. This inverse is unique, as U (F) is a group. Therefore, if 0 \u0004= b ∈ S, then b−1 ∈ S. Since S is a subring, we get (3) as well. Conversely, suppose that (1)–(3) hold. In view of (1) and (2), we see that 0 = 1−1 ∈ S. Take any a, b ∈ S. By (2), a−b ∈ S. If b = 0, then ab = 0 ∈ S. Otherwise, we have b−1 = 1b−1 ∈ S, and therefore ab = a(b−1 )−1 ∈ S. By Theorem 8.5, S is a subring of F. It certainly has an identity 1 \u0004= 0, and it is commutative, since F is. Thus, it remains only to check that every nonzero element has an inverse in S. But we just did that! If 0 \u0004= b ∈ S, then b−1 = 1b−1 ∈ S. Therefore, S is indeed a subfield of F. \u0002 A small word of caution. It is not sufficient to replace (1) with the condition that S is not empty; indeed, if we did so, then we would accept {0} as a field, which is wrong. It would be sufficient to assume that S contains a nonzero element b, for then (3) would give 1 = bb−1 ∈ S. √ We claim that F is a subfield of R. Example 8.30. Let F = {a + b 2 : a, b ∈ Q}. √ 2 ∈ F, √ so (1) holds. If ai , bi ∈ Q, Let us check the conditions. Certainly 1 = 1 + 0 √ √ then (a1 + b1 2) − (a2 + b2 2) = (a1 − a2 ) + (b1 − b2 ) 2 ∈ F, and we have (2). Let us check the final condition. √ To begin √with, we shall show that F is closed under √ multiplication. But (a1 + b1 2)(a2 + b2 2) = (a1 a2 + 2b1 b2 ) + (a1 b2 + a2 b1 ) 2 ∈\n\n146\n\n8 Introduction to Rings\n\nF. Thus, if we can show that every nonzero element of F has √ an inverse in F, then we will be done, as we can obtain (3). Take 0 \u0004= a + b 2 ∈ F. If b = 0, then 0 √ \u0004= a ∈ Q,√and certainly a −1 ∈ Q ⊆ F. Assume that b \u0004= 0. Notice that 2 2 2 (a + b 2)(a − b 2) = a 2 − 2b √ ∈ Q. Also, a − 2b \u0004= 0. Otherwise, we2 would2 −1 2 rational, which have (ab ) = 2, meaning that 2 is√ √ is not the case. Thus, a − 2b has an√inverse c ∈ Q. But then (a + b 2)(ac − bc 2) = (a 2 − 2b2 )c = 1. Hence, a + b 2 has an inverse in F, and F is a subfield of R. Exercises 8.27. Let R = {a + bi : a, b ∈ Q}. Show that R is a subfield of C. 8.28. For each of the following rings, which elements are units? Which are zero divisors? 1. Z18 2. Z3 ⊕ Z9 8.29. Let R and S be rings with identity. Show that U (R ⊕ S) = U (R) × U (S). 8.30. Let n ≥ 2 be a positive integer. Show that U (Zn ) = U (n). 8.31. Show that every integral domain contains exactly two elements a satisfying a 2 = a. 8.32. Let R and S be rings. Under precisely what circumstances is R ⊕ S an integral domain? 8.33. Let F be a field with subfields K and L. Show that K ∩ L is a subfield of F. Extend this to show that the intersection of any collection of subfields is a subfield. 8.34. Let p be a prime and F a field with p 2 elements. Show that F cannot have more than one proper subfield. 8.35. Let R be an integral domain. Suppose that we have a, b ∈ R such that a 13 = b13 and a 10 = b10 . Show that a = b. 8.36. Let R be a finite commutative ring having no zero divisors. Show that R is {0} or an integral domain.\n\n8.5 The Characteristic of a Ring One rather important property of a ring is its characteristic. Letting R be a ring, recall that using additive notation, if we have a ∈ R and some positive integer n, then na = a\u0010 + a + \u0011\u0012· · · + a\u0013 . n times\n\n8.5 The Characteristic of a Ring\n\n147\n\nDefinition 8.13. Let R be a ring. Then the characteristic of R, denoted char R, is the smallest positive integer n such that na = 0 for all a ∈ R. If no such n exists, then char R = 0. Example 8.31. The characteristic of Zn is n, as clearly na = 0 for any a ∈ Zn , whereas no smaller value than n will work if we take a = 1. Example 8.32. The ring of integers has characteristic zero. In fact, for rings with identity, we only need to look at the identity. Theorem 8.13. Let R be a ring with identity. Regarding R as an additive group, if the order of 1 is n < ∞, then R has characteristic n. If 1 has infinite order, then R has characteristic zero. Proof. If 1 has infinite order, then there is no positive integer n such that n1 = 0, and therefore char R = 0. Suppose 1 has order n < ∞. Then no number 1 ≤ m < n can be the characteristic, as m1 \u0004= 0. But on the other hand, if a ∈ R, then na = a\u0010 + a + \u0010 + 1a + \u0010+1+ \u0011\u0012· · · + a\u0013 = 1a \u0011\u0012 · · · + 1a\u0013 = (1 \u0011\u0012· · · + 1\u0013)a = 0a = 0. n times\n\nn times\n\nn times\n\n\u0002\n\nThus, n is the characteristic.\n\nCorollary 8.2. Let R be a ring with identity. Then every unital subring of R has the same characteristic as R. Proof. The same identity has the same order.\n\n\u0002\n\nThe corollary does not apply to subrings that are not unital. For instance, if R = Z6 , then char R = 6, but taking the subring S = {0, 2, 4}, we see that char S = 3. In a commutative ring of prime characteristic, we have the following interesting fact. Theorem 8.14 (Freshman’s Dream). Let R be a commutative ring of prime characteristic p. Then for any a, b ∈ R, we have (a + b) p = a p + b p . Proof. Let us apply the Binomial Theorem. (We are really only familiar with it for real numbers, but the proof in any commutative ring is the same.) We have \u0002 \u0003 \u0002 \u0003 \u0002 \u0003 p p−1 p p−2 2 p a b+ a b + ··· + ab p−1 + b p . (a + b) = a + 1 2 p−1 p\n\np\n\nNow, if 1 < k < p, then\n\n\u0002 \u0003 p p! = . k ( p − k)!k!\n\n148\n\n8 Introduction to Rings\n\nNotice that the numerator is divisible by p. However, p does not divide any of the terms in the denominator and, therefore, it does not divide the denominator. Thus, p \u0014 \u0015 divides each kp , with 1 < k < p. As our ring has characteristic p, multiplying any element by p, and hence by any multiple of p, gives 0. We have our result. \u0002 We tend to encounter commutative rings with prime characteristic a lot in the context of integral domains. Theorem 8.15. The characteristic of an integral domain is either zero or a prime. Proof. Let R be an integral domain. There is nothing to do if char R = 0, so let char R = n > 0. We cannot have n = 1, for then 1 has additive order 1, but only 0 has that order. The only remaining problem is if n is composite. Suppose that n = kl, with 1 < k, l < n. Then we have 0 = n1 = (kl)1 = \u00101 + ·\u0011\u0012 · · + 1\u0013 = (1\u0010 + ·\u0011\u0012 · · + 1\u0013)(1 · · + 1\u0013) = (k1)(l1). \u0010 + ·\u0011\u0012 kl times\n\nk times\n\nl times\n\nSince R is an integral domain, k1 = 0 or l1 = 0. But the additive order of 1 is n, and we have a contradiction. \u0002 Exercises 8.37. Find the characteristic of each of the following rings. 1. 3Z21 = {0, 3, . . . , 18} 2. R[x] 8.38. Find the characteristic of each of the following rings. 1. Z4 ⊕ Z10 2. M2 (Z3 ) 8.39. Show that a finite integral domain R must have order p n for some prime p and positive integer n. 8.40. Let F be a field of prime characteristic p. Show that for every positive integer n n, {a ∈ F : a p = a} is a subfield of F. 8.41. Let R be a commutative ring with identity, and suppose that a ∈ R satisfies a n = 0 for some positive integer n. 1. Show that 1 + a ∈ U (R). 2. If char R is prime, show that 1 + a has finite order in U (R). 8.42. Let F = {0, 1, a, b} be a field with four elements. Write the addition and multiplication tables for F.\n\nChapter 9\n\nIdeals, Factor Rings and Homomorphisms\n\nWe saw in Chapter 4 that for some purposes, subgroups are not quite good enough; we needed to consider normal subgroups. There is a similar concept in ring theory. In this chapter, we introduce the notion of an ideal, which is a subring with one additional condition imposed. We can then define factor rings and discuss ring homomorphisms and isomorphisms. Along the way, we will mention several important sorts of ideals, including principal, maximal and prime ideals.\n\n9.1 Ideals When we discussed normal subgroups of a group, our main concern was to find a condition that we could impose in order to make the group operation on a factor group well-defined. Our motivation is the same here. Of course, a subring is necessarily an additive subgroup of a ring, and as the additive group is abelian, we do not have to worry about normality. However, we need an additional condition to make the multiplication operation work properly. Definition 9.1. Let R be a ring. Then a subring I of R is said to be an ideal if ir, ri ∈ I for all i ∈ I and r ∈ R. We call this the absorption property. Note that closure under multiplication is not enough. We need to be able to multiply an element of the ideal by any element of the ring and stay within the ideal. Combining the definition with Theorem 8.5, we immediately obtain the following. Theorem 9.1. Let R be a ring and I a subset of R. Then I is an ideal if and only if 1. 0 ∈ I ; 2. i − j ∈ I for all i, j ∈ I ; and 3. ir, ri ∈ I for all i ∈ I , r ∈ R. © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_9\n\n149\n\n150\n\n9 Ideals, Factor Rings and Homomorphisms\n\nExample 9.1. Let n be any integer. Then nZ is an ideal of Z. Indeed, we already know that it is a subring. But also, if nk ∈ nZ, then for any integer r , r (nk) = n(r k) ∈ nZ. Example 9.2. Let I be the set of all polynomials f (x) ∈ R[x] such that f (0) = 0. We claim that I is an ideal in R[x]. Certainly I contains the zero polynomial. Also, if f (0) = g(0) = 0, then ( f −g)(0) = f (0)−g(0) = 0, hence f (x)−g(x) ∈ I . Also, if f (0) = 0 and h(x) ∈ R[x], then h(0) f (0) = h(0)0 = 0. Hence, h(x) f (x) ∈ I . Example 9.3. Let I be the set of all polynomials in Z[x] whose constant term is a multiple of 5. Then I is an ideal. See Exercise 9.2. Example 9.4. For any ring R, {0} and R are ideals of R. Example 9.5. In Q, the ring of integers is a subring but not an ideal. Indeed, 3 ∈ Z, but 3(1/5) ∈ / Z. Thus, Z does not have the absorption property. Actually, this last example is not particularly surprising. Fields do not have interesting ideals, as the following results illustrate. Theorem 9.2. Let R be a ring with identity. If an ideal I of R contains a unit, then I = R. Proof. Let u ∈ I be a unit. Then by the absorption property, 1 = uu −1 ∈ I . But then for any a ∈ R, we also have a = 1a ∈ I . Thus, I = R. \u0002 Corollary 9.1. Let F be a field. Then the only ideals of F are {0} and F. Proof. Let I be an ideal of F. If I = {0}, there is nothing to do, so assume that 0 \u0003= a ∈ F. Then a is a unit, and by the preceding theorem, I = F. \u0002 There are a number of ways in which we can obtain new ideals from old ones. For instance, let I and J be ideals of a ring R. Then we write I + J = {i + j : i ∈ I, j ∈ J }. Example 9.6. In Z, if we let I = 4Z and J = 6Z, then I + J = 2Z. Indeed, if m ∈ I + J , then m = 4a + 6b = 2(2a + 3b), for some integers a and b. In particular, I + J ⊆ 2Z. On the other hand, for any c ∈ Z, 2c = 4(−c) + 6c ∈ I + J , and hence 2Z ⊆ I + J . We can say something more general here. See Exercise 9.4. Theorem 9.3. If I and J are ideals of a ring R, then so is I + J . Proof. As I and J are subgroups of the abelian additive group R, and hence normal, we know from Theorem 4.5 that I + J is an additive subgroup of R. It remains only to check the absorption property. Take i ∈ I , j ∈ J and r ∈ R. Then r (i + j) = ri +r j. Now, ri ∈ I and r j ∈ J . Thus, r (i + j) ∈ I + J . Similarly, (i + j)r = ir + jr ∈ I + J. \u0002 Also, we can define I J = {i 1 j1 + i 2 j2 + · · · + i n jn : i k ∈ I, jk ∈ J, n ∈ N}. (We cannot simply take terms of the form i j with i ∈ I and j ∈ J , as sums of terms of that form cannot necessarily be written in the same form.)\n\n9.1 Ideals\n\n151\n\nTheorem 9.4. Let I and J be ideals in a ring R. Then I J is also an ideal. Proof. Clearly 0 = 0 · 0 ∈ I J . If we have i k ∈ I , jk ∈ J , then (i 1 j1 + · · · + i m jm ) − (i m+1 jm+1 + · · · + i n jn ) = i 1 j1 + · · · + i m jm + (−i m+1 ) jm+1 + · · · + (−i n ) jn ∈ I J. Also, for any r ∈ R, r (i 1 j1 + · · · + i m jm ) = (ri 1 ) j1 + · · · + (ri m ) jm ∈ I J since ri 1 , . . . , ri m ∈ I and, similarly, (i 1 j1 + · · · + i m jm )r = i 1 ( j1r ) + · · · + i m ( jm r ) ∈ I J. \u0002 Notice that I and J are both subsets of I + J , as we can take i + 0 and 0 + j for any i ∈ I , j ∈ J . But by the absorption property, I J ⊆ I ∩ J . One type of ideal is particularly important. Definition 9.2. Let R be a commutative ring with identity and a ∈ R. Then the principal ideal generated by a, denoted (a), is the set {ra : r ∈ R}. Example 9.7. In Z, we have (n) = nZ for any n ∈ Z. Example 9.8. The ideal from Example 9.2 is (x). Indeed, if f (x) ∈ R[x], then f (0) = 0 if and only if the constant term is 0; that is, if and only if the polynomial is a multiple of x. Theorem 9.5. If R is a commutative ring with identity, and a ∈ R, then (a) is an ideal of R; indeed, it is the intersection of all ideals of R containing a. Proof. We have 0 = 0a ∈ (a). If r, s ∈ R, then ra − sa = (r − s)a ∈ (a). Also, if ra ∈ (a) and s ∈ R, then s(ra) = (ra)s = (sr )a ∈ (a). Furthermore, if I is an ideal of R containing a, then by the absorption property, ra ∈ I for all r ∈ R. Thus, (a) is a subset of every ideal containing a. As (a) is an ideal containing a, our result is proved. \u0002 Notice that the preceding proof does not work if R is not a commutative ring with identity. Indeed, if R is not commutative, then {ra : r ∈ R} need not be an ideal; if R does not have an identity, then it may not contain a. See Exercise 9.5. In a similar fashion, if R is a commutative ring with identity, and a1 , . . . , an ∈ R, we can construct the ideal generated by a1 , . . . , an , namely, the set of all elements r1 a1 + · · · + rn an , with ri ∈ R. Exercises 9.1. List all of the elements in each of the following principal ideals of Z4 ⊕ Z6 .\n\n152\n\n9 Ideals, Factor Rings and Homomorphisms\n\n1. ((0, 5)) 2. ((2, 3)) 9.2. Let I be the set of all polynomials in Z[x] whose constant term is a multiple of 5. Show that I is an ideal of Z[x]. 9.3. Show that the intersection of ideals I and J in a ring R is also an ideal. Extend this to the intersection of an arbitrary collection of ideals. 9.4. Let m and n be positive integers. Show that mZ + nZ = (m, n)Z. 9.5. Given a ring R and an element a, let S = {ra : r ∈ R}. Show by example that 1. S need not contain a, even if R is commutative; and 2. S need not be an ideal, even if R has an identity. 9.6. Let R be a commutative ring. If I = {a ∈ R : a n = 0 for some n ∈ N}, show that I is an ideal of R. 9.7. Find ideals I and J in a ring R such that I J \u0003= I ∩ J . 9.8. Let R be a commutative ring with identity having exactly two ideals. Show that R is a field. 9.9. Consider the additive group G and subgroup H from Exercise 3.42. Define a multiplication operation on G via (a1 , a2 , . . .)(b1 , b2 , . . .) = (a1 b1 , a2 b2 , . . .). Show that G is a ring and H is an ideal. 9.10. In the preceding exercise, show that H is not a principal ideal.\n\n9.2 Factor Rings Let R be a ring and I an ideal. Then R is an abelian group under addition, and I is a subgroup. Thus, we can consider the left cosets a + I , for all a ∈ R. We use these to form a factor ring. Remember that a + I = b + I if and only if a − b ∈ I . Definition 9.3. Let R be a ring and I an ideal of R. Then the factor ring (or quotient ring), R/I , is the set of all left cosets {a + I : a ∈ R} together with the operations (a + I ) + (b + I ) = a + b + I and (a + I )(b + I ) = ab + I , for all a, b ∈ R. Theorem 9.6. For any ring R and ideal I , the factor ring R/I is a ring. Proof. Since R is an abelian group under addition, I is necessarily a normal subgroup. Thus, we know from Theorem 4.6 that R/I is a group under addition. By Theorem 4.7, it is abelian. Let us show that the multiplication operation is well-defined. Suppose that a1 +I = a2 + I and b1 + I = b2 + I . Then notice that\n\n9.2 Factor Rings\n\n153\n\na1 b1 − a2 b2 = a1 b1 − a1 b2 + a1 b2 − a2 b2 = a1 (b1 − b2 ) + (a1 − a2 )b2 . Now, b1 − b2 ∈ I . Thus, by absorption, a1 (b1 − b2 ) ∈ I . Similarly, a1 − a2 ∈ I , and hence (a1 − a2 )b2 ∈ I . Thus, a1 b1 − a2 b2 ∈ I , and therefore a1 b1 + I = a2 b2 + I . That is, the multiplication operation is well-defined. We must check the remaining properties of a ring. Take any a, b, c ∈ R. Then since ab + I ∈ R/I , we have closure under multiplication. Also, (a + I )((b + I )(c + I )) = (a + I )(bc + I ) = a(bc) + I = (ab)c + I = (ab + I )(c + I ) = ((a + I )(b + I ))(c + I ), and associativity holds. Similarly, (a + I )((b + I ) + (c + I )) = (a + I )(b + c + I ) = a(b + c) + I = ab + ac + I = (ab + I ) + (ac + I ) = (a + I )(b + I ) + (a + I )(c + I ). The other distributive law is proved in the same fashion.\n\n\u0002\n\nLet us discuss a few examples of factor rings. Example 9.9. Let R = Z and I = (5) = 5Z. Then R/I = {0 + I, 1 + I, 2 + I, 3 + I, 4 + I } and, for instance, (2 + I ) + (4 + I ) = 6 + I = 1 + I and (3 + I )(4 + I ) = 12 + I = 2 + I . Example 9.10. Let R = M2 (Z) and let I be the ideal consisting of all matrices whose entries are even. Then notice that for any ai j ∈ Z, we have \u0002 \u0003 \u0002 \u0003 a11 a12 b b + I = 11 12 + I, a21 a22 b21 b22 where bi j is 0 if a\u0002i j is even \u0003 and 1 if ai j is odd. Thus, R/I consists of the sixteen b11 b12 +I , bi j ∈ {0, 1}. We perform arithmetic in the following different elements b21 b22 fashion: \u0002\u0002 \u0003 \u0003 \u0002\u0002 \u0003 \u0003 \u0002 \u0003 \u0002 \u0003 10 11 21 01 +I + +I = +I = +I 11 01 12 10 and\n\n154\n\n9 Ideals, Factor Rings and Homomorphisms\n\n\u0002\u0002 \u0003 \u0003 \u0002\u0002 \u0003 \u0003 \u0002 \u0003 \u0002 \u0003 10 11 11 11 +I +I = +I = + I. 11 01 12 10 Example 9.11. Let R = R[x] and I = (x 2 + 3). Readers familiar with polynomial long division will know that if f (x) ∈ R, then f (x) = (x 2 + 3)q(x) + r (x), where q and r are polynomials, with r (x) = a + bx, for some a, b ∈ R. (Those who are unfamiliar with polynomial long division can peek ahead to Section 10.1, where it will be discussed in more generality.) Since (x 2 + 3)q(x) ∈ I by absorption, we know that elements of R/I are of the form a + bx + I , with a, b ∈ R. Addition behaves as expected; for instance, (2 + 3x + I ) + (7 − 4x + I ) = 9 − x + I . To deal with multiplication, observe that x 2 − (−3) ∈ I ; thus, x 2 + I = −3 + I . Therefore, we have calculations such as (5 + 4x + I )(−7 + 2x + I ) = −35 − 18x + 8x 2 + I = −35 − 18x + 8(−3) + I = −59 − 18x + I. Let us also record a few basic facts about factor rings. Theorem 9.7. Let R be a ring and I an ideal. Then 1. if R is commutative, then so is R/I ; 2. if R has an identity, then so does R/I ; and 3. if u is a unit of R, then u + I is a unit of R/I . Proof. (1) If a, b ∈ R, then (a + I )(b + I ) = ab + I = ba + I = (b + I )(a + I ). (2) If a ∈ R, then (1 + I )(a + I ) = a + I = (a + I )(1 + I ). Hence, 1 + I is the identity of R/I . (3) Observe that (u + I )(u −1 + I ) = 1 + I = (u −1 + I )(u + I ); thus, (u + I )−1 = \u0002 u −1 + I . Theorem 9.8. Let R be a ring with ideals I and J , such that I ⊆ J . Then J/I is an ideal of R/I . Proof. We see that I is a subring of J , and since I enjoys the absorption property in R, it enjoys it in J as well. Thus, I is an ideal of J , and so J/I makes sense. Now, 0 ∈ J , and therefore 0+I ∈ J/I . If j1 , j2 ∈ J , then ( j1 +I )−( j2 +I ) = ( j1 − j2 )+I ∈ J/I . Also, if j ∈ J and r ∈ R, then (r + I )( j + I ) = r j + I ∈ J/I , since r j ∈ J . Similarly, ( j + I )(r + I ) = jr + I ∈ J/I . The proof is complete. \u0002 Exercises 9.11. Let R = Z and I = (5). Write the addition and multiplication tables for R/I . 9.12. Let R = Z6 ⊕ Z4 and I = ((4, 2)). Write the addition and multiplication tables for R/I .\n\n9.2 Factor Rings\n\n155\n\n9.13. Let R = R[x] and f (x) = x 3 + 6x 2 + 2. If I = ( f (x)), calculate the product (4x 2 + 3x + 1 + I )(2x 2 − x + 2 + I ) in R/I . Reduce the answer to the form ax 2 + bx + c + I , for some a, b, c ∈ R. 9.14. Let R be a ring and I an ideal. Show that R/I is commutative if and only if ab − ba ∈ I for all a, b ∈ R. 9.15. Let I and J be ideals in a ring R. Show that R/I and R/J are both commutative rings if and only if R/(I ∩ J ) is commutative. 9.16. Let R be a ring and I a proper ideal. 1. If R is an integral domain, does it follow that R/I is an integral domain? Prove that it does, or find a counterexample. 2. If R/I is an integral domain, does it follow that R is an integral domain? Prove that it does, or find a counterexample. 9.17. If F is a field of order 81, what are the possible orders of F/I , where I is an ideal of F? \u0004 9.18. Let I1 ⊆ I2 ⊆ I3 ⊆ · · · be ideals of R. Let I = ∞ n=1 In . 1. Show that I is an ideal of R. 2. Suppose that R/I is commutative. Show that for every a, b ∈ R, there exists an n ∈ N such that ab − ba ∈ In . 9.19. Let R and I be as in Exercise 9.6. Define the analogous ideal for R/I , namely {a + I ∈ R/I : (a + I )n = 0 + I for some n ∈ N}. Show that this ideal is {0 + I }. 9.20. If I is an ideal of a ring R, show that the subrings of R/I are precisely of the form S/I , where S is a subring of R containing I . Further show that S/I is an ideal of R/I if and only if S is an ideal of R.\n\n9.3 Ring Homomorphisms We recall that a group homomorphism is a function from one group to another that respects the group operation. There is a similar concept for rings, but both of the ring operations must be respected. Definition 9.4. Let R and S be rings. Then a ring homomorphism (or, simply, homomorphism) from R to S is a function α : R → S satisfying α(r1 + r2 ) = α(r1 ) + α(r2 ) and α(r1r2 ) = α(r1 )α(r2 ) for all r1 , r2 ∈ R.\n\n156\n\n9 Ideals, Factor Rings and Homomorphisms\n\nThus, a ring homomorphism is a homomorphism of additive groups, with the additional property that it respects the multiplication operation. The kernel is the same as the kernel of the additive group homomorphism. Definition 9.5. Let α : R → S be a ring homomorphism. Then the kernel of α is ker(α) = {r ∈ R : α(r ) = 0}. Example 9.12. Let n ≥ 2 be a positive integer. Then α : Z → Zn given by α(a) = [a] (where we insert the square brackets for clarity) is a homomorphism. Indeed, by Example 4.10, it respects the addition operation. Also, for any a, b ∈ Z, α(ab) = [ab] = [a][b] = α(a)α(b). Furthermore, by Example 4.10, ker(α) = nZ. Example 9.13. Define α : C → C via α(a + bi) = a − bi, for all a, b ∈ R. Then notice that α((a + bi) + (c + di)) = α((a + c) + (b + d)i) = a + c − (b + d)i = (a − bi) + (c − di) = α(a + bi) + α(c + di), for all a, b, c, d ∈ R. Similarly, α((a + bi)(c + di)) = α((ac − bd) + (ad + bc)i) = ac − bd − (ad + bc)i = (a − bi)(c − di) = α(a + bi)α(c + di). Thus, α is a homomorphism. Also, if α(a + bi) = 0, then a − bi = 0, and hence a = b = 0. Thus, ker(α) = {0}. Example 9.14. If R and S are any rings, then α : R → S given by α(r ) = 0 for all r ∈ R is a homomorphism. Indeed, by Example 4.13, it is an additive group homomorphism, and α(r1r2 ) = 0 = 0 · 0 = α(r1 )α(r2 ), for all r1 , r2 ∈ R. The kernel of α is R. Let us record a few basic properties of ring homomorphisms. If α : R → S is a ring homomorphism, then as with group homomorphisms, we write α(M) = {α(m) : m ∈ M} and α −1 (N ) = {r ∈ R : α(r ) ∈ N }, for any subring M of R, and any subring N of S. Theorem 9.9. Let α : R → S be a ring homomorphism. Then 1. 2. 3. 4.\n\nker(α) is an ideal of R; α is one-to-one if and only if ker(α) = {0}; α(0) = 0; and if R is a ring with identity, then so is α(R), and α(1) is the identity of α(R).\n\n9.3 Ring Homomorphisms\n\n157\n\nProof. (1) By Theorem 4.11, ker(α) is an additive subgroup of R. It remains to check the absorption property. But if k ∈ ker(α) and r ∈ R, then α(r k) = α(r )α(k) = α(r )0 = 0, and hence r k ∈ ker(α). Similarly, kr ∈ ker(α). (2) See Theorem 4.11. (3) This follows immediately from Theorem 4.10. (4) If r ∈ R, then α(1)α(r ) = α(1r ) = α(r ) = α(r 1) = α(r )α(1). \u0002 We must be a bit careful with the final part of the last theorem. It certainly does not follow that α(1) is the identity of S, as the following example illustrates. Example 9.15. Define α : R → M2 (R) via \u0002 \u0003 r 0 α(r ) = 00 for all r ∈ R. It is easy to verify that α is a homomorphism. Now, α(1) is the identity of α(R), but not of M2 (R). Theorem 9.10. Let α : R → S be a ring homomorphism. Let M be a subring of R and N a subring of S. Then 1. 2. 3. 4.\n\nα(M) is a subring of S; if M is an ideal of R, then α(M) is an ideal of α(R); α −1 (N ) is a subring of R; and if N is an ideal of S, then α −1 (N ) is an ideal of R.\n\nProof. (1) By Theorem 4.12, α(M) is an additive subgroup of S. If m 1 , m 2 ∈ M, then α(m 1 )α(m 2 ) = α(m 1 m 2 ) ∈ α(M), since m 1 m 2 ∈ M. (2) By (1), it remains only to check the absorption property. If m ∈ M, r ∈ R, then α(r )α(m) = α(r m) ∈ α(M), since r m ∈ M. Similarly, α(m)α(r ) ∈ α(M). (3) By Theorem 4.12, α −1 (N ) is an additive subgroup of R. If r1 , r2 ∈ α −1 (N ), then α(r1r2 ) = α(r1 )α(r2 ) ∈ N , since α(r1 ), α(r2 ) ∈ N . Thus, r1r2 ∈ α −1 (N ). (4) In view of (3), we only need to check the absorption property. Take a ∈ α −1 (N ) and r ∈ R. Then α(ra) = α(r )α(a) ∈ N , since α(a) ∈ N , and N is an ideal of S. \u0002 Therefore, ra ∈ α −1 (N ). Similarly, ar ∈ α −1 (N ). Once again, we note that the second part of the preceding theorem does not say that α(M) is necessarily an ideal of S. One more homomorphism will prove useful later. Theorem 9.11. Let R be a ring with identity of characteristic n. Then there is a homomorphism α : Z → R with kernel (n). Proof. Define α : Z → R via α(k) = k1, for all k ∈ Z. Let us check that α is a homomorphism. If j, k ∈ Z, then α( j +k) = ( j +k)1 = j1+k1 = α( j)+α(k). (This is Theorem 3.6, using additive notation.) Also, α( jk) = ( jk)1, whereas α( j)α(k) = ( j1)(k1). Again by Theorem 3.6, ( jk)1 = j (k1). If j > 0, then\n\n158\n\n9 Ideals, Factor Rings and Homomorphisms\n\n( j1)(k1) = (1 · · + 1\b)(k1) = k1 · · + k1\b = j (k1). \u0005 + ·\u0006\u0007 \u0005 + ·\u0006\u0007 j times\n\nj times\n\nIf j < 0, then ( jk)1 = ((− j)(−k))1 and ( j1)(k1) = (− j1)(−k1), and we can use the j > 0 argument. If j = 0, then ( jk)1 = ( j1)(k1) = 0. Thus, in any case, α( jk) = α( j)α(k), and α is a homomorphism. We note that k ∈ ker(α) if and only if k1 = 0. If n = 0, then by Theorem 8.13, 1 has infinite additive order. Therefore, the only solution is k = 0; thus, ker(α) = (0). If n > 0, then by Theorem 8.13, n is the additive order of 1. Furthermore, by Corollary 3.2, k1 = 0 if and only if the additive order of 1 divides k; that is, if and only if n divides k. In other words, the kernel is the set of multiples of n. The proof is complete. \u0002 Exercises 9.21. Decide if each of the following functions is a ring homomorphism. 1. α : Z → R, α(a) = 2a. 2. α : R[x] → R, α( f (x)) = f (2). 9.22. Decide if each of the following functions is a ring homomorphism. \u0002 \u0003 \u0002 \u0003 11 1 −1 A . 1. α : M2 (R) → M2 (R), α(A) = 01 0 1 2. α : M2 (R) → R, α(A) = det(A). 9.23. Let α : R → S and β : S → T be ring homomorphisms. Show that βα : R → T is also a ring homomorphism. 9.24. Let α : R → S and β : S → T be ring homomorphisms. Show that ker(α) ⊆ ker(βα). If β is one-to-one, show that ker(βα) = ker(α). 9.25. Define α : Z ⊕ Z → Z ⊕ Z via α((a, b)) = (a, 0). Is this a ring homomorphism? If so, find ker(α) and α −1 (2Z ⊕ 3Z). 9.26. Define α : Z8 → Z16 via α([a]) = [a], for all a ∈ {0, 1, . . . , 7}, where the square brackets represent the congruence classes. Is this a ring homomorphism? If so, find ker(α) and α −1 (). 9.27. Let R be a ring and I an ideal. Show that there exist a ring S and a homomorphism α : R → S such that ker(α) = I . 9.28. Let R be a commutative ring with prime characteristic p. Show that α : R → R given by α(a) = a p is a ring homomorphism. 9.29. Let F be a field of order 16 and K a field of order 4. Find all homomorphisms from F to K . 9.30. Find all homomorphisms from Z to Q.\n\n9.4 Isomorphisms and Automorphisms\n\n159\n\n9.4 Isomorphisms and Automorphisms As with groups, we use isomorphisms to establish if two rings have the same structure. Definition 9.6. Let R and S be rings. Then a ring isomorphism (or, simply, isomorphism) is a bijective homomorphism from R to S. When such an isomorphism exists, we say that R and S are isomorphic rings. Example 9.16. Consider the function α : Z15 → Z3 ⊕Z5 given by α(a) = (a, a) for all a. By Example 4.16, this is an isomorphism of additive groups. We claim that it is, in fact, a ring isomorphism. All that remains is to show that α respects multiplication. But if a, b ∈ Z15 , then α(ab) = (ab, ab) = (a, a)(b, b) = α(a)α(b). Thus, Z15 and Z3 ⊕ Z5 are isomorphic rings. We must not, however, make the mistake of thinking that rings that are isomorphic as additive groups are necessarily isomorphic as rings! Example 9.17. Let R = Z and S = 5Z. As additive groups, these are isomorphic, since 5Z is infinite cyclic, being generated by 5, and we can apply Theorem 4.14. However, as rings, there cannot even be an onto homomorphism from R to S. Why not? If there were, by Theorem 9.9, 1 would have to map to the identity of S, which is sadly lacking. Thus, R and S are not isomorphic rings. Example 9.18. Define α : C → M2 (R) via α(a + bi) =\n\n\u0002 \u0003 a −b , b a\n\nfor all a, b ∈ R. Let us check that α is a homomorphism. If a, b, c, d ∈ R, then α((a + bi) + (c + di)) = α((a + c) + (b + d)i) \u0002 \u0003 a + c −(b + d) = b+d a+c = α(a + bi) + α(c + di). Also, α((a + bi)(c + di)) = α((ac − bd) + (ad + bc)i) = whereas α(a + bi)α(c + di) =\n\n\u0002 \u0003 ac − bd −(ad + bc) ad + bc ac − bd\n\n\u0002 \u0003\u0002 \u0003 a −b c −d , b a d c\n\nand these are the same. Clearly, ker(α) = {0}, so α is one-to-one. Now, α is not onto, but we see that C is isomorphic to the image α(C), namely\n\n160\n\n9 Ideals, Factor Rings and Homomorphisms\n\n\u0002 \u0003\n\na −b : a, b ∈ R . b a Let us discuss a few properties of isomorphisms. Theorem 9.12. On any collection of rings, isomorphism is an equivalence relation. Proof. Reflexivity: The function α : R → R given by α(a) = a for all a is an isomorphism. Symmetry: Let α : R → S be an isomorphism. By Theorem 4.13, the inverse α −1 : S → R is an isomorphism of additive groups. We only need to check that it respects multiplication. Take any s1 , s2 ∈ S, and suppose that α −1 (si ) = ri . Then α(r1r2 ) = α(r1 )α(r2 ) = s1 s2 ; that is, α −1 (s1 s2 ) = r1r2 = α −1 (s1 )α −1 (s2 ). Transitivity: Let α : R → S and β : S → T be isomorphisms. By Theorem 4.13, β ◦ α : R → T is an isomorphism of additive groups. Again, we must check that it respects multiplication. Take r1 , r2 ∈ R. Then (β ◦ α)(r1r2 ) = β(α(r1r2 )) = β(α(r1 )α(r2 )) = β(α(r1 ))β(α(r2 )). The proof is complete.\n\n\u0002\n\nTheorem 9.13. Let α : R → S be a ring isomorphism. Then 1. 2. 3. 4.\n\nif if if if\n\nR is commutative, then so is S; R has an identity, then so does S; R is an integral domain, then so is S; and R is a field, then so is S.\n\nProof. (1) Take s1 , s2 ∈ S. Then si = α(ri ), for some ri ∈ R. Thus, s1 s2 = α(r1 )α(r2 ) = α(r1r2 ) = α(r2 r1 ) = α(r2 )α(r1 ) = s2 s1 . (2) Use Theorem 9.9 and the fact that α is onto. (3) By (1) and (2), S is a commutative ring with identity. If 1 = 0 in S, then S = {0}. As α is bijective, R = {0}, which is impossible. It remains only to check for zero divisors. Suppose that s1 and s2 are nonzero elements of S with s1 s2 = 0. Let us say α(ri ) = si . Then 0 = s1 s2 = α(r1 )α(r2 ) = α(r1r2 ). As α is one-to-one, r1r2 = 0. But R is an integral domain, so either r1 = 0 or r2 = 0, which means that s1 = 0 or s2 = 0, giving us a contradiction. (4) Once again, S is a commutative ring with identity and 1 \u0003= 0. Suppose that 0 \u0003= s ∈ S. Then s = α(r ), for some r ∈ R. Now, r \u0003= 0, so r has an inverse in R. Since α is onto, we know that α(1) = 1. Thus, 1 = α(1) = α(rr −1 ) = α(r )α(r −1 ) = sα(r −1 ). That is, α(r −1 ) = s −1 , and every nonzero element of S has an inverse.\n\n\u0002\n\n9.4 Isomorphisms and Automorphisms\n\n161\n\nExample 9.19. The rings 2Z, M2 (R), Z6 , Z and Q are all nonisomorphic. Indeed, 2Z does not have an identity, so it cannot be isomorphic to any of the others, which do have an identity. Also, all of the rings are commutative except for M2 (R), so it is ruled out. Next, Z6 is not an integral domain, but the remaining two are. (Also, it is finite and the others are infinite.) Finally, Q is a field but Z is not. Given a ring R, we might well ask if it is a subring of a field. For example, Z is a subring of Q. But not every ring can be a subring of a field. Indeed, a noncommutative ring or a ring containing zero divisors cannot exist inside a field. So, it seems that integral domains are a good place to start. In fact, if R is an integral domain, then we can construct a field F containing an isomorphic copy of R. The method we will use may seem somewhat familiar; actually, it is exactly the way in which Q is constructed from Z. We will need something comparable to numerators and denominators. Also, the denominator must not be zero. Furthermore, we need some way of recognizing that 10/25 = 8/20, for instance. This gives us an idea of how to proceed. Let R be an integral domain. Then let S be the Cartesian product R × (R\\{0}). Note that S is only a set, not a ring. (If R = Z, for instance, when we look at (10, 25) ∈ S, we are thinking of the fraction 10/25.) Let us define a relation ∼ on S via (a, b) ∼ (c, d) if and only if ad = bc. (Continuing our parenthetical thought, (10, 25) ∼ (8, 20).) We claim that ∼ is an equivalence relation. Reflexivity: As R is commutative, we see that (a, b) ∼ (a, b). Symmetry: Suppose that (a, b) ∼ (c, d). Then ad = bc and again, by commutativity, (c, d) ∼ (a, b). Transitivity: Suppose that (a, b) ∼ (c, d) and (c, d) ∼ (e, f ). Then ad = bc, and hence ad f = bc f . Also, c f = de, and hence bc f = bde. Thus, ad f = bde. Now, d \u0003= 0 and R is an integral domain. Thus, we have a f = be, and hence (a, b) ∼ (e, f ). For the sake of simplicity, write [a, b] for the equivalence class of (a, b). (In our example in Z, we have [10, 25] = [8, 20], and this is the set of all pairs (a, b), with a, b ∈ Z, b \u0003= 0 and 10b = 25a.) Let us write F for the set of all equivalence classes of S. The addition and multiplication operations on F work precisely as we would expect with fractions. Specifically, [a, b] + [c, d] = [ad + bc, bd] and [a, b][c, d] = [ac, bd]. We must verify that these operations are well-defined. Suppose that [a1 , b1 ] = [a, b] and [c1 , d1 ] = [c, d]. Then [a1 , b1 ] + [c1 , d1 ] = [a1 d1 + b1 c1 , b1 d1 ]. But (ad + bc)(b1 d1 ) − (a1 d1 + b1 c1 )bd = dd1 (ab1 − a1 b) + bb1 (cd1 − c1 d) = dd1 (0) + bb1 (0) = 0;\n\n162\n\n9 Ideals, Factor Rings and Homomorphisms\n\nthus, [a, b] + [c, d] = [a1 , b1 ] + [c1 , d1 ]. Similarly, [a1 , b1 ][c1 , d1 ] = [a1 c1 , b1 d1 ], and acb1 d1 − a1 c1 bd = (acb1 d1 − a1 cbd1 ) + (a1 cbd1 − a1 c1 bd) = cd1 (ab1 − a1 b) + a1 b(cd1 − c1 d) = cd1 (0) + a1 b(0) = 0; thus, [a, b][c, d] = [a1 , b1 ][c1 , d1 ]. Also, we must note that if b and d are nonzero, then so is bd, since R is an integral domain. Definition 9.7. Let R be an integral domain. Then the field of fractions (or field of quotients) of R is the field F constructed above. Of course, the fact that F is indeed a field needs proving! Theorem 9.14. Let R be an integral domain. Then the field of fractions, F, is a field. Proof. The proof of this theorem is not difficult. However, there are many steps to complete, as we must verify that F is an abelian group under addition, then that it has all of the remaining properties of a ring, and finally that it is a field. We will prove a few selected properties, and leave the rest to the reader1 as Exercise 9.38. Take any a, b, c, d, e, f ∈ R with b, d and f all nonzero. Let us show that the addition operation on F is associative. We have ([a, b] + [c, d]) + [e, f ] = [ad + bc, bd] + [e, f ] = [ad f + bc f + bde, bd f ] = [a, b] + [c f + de, d f ] = [a, b] + ([c, d] + [e, f ]), as required. Next, let us prove a distributive law. Observe that [a, b]([c, d] + [e, f ]) = [a, b][c f + de, d f ] = [ac f + ade, bd f ] whereas [a, b][c, d] + [a, b][e, f ] = [ac, bd] + [ae, b f ] = [abc f + abde, b2 d f ]. But as (ac f + ade)b2 d f = (abc f + abde)bd f , these are equal. Notice that [0, 1] is the additive identity of F and [1, 1] is the multiplicative identity. Let us show that every nonzero element of F has an inverse. Take a nonzero [a, b] ∈ F. Note that [0, b] = [0, 1], so a \u0003= 0. But then [b, a] ∈ F as well, and \u0002 [a, b][b, a] = [ab, ab] = [1, 1]. Thus, [b, a] = [a, b]−1 . 1 Aren’t\n\nyou lucky!\n\n9.4 Isomorphisms and Automorphisms\n\n163\n\nWhat is the connection between R and F? The idea is that F contains a copy of R and it is, in a sense, the smallest field that does. Theorem 9.15. Let R be an integral domain and F its field of fractions. Then F has a subring isomorphic to R. Furthermore, if K is any field having R as a subring, then K has a subfield isomorphic to F, and this subfield contains R. Proof. Define α : R → F via α(r ) = [r, 1], for all r ∈ R. We claim that α is a homomorphism. If r1 , r2 ∈ R, then α(r1 + r2 ) = [r1 + r2 , 1] = [r1 , 1] + [r2 , 1] = α(r1 ) + α(r2 ) and α(r1r2 ) = [r1r2 , 1] = [r1 , 1][r2 , 1] = α(r1 )α(r2 ), proving the claim. Furthermore, if r ∈ ker(α), then [r, 1] = [0, 1], and therefore r = 0. Thus, α is one-to-one, and hence R is isomorphic to α(R) which, by Theorem 9.10, is a subring of F. Also, define β : F → K via β([a, b]) = ab−1 , for all a, b ∈ R with b \u0003= 0. (Since K is a field, b has an inverse in K . But we still need to check that β is well-defined. Suppose that [a, b] = [a1 , b1 ]. Then ab1 = a1 b, and hence ab−1 = a1 b1−1 .) Now, let us show that β is a homomorphism. But β([a, b] + [c, d]) = β([ad + bc, bd]) = (ad + bc)(bd)−1 = ab−1 + cd −1 = β([a, b]) + β([c, d]). Similarly, β([a, b][c, d]) = β([ac, bd]) = ac(bd)−1 = ab−1 cd −1 = β([a, b])β([c, d]). Thus, β is a homomorphism. Now, ker(β) is an ideal of F. By Corollary 9.1, ker(β) = {0} or F. But β([1, 1]) = 1 \u0003= 0, and therefore β is one-to-one. Thus, K has a subfield β(F), which is isomorphic to F. Also, for any r ∈ R, we have r = β([r, 1]) ∈ β(F). Thus, R is a subring of the isomorphic copy of the field of fractions of R contained in K . \u0002 Example 9.20. As we mentioned above, the field of fractions of Z is Q. √ Example 9.21. Let R = {a + b 2 : a, b ∈√Z}. Then R is an integral domain, and its field of fractions is isomorphic to {a + b 2 : a, b ∈ Q}. See Exercise 9.36. One particular type of isomorphism deserves special mention. Definition 9.8. Let R be a ring. Then an automorphism of R is an isomorphism from R to R.\n\n164\n\n9 Ideals, Factor Rings and Homomorphisms\n\nExample 9.22. For any ring R, the function α : R → R given by α(a) = a for all a is an automorphism. Example 9.23. The function α : C → C given by α(a +bi) = a −bi for all a, b ∈ R is an automorphism. Indeed, we saw in Example 9.13 that it is a homomorphism. It is immediately obvious that α is one-to-one, and if a +bi ∈ C, then α(a −bi) = a +bi, and therefore α is onto. We have something similar to an inner automorphism of a group as well. Theorem 9.16. Let R be a ring with identity and u a unit of R. Then α : R → R, given by α(a) = u −1 au for all a ∈ R, is an automorphism of R. Proof. Take a, b ∈ R. Then α(a + b) = u −1 (a + b)u = u −1 au + u −1 bu = α(a) + α(b). Also,\n\nα(ab) = u −1 abu = u −1 auu −1 bu = α(a)α(b).\n\nThus, α is a homomorphism. If α(a) = 0, then u −1 au = 0, and therefore a = uu −1 auu −1 = u0u −1 = 0. Therefore, α is one-to-one. Finally, take any a ∈ R. Then α(uau −1 ) = u −1 uau −1 u = a. Thus, α is onto, and the proof is complete. \u0002 Exercises 9.31. Explain why the following pairs of rings are not isomorphic. 1. Z4 ⊕ Z4 and Z4 ⊕ Z2 ⊕ Z2 2. Z[x] and 2Z[x] 9.32. Explain why the following pairs of rings are not isomorphic. 1. R and M2 (R) 2. R and R ⊕ R 9.33. Show that if two rings are isomorphic, then their centres are isomorphic. 9.34. Let R and S be any rings. Show that R ⊕ S is isomorphic to S ⊕ R. 9.35. Let F be a field. Show that F is isomorphic to its field of fractions by constructing an explicit isomorphism. √ 9.36. Let R = {a + b 2 : a, b ∈ Z}. Show √ that R is an integral domain, and that its field of fractions is isomorphic to {a + b 2 : a, b ∈ Q}. 9.37. Let R and S be integral domains. If the fields of fractions of R and S are isomorphic, does it follow that R and S are isomorphic? Prove that it does, or give an explicit counterexample.\n\n9.4 Isomorphisms and Automorphisms\n\n165\n\n9.38. Complete the proof of Theorem 9.14. 9.39. Let R be a ring. An involution on R is a function α : R → R such that, for all ri ∈ R, we have α(r1 + r2 ) = α(r1 ) + α(r2 ), α(r1r2 ) = α(r2 )α(r1 ) and α(α(r1 )) = r1 . Show that the following functions α are involutions on M2 (R). \u0002\u0002 \u0003\u0003 \u0002 \u0003 ab a c 1. α = (called the transpose involution) cd bd \u0002\u0002 \u0003\u0003 \u0002 \u0003 ab d −b 2. α = (called the symplectic involution) cd −c a 9.40. Let R be a ring. Using the definition of an involution from the preceding question, 1. determine under what circumstances an involution on R is an automorphism; and 2. show that the composition of two involutions on R is an automorphism.\n\n9.5 Isomorphism Theorems for Rings We recall that the three isomorphism theorems for groups were presented in Section 4.5. Let us now state the analogues for rings. The first is certainly the most important. Theorem 9.17 (First Isomorphism Theorem for Rings). Let α : R → S be a ring homomorphism. Then R/ ker(α) is isomorphic to α(R). Proof. Let K = ker(α). Define β : R/K → α(R) via β(a + K ) = α(a). From the proof of Theorem 4.18, we see that β is an isomorphism of additive groups. Thus, it remains only to check that β respects multiplication. Take any a, b ∈ R. Then β((a + K )(b + K )) = β(ab + K ) = α(ab) = α(a)α(b) = β(a + K )β(b + K ), as required.\n\n\u0002\n\nWhenever we are asked to show that a ring modulo an ideal is isomorphic to some other ring, it is usually a good indication that we should employ the First Isomorphism Theorem. Example 9.24. We already know that for any n ≥ 2, the additive groups Z/nZ and Zn are isomorphic. Indeed, in Example 4.21, we showed that α : Z → Zn , given by α(a) = [a], is an onto group homomorphism with kernel nZ. But, in fact, we also have α(ab) = [ab] = [a][b] = α(a)α(b), for all a, b ∈ Z. Thus, α is actually an onto ring homomorphism, and we now know that Z/nZ and Zn are isomorphic rings.\n\n166\n\n9 Ideals, Factor Rings and Homomorphisms\n\nExample 9.25. Let us show that R[x]/(x) is isomorphic to R. To this end, let us define α : R[x] → R via α( f (x)) = f (0). Now, if f (x), g(x) ∈ R[x], then α( f (x) + g(x)) = f (0) + g(0) = α( f (x)) + α(g(x)). Furthermore, α( f (x)g(x)) = f (0)g(0) = α( f (x))α(g(x)). Thus, α is a homomorphism. Also, if r ∈ R, then simply regarding r as a constant polynomial, we have α(r ) = r ; hence, α is onto. The kernel of α is the set of all polynomials f (x) satisfying f (0) = 0. But f (0) is the constant term of the polynomial. Thus, ker(α) is the set of all polynomials with zero constant term, that is, the set of all polynomials that are multiples of x. Now apply Theorem 9.17. A couple of rather interesting consequences follow. Corollary 9.2. Let R be a ring with identity of characteristic n. If n = 0, then R has a subring isomorphic to Z. If n ≥ 2, then R has a subring isomorphic to Zn . Proof. By Theorem 9.11, there is a homomorphism α : Z → R with kernel nZ. Now, Theorem 9.10 says that α(Z) is a subring of R, and Theorem 9.17 tells us that this subring is isomorphic to Z/nZ. If n = 0, then nZ = {0}, and there is nothing to do. Otherwise, we use Example 9.24. \u0002 Corollary 9.3. Let F be a field. If F has characteristic 0, then F has a subfield isomorphic to Q. If F has prime characteristic p, then F has a subfield isomorphic to Z p . Proof. If char F = p > 0, then we use the preceding corollary. If char F = 0, then we note that F has a subring isomorphic to Z. By Theorem 9.15, F also has a subfield isomorphic to the field of fractions of Z, namely, Q. \u0002 The subfield discussed in Corollary 9.3 (either Q or Z p ) is the smallest subfield of F, and it is called the prime subfield. Theorem 9.18 (Second Isomorphism Theorem for Rings). Let R be a ring with ideals I and J . Then I /(I ∩ J ) is isomorphic to (I + J )/J . Proof. Define α : I → (I + J )/J via α(i) = i + J , for all i ∈ I . Consulting the proof of Theorem 4.19, we see that α is an onto homomorphism of additive groups with kernel I ∩ J . In view of the First Isomorphism Theorem for Rings, it suffices to show that α respects multiplication. But for any i 1 , i 2 ∈ I , we have \u0002 α(i 1 i 2 ) = i 1 i 2 + J = (i 1 + J )(i 2 + J ) = α(i 1 )α(i 2 ). The proof is complete. Example 9.26. Let R = Z, I = (4) and J = (6). Then the preceding theorem tells us that (4)/((4) ∩ (6)) is isomorphic to ((4) + (6))/(6). That is, (4)/(12) is isomorphic to (2)/(6).\n\n9.5 Isomorphism Theorems for Rings\n\n167\n\nTheorem 9.19 (Third Isomorphism Theorem for Rings). Let R be a ring, and let I and J be ideals of R with I ⊆ J . Then (R/I )/(J/I ) is isomorphic to R/J . Proof. Define α : R/I → R/J via α(a + I ) = a + J , for all a ∈ R. The proof of Theorem 4.20 shows us that α is an onto additive group homomorphism with kernel J/I . It remains only to show that α respects multiplication, for then we can apply Theorem 9.17. Take a, b ∈ R. Then α((a + I )(b + I )) = α(ab + I ) = ab + J = (a + J )(b + J ) = α(a + I )α(b + I ). We are done.\n\n\u0002\n\nExercises 9.41. Let R and S be rings. Show that (R ⊕ S)/(R ⊕ {0}) is isomorphic to S. 9.42. Let m and n be positive integers, both greater than 1. Show that the rings (Z ⊕ Z)/((m) ⊕ (n)) and Zm ⊕ Zn are isomorphic. 9.43. Let I be the set of all polynomials f (x) ∈ Z[x] such that the constant term of f (x) is a multiple of 5. Show that Z[x]/I is isomorphic to Z5 . 9.44. Let I be the set of all matrices in M2 (Z) in which every entry is even. Show that M2 (Z)/I is isomorphic to M2 (Z2 ). 9.45. Show that the rings (3Z/60Z)/(12Z/60Z) and 3Z/12Z are isomorphic. Then show that both are isomorphic to Z4 . 9.46. Let I and J be ideals in a ring R such that I + J = R. Show that R/(I ∩ J ) is isomorphic to (R/I ) ⊕ (R/J ).\n\n9.6 Prime and Maximal Ideals We conclude this chapter by discussing two special sorts of ideals. Definition 9.9. Let R be a ring. An ideal M of R is said to be maximal if 1. M \u0003= R; and 2. if I is an ideal of R containing M, then I = M or I = R. Example 9.27. Let R = Z and let n be a nonnegative integer. Then we claim that (n) is a maximal ideal of R if and only if n is prime. Indeed, (0) is certainly not maximal, as (0) \u0002 (2) \u0002 R. Also, (1) is not maximal, since (1) = R. If n is composite, say n = kl, with 1 < k, l < n, then we note that (n) \u0002 (k) \u0002 R, so (n) is not maximal. Finally, let n be prime. Suppose that I is an ideal of R with (n) \u0002 I \u0002 R. Take a ∈ I \\(n). Since a is not divisible by n, and n is prime, we know that (a, n) = 1. Thus, by Corollary 2.1, we can find integers u and v such that au + nv = 1. But as a, n ∈ I , this implies that 1 ∈ I , hence I = R, giving us a contradiction. (As we shall see shortly, there is another way to prove this.)\n\n168\n\n9 Ideals, Factor Rings and Homomorphisms\n\nExample 9.28. In any field, the ideal {0} is maximal! Remember, by Corollary 9.1, a field only has two ideals. In a commutative ring with identity, there is a nice test for maximality of ideals. Theorem 9.20. Let R be a commutative ring with identity, and M an ideal of R. Then M is maximal if and only if R/M is a field. Proof. Suppose that M is a maximal ideal. By Theorem 9.7, R/M is a commutative ring with identity. Furthermore, as M \u0003= R, we know that R/M consists of more than one additive left coset. But the only ring in which 0 = 1 is the ring consisting only of zero; thus, 0 + M \u0003= 1 + M. It remains to show that every nonzero element of R/M has an inverse. Let a + M \u0003= 0 + M. Now, define I = {m + ra : m ∈ M, r ∈ R}. We claim that I is an ideal of R. Taking r = 0, we note that m ∈ I for all m ∈ M; thus, M ⊆ I and, in particular, 0 ∈ I . If m i ∈ M, ri ∈ R, then (m 1 + r1 a) − (m 2 + r2 a) = (m 1 − m 2 ) + (r1 − r2 )a ∈ I. Also, for any s ∈ R, s(m 1 + r1 a) = sm 1 + sr1 a. As sm 1 ∈ M and sr1 ∈ R, we see that I has the absorption property and is, therefore, an ideal. But we noted above that M ⊆ I . Furthermore, a = 0 + 1a ∈ I \\M. By the maximality of M, we have I = R. In particular, 1 ∈ I , so there exist m ∈ M and r ∈ R such that m + ra = 1. But then (r + M)(a + M) = 1 − m + M = 1 + M, since m ∈ M. That is, r + M is the inverse of a + M, and R/M is a field. Conversely, suppose that R/M is a field. We must show that M is maximal. If M = R, then R/M consists only of a single additive left coset, contradicting the fact that a field must have a distinct 0 and 1. Thus, M \u0003= R. Suppose that I is an ideal of R with M \u0002 I \u0002 R. Take a ∈ I \\M. Now, a + M \u0003= 0 + M, so a + M has an inverse, say b+ M. Then (a + M)(b+ M) = 1+ M; in other words, 1−ab ∈ M ⊆ I . But also a ∈ I , which means that ab ∈ I by absorption, and therefore 1 = (1 − ab) + ab ∈ I . By Theorem 9.2, I = R, giving us a contradiction and completing the proof. \u0002 Example 9.29. This gives us another way to deal with Example 9.27. If n = 0, then we note that Z/{0} is simply Z, which is not a field. Thus, (0) is not maximal. If n = 1, then observe that Z/Z is the ring with one element, which is not a field; hence, (1) is not maximal. For any n ≥ 2, we see from Example 9.24 that Z/nZ is isomorphic to Zn . But by Theorem 8.11, Zn is a field if and only if n is prime. Thus, (n) is maximal if and only if n is prime. Example 9.30. By Example 9.25, R[x]/(x) is isomorphic to R, which is a field. Thus, (x) is a maximal ideal of R[x]. Example 9.31. In the same manner as Example 9.30, we see that Z[x]/(x) is isomorphic to Z. But Z is not a field, and hence (x) is not maximal. In fact, we can see this by noting that (x) is properly contained in the ideal M consisting of those polynomials whose constant terms are multiples of 5. We can use Theorem 9.20 to show that M is maximal. See Exercise 9.43.\n\n9.6 Prime and Maximal Ideals\n\n169\n\nIt is worth mentioning that Theorem 9.20 only applies when R is a commutative ring with identity. For instance, the ideal containing only the zero matrix is maximal in M2 (R)! See Exercise 9.54. Definition 9.10. Let R be a commutative ring and P an ideal of R. Then we say that P is a prime ideal2 if 1. P \u0003= R; and 2. if a, b ∈ R and ab ∈ P, then either a ∈ P or b ∈ P. Example 9.32. In any integral domain, {0} is a prime ideal. If ab = 0, then a = 0 or b = 0. Example 9.33. Let us consider R = Z. By the preceding example, {0} is prime, so we know immediately that maximal and prime are not the same thing. Of course, (1) = R, so (1) is not prime. Suppose that n ≥ 2. If n is composite, say n = kl with 1 < k, l < n, we see that kl ∈ (n) but neither k nor l lies in (n). Thus, (n) is not prime. But if n is prime, then (n) is a prime ideal. Indeed, if ab ∈ (n), then n|ab. Thus, by Theorem 2.7, n|a or n|b, and hence a or b is in (n). Once again, there is another way to handle this last example. Theorem 9.21. Let R be a commutative ring with identity and P an ideal. Then P is prime if and only if R/P is an integral domain. Proof. Suppose that P is prime. Since R is a commutative ring with identity, so is R/P, by Theorem 9.7. Also, as P \u0003= R, R/P has more than one element, and therefore 0 + P \u0003= 1 + P. Thus, it remains to show that R/P has no zero divisors. Suppose that (a + P)(b + P) = 0 + P. Then ab ∈ P, and hence a ∈ P or b ∈ P. That is, a + P = 0 + P or b + P = 0 + P, and R/P is an integral domain. Conversely, let R/P be an integral domain. As R/P cannot be the ring with one element, P \u0003= R. Suppose that ab ∈ P. Then (a + P)(b + P) = 0 + P. Since R/P has no zero divisors, a + P = 0 + P or b + P = 0 + P. That is, a ∈ P or b ∈ P, and P is prime. \u0002 Example 9.34. Let us look at Example 9.33 again. We know that Z/(0) is just Z, which is an integral domain, and hence (0) is a prime ideal. If n ≥ 2, then Z/(n) is isomorphic to Zn , by Example 9.24, and Theorem 8.11 tells us that this is an integral domain if and only if n is prime. Thus, for a nonnegative integer n, (n) is a prime ideal of Z if and only if n is 0 or prime. Example 9.35. Refer to Example 9.31. We see that (x) is a prime ideal in Z[x], since Z[x]/(x) is isomorphic to Z. Example 9.36. Naturally, (x) is prime in R[x], because we saw in Example 9.30 that R[x]/(x) is isomorphic to R, which is a field, hence an integral domain. 2 Please note that for noncommutative rings, the definition of a prime ideal is different. We will only\n\nconcern ourselves with prime ideals in commutative rings in this book.\n\n170\n\n9 Ideals, Factor Rings and Homomorphisms\n\nOf course, this last example can be generalized. Theorem 9.22. Let R be a commutative ring with identity. Then every maximal ideal of R is also a prime ideal. Proof. Use the last two theorems and the fact that every field is an integral domain. \u0002 As we have already seen, not every prime ideal is maximal. Also, this last theorem only applies to commutative rings with identity. In some commutative rings without an identity, it is possible to find maximal ideals that are not prime, and Exercise 9.51 asks for an example of this phenomenon. Exercises 9.47. Let R be the ring from Exercise 8.17. Is the ideal (2) prime? Is it maximal? 9.48. Find all prime ideals in each of the following rings. 1. Z10 2. Z50 9.49. Let R be a finite commutative ring with identity. Show that every prime ideal of R is maximal. 9.50. Find every maximal ideal of Z7 ⊕ Z7 . 9.51. Find an example of a commutative ring having an ideal that is maximal but not prime. 9.52. Suppose that R is a commutative ring with identity in which the elements of R that are not units form an ideal. Show that this ideal is the unique maximal ideal of R. 9.53. Show that every field has the property described in the preceding exercise. Also show that Z pn has this property, for every prime p and positive integer n. 9.54. Show that the ideal containing only the zero matrix is maximal in M2 (R). 9.55. Let R be a commutative ring with identity having a prime ideal I . Find a prime ideal in R ⊕ R. 9.56. Let R \u0003= {0} be a commutative ring with identity. Suppose that every proper ideal of R is prime. Show that R is an integral domain, and then use this information to show that R is, in fact, a field.\n\nChapter 10\n\nSpecial Types of Domains\n\nIn this chapter, we begin with a specific and rather familiar sort of integral domain, and then generalize slightly in each section. First, we define a polynomial ring over a field, and show that we have a division algorithm in such a ring. As a result, this polynomial ring is a special type of ring called a Euclidean domain. Subsequently, we demonstrate that Euclidean domains are principal ideal domains; that is, every ideal is principal. Finally, we prove that principal ideal domains are examples of unique factorization domains, in which we have something similar to the Fundamental Theorem of Arithmetic.\n\n10.1 Polynomial Rings We are certainly familiar with polynomials having real coefficients. There is no reason why we cannot consider coefficients in other rings. Definition 10.1. Let R be a ring. Then a polynomial with coefficients in R is a formal expression a0 + a1 x + a2 x 2 + · · · + an x n , where ai ∈ R and n is a nonnegative integer. Suppose that b0 + b1 x + · · · + bm x m is also a polynomial with coefficients in R. Without loss of generality, let us say that n ≤ m. Then these polynomials are equal if and only if ai = bi for all i ≤ n and bi = 0 for all i > n. The set of all polynomials with coefficients in R is denoted R[x]. Example 10.1. Let R = Z5 . Then (inserting congruence class brackets for clarity), an example of a polynomial in R[x] would be f (x) = + x + x 2 . As part of the above definition, we observe that f (x) = g(x), where g(x) = + x + x 2 + x 3 . © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_10\n\n171\n\n172\n\n10 Special Types of Domains\n\n10.1 Polynomial Rings\n\n173\n\na0 (bi + ci ) + a1 (bi−1 + ci−1 ) + · · · + ai (b0 + c0 ). But this is (a0 bi + · · · + ai b0 ) + (a0 ci + · · · + ai c0 ), which is the coefficient of x i in f (x)g(x)+ f (x)h(x). The other distributive law is proved similarly. Finally, we must check that multiplication is associative. But by repeated application of the distributive laws, we see that we may reduce to proving that (au x u bv x v )cw x w = au x u (bv x v cw x w ), for all au , bv , cw ∈ R and all u, v, w ≥ 0. However, both sides of this equation are \u0002 equal to au bv cw x u+v+w , and the proof is complete. Corollary 10.1. Let R be a ring. Then 1. if R has an identity, then so does R[x]; and 2. if R is commutative, then so is R[x]. Proof. (1) The constant polynomial 1 is the identity. (2) Repeatedly applying the distributive laws, we see that we need only check that ai x i and b j x j commute, where ai , b j ∈ R and i, j ≥ 0. But ai x i b j x j = ai b j x i+ j and b j x j ai x i = b j ai x i+ j . Since R is commutative, these are equal. \u0002 When our ring is an integral domain, degrees of polynomials behave in a way we would expect. Theorem 10.2. Let R be an integral domain, and let f (x) and g(x) be nonzero polynomials in R[x], of degree m and n respectively. Then 1. deg( f (x) + g(x)) is at most the larger of m and n (or f (x) + g(x) = 0); and 2. deg( f (x)g(x)) = m + n. Proof. (1) This is clear from the definition of polynomial addition. (2) Let f (x) = a0 + · · · + am x m and g(x) = b0 + · · · + bn x n . Then we see from the definition of polynomial multiplication that the only term of highest degree in f (x)g(x) is am bn x m+n . Furthermore, am \u0004= 0 \u0004= bn and, since R is an integral \u0002 domain, am bn \u0004= 0. Thus, deg( f (x)g(x)) = m + n. Note that the second part of the theorem fails if R is not an integral domain. For instance, in Z6 [x], we have (2 + 3x)(1 + 2x) = 2 + x, which does not have degree 2. Corollary 10.2. If R is an integral domain, then so is R[x]. Proof. By Corollary 10.1, R[x] is a commutative ring with identity. Furthermore, 1 \u0004= 0. By the preceding theorem, the product of nonzero polynomials cannot be the zero polynomial. \u0002 Why are we so interested in polynomial rings? We now know that if F is a field, then F[x] is an integral domain. But it has another attractive property. Indeed, we have an analogue of the division algorithm with which we are familiar for the integers. Readers who have seen polynomial long division for real polynomials will find the procedure very similar.\n\n174\n\n10 Special Types of Domains\n\nTheorem 10.3. (Division Algorithm for Polynomials). Let F be a field, and let f (x), g(x) ∈ F[x], with g(x) \u0004= 0. Then there exist unique q(x), r (x) ∈ F[x] such that f (x) = g(x)q(x) + r (x), with either r (x) = 0 or deg(r (x)) < deg(g(x)). Proof. Let us verify the existence of q(x) and r (x). If f (x) = 0, there is nothing to do; indeed, we let q(x) = r (x) = 0. Therefore, assume that f (x) is not the zero polynomial. We proceed by strong induction on deg( f (x)). Suppose that deg( f (x)) = 0. If deg(g(x)) > 0, then use q(x) = 0 and r (x) = f (x). On the other hand, if deg(g(x)) = 0, then g(x) = b is a nonzero constant in F. As F is a field, we have b−1 ∈ F, and we can use q(x) = b−1 f (x) and r (x) = 0. Thus, suppose that deg( f (x)) = n > 0 and that our result holds for polynomials of smaller degree. Let us write f (x) = a0 + a1 x + · · · + an x n . Also suppose that deg(g(x)) = m, and write g(x) = b0 + b1 x + · · · + bm x m . If n < m, then we can use q(x) = 0 and r (x) = f (x). Otherwise, notice that in f (x) − g(x)bm−1 an x n−m , no term of degree greater than n appears, and the coefficient of x n is an − bm bm−1 an = 0; thus, either f (x) − g(x)bm−1 an x n−m is the zero polynomial, or it has degree strictly smaller than f (x). By our inductive hypothesis, there exist q(x), r (x) ∈ F[x] such that f (x) − g(x)bm−1 an x n−m = g(x)q(x) + r (x), with r (x) = 0 or deg(r (x)) < deg(g(x)). But then f (x) = g(x)(q(x) + bm−1 an x n−m ) + r (x), as required. Now for uniqueness. Suppose that f (x) = g(x)q(x)+r (x) = g(x)q1 (x)+r1 (x), with q(x), q1 (x), r (x), r1 (x) ∈ F[x] and each of r (x) and r1 (x) either is 0 or has degree smaller than that of g(x). Then g(x)(q(x) − q1 (x)) = r1 (x) − r (x). Suppose that q(x) \u0004= q1 (x). By Theorem 10.2, deg(g(x)(q(x) − q1 (x))) ≥ deg(g(x)), but r1 (x) − r (x) cannot possibly have a degree that large. Thus, q(x) = q1 (x) and hence \u0002 r (x) = r1 (x). The proof also shows us how to construct q(x) and r (x). We look only at the leading terms of f (x) and g(x) (say, respectively, an x n and bm x m ). Assuming that n ≥ m, we subtract bm−1 an x n−m g(x) from f (x) and obtain either the zero polynomial or a polynomial of degree smaller than deg( f (x)). Then repeat. Example 10.4. Let us apply the division algorithm in Q[x] with f (x) = 8x 4 −4x 3 + 2x 2 + x + 1 and g(x) = 2x 2 + 3x + 7. We take 2−1 · 8x 4−2 = 4x 2 , multiply by g(x) and subtract from f (x). \u0002 2x + 3x + 7 2\n\n4x 2 8x − 4x + 2x 2 − 8x 4 − 12x 3 − 28x 2 4\n\n3\n\n− 16x 3 − 26x 2\n\n+x +1 +x\n\nNext, take 2−1 (−16)x 3−2 = −8x, multiply by g(x) and subtract.\n\n10.1 Polynomial Rings\n\n\u0002 2x 2 + 3x + 7\n\n175\n\n4x 2 − 8x 8x 4 − 4x 3 + 2x 2 + x + 1 − 8x 4 − 12x 3 − 28x 2 − 16x 3 − 26x 2 + x 16x 3 + 24x 2 + 56x − 2x 2 + 57x + 1\n\nFinally, take 2−1 (−2)x 2−2 = −1, multiply by g(x) and subtract. \u0002 2x + 3x + 7 2\n\n4x 2 − 8x − 1 8x − 4x + 2x 2 − 8x 4 − 12x 3 − 28x 2 4\n\n3\n\n+x +1\n\n− 16x 3 − 26x 2 + x 16x 3 + 24x 2 + 56x − 2x 2 + 57x + 1 2x 2 + 3x + 7 60x + 8 We now have a remainder with degree smaller than deg(g(x)), so we are done. Indeed, f (x) = g(x)(4x 2 − 8x − 1) + (60x + 8). Note that it is not sufficient to work in R[x], where R is an integral domain. By Corollary 10.2, R[x] is also an integral domain, but we cannot implement the division algorithm if we are unable to take the inverse of the leading coefficient of g(x). Indeed, if we worked in Z[x], we would be immediately stymied if we tried to perform the division algorithm using f (x) = 2x 2 + 3x + 5 and g(x) = 3x + 7. In fact, a polynomial ring over a field is a nice example of a special type of integral domain that we can now discuss. Exercises 10.1. In Z11 [x], let f (x) = 2x 3 + 4x 2 + 2x + 5 and g(x) = 2x 4 + 5x 3 + 7x + 1. Find f (x) − g(x) and f (x)g(x). 10.2. Let f (x) = 3x 5 + x 4 + x 3 + 3x 2 + 2x + 4 and g(x) = 2x 3 + 3x 2 + x + 1 be polynomials in Z5 [x]. Find q(x), r (x) ∈ Z5 [x], with deg(r (x)) < 3, such that f (x) = g(x)q(x) + r (x). 10.3. Let f (x) = 3x 5 + 6x 4 + x 3 + 3x 2 + 2x + 4 and g(x) = 2x 3 + 3x 2 + x + 1 be polynomials in Z7 [x]. Find q(x), r (x) ∈ Z7 [x], with deg(r (x)) < 3, such that f (x) = g(x)q(x) + r (x). 10.4. Let R be an integral domain. Show that the units of R[x] are precisely the constant polynomials a, where a ∈ U (R). 10.5. If F is a field, is F[x] a field?\n\n176\n\n10 Special Types of Domains\n\n10.6. Show that 2x + 1 is a unit in Z4 [x]. Then, for any prime p, find a unit in Z p2 [x] that is not a constant polynomial. 10.7. For any ring R, show that R and R[x] have the same characteristic. 10.8. If R and S are isomorphic rings, show that R[x] and S[x] are also isomorphic. 10.9. Let S be a subring of R. Show that S[x] is a subring of R[x]. In particular, if S is an ideal of R, show that S[x] is an ideal of R[x]. 10.10. Let R be a commutative ring with identity and P a prime ideal of R. Show that P[x] is a prime ideal of R[x].\n\n10.2 Euclidean Domains A Euclidean domain is an integral domain having an additional property. Definition 10.3. Let R be an integral domain. Then a Euclidean function is a function ε from the set of nonzero elements of R to the nonnegative integers such that, for all nonzero a, b ∈ R, we have 1. ε(a) ≤ ε(ab); and 2. there exist q, r ∈ R such that a = bq + r , and either r = 0 or ε(r ) < ε(b). Definition 10.4. A Euclidean domain is an integral domain having a Euclidean function. We have already seen several examples of Euclidean domains. Example 10.5. The integers form a Euclidean domain. We already know that Z is an integral domain. Define ε(a) = |a|. If a and b are nonzero integers, then |ab| = |a||b| ≥ |a|. Furthermore, by the division algorithm, there exist q, r ∈ Z such that a = |b|q + r , with 0 ≤ r < |b|. If b > 0, we are done. Otherwise, simply note that a = b(−q) + r . Example 10.6. Any field is a Euclidean domain. See Exercise 10.12. Example 10.7. If F is a field, then F[x] is a Euclidean domain. Indeed, Corollary 10.2 tells us that it is an integral domain. For any 0 \u0004= f (x) ∈ F[x], let ε( f (x)) = deg( f (x)). If 0 \u0004= g(x) ∈ F[x], then by Theorem 10.2, deg( f (x)g(x)) = deg( f (x)) + deg(g(x)) ≥ deg( f (x)). The division algorithm for polynomials completes the proof. Let us construct a new Euclidean domain.\n\n10.2 Euclidean Domains\n\n177\n\nExample 10.8. Let R = {a + bi : a, b ∈ Z}. We call this the ring of Gaussian integers. By Exercises 8.17 and 8.27, R is a subring of F = {a + bi : a, b ∈ Q} which, in turn, is a subfield of C. We claim that R is a Euclidean domain. It is surely an integral domain, since it is a unital subring of a field and therefore has no zero divisors. It remains to construct a Euclidean function. Define ε : F → Q via ε(a + bi) = a 2 + b2 . In particular, if 0 \u0004= a + bi ∈ R, then ε(a + bi) ∈ N. If a, b, c, d ∈ Q, then ε((a + bi)(c + di)) = ε((ac − bd) + (ad + bc)i) = (ac − bd)2 + (ad + bc)2 = a 2 c2 + b2 d 2 + a 2 d 2 + b2 c2 = (a 2 + b2 )(c2 + d 2 ) = ε(a + bi)ε(c + di). In particular, if a + bi and c + di are nonzero elements of R, then ε(a + bi) ≤ ε((a + bi)(c + di)). Take any nonzero u, v ∈ R. Then as F is a field, uv−1 ∈ F. Let us write uv−1 = s + ti, with s, t ∈ Q. Choose integers m and n such that |s − m| ≤ 21 and |t − n| ≤ 21 . Then u − v(m + ni) = u − v((s + ti) + ((m − s) + (n − t)i)) = u − v(uv−1 ) + v((s − m) + (t − n)i) = v((s − m) + (t − n)i). Now, ε((s − m) + (t − n)i) = (s − m)2 + (t − n)2 ≤\n\n1 . 2\n\nTherefore, ε(u − v(m + n)i) = ε(v)ε((s − m) + (t − n)i) < ε(v). Letting q = m + ni and r = u − v(m + ni), we have u = vq + r and we are done. What is so special about Euclidean domains? Let us begin with some definitions. Definition 10.5. Let R be a commutative ring with identity. If a, b ∈ R, then we say that a divides b, and write a|b, if there exists a c ∈ R such that b = ac. Of course, this agrees with our definition of divisibility in Z. We are very much interested in extending the notion of a greatest common divisor as well. For an arbitrary ring, this is problematic, as there is no particular notion of ordering. But for a Euclidean domain, we have ε!\n\n178\n\n10 Special Types of Domains\n\nDefinition 10.6. Let R be a Euclidean domain, and let a, b ∈ R, not both zero. Then a nonzero element d of R is said to be a greatest common divisor (or gcd) if 1. d|a and d|b; and 2. whenever c is an element of R satisfying c|a and c|b, we have ε(c) ≤ ε(d). Certainly a gcd must exist. Indeed, 1 is a common divisor of any two elements, so the set of common divisors is not empty. Furthermore, by definition of a Euclidean function, if c|a and a \u0004= 0, then ε(c) ≤ ε(a). Thus, there is an upper bound on the ε values of the common divisors, so we can select one having the largest possible value. Notice that we called d “a gcd”, not “the gcd”. Indeed, this definition does not produce a unique gcd. In particular, in Z, we see that both 5 and −5 would meet the description of “a gcd” of 10 and 35. However, when we say “the gcd”, we will still mean the positive one; that is, (10, 35) = 5, not −5. Similarly, if F is a field, suppose that d(x) is a gcd of f (x) and g(x). If u is a nonzero element of F, we see immediately that ud(x) also divides both f (x) and g(x), and that deg(ud(x)) = deg(d(x)). Thus, ud(x) is also a gcd. But again, we can choose a specific gcd here. Definition 10.7. Let F be a field and let f (x) and g(x) be polynomials in F[x], not both the zero polynomial. By the gcd of f (x) and g(x) we mean a monic gcd. When we write ( f (x), g(x)), we mean specifically this monic gcd. For more general Euclidean domains, we cannot easily single out a particular gcd in this manner. But we will see that, in fact, the gcds are all related to each other in a nice way. While proving this, we can produce some other interesting results. For instance, the Euclidean domain is so named because there is a Euclidean algorithm just like in Z. Theorem 10.4 (Euclidean Algorithm for Euclidean Domains). Let R be a Euclidean domain. Take a, b ∈ R with b \u0004= 0. If b|a, then b is a gcd of a and b. Otherwise, apply the division algorithm repeatedly. To wit, write a = bq1 + r1 b = r 1 q2 + r 2 r 1 = r 2 q3 + r 3 .. . rk−2 = rk−1 qk + rk rk−1 = rk qk+1 + 0, where all qi , ri ∈ R and ri \u0004= 0, with ε(r1 ) < ε(b) and ε(r j ) < ε(r j−1 ) for all j ≥ 2. Then rk is a gcd of a and b.\n\n10.2 Euclidean Domains\n\n179\n\nProof. If b|a, then b is a common divisor of a and b. Also, if c|b, then ε(c) ≤ ε(b), and so b is a gcd. Assume that b does not divide a, and we perform the division algorithm repeatedly, as indicated. Note, first of all, that this process must end, as the ε(ri ) are strictly decreasing integers and cannot be negative. Suppose that c|a and c|b, say a = ca1 and b = ca2 , with ai ∈ R. Then r1 = c(a1 − a2 q1 ), and hence c|r1 . Similarly, any common divisor of b and r1 must also divide a. Thus, the set of common divisors of a and b is precisely the same as the set of common divisors of b and r1 . In particular, they have the same set of gcds. By the same argument, the gcds of b and r1 are the same as those of r1 and r2 . We then repeat this and find that the gcds of a and b are the same as the gcds of rk and 0. But as everything divides 0, we are looking only for the largest value of ε among the divisors of rk . However, as if u|v and v \u0004= 0, then ε(u) ≤ ε(v), we see that rk is \u0002 a gcd of rk and 0, as required. Corollary 10.3. Let R be a Euclidean domain. Take a, b ∈ R with b \u0004= 0. Let d be the gcd of a and b found in the preceding theorem. Then there exist u, v ∈ R such that d = au + bv. Proof. If b|a, then d = b = a(0) + b(1). Assume otherwise. We have d = rk = rk−2 + rk−1 (−qk ), a multiple of rk−2 plus a multiple of rk−1 . But the preceding equation is rk−3 = rk−2 qk−1 + rk−1 . Thus, d = rk−2 + (rk−3 + rk−2 (−qk−1 ))(−qk ). We have written d as a multiple of rk−3 plus a multiple of rk−2 . Now move backwards through the equations, and we will eventually write d as a multiple of a plus a multiple of b. \u0002 Example 10.9. Let us apply the Euclidean algorithm and its corollary in Z7 [x], starting with f (x) = 2x 3 + 4x 2 + x + 1 and g(x) = 6x 3 + 4x 2 + 4x + 5. We write 2x 3 + 4x 2 + x + 1 = (6x 3 + 4x 2 + 4x + 5)(5) + (5x 2 + 2x + 4) 6x 3 + 4x 2 + 4x + 5 = (5x 2 + 2x + 4)(4x + 2) + (5x + 4) 5x 2 + 2x + 4 = (5x + 4)(x + 1) + 0. Thus, 5x + 4 is a gcd of f (x) and g(x). Now let us apply the method discussed in the proof of the preceding corollary. We have 5x + 4 = g(x) − (5x 2 + 2x + 4)(4x + 2) = g(x) − ( f (x) − g(x)(5))(4x + 2) = f (x)(3x + 5) + g(x)(6x + 4). If we want to use ( f (x), g(x)), we must make it monic. Now, 5−1 = 3, and therefore ( f (x), g(x)) = 3(5x + 4) = x + 5. Then we get\n\n180\n\n10 Special Types of Domains\n\nx + 5 = 3(5x + 4) = f (x)(2x + 1) + g(x)(4x + 5). Corollary 10.4. Let R be a Euclidean domain, and let a, b ∈ R with b \u0004= 0. Let d be the gcd of a and b found in Theorem 10.4. Then if c ∈ R is a divisor of both a and b, then c|d. Proof. We have d = au + bv, for some u, v ∈ R. If c|a and c|b, then c|d.\n\n\u0002\n\nLet us now discuss how the gcds of two elements of a Euclidean domain relate to each other. Definition 10.8. Let R be a commutative ring with identity. If a, b ∈ R, then we say that a and b are associates if there exists a unit u of R such that b = au. Note that if b = au, where u is a unit, then a = bu −1 . Thus, if a is an associate of b, then b is an associate of a. Example 10.10. In Z, the only units are 1 and −1, so the only associates of a are a and −a. Example 10.11. Let F be a field. The units in F[x] are the nonzero constants. (See Exercise 10.4.) Thus, the associates of f (x) are of the form a f (x), where 0 \u0004= a ∈ F. Lemma 10.1. Let R be an integral domain. Then a and b are associates in R if and only if a|b and b|a. Proof. If a and b are associates, the fact that a|b and b|a follows from the definition. Suppose that a|b and b|a, say b = ar and a = bs, with r, s ∈ R. Then a = ar s. If a = 0, then b = 0, so a = b · 1. Otherwise, by cancellation, r s = 1, and hence r is a unit. \u0002 Theorem 10.5. Let R be a Euclidean domain. Take a, b ∈ R, not both 0. Let d be any gcd of a and b. Then c ∈ R is a gcd of a and b if and only if c and d are associates. Proof. Suppose that c is a gcd of a and b. Let g be the gcd of a and b found in Theorem 10.4. By Corollary 10.4, c and d divide g. Applying the division algorithm, we have c = gq + r , where q, r ∈ R and either r = 0 or ε(r ) < ε(g). Suppose the latter. Now, c|g, and therefore c|r = c − gq. But then ε(c) ≤ ε(r ). However, if c and g are both gcds, we must have ε(c) = ε(g), giving us a contradiction. Therefore, r = 0 and g|c. By the preceding lemma, g and c are associates, say c = gu, with u a unit in R. By the same argument, d = gv, where v is a unit in R. Then c = duv−1 , where uv−1 is a unit in R, and hence c and d are associates. Conversely, let c and d be associates. Then since d|a and d|b, we have c|a and c|b as well. Furthermore, since c|d and d|c we can only have ε(c) = ε(d). Therefore, c is a gcd of a and b. \u0002\n\n10.2 Euclidean Domains\n\n181\n\nWe can now feel better about Definition 10.7, where we referred to “the” monic gcd of f (x) and g(x) in F[x]. As any two gcds are associates, and the only units are nonzero elements of F, there can only be one gcd that is monic. Time to tidy up! We can strengthen Corollary 10.3. It actually applies to any gcd, not just the one found in Theorem 10.4. Theorem 10.6. Let R be a Euclidean domain. Take a, b ∈ R, not both 0. Let d be a gcd of a and b. Then there exist u, v ∈ R such that d = au + bv. Proof. Without loss of generality, assume that b \u0004= 0, and calculate the gcd g of a and b from Theorem 10.4. Then by Corollary 10.3, g = au + bv, for some u, v ∈ R. But by Theorem 10.5, d = gw, for some unit w of R. Thus, d = auw + bvw. \u0002 We conclude by strengthening Corollary 10.4. Theorem 10.7. Let R be a Euclidean domain. Take a, b ∈ R, not both 0. Then the following are equivalent for an element d of R: 1. d is a gcd of a and b; and 2. d|a, d|b, and if c|a and c|b, then c|d. Proof. Suppose (1) holds. Without loss of generality, assume that b \u0004= 0. By definition, d|a and d|b. Suppose that c|a and c|b. If g is the gcd of a and b found in Theorem 10.4, then by Corollary 10.4, c|g. But Theorem 10.5 tells us that g|d. Thus, c|d. Conversely, suppose that (2) holds. Then d is a common divisor of a and b. Suppose that c is another common divisor of a and b. Then by assumption, c|d. But this means that ε(c) ≤ ε(d); hence, d is a gcd. \u0002 A nice feature of Theorem 10.7 is that it shows that gcds in a Euclidean domain do not depend upon the particular Euclidean function that is used. Exercises 10.11. In an integral domain, if a and ab are associates, show that a = 0 or b is a unit. 10.12. Show that every field is a Euclidean domain. 10.13. Let R be a Euclidean domain. Let n be the smallest value of ε(s), for all 0 \u0004= s ∈ R. Show that for each 0 \u0004= a ∈ R we have ε(a) = n if and only if a is a unit. 10.14. Find all units in the ring of Gaussian integers. 10.15. In Q[x], let f (x) = 3x 4 + 7x 3 + 13x 2 + 7x + 6 and g(x) = 2x 4 + 7x 3 + 13x 2 + 11x + 3. Find ( f (x), g(x)). 10.16. In Z5 [x], let f (x) = 3x 4 + 3x 3 + x + 1 and g(x) = 2x 3 + 4x 2 + x + 1. Find ( f (x), g(x)).\n\n182\n\n10 Special Types of Domains\n\n10.17. Taking f (x) and g(x) as in Exercise 10.15, find u(x), v(x) ∈ Q[x] such that ( f (x), g(x)) = f (x)u(x) + g(x)v(x). 10.18. Taking f (x) and g(x) as in Exercise 10.16, find u(x), v(x) ∈ Z5 [x] such that ( f (x), g(x)) = f (x)u(x) + g(x)v(x). 10.19. Find a gcd for 5 + 7i and 1 + 3i in the ring of Gaussian integers. 10.20. Let R be a Euclidean domain having the following additional property: for every a, b ∈ R such that a, b and a + b are all nonzero, ε(a + b) is no bigger than the larger of ε(a) and ε(b). (For example, if F is a field, the degree function on F[x]\\{0} has this property.) Show that in the second part of the definition of a Euclidean function, the elements q and r are uniquely determined.\n\n10.3 Principal Ideal Domains Let us discuss another sort of integral domain with a nice property. Definition 10.9. A principal ideal domain (or PID) is an integral domain in which every ideal is principal. A field F is an obvious example of a PID; indeed, its only ideals are (0) and F = (1). But we can obtain others through the following theorem. Theorem 10.8. Every Euclidean domain is a PID. Proof. Let R be a Euclidean domain with Euclidean function ε, and I an ideal of R. If I = {0}, then I = (0), and there is nothing to do. Assume that I \u0004= {0}. Among the nonzero elements of I , choose b so that ε(b) is as small as possible. (Since ε takes on values that are nonnegative integers, there must be a smallest such value.) We claim that I = (b). Take a ∈ I . As ε is a Euclidean function, we have a = bq + r , where q, r ∈ R and either r = 0 or ε(r ) < ε(b). If r = 0, then b|a, as required. Otherwise, we note that a, b ∈ I , and since I is an ideal, r = a − bq ∈ I . But by the minimality of ε(b), this is impossible. \u0002 Example 10.12. Since Z is a Euclidean domain, it is a PID. Example 10.13. Let F be a field. Since F[x] is a Euclidean domain, it is a PID. Proving that an integral domain is not a Euclidean domain can be a bit tricky; it is often simpler to show that is not a PID, from which it follows that it is not a Euclidean domain. Example 10.14. We claim that Z[x] is not a PID, and hence not a Euclidean domain. To prove this, consider the set I of all f (x) ∈ Z[x] whose constant terms are divisible by 5. We saw in Exercise 9.2 that I is an ideal. But it is not principal. Indeed, suppose\n\n10.3 Principal Ideal Domains\n\n183\n\nthat I = ( f (x)). Then as the constant polynomial 5 is in I , we see that f (x)|5. In view of Theorem 10.2, f (x) is a constant polynomial. As it divides 5, the constant must be in {±1, ±5}. However, (1) = (−1) = Z[x], whereas (5) = (−5) = 5Z[x], which does not include x. But x ∈ I , and therefore 5Z[x] \u0004= I . We might, at this point, ask if every PID is a Euclidean domain. The answer is no, but this is not obvious. Theodore S. Motzkin showed that there is a subring of the complex numbers that is a PID but not a Euclidean domain. We will not use this fact, but the interested reader can find an accessible proof in the paper of Wilson . Let us explore a couple of other properties of PIDs. The following theorem shows that a PID has the ascending chain condition. Theorem 10.9. Let R be a PID. Suppose that R has ideals Ik , k ∈ N, such that I1 ⊆ I2 ⊆ I3 ⊆ · · · . Then there exists a positive integer n such that Ik = In for all k ≥ n. \u0003 Proof. Let I = ∞ k=1 Ik . We claim that I is an ideal. Certainly 0 ∈ I1 ⊆ I . If a, b ∈ I , then there exist positive integers k and l such that a ∈ Ik and b ∈ Il . Let m be the larger of k and l. Then a, b ∈ Im , and hence a − b ∈ Im ⊆ I . Similarly, if a ∈ I , say a ∈ Ik , and r ∈ R, then ra ∈ Ik ⊆ I . Thus, I is an ideal. As R is a PID, we must have I = (c) for some c ∈ I . But then c ∈ In , for some positive integer n. It now follows that I = (c) ⊆ In . That is, I = In , and hence Ik = In for all k ≥ n. \u0002 We are familiar with the notion of a prime positive integer. Let us extend the idea. Definition 10.10. Let R be an integral domain. Then an element p of R is prime if it is not zero, not a unit, and if p|ab, with a, b ∈ R, then p|a or p|b. We observe that the definition of a prime positive integer that we introduced in Chapter 2 is different. However, Theorem 2.7 assures us that the definitions are equivalent, for positive integers. Of course, the positive integers do not form a ring, so in Z, we see that the primes are ±2, ±3, ±5, . . .. (Note that 1 and −1 are units, so we exclude them.) We have an easy lemma. Lemma 10.2. Let R be an integral domain, and take 0 \u0004= p ∈ R. Then p is prime if and only if ( p) is a prime ideal. Proof. Let p be prime. If ( p) = R, then there exists an r ∈ R such that r p = 1; hence, p is a unit. But primes cannot be units, so this is impossible. If ab ∈ ( p), then p|ab, and hence p|a or p|b. Thus, a ∈ ( p) or b ∈ ( p), and ( p) is a prime ideal. Conversely, suppose that ( p) is a prime ideal and p|ab. Then ab ∈ ( p), and hence a ∈ ( p) or b ∈ ( p). That is, p|a or p|b. Furthermore, if p is a unit, then by Theorem 9.2, ( p) = R, which contradicts the assumption that ( p) is a prime ideal. Thus, p is prime. \u0002\n\n184\n\n10 Special Types of Domains\n\nDefinition 10.11. Let R be an integral domain, and take p ∈ R. We say that p is irreducible if it is not zero, not a unit, and if p = ab, with a, b ∈ R, then either a or b must be a unit. This is, essentially, the definition we used for a prime positive integer. As we noted above, in the integers, these concepts are equivalent. What is the general situation? Theorem 10.10. Let R be an integral domain. Then every prime in R is irreducible. Proof. Let p be a prime, and suppose that p = ab, with a, b ∈ R. Then p|ab, so p|a or p|b. Without loss of generality, say p|a. But a| p as well. By Lemma 10.1, p and a are associates. Thus, by Exercise 10.11, b is a unit, as required. \u0002 Unfortunately, the converse is not true in general. √ Example 10.15. Let R = {a + b 5i : a, b ∈ Z}. It is easy to check that R is a unital subring of C, and hence √ an integral domain. We can define a function called a norm on R via N (a + b 5i) = a 2 + 5b2 . If u, v ∈ R, then N (uv) = N (u)N (v). (This is the same calculation as in Example 10.8.) We claim that 3 is irreducible in R. If 3 = uv, then 9 = N (3) = N (u)N (v). Noting that the norms of elements of R are nonnegative integers, we can only have N (u) = N (v) = 3 or, without loss of generality, N (u) = 1 and N (v) = 9. But the equation a 2 + 5b2 = 3 has no solution in the integers, so N (u) = N (v) = 3 is impossible. Also, the only solutions to a 2 + 5b2 = 1 are a ∈ {1, −1} and b = 0. However, 1 and −1 are units in R. Also, 3 is clearly not a unit, and √ is proved. Nevertheless, 3 is not prime. To see √ the claim this, √ we note that√(2 + 5i)(2 − 5i) = 9. Of course, 3|9, but 3 does not divide 2 + 5i or 2 − 5i. The good news, however, is that in a PID, primeness and irreducibility are equivalent. Theorem 10.11. Let R be a PID and p ∈ R. Then p is prime if and only if p is irreducible. Proof. In view of Theorem 10.10, we only need to show the converse. Let p be irreducible, and let I = ( p). We claim that I is a maximal ideal of R. If not, suppose that J is an ideal of R with I \u0002 J \u0002 R. Since R is a PID, we have J = (a), for some a ∈ J . Now, p ∈ I ⊆ J , so p = ab, for some b ∈ R. As p is irreducible, either a or b is a unit. If a is a unit, then by Theorem 9.2, J = R, which is not permitted. Therefore, b is a unit. But then a = pb−1 ∈ I . Thus, J ⊆ I , which is also not allowed. On the other hand, if I = R, then p is a unit, which is impossible. Our claim is proved. By Theorem 9.22, a maximal ideal is necessarily prime. Lemma 10.2 completes the proof. \u0002\n\n10.3 Principal Ideal Domains\n\n185\n\nExercises √ 10.21. With R as in Example 10.15, show that 1 + 2 5i is irreducible, but not prime. √ √ 10.22. Let S be {a + b 3i : a, b ∈ Z}, a subring of C. Show that 1 + 3i is irreducible, but not prime. 10.23. Show that R and S from the preceding two exercises are not PIDs. 10.24. Let R be an integral domain. Show that an associate of an irreducible element is irreducible, and an associate of a prime element is prime. 10.25. If R is a Euclidean domain, does it follow that R[x] is a Euclidean domain? Prove that it does, or give an explicit counterexample. 10.26. Let R be a PID. Show that every proper ideal of R is a subset of a maximal ideal of R. 10.27. Let R be an integral domain and p a prime in R. If p|a1 a2 · · · an , with ai ∈ R, show that some ai is divisible by p. 10.28. Let R be a PID and 0 \u0004= a ∈ R. Show that a is irreducible if and only if (a) is a maximal ideal. 10.29. Let R be an integral domain, but not a field. Show that there exist infinitely many ideals I1 , I2 , . . . of R such that In+1 is a proper subset of In for all n. 10.30. Let R be an integral domain. If R[x] is a PID, show that R is a field.\n\n10.4 Unique Factorization Domains We now reach our main conclusion, which is that every PID has an analogue of the Fundamental Theorem of Arithmetic. Definition 10.12. Let R be an integral domain. We say that R is a unique factorization domain (or UFD) if 1. every nonzero, nonunit element of R can be written as a product of one or more irreducibles; and 2. the product is unique up to order and associates; that is, if p1 p2 · · · pk = q1 q2 · · · ql , for some irreducibles pi and qi , then k = l and, after rearranging, each pi is an associate of qi . Theorem 10.12. Every PID is a UFD.\n\n186\n\n10 Special Types of Domains\n\nProof. Let R be a PID. We shall prove that R satisfies the first part of the definition of a UFD. Take any nonzero nonunit a1 ∈ R, and suppose that a1 is not a product of irreducibles. If a1 is irreducible then we have an immediate contradiction. Therefore, we may write a1 = a2 b2 , where a2 and b2 are nonunits in R. If a2 and b2 are both products of irreducibles then again, we have a contradiction, as a1 is then a product of irreducibles. Without loss of generality, let us say that a2 is not a product of irreducibles. In particular, it is not irreducible, so write a2 = a3 b3 , where a3 and b3 are nonunits, and so forth. Then we have ai+1 |ai for all positive integers i. Furthermore, as bi+1 is not a unit, we see that ai and ai+1 are not associates. By Lemma 10.1, ai does not divide ai+1 . In particular, ai ∈ (ai+1 ), but ai+1 ∈ / (ai ), so (ai ) \u0002 (ai+1 ) for all positive integers i. But this contradicts Theorem 10.9, and we see that each nonzero nonunit is a product of irreducibles. Now let us verify the uniqueness. Suppose that p1 · · · pk = q1 · · · ql , where the pi and qi are irreducible, and k ≤ l. Then p1 |q1 · · · ql . By Theorem 10.11, p1 is prime. Thus, p1 divides one of the terms in the product. After rearranging, we may assume that p1 |q1 . Let us write q1 = p1 u 1 , with u 1 ∈ R. As q1 is irreducible and p1 is not a unit, we see that u 1 is a unit, and hence p1 and q1 are associates. Thus, p1 p2 · · · pk = u 1 p1 q2 · · · ql . Cancelling, we have p2 · · · pk = u 1 q2 · · · ql . Now, p2 |u 1 q2 · · · ql , and since p2 is prime, it divides a term in the product. Since a divisor of a unit is a unit, we cannot have p2 |u 1 , and therefore p2 |qi , for some i ≥ 2. Rearranging, we have p2 |q2 . Just as before, we see that q2 = p2 u 2 , for some unit u 2 . Repeating, we find that pi and qi are associates, 1 ≤ i ≤ k. If k = l, we are done. Otherwise, we have 1 = u 1 · · · u k qk+1 · · · ql . But nonunits cannot divide 1, so we have a contradiction. \u0002 Our examples of UFDs will largely be PIDs. Example 10.16. As we already knew from the Fundamental Theorem of Arithmetic, Z is a UFD. Example 10.17. For any field F, F[x] is a UFD. There are also UFDs that are not PIDs. In fact, Z[x] is such a ring. We opt to postpone the proof of this until Section 11.2. What sort of integral domains are not UFDs? Either of the two conditions could fail. Let us first consider one where nonzero nonunit elements are not necessarily products of irreducibles. Example 10.18. Let R be the subset of Q[x] consisting of all polynomials with an integer constant term. It is easy to see that R is a unital subring of Q[x]. As Q[x] is an integral domain, so is R. We claim that the only units of R are the constant polynomials 1 and −1. Indeed, a unit is necessarily a unit in Q[x] as well. By Exercise 10.4, our unit is a nonzero constant a. But as the constant term of an element of R must be an integer, we see that if a f (x) = 1, then a can only be ±1, proving the claim. In particular, x is a nonzero nonunit. If we write x = p1 (x) · · · pk (x), a product of irreducibles, then all but one of the pi (x) (say p1 (x)) are integers and p1 (x) = q x,\n\n10.4 Unique Factorization Domains\n\n187\n\n\u0004 \u0002 for some 0 \u0004= q ∈ Q. But q x is not irreducible; indeed, q x = 2 q2 x , and neither 2 nor q2 x is a unit. Thus, x is not a product of irreducibles, and R is not a UFD. Even if every nonzero nonunit is a product of irreducibles, this product may not be unique. √ Example 10.19. Consider the ring R = {a + b 5i : a, b ∈ Z} from Example 10.15. We noted in √ that example√ that 3 is irreducible. Applying a similar argument, we can see that 2 + 5i and 2 − 5i are irreducible. (We do have to check that they are not units, but if uv = 1, then N (u)N (v) = N (1) = 1, and as we noted in Example 10.15, √ √ this must mean that u = v = ±1.) Thus, we can write 9 = 3·3 = (2+ 5i)(2− 5i), giving two different products As the only units are ±1, we see that 3 √ √ of irreducibles. is not an associate of 2 + 5i or 2 − 5i. Therefore, our factorization is not unique. We close with a few remarks concerning divisibility in a UFD. Theorem 10.13. Let R be a UFD, and let a and b be nonzero nonunit elements of R. Then there exist irreducibles p1 , . . . , pk , none of which are associates, such that a = up1m 1 · · · pkm k and b = v p1n 1 · · · pkn k , for some units u, v ∈ R and some nonnegative integers m i and n i . Furthermore, a|b if and only if m i ≤ n i for all i. Proof. Write each of a and b as a product of irreducibles. List all of the irreducibles that appear, and if some are associates, say q1 , q2 , . . ., then delete all but one. Let p1 , . . . , pk be the irreducibles that remain. Then a can be written as a product of irreducibles, each of which is an associate of some pi , and so can be written as a product of a unit and pi . Gathering the units together, we obtain our expression for a, and similarly for b. If m i ≤ n i for all i, then we see that b = avu −1 p1n 1 −m 1 · · · pkn k −m k ; hence, a|b. Conversely, without loss of generality, suppose that m 1 > n 1 . If a|b, then write b = ac. As R is an integral domain, we can use cancellation, and obtain p1m 1 −n 1 p2m 2 · · · pkm k c = u −1 v p2n 2 · · · pkn k . Here, c is either a unit or a product of irreducibles. By unique factorization, p1 must be an associate of one of u −1 v p2 , p3 , . . . , pk . But by our choice of the pi , this is impossible. \u0002 A UFD does not necessarily have anything comparable to a Euclidean function, so we cannot order elements in any logical way. However, we can obtain the equivalent form of a gcd given in Theorem 10.7. Theorem 10.14. Let R be a UFD. Take any nonzero nonunits a, b ∈ R, and write them in the form a = up1m 1 · · · pkm k , b = v p1n 1 · · · pkn k , as in Theorem 10.13. Let d = p1l1 · · · pklk , where li is the smaller of m i and n i , for all i. Then d|a, d|b, and if c|a and c|b, then c|d.\n\n188\n\n10 Special Types of Domains\n\nProof. Theorem 10.13 tells us that d|a and d|b. Suppose that c|a and c|b. If c is a unit, then surely c|d. Suppose it is not. Then write a = cr , with r ∈ R. Now c can be written as a product of irreducibles, and r is a unit or a product of irreducibles. By unique factorization, all of these irreducibles must be associates of the pi . Using j j Theorem 10.13 again, we can write c = w p11 · · · pkk , where w is a unit and jk ≤ m k , for all k. By the same argument, as c|b, we have jk ≤ n k , and hence jk ≤ lk , for all k. Therefore, c|d. \u0002 Exercises 10.31. Show that 1 + i is prime in the ring R of Gaussian integers. 10.32. In the ring of Gaussian integers, which of the numbers 3, 5 and 7 are irreducible? 10.33. Must a unital subring of a UFD be a UFD? Prove that it must, or give an explicit counterexample. 10.34. Let R be a UFD. Suppose that a and b are nonzero nonunit elements of R. If d1 and d2 are gcds of a and b (in the sense discussed in the second part of Theorem 10.7), show that d1 and d2 are associates. √ 10.35. Let R = {a +b 6i : a, b ∈ Z}. Find a, b, c, d ∈ R such that 10 = ab = cd, but a, b, c and d are all irreducible and neither of {a, b} is an associate of either of {c, d}. Conclude that R is not a UFD. 10.36. Let R be a UFD, and let p be an irreducible element of R. If a and b are nonzero nonunits of R, and p|ab, writing both a and b as products of irreducibles, show that p is an associate of at least one of the irreducibles appearing in at least one of these products. 10.37. Show that every irreducible in a UFD is prime. 10.38. Let R be a UFD. Suppose that there exist a1 , a2 , . . . ∈ R such that (a1 ) ⊆ (a2 ) ⊆ · · · . Show that there exists an i such that (ai ) = (ai+1 ).\n\nReference 1. Wilson, J.C.: A principal ideal ring that is not a Euclidean ring. Math. Mag. 46, 34–38 (1973)\n\nPart IV\n\nFields and Polynomials\n\nChapter 11\n\nIrreducible Polynomials\n\nLet F[x] be the polynomial ring over a field F. If f (x) ∈ F[x], we can now discuss some conditions under which f (x) is irreducible.\n\n11.1 Irreducibility and Roots For any field F, we recall that the polynomial ring F[x] is a UFD (see Example 10.17). Also, by Exercise 10.4, the units in F[x] are the nonzero elements of F. Thus, every polynomial of degree greater than 0 is a product of one or more irreducibles. Here, a polynomial f (x) of degree greater than 0 is irreducible over F if, whenever f (x) = g(x)h(x) for some g(x), h(x) ∈ F[x], either g(x) or h(x) is an element of F. Otherwise, f (x) is reducible. Note that irreducibility depends very much upon the particular field. 2 Example 11.1. The polynomial √ √x − 2 is irreducible over Q, but reducible over R, 2 since x − 2 = (x − 2)(x + 2).\n\nLet f (x) = a0 + a1 x + · · · + an x n ∈ F[x]. If r ∈ F, we can evaluate f (x) at r , and obtain f (r ) = a0 + a1 r + a2 r 2 + · · · + an r n . In this way, we obtain a function (not a homomorphism!) α : F → F given by α(r ) = f (r ). In dealing with polynomials in R[x], we are accustomed to identifying the polynomial f (x) with this function α. But over a more general field, we cannot do this. Indeed, two different polynomials can induce the same function. Example 11.2. In Z5 [x], the polynomials f (x) = x 3 + x +1 and g(x) = x 5 + x 3 +1 induce the same function. That is, f (r ) = g(r ) for all r ∈ Z5 . (There are only five elements in Z5 , so this is easily checked.) It is worth mentioning that we do obtain a homomorphism if we fix an element r of the field and consider evaluating polynomials at r . © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_11\n\n191\n\n192\n\n11 Irreducible Polynomials\n\nLemma 11.1. Let R be a commutative ring and fix r ∈ R. Then the function α : R[x] → R given by α( f (x)) = f (r ) is a homomorphism. Proof. Let f (x) = a0 + · · · + an x n and g(x) = b0 + · · · + bn x n be arbitrary polynomials in R[x] (adding in terms with coefficient zero if necessary). Then α( f (x) + g(x)) = a0 + b0 + a1r + b1 r + · · · + an r n + bn r n = (a0 + · · · + an r n ) + (b0 + · · · + bn r n ) = α( f (x)) + α(g(x)). Also, writing f (x)g(x) = c0 + · · · + c2n x 2n , where ci = a0 bi + · · · + ai b0 , we have α( f (x)g(x)) = c0 + · · · + c2n r 2n , whereas α( f (x))α(g(x)) = (a0 + · · · + an r n )(b0 + · · · + bn r n ). But for any i, a0 bi r i + a1r bi−1r i−1 + · · · + ai r i b0 = ci r i , and so α( f (x)g(x)) = α( f (x))α(g(x)).\n\n\u0002\n\nWe can now use the division algorithm to write a polynomial over a field F as a multiple of x − a, for any a ∈ F, plus a constant. Theorem 11.1 (Remainder Theorem). Let F be a field and f (x) ∈ F[x]. Take any a ∈ F. Then there exists a q(x) ∈ F[x] such that f (x) = (x − a)q(x) + f (a). Proof. By the division algorithm for polynomials, f (x) = (x − a)q(x) + r (x), where q(x), r (x) ∈ F[x], and either r (x) is the zero polynomial, or deg(r (x)) < deg(x − a) = 1. That is, r (x) is some constant, b ∈ F. By the preceding lemma, f (a) = (a − a)q(a) + b = b.\n\n\u0002\n\nIt is crucial for us to know if a polynomial has any roots. Definition 11.1. Let F be a field and f (x) ∈ F[x]. If a ∈ F, then we say that a is a root of f (x) if f (a) = 0. Example 11.3. The polynomial x 2 −√2 has no√ roots in Q. However, if we regard it as a polynomial over R, we see that 2 and − 2 are roots.\n\n11.1 Irreducibility and Roots\n\n193\n\nRecall that if f (x), g(x) ∈ F[x], we say that f (x) divides g(x), and write f (x)|g(x), if there exists an h(x) ∈ F[x] such that g(x) = f (x)h(x). Theorem 11.2 (Factor Theorem). Let F be a field and f (x) ∈ F[x]. Take any a ∈ F. Then a is a root of f (x) if and only if (x − a)| f (x). Proof Suppose that a is a root of f (x). By the Remainder Theorem, we have f (x) = (x − a)q(x), and hence (x − a)| f (x). Conversely, suppose that (x − a)| f (x). Then f (x) = (x − a)g(x), for some g(x) ∈ F[x]. In this case, f (a) = (a − a)g(a) = 0, and hence a is a root. \u0002 Example 11.4. In Z7 [x], let f (x) = 3x 3 +5x 2 +4x +4. We note that 2 is a root. Thus, x −2 (in other words, x +5) must divide f (x). In fact, f (x) = (x −2)(3x 2 +4x +5). Corollary 11.1. Let F be a field and f (x) ∈ F[x]. If deg( f (x)) > 1 and f (x) has a root in F, then f (x) is reducible over F. Proof. Let a be a root of f (x). By the Factor Theorem, f (x) = (x − a)g(x), for some g(x) ∈ F[x]. Since deg( f (x)) > 1, we note that g(x) is not a constant. Thus, f (x) is reducible. \u0002 The converse is false! Example 11.5. In R[x], let f (x) = x 4 + 2x 2 + 1. For any a ∈ R, we have f (a) ≥ 1; thus, f (x) has no real roots. However, f (x) = (x 2 + 1)2 . Thus, f (x) is reducible. However, for polynomials of degree 2 and 3, the converse does hold. Corollary 11.2. Let F be a field and f (x) ∈ F[x]. Then 1. if deg( f (x)) = 1, then f (x) is irreducible over F; and 2. if f (x) has degree 2 or 3, then f (x) is irreducible over F if and only if it has no roots in F. Proof. (1) If f (x) = g(x)h(x), then by Theorem 10.2, either g(x) or h(x) has degree 0. (2) If f (x) is irreducible, then the preceding corollary tells us that f (x) has no roots. Suppose that f (x) is reducible, say f (x) = g(x)h(x) for some nonconstant polynomials g(x) and h(x) in F[x]. As the sum of their degrees is 2 or 3, either g(x) or h(x) must have degree 1. Without loss of generality, say g(x) = ax +b, with a, b ∈ F and a \u0006= 0. But then notice that f (−a −1 b) = (a(−a −1 b) + b)h(−a −1 b) = 0. Thus, \u0002 −a −1 b is a root of f (x). We can also put a limit on the number of roots of a polynomial. Corollary 11.3. Let F be a field and f (x) ∈ F[x] a nonzero polynomial. If f (x) has degree n, then f (x) has at most n roots in F.\n\n194\n\n11 Irreducible Polynomials\n\nProof. We proceed by induction on n. If n = 0, then f (x) is a nonzero constant polynomial, which clearly has no roots. Assume that our result is true for n, and let deg( f (x)) = n + 1. If f (x) has no roots, then we are done. Otherwise, let a be a root. By Theorem 11.2, f (x) = (x − a)g(x), for some g(x) ∈ F[x]. Furthermore, by Theorem 10.2, deg(g(x)) = n. Thus, our inductive hypothesis tells us that g(x) has at most n roots. Let b be any root of f (x). Then 0 = f (b) = (b − a)g(b). Therefore, either b − a = 0 (and b = a) or g(b) = 0 (and b is among the at most n roots of g(x)). Thus, f (x) has at most n + 1 roots, as required. \u0002 Exercises 11.1. Are the following polynomials irreducible in Z7 [x]? 1. x 3 + 5x 2 + 4x + 3 2. x 3 + x 2 + 1 3. x 4 + x 2 + 2 11.2. Write each of the following as products of irreducibles in Z5 [x]. 1. x 3 + 3x 2 + 3x + 2 2. x 3 + 2x 2 + 4x + 2 3. x 4 + 2x 3 + 4x + 3 11.3. Find every irreducible polynomial of degree 3 over Z2 . 11.4. If we divide 3x 59 + 4x 16 + 2 by x + 5 in Z7 [x], what is the remainder? (The answer must be in {0, 1, . . . , 6}.) 11.5. Let F be an infinite field. If f (x), g(x) ∈ F[x], and f (a) = g(a) for all a ∈ F, show that f (x) = g(x). 11.6. Let p be a prime. Find infinitely many polynomials f 1 (x), f 2 (x), . . . in Z p [x] such that f i (a) = 0 for all a ∈ Z p and all positive integers i. 11.7. Is Lemma 11.1 still true for noncommutative rings? 11.8. Let R be an integral domain. Show that U (R) has at most n elements of order n, for every positive integer n. Also give an example of a commutative ring R with identity which is not an integral domain for which this is not true. 11.9. Let p be a prime number. Show that the following are equivalent: 1. x 2 + 1 is reducible in Z p [x]; and 2. there exist nonnegative integers m and n such that p = m + n and p|(mn − 1). 11.10. Show that Theorems 11.1 and 11.2 remain true if F is replaced with an integral domain.\n\n11.2 Irreducibility over the Rationals\n\n195\n\n11.2 Irreducibility over the Rationals If we have a polynomial f (x) ∈ Q[x], then by multiplying by a suitable positive integer, we obtain a polynomial in Z[x]. It is often simpler to start with a polynomial with integer coefficients. As we noted in the preceding section, a polynomial of degree greater than 1 in Q[x] is necessarily reducible if it has a root. Of course, there are infinitely many possible roots, so testing them all is impossible. However, we can narrow the possible roots down to a finite set of rational numbers. Theorem 11.3 (Rational Roots Theorem). Let f (x) = a0 + a1 x + · · · + an x n ∈ Z[x], with an \u0006= 0. Suppose that q ∈ Q is a root of f (x). If q = rs , with r, s ∈ Z and (r, s) = 1, then r |a0 and s|an . Proof. We have 0= f\n\n\u0002r \u0003 s\n\n= a0 +\n\na1 r an r n an−1 r n−1 + n . + ··· + n−1 s s s\n\nMultiplying through by s n , we obtain a0 s n + a1r s n−1 + · · · + an−1 r n−1 s + an r n = 0. As s divides every term except an r n , it also divides an r n . Since (r, s) = 1, Corollary 2.2 tells us that s|an . Similarly, r divides every term except a0 s n , so it also divides \u0002 a0 s n . Since (r, s) = 1, we see that r |a0 . Example 11.6. Let f (x) = 3x 3 + 2x 2 − 2x − 8. In view of the Rational Roots Theorem, the only possible rational roots of f (x) are ±1, ±2, ±4, ±8, ± 13 , ± 23 , ± 43 and ± 83 . Trying them all, we see that the only rational root of f (x) is 43 . Of course, polynomials can be reducible without having roots. If we wish to restrict our attention to polynomials in Z[x], we must be sure that it makes sense to do so. At first blush, it seems conceivable that we could have a polynomial in Z[x] that factors into a product of polynomials of lower degree in Q[x], but not in Z[x]. In fact, this does not happen. Let us see why. Definition 11.2. If f (x) is a nonzero polynomial in Z[x], then the content of f (x) is the largest positive integer that divides every coefficient of f (x). We say that f (x) is primitive if its content is 1. Example 11.7. The polynomial 6x 3 − 15x 2 + 81x − 12 has content 3, whereas 5x 2 + 14x − 2 is primitive. We can now present a famous result due to Carl F. Gauss. Lemma 11.2 (Gauss’s Lemma). The product of two primitive polynomials in Z[x] is also primitive.\n\n196\n\n11 Irreducible Polynomials\n\nProof. Let f (x) = a0 + · · · + an x n and g(x) = b0 + · · · + bm x m be primitive. Suppose that f (x)g(x) is not primitive. Let p be a prime dividing the content of f (x)g(x). As p cannot divide all of the coefficients of f (x), let i be the smallest nonnegative integer such that p does not divide ai . Similarly, let j be the smallest nonnegative integer such that b j is not divisible by p. Then the coefficient of x i+ j in f (x)g(x) is a0 bi+ j + a1 bi+ j−1 + · · · + ai−1 b j+1 + ai b j + ai+1 b j−1 + · · · + ai+ j−1 b1 + ai+ j b0 , where we add terms with coefficient zero if necessary. Now, this coefficient must be divisible by p. Also, p divides ak , 0 ≤ k < i, and p divides bl , 0 ≤ l < j. Thus, every term in the sum is divisible by p except ai b j , which means that p|ai b j as well. But this contradicts Theorem 2.7. \u0002 As a consequence, we can see that if a polynomial in Z[x] is reducible in Q[x], then it is reducible in Z[x] as well. Theorem 11.4. Let f (x) be a polynomial in Z[x], and suppose that f (x) = g(x)h(x), where g(x), h(x) ∈ Q[x]. Then there is a positive rational number q such that qg(x) and q1 h(x) lie in Z[x]. Proof. Assume, first of all, that f (x) is primitive. Choose positive integers a and b such that ag(x), bh(x) ∈ Z[x]. Then ab f (x) = (ag(x))(bh(x)). Let c be the content of ag(x) and d the content of bh(x). Then ac g(x), db h(x) ∈ Z[x], and both are primitive polynomials. By Gauss’s lemma, their product, ab f (x), cd is also primitive. Thus, the content of ab f (x) is cd. But as f (x) is primitive, the content of ab f (x) is also ab. Thus, ab = cd, and hence letting q = ac , we see that b = q1 . d Suppose that f (x) is not primitive. If it is the zero polynomial, then either g(x) or h(x) must be as well. Without loss of generality, say that h(x) is the zero polynomial. Then let q be a positive integer such that qg(x) ∈ Z[x]. On the other hand, if f (x) is not the zero polynomial, then\u0004 let k be \u0005 its content. Writing f (x) = k f 1 (x), with f 1 (x) ∈ Z[x], we have f 1 (x) = k1 g(x) h(x). By the argument above, there exists a positive rational number q such that qk g(x), q1 h(x) ∈ Z[x]. But then qg(x), q1 h(x) ∈ Z[x] as well. \u0002 Example 11.8. The polynomial f (x) = 3x 3 + 2x 2 − 2x − 8 has 43 as a rational root. Thus, by Theorem 11.2, g(x) = x − 43 is a divisor of f (x) in Q[x]. Performing polynomial long division, we see that f (x) = g(x)h(x), where h(x) = 3x 2 +6x +6. Using q = 3 in the above theorem, we find that f (x) = (3x − 4)(x 2 + 2x + 2), and we have a factorization in Z[x]. Even if a polynomial has coefficients in Z, it can still be difficult to tell if it is irreducible over Q. One nice result that can be rather helpful is attributed to F. Gotthold M. Eisenstein, although a proof was first published by Theodor Schönemann.\n\n11.2 Irreducibility over the Rationals\n\n197\n\nTheorem 11.5 (Eisenstein’s Criterion). Let f (x) = a0 + a1 + · · · + an x n ∈ Z[x], with n ≥ 1 and an \u0006= 0. Suppose that there exists a prime p such that p|ai , 0 ≤ i < n, but p \u0002 an and p 2 \u0002 a0 . Then f (x) is irreducible in Q[x]. Proof. If f (x) is reducible, then by Theorem 11.4, there exist nonconstant polynomials g(x) = b0 + · · · + bl x l and h(x) = c0 + · · · + cm x m in Z[x], with bl \u0006= 0 \u0006= cm and f (x) = g(x)h(x). Now, p divides a0 = b0 c0 , but p 2 does not. Thus, p divides exactly one of {b0 , c0 }. Without loss of generality, say p|b0 . But p does not divide an = bl cm . Thus, p divides neither bl nor cm . Let i be the smallest positive integer such that p \u0002 bi . Then ai = b0 ci + b1 ci−1 + · · · + bi−1 c1 + bi c0 . Now, p|b j , 0 ≤ j < i. Furthermore, as i ≤ l < n, we know that p|ai . Thus, p|bi c0 . \u0002 But p divides neither bi nor c0 , and we have a contradiction. Example 11.9. The polynomial 13x 3 − 42x 2 + 81x − 15 is irreducible over Q, using Eisenstein’s criterion with p = 3. Example 11.10. For any positive integer n and any prime p, we observe that x n − p is irreducible over Q. Note that if F is a subfield of K , and f (x) is a reducible polynomial in F[x], then it is also necessarily reducible in K [x] (just using the same factorization). Of course, the fact that it is reducible in K [x] does not imply that it is reducible in F[x], as we illustrated in Example 11.3. But the relationship between Z[x] and Q[x] is backwards. Indeed, we have seen that if a polynomial in Z[x] is reducible in Q[x], then it is also reducible in Z[x]. The other direction does not work! Example 11.11. Let f (x) = 2x − 6. Then by Corollary 11.2, f (x) is irreducible in Q[x]. But f (x) is reducible in Z[x]; indeed, f (x) = 2(x − 3), and neither 2 nor x − 3 is a unit in Z[x]. The problem, then, is that the nonzero constants are not necessarily units in Z[x], and this affects irreducibility. Lemma 11.3. Let f (x) ∈ Z[x]. Then f (x) is irreducible in Z[x] if and only if either 1. f (x) is a (positive or negative) prime in Z; or 2. f (x) is a primitive polynomial that is irreducible in Q[x]. Proof. Note that a unit in Z[x] is also a unit in Q[x], and hence a constant. But the only constants having inverses in Z[x] are ±1, so those are the only units. Suppose that f (x) is a constant c ∈ Z. If c is prime, then its only factorizations are 1 · c and (−1)(−c), so f (x) is irreducible. Otherwise, c has some other factorization, and f (x) is not irreducible.\n\n198\n\n11 Irreducible Polynomials\n\nSo, let deg( f (x)) ≥ 1. Suppose that f (x) is irreducible \u0005 in Z[x]. If f (x) has \u0004 content d > 1, then f (x) has a factorization d d1 f (x) , and therefore f (x) is reducible. So, we may assume that f (x) is primitive. If it is reducible in Q[x], then by Theorem 11.4, it is reducible in Z[x] as well. Conversely, assume that f (x) is irreducible in Q[x] and primitive. If f (x) = g(x)h(x), with g(x), h(x) ∈ Z[x], then we have a factorization in Q[x] as well, which would make f (x) reducible over Q, unless either g(x) or h(x) is a constant. Without loss of generality, let g(x) = e \u0006= 0. If e = ±1, then g(x) is a unit in Z[x]. If not, then f (x) has content |e| times the content of h(x), contradicting the assumption that f (x) is primitive. Thus, f (x) is irreducible in Z[x] in this case. \u0002 Let us now present the counterexample promised in Section 10.4. We already know that Z[x] is not a PID. But we have the following. Theorem 11.6. The ring Z[x] is a UFD. Proof. Let f (x) ∈ Z[x] be a nonzero nonunit. We will show that f (x) is a product of irreducibles. First, suppose that deg( f (x)) = n ≥ 1. We claim that f (x) is a product of polynomials in Z[x] that are irreducible in Q[x]. Our proof is by strong induction on n. If n = 1, then f (x) is irreducible in Q[x] and there is nothing to do. Let n ≥ 2, and suppose that our claim holds for polynomials of smaller degree. If f (x) is irreducible in Q[x], then again, there is nothing to do. Otherwise, we know that f (x) = g(x)h(x), where g(x) and h(x) are polynomials of degree less than n in Q[x]. By Theorem 11.4, we may choose g(x) and h(x) to be in Z[x]. Then our inductive hypothesis tells us that g(x) and h(x) are products of polynomials in Z[x] that are irreducible in Q[x], and hence, so is f (x), proving the claim. If f (x) = f 1 (x) · · · f k (x), where each f i (x) is irreducible in Q[x], then let ci be the content of f i (x). We now have \u0006 f (x) = (c1 · · · ck )\n\n\u0007 \u0006 \u0007 1 1 f 1 (x) · · · f k (x) , c1 ck\n\nwhere each c1i f i (x) is irreducible in Z[x], by the preceding lemma. Thus, bringing the deg( f (x)) = 0 possibility back into consideration, we see that f (x) is either an integer not in {0, ±1}, or a nonzero integer multiplied by a product of irreducibles in Z[x]. It remains only to consider the case of an integer. But the Fundamental Theorem of Arithmetic tells us that any integer not in {0, ±1} is a product of (positive or negative) primes, which are certainly irreducible in Z[x]. We do still have to deal with f (x) = (−1)g1 (x) · · · gk (x), where each gi (x) is irreducible, but then this is (−g1 (x))g2 (x) · · · gk (x), and −g1 (x) is irreducible as well. Let us verify the uniqueness. Suppose that f (x) = p1 · · · pk g1 (x) · · · gl (x) = q1 · · · qm h 1 (x) · · · h n (x),\n\n11.2 Irreducibility over the Rationals\n\n199\n\nwhere each pi and qi is a (positive or negative) prime in Z, and each gi (x) and h i (x) is a primitive polynomial which is irreducible in Q[x]. (We allow the possibility that k, l, m or n may be zero.) By Gauss’s lemma, the product of primitive polynomials is primitive. Thus, the content of f (x) is | p1 · · · pk | = |q1 · · · ql |. By the Fundamental Theorem of Arithmetic, k = l and after rearranging, each pi = ±qi . Cancelling, we have g1 (x) · · · gm (x) = ±h 1 (x) · · · h n (x). But these are products of irreducible polynomials in Q[x]. As Q[x] is a UFD, m = n and, after rearranging, each gi (x) = qi h i (x), for some qi ∈ Q. Write qi = rsii , with ri , si ∈ Z and si \u0006= 0. Then si gi (x) = ri h i (x). As gi (x) and h i (x) are primitive, looking at the content of each side of the \u0002 equation, we have |si | = |ri |, and hence qi ∈ {1, −1}. We are done. Exercises 11.11. Find all rational roots of each of the following polynomials. 1. x 3 − 7x 2 + 5x + 2 2. 6x 4 − x 3 + 4x 2 − x − 2 11.12. Are the following polynomials irreducible over Q? 1. 3x 4 + 15x 3 − 25x 2 + 45x + 10 2. 2x 3 + 5x 2 + x + 7 3. x 14 − 75 11.13. Write each of the following polynomials as a product of irreducibles in Q[x]. 1. x 4 − 10x 3 + 35x 2 − 48x + 18 2. x 4 + 2x 3 + x 2 + 3x + 2 11.14. Write each of the following polynomials as a product of irreducibles in Z[x]. 1. 6x 4 + 84x 3 − 126x 2. 6x 4 − 3x 3 + 18x 2 − 3x − 3 11.15. Let F be a field, a ∈ F and f (x) ∈ F[x]. Show that f (x) is irreducible if and only if f (x + a) is irreducible. 11.16. Modify Eisenstein’s criterion as follows, namely, insist that p|ai , 1 ≤ i ≤ n, but p \u0002 a0 and p 2 \u0002 an . Show that the result still holds. 11.17. Is 7x 6 + 21x 5 − 49x 3 + 14x 2 + 7x + 2 reducible or irreducible over Q? 11.18. Let R be a Euclidean domain. If f (x) ∈ R[x] is a nonzero polynomial, let us say that it is primitive if the only common divisors of its coefficients are the units of R. Show that Gauss’s lemma holds in R[x].\n\n200\n\n11 Irreducible Polynomials\n\n11.3 Irreducibility over the Real and Complex Numbers While the real numbers may seem like a more natural field with which to work, the complex numbers have a more attractive algebraic structure. Indeed, we wish to consider the complex numbers, because there are nonconstant real polynomials, such as x 2 + 1, having no real roots. The complex numbers do not have this problem. Indeed, this is a famous result known as the Fundamental Theorem of Algebra. There are many different proofs of this theorem. Curiously, to the best of the author’s knowledge, all of these proofs require results from outside of algebra. A proof that is mostly algebraic can be found in the advanced textbook of Dummit and Foote . (Sadly, the algebra involved is somewhat beyond the scope of this course.) Theorem 11.7 (Fundamental Theorem of Algebra). Let f (x) be a nonconstant polynomial in C[x]. Then f (x) has a root in C. We say that the field of complex numbers is algebraically closed. Corollary 11.4. If f (x) ∈ C[x], then f (x) is irreducible if and only if deg( f (x)) = 1. Proof. Combine Theorem 11.7 with Corollaries 11.1 and 11.2.\n\n\u0002\n\nCorollary 11.5. Let f (x) ∈ C[x] be a nonconstant polynomial. Then there exist a, c1 , c2 , . . . , cn ∈ C such that f (x) = a(x − c1 )(x − c2 ) · · · (x − cn ). Proof. We proceed by induction on deg( f (x)) = n. If n = 1, then f (x) = ax + b = a(x − (−a −1 b)), for some a, b ∈ C, with a \u0006= 0. Suppose that the result is true for n, and let deg( f (x)) = n + 1. By Theorem 11.7, f (x) has a root, cn+1 ∈ C. But then Theorem 11.2 tells us that f (x) = g(x)(x − cn+1 ), where deg(g(x)) = n, by Theorem 10.2. Now apply our inductive hypothesis to g(x). \u0002 Thus, complex polynomials behave as nicely as we could possibly wish. What about real polynomials? The situation there is slightly more complicated. Lemma 11.4. Let f (x) ∈ R[x]. If c, d ∈ R, and c + di is a complex root of f (x), then so is c − di. Proof. Write c + di = c − di. Let f (x) = a0 + · · · + an x n , ai ∈ R. Then if z = c + di, we have f (¯z ) = a0 + a1 z¯ + a2 (¯z )2 + · · · + an (¯z )n . But by Example 9.13, the function mapping z to z¯ is a homomorphism. Thus,\n\n11.3 Irreducibility over the Real and Complex Numbers\n\n201\n\nf (¯z ) = a0 + a1 z¯ + a2 z 2 + · · · + an z n = a0 + a1 z¯ + a2 z 2 + · · · + an z n = a0 + a1 z + a2 z 2 + · · · + an z n = f (z) = 0¯ = 0, making using of the fact that each ai = ai , since ai ∈ R.\n\n\u0002\n\nWe can use this to classify the irreducible real polynomials. Theorem 11.8. Let f (x) ∈ R[x]. Then f (x) is irreducible over R if and only if either 1. deg( f (x)) = 1; or 2. f (x) = ax 2 + bx + c, where a \u0006= 0 and b2 < 4ac. Proof. Since R is a field, constant polynomials are either 0 or a unit, and therefore need not be considered. If deg( f (x)) = 1, then Corollary 11.2 tells us that f (x) is indeed irreducible. Therefore, let f (x) have degree at least 2. Suppose that f (x) is irreducible. By Theorem 11.7, f (x) has a root z = a+bi ∈ C. If z ∈ R, then by Corollary 11.1, we have a contradiction. Assume otherwise. By Lemma 11.4, a − bi is also a root. Expressing f (x) as in Corollary 11.5, we see that (x − (a + bi))(x − (a − bi))| f (x) in C[x]. But (x − (a + bi))(x − (a − bi)) = x 2 − 2ax + (a 2 + b2 ) ∈ R[x]. Thus, applying the division algorithm, we see that there exist q(x), r (x) ∈ R[x] such that f (x) = (x 2 − 2ax + (a 2 + b2 ))q(x) + r (x), and r (x) = 0 or deg(r (x)) < 2. By the uniqueness of the division algorithm in C[x], we must have r (x) = 0 and x 2 − 2ax + (a 2 + b2 ) divides f (x) in R[x]. In particular, if deg( f (x)) > 2, then f (x) must be reducible. Thus, we may assume that f (x) = ax 2 + bx + c, with a \u0006= 0. By Corollary 11.2, such a polynomial is irreducible over R if and only if it has no roots in R. But the quadratic formula tells us that this happens if and only if b2 − 4ac < 0. We are done. \u0002 We can use this to recover a well-known fact from calculus. Corollary 11.6. Let f (x) ∈ R[x] be a polynomial of odd degree. Then f (x) has a real root. Proof. We know that R[x] is a UFD. Thus, write f (x) as a product of irreducible polynomials. By the preceding theorem, each such irreducible has degree 1 or 2. Since f (x) has odd degree, at least one of these irreducible polynomials has degree 1. Therefore, there exist a, b ∈ R, with a \u0006= 0, such that ax + b divides f (x). But then −a −1 b is a root of f (x). \u0002\n\n202\n\n11 Irreducible Polynomials\n\nExercises 11.19. Given that a is a root of f (x), find all complex roots of f (x). 1. f (x) = x 3 − 11x 2 + 41x − 91, a = 2 + 3i 2. f (x) = x 4 + x 2 − 2x + 6, a = 1 − i 11.20. Given that a is a root of f (x), find all complex roots of f (x). 1. f (x) = x 3 + x 2 − 5x − 21, a = 3 2. f (x) = x 4 − 6x 3 + 33x 2 − 84x + 136, a = 1 + 4i 11.21. Write each of the following polynomials as a product of irreducibles in Q[x], R[x] and C[x]. 1. x 4 − 10 2. x 3 + x 2 + 5x − 22 11.22. Write each of the following polynomials as a product of irreducibles in Q[x], R[x] and C[x]. 1. x 3 + 12 2. x 4 + 4x 2 + 4 11.23. Find a nonzero polynomial in R[x] having 2 − 5i, 4 + i and 6 as roots. 11.24. Let f (x) and g(x) be nonzero polynomials in Q[x]. Consider the gcds of f (x) and g(x) in Q[x], R[x] and C[x]. Must these gcds be the same, or can they be different?\n\n11.4 Irreducibility over Finite Fields When our field is finite, we have the luxury of taking a brute force approach to factoring polynomials. That is, we can simply list all of the polynomials of suitable degrees, and see if the products work. Of course, we can save ourselves some effort by narrowing the possibilities first. Example 11.12. Let f (x) = x 4 + x 3 + x 2 + x + 1 ∈ Z2 [x]. We claim that f (x) is irreducible over Z2 . If not, there are two possibilities. First, f (x) could be a product of a degree 1 polynomial and a degree 3 polynomial. But if it has a degree 1 polynomial as a factor, then it has a root. There are only two possible roots in Z2 , namely, 0 and 1, and neither works. Second, f (x) could be a product of two polynomials of degree 2. Now, the only possible coefficients are 0 and 1. Furthermore, the leading coefficients and the constant terms must multiply to give 1. Thus, the only possible factors are x 2 + 1 and x 2 + x + 1. But x 2 + 1 has 1 as a root, and f (x) does not, so\n\n11.4 Irreducibility over Finite Fields\n\n203\n\nonly x 2 + x + 1 remains. However, (x 2 + x + 1)2 = x 4 + x 2 + 1 \u0006= f (x). Thus, f (x) is indeed irreducible. Example 11.13. Let f (x) = 3x 5 + x 4 + 4x 3 + 4x 2 + 3x + 2 ∈ Z5 [x]. We would like to write f (x) as a product of irreducibles. The first thing we should do is check for roots. We run through the five elements of Z5 and find that 3 is a root. Thus, x − 3 (or, equivalently, x + 2) divides f (x). Performing polynomial long division, we find that f (x) = (x + 2)(3x 4 + 4x 2 + x + 1). Let g(x) = 3x 4 + 4x 2 + x + 1. Evaluating g(x) at each element of Z5 , we see that g(x) has no roots. Thus, if it is to be factored, it must be as a product of two polynomials of degree 2. Up to a unit in Z5 , these factors would have to be x 2 + ax + b and 3x 2 + cx + d, for some a, b, c, d ∈ Z5 . Furthermore, looking at the constant terms, we have bd = 1. Thus, once b is decided, d = b−1 . Looking at the coefficients of x 3 , we have 3a + c = 0. Thus, once a is decided, we have c = 2a. Trying the various possibilities for a and b, we have g(x) = (x 2 + 2x + 3)(3x 2 + 4x + 2). Since g(x) has no roots, this cannot be factored any further. Thus, f (x) = (x + 2)(x 2 + 2x + 3)(3x 2 + 4x + 2) is a product of irreducibles in Z5 [x]. Our ability to handle polynomials over finite fields can be helpful when we consider polynomials in Q[x]. Theorem 11.9. Let f (x) = a0 + a1 x + · · · + an x n ∈ Z[x]. Let p be a prime such that p \u0002 an . Reducing all of the coefficients modulo p, if [a0 ] + [a1 ]x + · · · + [an ]x n is irreducible in Z p [x], then f (x) is irreducible in Q[x]. Proof. Suppose f (x) is reducible in Q[x]. Then by Theorem 11.4, we must have f (x) = g(x)h(x), where g(x) = b0 + · · · + bk x k and h(x) = c0 + · · · + cm x m are polynomials in Z[x], with k, m > 0 and bk \u0006= 0 \u0006= cm . Now, we have ai = b0 ci + b1 ci−1 + · · · + bi c0 , for each i. By Example 9.12, the function from Z to Z p sending d to [d] is a ring homomorphism. Thus, [ai ] = [b0 ][ci ] + [b1 ][ci−1 ] + · · · + [bi ][c0 ]. It now follows that [a0 ] + · · · + [an ]x n = ([b0 ] + · · · + [bk ]x k )([c0 ] + · · · + [cm ]x m ). That is, [a0 ] + · · · + [an ]x n is reducible, unless one of the factors is a constant polynomial. But as p \u0002 an , we see that the degree of [a0 ] + · · · + [an ]x n is n = k + m. The only way the product will have the correct degree is if [bk ] \u0006= \u0006= [cm ]. This contradiction completes the proof. \u0002\n\n204\n\n11 Irreducible Polynomials\n\nNote that the condition that p does not divide the leading coefficient is important. Indeed, 3x 2 + x − 4 is reducible in Q[x], as it is (x − 1)(3x + 4). But if we tried to use p = 3, we would obtain x + 2 ∈ Z3 [x], which is certainly irreducible. Also, the converse of the theorem is not true. For instance, x 2 + 1 is irreducible over Q (or, for that matter, R), but in Z5 [x], we have x 2 + 1 = (x + 2)(x + 3). Example 11.14. We claim that 15x 4 − 29x 3 + 13x 2 + 33x − 201 is irreducible over Q[x]. Use Theorem 11.9 with p = 2. In Z2 [x], we obtain x 4 + x 3 + x 2 + x + 1 which, by Example 11.12, is irreducible. Sometimes, we might have to try more than one prime. Example 11.15. Let f (x) = 5x 3 + 3x 2 + x + 1. If we use p = 2, we obtain x 3 + x 2 + x + 1 ∈ Z2 [x]. But this polynomial has 1 as a root, so it is reducible. No help here! Let us try p = 3. Then we get 2x 3 + x + 1 ∈ Z3 [x]. By Corollary 11.2, it is irreducible if it has no roots. But trying 0, 1 and 2, we see that it has no roots in Z3 . Thus, f (x) is irreducible in Q[x]. Exercises 11.25. Are the following polynomials reducible or irreducible over the rationals? 1. f (x) = x 3 + 5x 2 + 2x + 16 2. f (x) = 22x 4 − 9x 3 + 16x 2 + 18x + 20 11.26. Are the following polynomials reducible or irreducible over the rationals? 1. f (x) = 9x 4 − 15x 3 + 8x 2 − 6x + 25 2. f (x) = 2x 4 + 11x 3 + 16x 2 + 5x + 6 11.27. Let F be a finite field with n elements. How many monic irreducible polynomials of degree 2 are there in F[x]? 11.28. Write each of the following polynomials as a product of irreducibles in Z11 [x]. 1. 2x 3 + 3x 2 + 9x + 10 2. x 4 + 4x 3 + 5x 2 + x + 7 11.29. Let p be an odd prime. Show that x 4 + 1 is reducible over Z p in each of the following cases: 1. there exists an a ∈ Z p such that a 2 = p − 1; 2. there exists an a ∈ Z p such that a 2 = p − 2; or 3. there exists an a ∈ Z p such that a 2 = 2.\n\n11.4 Irreducibility over Finite Fields\n\n205\n\n11.30. Show that x 4 + 1 is irreducible in Q[x] but reducible in Z p [x] for every prime p. (Thus, the converse of Theorem 11.9 is wildly false!)\n\nReference 1. Dummit, D.S., Foote, R.M.: Abstract Algebra, 3rd edn. Wiley, Hoboken (2004)\n\nChapter 12\n\nVector Spaces and Field Extensions\n\nWe begin this chapter with some basic facts about vector spaces. These will be familiar (at least in the case of real vector spaces) to those readers who have studied linear algebra. We then focus our attention on the particular case of a field extension. A number of properties of field extensions are discussed. Let F be a field and f (x) ∈ F[x] a nonconstant polynomial. We demonstrate how to create a field extension in which f (x) splits into a product of polynomials of degree 1. This leads to a classification of all finite fields.\n\n12.1 Vector Spaces We begin with the definition of a vector space. In most linear algebra courses, vector spaces are defined over R or, occasionally, C. But we can do the same thing over any field. If F is a field and V is a set, then a scalar multiplication on V is a function from F × V to V . If a ∈ F, v ∈ V , then we write av for the image of (a, v) under such a function. Definition 12.1. Let F be a field. Then a vector space over F is a set V having an addition operation and a scalar multiplication such that 1. 2. 3. 4. 5. 6.\n\nV is an abelian group under addition; av ∈ V for all a ∈ F and all v ∈ V ; (a + b)v = av + bv for all a, b ∈ F and all v ∈ V ; a(u + v) = au + av for all a ∈ F and all u, v ∈ V ; a(bv) = (ab)v for all a, b ∈ F and all v ∈ V ; and 1v = v for all v ∈ V .\n\n© Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_12\n\n207\n\n208\n\n12 Vector Spaces and Field Extensions\n\nOf course, condition (2) is redundant, given the definition of a scalar multiplication, but we include it, because it must be checked. Certainly the most familiar vector space over R is Rn . We can generalize this. Example 12.1. Let F be a field. For any positive integer n, let F n = \u0002F⊕F⊕\u0003\u0004· · · ⊕F\u0005. n times\n\nThen F n is a vector over F with the usual addition operation and scalar multiplication a(b1 , . . . , bn ) = (ab1 , . . . , abn ), for any a, b1 , . . . , bn ∈ F. Example 12.2. Let F be any field. Then F[x] is a vector space over F with the usual polynomial addition and a(b0 + b1 x + · · · + bn x n ) = ab0 + ab1 x + · · · + abn x n , for any a, b0 , . . . , bn ∈ F. Example 12.3. Let m and n be any positive integers, and let V be the set of m × n matrices with entries in a field F. Then V is a vector space over F using matrix addition and scalar multiplication. The least exciting example of a vector space is the following. Example 12.4. Let F be any field and V the trivial additive group, {0}. Then V is a vector space using the only available addition and scalar multiplication options, 0 + 0 = 0 and a0 = 0, for all a ∈ F. The most important example for our purposes is the following. Definition 12.2. If F and K are fields, with F a subfield of K , then we say that K is an extension field of F. Example 12.5. Any extension field K of F is a vector space over F, using the addition operation in K and multiplication in K as the scalar multiplication. (All the properties are immediate, except that 1v = v for all v ∈ K . To be sure of that, we must know that the identity of F is the identity of K . But this follows from Theorem 8.12.) For example, R and C are vector spaces over Q. Let us mention a few basic properties of vector spaces. Theorem 12.1. Let V be a vector space over F. Then 1. a0 = 0 for all a ∈ F; 2. 0v = 0 for all v ∈ V ; and 3. (−1)v = −v for all v ∈ V . Proof. (1) Note that a0 = a(0 + 0) = a0 + a0. Adding −(a0) to both sides, we see that a0 = 0. (2) We have 0v = (0 + 0)v = 0v + 0v. Adding −0v to both sides, we obtain the desired conclusion. (3) Observe that v + (−1)v = 1v + (−1)v = (1 − 1)v = 0v = 0. Thus, (−1)v = −v. \u0002\n\n12.1 Vector Spaces\n\n209\n\nWe do have to be a bit careful about which 0 we are using. For example, when we write 0v = 0 in the theorem above, the first 0 is in F and the second is in V . Definition 12.3. Let V be a vector space over a field F. Then a subset W of V is said to be a subspace of V if it is a vector space over F using the same addition and scalar multiplication. Example 12.6. If F is a subfield of K , and K is a subfield of L, then L is a vector space over F having K and F as subspaces. Example 12.7. Regarding R[x] as a vector space over Q, we note that Q[x] is a subspace. There is a simple test for a subspace. Theorem 12.2. Let F be a field and V a vector space over F. Then a subset W of V is a subspace if and only if 1. 0 ∈ W ; 2. w1 + w2 ∈ W for all w1 , w2 ∈ W (closure under addition); and 3. aw ∈ W for all a ∈ F and w ∈ W (closure under scalar multiplication). Proof. If W is a subspace then, in particular, it is an additive subgroup, so (1) and (2) hold. Part (3) is one of the conditions for a vector space. Conversely, suppose that (1), (2) and (3) hold. Noting that (3) tells us that −w = (−1)w ∈ W , for all w ∈ W , we see from Theorem 3.10 that W is an additive subgroup of V . We are given closure under scalar multiplication. The remaining vector space properties hold in V , and therefore in any subset of V . \u0002 Note that in the preceding theorem, condition (1) could be replaced with the condition that W is not the empty set, for if w ∈ W , then −w ∈ W , and therefore 0 = w + (−w) ∈ W . Example 12.8. Let V = R4 , which is a vector space over R. We claim that W = {(a, b, 2a − b + 3c, c) : a, b, c ∈ R} is a subspace of V . Letting a = b = c = 0, we see that (0, 0, 0, 0) ∈ W . To check closure under addition, take ai , bi , ci ∈ R. Then (a1 , b1 , 2a1 − b1 + 3c1 , c1 ) + (a2 , b2 , 2a2 − b2 + 3c2 , c2 ) = (a1 + a2 , b1 + b2 , 2(a1 + a2 ) − (b1 + b2 ) + 3(c1 + c2 ), c1 + c2 ) ∈ W. Similarly, if a ∈ R, then a(a1 , b1 , 2a1 − b1 + 3c1 , c1 ) = (aa1 , ab1 , 2aa1 − ab1 + 3ac1 , ac1 ) ∈ W. Thus, we have closure under scalar multiplication, and the claim is proved.\n\n210\n\n12 Vector Spaces and Field Extensions\n\nExercises 12.1. Let F be a field and n a positive integer. If V is the set of all polynomials of degree n in F[x], together with the zero polynomial, is V a subspace of F[x]? 12.2. Let F be a field and n a positive integer. If V is the set of all polynomials of degree at most n in F[x], together with the zero polynomial, show that V is a subspace of F[x]. 12.3. Let V be a vector space having subspaces U and W . Show that U ∩ W is a subspace of V . Extend this to the intersection of an arbitrary collection of subspaces. 12.4. Let V be a vector space having subspaces U and W . Show that U + W (regarding U and W as additive subgroups of V ) is a subspace of V . 12.5. Let V and W be vector spaces over a field F. A function α : V → W is said to be a linear transformation if α(v1 + v2 ) = α(v1 ) + α(v2 ) and α(av1 ) = aα(v1 ) for all a ∈ F, v1 , v2 ∈ V . If U is a subspace of V , show that α(U ) is a subspace of W . 12.6. Let F, V , W and α be as in the preceding exercise. Show that the kernel of α (regarding α as a homomorphism of additive groups) is a subspace of V . Further show that α is one-to-one if and only if the kernel is {0}. 12.7. Let F = Z11 and V = F 3 . If W = {(a, b, c) ∈ V : 2a + 3b + 7c = 0}, is W a subspace of V ? 12.8. Let F be a field with vector spaces V and W . Let U = V × W be the direct product of the additive groups V and W . Define a scalar multiplication on U via a(v, w) = (av, aw) for all a ∈ F, v ∈ V and w ∈ W . Is U a vector space over F? 12.9. Let F be a field of characteristic 3 and V a vector space over F. Show that v + v + v = 0 for all v ∈ F. 12.10. Suppose that V is a vector space over an infinite field F. Show that V is not the union of a finite number of proper subspaces.\n\n12.2 Basis and Dimension In order to define a basis for a vector space, we must first discuss linear combinations of vectors. Definition 12.4. Let V be a vector space over a field F. If v1 , v2 , . . . , vk ∈ V , then a vector v ∈ V is said to be a linear combination of the vi if v = a1 v1 + · · · + ak vk , for some ai ∈ F. Example 12.9. Let F = Q and V = F 3 . If v1 = (2, −3, 7) and v2 = (4, 0, 1), then (24, −6, 19) is a linear combination of v1 and v2 , since (24, −6, 19) = 2v1 + 5v2 .\n\n12.2 Basis and Dimension\n\n211\n\nDefinition 12.5. Let F be a field and V a vector space over F. Let v1 , v2 , . . . , vk ∈ V . We say that the vi are linearly dependent if there exist a1 , . . . , ak ∈ F, not all zero, such that a1 v1 + · · · + ak vk = 0. Otherwise, the vi are linearly independent. Example 12.10. Let F = Z5 and V = F 3 . The vectors (2, 1, 3), (1, 3, 0) and (2, 1, 4) are linearly dependent, since 3(2, 1, 3) + (1, 3, 0) + 4(2, 1, 4) = (0, 0, 0). On the other hand, (1, 0, 4), (3, 2, 1) and (2, 0, 2) are linearly independent. Indeed, if a1 (1, 0, 4) + a2 (3, 2, 1) + a3 (2, 0, 2) = (0, 0, 0), then looking at the middle entry, we see immediately that a2 = 0. Then a1 + 2a3 = 4a1 + 2a3 = 0. This yields 3a1 = 0, and hence a1 = 0 and, finally, a3 = 0. Here is a handy test for linear dependence. Theorem 12.3. Let V be a vector space over a field F and v1 , . . . , vk ∈ V . Then the vi are linearly dependent if and only if either 1. v1 = 0; or 2. there exists an m ≥ 2 such that vm is a linear combination of v1 , . . . , vm−1 . Proof. Suppose that the vi are linearly dependent. Choose ai ∈ F, not all zero, such that a1 v1 + · · · + ak vk = 0. Let m be the largest positive integer such that am \u0007= 0. Then a1 v1 + · · · + am vm = 0. If m = 1, then a1 v1 = 0, with a1 \u0007= 0. Thus, v1 = a1−1 0 = 0, giving case (1). If m > 1, then vm = −am−1 a1 v1 −· · ·−am−1 am−1 vm−1 , and so vm is a linear combination of v1 , . . . , vm−1 , which proves case (2). Conversely, suppose that (1) or (2) is satisfied. If v1 = 0, then 1v1 + 0v2 + · · · + 0vk = 0, meaning that the vi are linearly dependent. If vm = b1 v1 + · · · + bm−1 vm−1 , for some bi ∈ F, then b1 v1 + · · · + bm−1 vm−1 − 1vm + 0vm+1 + · · · + 0vk = 0. Again, the vi are linearly dependent.\n\n\u0002\n\nLinear independence is most useful when combined with another property. Definition 12.6. Let V be a vector space over a field F, and let v1 , . . . , vk ∈ V . Then we say that the vi span V if every v ∈ V is a linear combination of the vi . Example 12.11. Regarding C as a vector space over R, we note that 1 and i span C, as a + bi = a1 + bi. Example 12.12. Let F = R and V = R3 . Then the vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) span V , since (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1). The following lemma describes a very nice relationship between linear independence and spanning. Lemma 12.1. Let V be a vector space over a field F. Suppose that v1 , . . . , vk span V . If w1 , . . . , wl ∈ V , and l > k, then the wi are linearly dependent.\n\n212\n\n12 Vector Spaces and Field Extensions\n\nProof. Since the vi span V , we know that w1 is a linear combination of the vi . Let us say that w1 = a1 v1 + · · · + ak vk , with ai ∈ F. If all of the ai are zero, then w1 is the zero vector. Thus, by Theorem 12.3, we are done. Therefore, we may assume that some ai is nonzero. Without loss of generality, say a1 \u0007= 0. We now observe that w1 , v2 , v3 , . . . , vk span V . Indeed, if v ∈ V , then v = b1 v1 + · · · + bk vk , for some bi ∈ F. But v1 = a1−1 w1 − a1−1 a2 v2 − · · · − a1−1 ak vk . Thus,\n\nv = b1 a1−1 w1 + (b2 − b1 a1−1 a2 )v2 + · · · + (bk − b1 a1−1 ak )vk ,\n\nproving the claim. Now consider w2 . It is a linear combination of w1 , v2 , v3 , . . . , vk . Let us say that w2 = c1 w1 + c2 v2 + c3 v3 + · · · + ck vk , with ci ∈ F. If ci = 0 for all i ≥ 2, then w2 is a linear combination of w1 , proving that the wi are linearly dependent. Thus, we may assume that there exists an i ≥ 2 with ci \u0007= 0. Without loss of generality, say c2 \u0007= 0. But then v2 is a linear combination of w1 , w2 , v3 , v4 , . . . , vk . And just as before, we now deduce that w1 , w2 , v3 , v4 , . . . , vk span V . Repeat this argument. We will conclude either that the wi are linearly dependent or, eventually, that w1 , . . . , wk span V . But then wk+1 is a linear combination of \u0002 w1 , . . . , wk . By Theorem 12.3, the wi are linearly dependent. What we really need is a basis for a vector space. Definition 12.7. Let V be a vector space over a field F. We say that v1 , . . . , vk ∈ V form a basis for V if they are linearly independent and span V . Example 12.13. Regarding C as a vector space over R, we can see that 1 and i form a basis for C. Example 12.14. For any field F and any positive integer n, the vectors (1, 0, 0, . . . , 0), (0, 1, 0, 0, . . . , 0), . . . , (0, 0, . . . , 0, 1) form a basis for F n . Example 12.15. Let F be any field and V = F[x]. Then V has no finite basis. Indeed, if v1 , . . . , vk ∈ V , then any linear combination of these vectors must have degree no larger than the maximum of the degrees of the vi . On the other hand, for any positive integer n, let W be the set of all polynomials having degree at most n (including the zero polynomial). By Exercise 12.2, W is a subspace of V , and the polynomials 1, x, x 2 , . . . , x n form a basis. Theorem 12.4. Let V be a vector space over a field F. If v1 , . . . , vk form a basis for V , then every element of V can be written uniquely in the form a1 v1 + · · · + ak vk , with ai ∈ F.\n\n12.2 Basis and Dimension\n\n213\n\nProof. Since a basis spans the space, only the uniqueness needs to be proved. Suppose that a1 v1 + · · · + ak vk = b1 v1 + · · · + bk vk , with ai , bi ∈ F. Then (a1 − b1 )v1 + · · · + (ak − bk )vk = 0. By linear independence, ai = bi for all i.\n\n\u0002\n\nBases are not unique. For instance, (1, 0) and (0, 1) form a basis for R2 over R, but so do (1, 3) and (5, 2). However, any two bases for a vector space must have the same number of vectors. Theorem 12.5. Let V be a vector space over a field F. If v1 , . . . , vk and w1 , . . . , wl are bases for V , then k = l. Proof. Suppose the theorem is false. Without loss of generality, say k < l. Then v1 , . . . , vk span V . Since k < l, Lemma 12.1 tells us that w1 , . . . , wl are linearly dependent. We have a contradiction. \u0002 Definition 12.8. Let V be a vector space over a field F. If v1 , . . . , vk is a basis for V , then we say that V has dimension k, and write dim V = k (or dim F V = k, if the field is unclear from the context). We also stipulate that dim{0} = 0. In either of these cases, V is finite-dimensional. If V has no finite basis, then V is infinitedimensional. Example 12.16. For any field F and positive integer n, dim F n = n. See Example 12.14. Example 12.17. The dimension of C over R is 2. See Example 12.13. Example 12.18. If F is any field, then F[x] is infinite-dimensional. The vector space consisting of the polynomials of degree at most n over F, including the zero polynomial, has dimension n + 1. See Example 12.15. In a finite-dimensional space, we can discard vectors from a spanning set to obtain a basis, or add vectors to a linearly independent set to obtain a basis. Theorem 12.6. Let V be any vector space over a field F, with V \u0007= {0}. Take v1 , . . . , vk ∈ V . Then 1. if v1 , . . . , vk span V , then some subset of {v1 , . . . , vk } is a basis for V ; and 2. if v1 , . . . , vk are linearly independent, and dim V = n < ∞, then there exist vk+1 , . . . , vn ∈ V such that v1 , . . . , vn form a basis for V .\n\n214\n\n12 Vector Spaces and Field Extensions\n\nProof. (1) We proceed by induction on k. If k = 1, then since v1 spans V , and V \u0007= {0}, we see that v1 \u0007= 0, and hence v1 is linearly independent. (If av1 = 0, and 0 \u0007= a ∈ F, then 0 = a −1 av1 = v1 .) Thus, v1 is a basis. Suppose the result is true for k, and let v1 , . . . , vk+1 span V . If they are linearly independent, there is nothing to do. Otherwise, refer to Theorem 12.3. If v1 = 0, then v2 , . . . , vk+1 span V as well. By our inductive hypothesis we are done. Otherwise, some vl is a linear combination of v1 , . . . , vl−1 . Without loss of generality, say l = k +1. Write vk+1 = a1 v1 +· · ·+ak vk . If v ∈ V , we know that v = b1 v1 + · · · + bk+1 vk+1 , for some bi ∈ F. But then v = (b1 + a1 bk+1 )v1 + · · · + (bk + ak bk+1 )vk . Thus, v1 , . . . , vk span V . Our inductive hypothesis completes the proof. (2) If v1 , . . . , vk span V , there is nothing to do. Otherwise, find vk+1 ∈ V which is not a linear combination of v1 , . . . , vk . Suppose that v1 , . . . , vk+1 are linearly dependent. Then a1 v1 + · · · + ak+1 vk+1 = 0, for some ai ∈ F. If ak+1 = 0, then v1 , . . . , vk are linearly dependent, which is not the case. Otherwise, −1 −1 a1 v1 − · · · − ak+1 ak vk ; vk+1 = −ak+1\n\nthat is, vk+1 is a linear combination of v1 , . . . , vk , giving us a contradiction. Therefore, v1 , . . . , vk+1 is a linearly independent set. Now repeat. This process must stop, because Lemma 12.1 tells us that V cannot have a linearly independent set with more than n vectors. \u0002 Example 12.19. Let F = Q and V = Q3 . The vectors (3, −7, 0) and (1, 2, 0) are easily seen to be linearly independent. Furthermore, (2, 5, 8) is not a linearly combination of these two vectors. Thus, since dim V = 3, we see that the vectors (3, −7, 0), (1, 2, 0), (2, 5, 8) form a basis for V . Exercises 12.11. Let F = R and V = R3 . Are the following sets of vectors in V linearly dependent or independent over F? 1. (1, 3, 5), (2, 1, 4), (7, 11, 23) 2. (1, 3, 4), (2, 2, 1), (3, 6, 3) 12.12. Let F = Z7 and V = M2 (F). Are the following sets of vectors in V linearly dependent or independent over F? \u0006 \u0007 \u0006 \u0007 \u0006 \u0007 23 60 40 1. , , 45 03 32 \u0006 \u0007 \u0006 \u0007 \u0006 \u0007 12 23 34 2. , , 34 45 56\n\n12.2 Basis and Dimension\n\n215\n\n12.13. Do the following vectors span Q3 (as a vector space over Q)? 1. (1, 0, 2), (2, 5, 3), (3, 5, 5) 2. (1, 0, 2), (2, 3, 5), (0, 0, 4) 12.14. matrices \u0006 \u0007 Do \u0006 the\u0007following \u0006 \u0007 \u0006 \u0007 span M2 (Z5 ) (as a vector space over Z5 ), namely 11 01 00 10 , , , ? 00 10 11 01 12.15. Let V = M2 (C). Find the dimension of V as a vector space over C, and as a vector space over R. 12.16. Let F = Z7 and V = {(a, b, c) ∈ F 3 : c = 3a + 5b}. Find the dimension of V over F. 12.17. Let F be a field and V a finite-dimensional vector space. If W is a subspace of V , show that dim W ≤ dim V , with equality if and only if W = V . (Do not assume, to begin with, that W is finite-dimensional.) 12.18. Suppose that a vector space V with dimension n has subspaces U and W with dimensions m and k, respectively. If m + k > n, show that U ∩ W \u0007= {0}. 12.19. Let F, V , W and α be as in Exercise 12.5. Suppose that v1 , . . . , vn ∈ V are linearly independent and α is one-to-one. Show that α(v1 ), . . . , α(vn ) are linearly independent. 12.20. Let F be a field and V a finite-dimensional vector space over F. Say dim V = n ∈ N. Show that there exists a bijective linear transformation (see Exercise 12.5 for the definition) α : V → F n .\n\n12.3 Field Extensions Let us now focus on our main vector space of interest: the field extension. Definition 12.9. Let K be a field extension of F. Then the degree of the extension is the dimension of K over F. We write [K : F] = dim F K . The extension is finite if [K : F] < ∞ and, in particular, quadratic if [K : F] = 2. Example 12.20. As we observed in Example 12.17, C is a quadratic extension of R. √ √ Example 12.21. Let K = {a + b 3 2 + c 3 4 : a, b, c ∈ Q}. We claim that K is a subfield of R and, therefore, an extension field of Q. All of the properties of a subfield are easy√to verify √ except, perhaps, that nonzero elements have inverses. Take 0 \u0007= a + b 3 2 + c 3 4 ∈ K . Then notice that (a + bx + cx 2 , x 3 − 2) divides x 3 − 2. But, by Example 11.10, x 3 − 2 is irreducible over Q. Thus, the gcd can only be a constant polynomial (in fact, 1, since we assume it to be monic). As Q[x] is a Euclidean domain, Theorem 10.6 guarantees that we can write\n\n216\n\n12 Vector Spaces and Field Extensions\n\n1 = u(x)(x 3 − 2) + v(x)(a + bx + c2 ), √ √ √ + b 3√ 2 + c 3 4). As it is easy to for some u(x), v(x) ∈ Q[x]. But then 1 = v( 3 2)(a √ √ 3 see that v( 3 2) ∈ K , we have an inverse for a + b 3 2 + c √ 4 in√K , as claimed. In fact, [K : Q] = 3. To see this, we observe that {1, 3 2, 3 4} is a basis for K over Q. Clearly, these numbers span K . If they√are linearly dependent, then there are rational numbers a, b, c, not all zero, such that 3 2 is a root of a +bx +cx 2 . But again, 1 = (a + bx + cx 2 , x 3 − 2), and we write√1 = u(x)(a + bx + cx 2 ) + v(x)(x 3 − 2), for some u(x), v(x) ∈ Q[x]. Evaluating at 3 2, we obtain 1 = 0, giving a contradiction and establishing that we have a basis. We are, in fact, engaging in a small abuse of notation here. If K is an extension field of F then, of course, F is also an additive subgroup of K . We could also use the notation [K : F] to mean the index of F in K as additive subgroup. This is not the same as the degree of the extension! For the remainder of the book, when we write [K : F], we will mean the degree of the extension. In the particular case of a finite field, we can illustrate the difference. By | . However, the Lagrange’s theorem, the index of the additive groups would be |K |F| degree is calculated as follows. Theorem 12.7. Let K be a field extension of F, such that K is a finite field. Then [K : F] = log|F| |K |. Proof. First, we note that K must be finite-dimensional over F. Indeed, the elements of K must span K , and by Theorem 12.6, we can obtain a finite basis. Let [K : F] = n, and suppose that {v1 , . . . , vn } is a basis for K over F. By Theorem 12.4, the elements of K are uniquely of the form a1 v1 +· · ·+an vn , with ai ∈ F. As there are |F| choices for each ai , the total number of elements of K is |F|n . Taking the base |F| logarithm, we obtain our result. \u0002 Degrees of extensions behave in a nice way. Theorem 12.8. Let K be a finite extension of F and L a finite extension of K . Then [L : F] = [L : K ][K : F]. Proof. Let {v1 , . . . , vn } be a basis for K over F, and let {w1 , . . . , wm } be a basis for L over K . We claim that {vi w j : 1 ≤ i ≤ n, 1 ≤ j ≤ m} is a basis for L over F. This will complete the proof. Take any l ∈ L. Then l = a1 w1 + · · · + am wm , for some ai ∈ K . But ai = bi1 v1 + · · · + bin vn , for some bi j ∈ F. Thus, l = b11 v1 w1 + b12 v2 w1 + · · · + b1n vn w1 + · · · + bm1 v1 wm + · · · + bmn vn wm . That is, the vi w j span L over F. Suppose that they are linearly dependent. Then there exist bi j ∈ F, not all zero, such that\n\n12.3 Field Extensions\n\n217\n\n0 = b11 v1 w1 + · · · + b1n vn w1 + · · · + bm1 v1 wm + · · · + bmn vn wm = (b11 v1 + · · · + b1n vn )w1 + · · · + (bm1 v1 + · · · + bmn vn )wm . As each bi1 v1 + · · · + bin vn ∈ K , and the wi are linearly independent over K , we have bi1 v1 + · · · + bin vn = 0, for all i. But the bi j ∈ F, and the v j are linearly \u0002 independent over F. Thus, all of the bi j are zero. The proof is complete. √ Example 12.22. Let K = √ {a+b 2 : a, b ∈ Q}. By Example 8.30, K is an extension then field of Q. Clearly, 1 and 2 span K over Q. If they were linearly dependent, √ √ 2 would lie in Q, which is not the case. Thus, [K : Q] = 2. Let L = {c + d 3 : c, d ∈ K }. We claim that L is a subfield of R and, hence, an extension of K . All of the subfield properties are easy to check the existence of inverses. √ except perhaps √ √ Let 0 \u0007= c + d 3 ∈√L. Then (c + d 3)(c − d 3) = c2 − 3d 2 . Suppose that giving √ us a contradiction. this is 0. Then √ c − d 3 = 0. If d = 0, then so is c, √ Otherwise, 3 = cd√−1 ∈ K . Thus, we can√write a + b 2 = 3, with a, b ∈ Q. Then\b a 2 +2b2 +2ab 2 = 3. If b = 0, then 3 = a ∈ Q, which is not true. If a = 0, = b ∈ Q. But then 2x 2 − 3 has a rational root which, by Theorem 11.5, √ is not the case. Thus,√ ab \u0007= 0, and 2 ∈ Q, giving us a contradiction. Therefore, √ √ 3 (c + d 3)−1 = cc−d 3 span L over K . If they were linearly 2 −3d 2 ∈ L. Now, 1 and √ dependent, then we would have 3 ∈ K which, as we have just seen, is not the case. Therefore, [L : K ] = 2. By the theorem above, [L : Q] = [L : K ][K : Q] = 4. then\n\n3 2\n\nOne particular type of extension is especially important. Definition 12.10. Let K be a field extension of F. If a ∈ K , then we write F(a) for the intersection of all subfields of K containing F and a. We say that K is a simple extension of F if K = F(a) for some a ∈ K . By Exercise 8.33, the intersection of some set of fields is a field. Thus, F(a) is always a field. Indeed, it is the smallest subfield of K containing F and a. √ Example 12.23. By Example 8.30, √ Thus, √ {a + b 2 : a, b ∈ Q} is a subfield of R. since any field including Q and 2 would surely contain this field, it is Q( 2). √ √ Example 12.24. In a similar manner, we note that Q( 3 2) would have to contain 3 2 √ √ √ √ 3 3 3 3 and ( 2)2 . Example 12.21 shows us that Q( 2) = {a + b 2 + c 4 : a, b, c ∈ Q}. Let us concentrate on simple extensions. In fact, we need to break them down into two types, depending upon one specific property of the element a. Definition 12.11. Let K be a field extension of F and a ∈ K . We say that a is algebraic over F if there exists a nonzero polynomial f (x) ∈ F[x] such that f (a) = 0. Otherwise, a is transcendental over F. √ Example 12.25. The number 3 2 is algebraic over Q, since it is a root of x 3 − 2 ∈ Q[x].\n\n218\n\nExample 12.26. The number x 4 − 10x 2 + 1.\n\n12 Vector Spaces and Field Extensions\n\n√ √ 2 + 3 is algebraic over Q, since it is a root of\n\nFinding examples of real numbers that are transcendental over Q is a bit tricky. As it happens, the constants e and π are both transcendental. (This is a difficult result. For a proof, see the advanced monograph of Baker .) Of course, the underlying field is important! If we let F = Q(π 2 ), then π is algebraic over F, as π is a root of x 2 − π 2 ∈ F[x]. We are primarily interested in algebraic elements. However, we can mention one important fact about transcendental elements. If F is a field, then F[x] is an integral domain, and so we can consider its field of fractions. Denote this field of fractions by F(x). Theorem 12.9. Let K be an extension field of F, and let a ∈ K be transcendental over F. Then F(a) is isomorphic to F(x). In particular, F(a) is of infinite degree over F. Proof. Define α : F[x] → K via α( f (x)) = f (a). By Lemma 11.1, α is a homomorphism. If f (x) ∈ ker(α), then f (a) = 0. Since a is transcendental, f (x) is the zero polynomial. Thus, α is one-to-one, and F[x] is isomorphic to α(F[x]). Also, f (a) ∈ F(a) for all f (x) ∈ F[x]; thus, α(F[x]) is a subring of F(a). By Theorem 9.15, there is a subfield L of F(a) such that L is isomorphic to F(x) and contains α(F[x]). Clearly α(b) = b for all b ∈ F and α(x) = a; thus, α(F[x]) contains both F and a. But F(a) is the smallest subfield of K containing both F and a; thus, F(a) = L. Suppose that [F(a) : F] = n < ∞. Then according to Lemma 12.1, the elements 1, a, a 2 , . . . , a n must be linearly dependent over F. But then there exist ci ∈ F, not all zero, such that a is a root of c0 + c1 x + · · · + cn x n . That is, a is algebraic, giving us a contradiction. \u0002 Now suppose that a is algebraic over F. We know that it satisfies a nonzero polynomial in F[x]. But one particular such polynomial is key. Definition 12.12. Let K be an extension field of F and let a ∈ K be algebraic over F. Then the minimal polynomial of a over F is the monic irreducible polynomial m(x) ∈ F[x] such that m(a) = 0. √ √ Example 12.27. The minimal polynomial of 3 2 over Q is x 3 − 2. Indeed, 3 2 is a root, and the polynomial is irreducible by Example 11.10. √ √ 4 2 Example 12.28. The minimal polynomial √ of 2 + 3 over Q is x − 10x + 1. √ As we noted in Example 12.26, 2 + 3 is a root. Suppose it were reducible over Q. The Rational Roots Theorem shows us that it has no roots in Q. Thus, it would have to factor as a product of two polynomials of degree 2. By Theorem 11.4, these polynomials may be assumed to be in Z[x]. Looking at the coefficients, we see immediately that (up to multiplying both factors by −1) the only possibilities are (x 2 + ax + 1)(x 2 − ax + 1) and (x 2 + ax − 1)(x 2 − ax − 1), for some a ∈ Z. But then 2 − a 2 = −10 or −2 − a 2 = −10. Neither of these has a solution in Z.\n\n12.3 Field Extensions\n\n219\n\nWe were a bit bold in our definition of the minimal polynomial. Indeed, we assumed that such a polynomial exists, and that there is only one. Fortunately, our presumptuousness was justified; in fact, we can say more. Theorem 12.10. Let K be an extension field of F, and let a ∈ K be algebraic over F. Then 1. the minimal polynomial m(x) of a over F exists, and is the unique monic polynomial of smallest degree in F[x] of which a is a root; and 2. if f (x) ∈ F[x], then f (a) = 0 if and only if m(x)| f (x). Proof. Let I = { f (x) ∈ F[x] : f (a) = 0}. We claim that I is an ideal of F[x]. Surely 0 ∈ I . If f (x), g(x) ∈ I , then f (a) − g(a) = 0, and hence f (x) − g(x) ∈ I . Also, if h(x) ∈ F[x], then f (a)h(a) = 0, and hence f (x)h(x) ∈ I , proving the claim. We know that F[x] is a Euclidean domain and hence, by Theorem 10.8, a PID. Thus, let I = (m(x)). Since a is algebraic, m(x) is not the zero polynomial. As (m(x)) = (cm(x)) if 0 \u0007= c ∈ F, we may as well assume that m(x) is monic. Now, f (x) ∈ I if and only if m(x)| f (x), as required by (2). As such, deg(m(x)) ≤ deg f (x), unless f (x) = 0. If deg(m(x)) = deg( f (x)), then f (x) is simply m(x) multiplied by an element of F. If f (x) is also monic, then f (x) = m(x). Thus, m(x) satisfies condition (1) as well. We must still establish that m(x) is actually the minimal polynomial of a over F. To demonstrate this, we must show that m(x) is irreducible. But if m(x) = f (x)g(x), with f (x), g(x) ∈ F[x], then 0 = m(a) = f (a)g(a). Thus, f (a) = 0 or g(a) = 0. Without loss of generality, say f (a) = 0. Then m(x)| f (x). But also f (x)|m(x). It now follows that deg( f (x)) = deg(m(x)), and hence g(x) is a constant polynomial. Thus, m(x) is irreducible, and hence a minimal polynomial for a. If g(x) is another minimal polynomial, then g(x) ∈ I , and hence m(x)|g(x). But g(x) is irreducible, and therefore g(x) = cm(x) for some c ∈ F. As m(x) and g(x) are both monic, m(x) = g(x), and the proof is complete. \u0002 We can use the minimal polynomial to describe the simple extension. Theorem 12.11. Let L be an extension field of F, and let a ∈ L be algebraic over F. If m(x) is the minimal polynomial of a over F, let n = deg(m(x)). Then 1. [F(a) : F] = n; 2. {1, a, a 2 , . . . , a n−1 } is a basis for F(a) over F; and 3. F(a) is isomorphic to F[x]/(m(x)). Proof. Of course, (1) follows immediately from (2), so let us prove (2). Suppose that 1, a, . . . , a n−1 are linearly dependent. Then there exist c0 , . . . , cn−1 ∈ F, not all zero, such that c0 + c1 a + · · · + cn−1 a n−1 = 0. That is, a is a root of c0 + c1 x + · · · + cn−1 x n−1 . But this polynomial has degree smaller than that of m(x), contradicting Theorem 12.10. Thus, 1, a, . . . , a n−1 are linearly independent. We claim that they span F(a). We know that F(a) is the smallest field containing F and a (and, hence, all of the a i ). Therefore, it is sufficient to show that K =\n\n220\n\n12 Vector Spaces and Field Extensions\n\n{c0 + c1 a + · · · + cn−1 a n−1 : ci ∈ F} is a field. Clearly, it contains 1 and is closed under subtraction. To show that it is closed under multiplication, it is enough to show that a i ∈ K for all positive integers i. Our proof is by strong induction upon i. If i < n, there is nothing to do, so let i ≥ n and suppose it is true for smaller exponents. Writing m(x) = b0 + · · · + bn−1 x n−1 + x n , we have a i = a n a i−n = (−b0 − b1 a − · · · − bn−1 a n−1 )a i−n . But this is a linear combination of terms of the form a j , with j < i. Thus, by our inductive hypothesis, a i ∈ K . Finally, we must check that every nonzero element of K has an inverse in K . But a nonzero element of K has the form f (a), for some 0 \u0007= f (x) ∈ F[x], with deg( f (x)) < n. Now, ( f (x), m(x))|m(x). As m(x) is irreducible, ( f (x), m(x)) is either 1 or an associate of m(x). However, deg( f (x)) < deg(m(x)). Thus, ( f (x), m(x)) = 1. By Theorem 10.6, there exist u(x), v(x) ∈ F[x] such that f (x)u(x) + m(x)v(x) = 1. Since m(a) = 0, we have f (a)u(a) = 1. Furthermore, as we noted above, u(a) ∈ K , so f (a) has an inverse in K . Therefore, K is a field, and (2) is proved. (3) Define α : F[x] → F(a) via α( f (x)) = f (a). By Lemma 11.1, α is a homomorphism. In view of (2), it is onto. The kernel is the set of all polynomials in F[x] of which a is a root. By Theorem 12.10, this is (m(x)). Apply the First Isomorphism Theorem. \u0002 Example 12.29. As x 2 + 1 is the minimal polynomial of i over R, we see that C = R(i) is isomorphic to R[x]/(x 2 + 1). √ √ Example 12.30. As we saw in Example √ 12.28,√ the minimal polynomial of 2 + 3 over Q is x 4 − 10x 2 + 1. Therefore, Q( 2 + 3) is isomorphic to Q[x]/(x 4 − 10x 2 + 1). √ √ Furthermore, letting a = 2 + 3, the elements of Q(a) are precisely c0 + c1 a + c2 a 2 + c3 a 3 , with ci ∈ Q. Addition works in the obvious way. To demonstrate multiplication, let us try (2 − 3a + 4a 2 )(5 + a − 6a 2 + 2a 3 ). We get 10 − 13a + 5a 2 + 26a 3 − 30a 4 + 8a 5 . Now, a 4 = 10a 2 − 1 and a 5 = 10a 3 − a. Thus, our product is 40 − 21a − 295a 2 + 106a 3 . Our last theorem has an interesting immediate consequence. Corollary 12.1. Let K be an extension field of F. If a, b ∈ K , and a and b have the same minimal polynomial over F, then F(a) is isomorphic to F(b). Proof. If m(x) is the minimal polynomial, then by Theorem 12.11, both F(a) and F(b) are isomorphic to F[x]/(m(x)). \u0002 Example 12.31. Let ω be a primitive cube root of unity in C. (That is, ω3 = 1 but √ √ 3 3 ω \u0007= 1.) Then 2 and ω 2 are both roots of x 3 − 2 ∈ Q[x]. As we have √ observed, √ irreducible over Q, so it is√the minimal polynomial of both 3 2 and ω√3 2. x 3 − 2 is√ to Q(ω 3 2). These fields are clearly distinct, as Q( 3 2) Thus, Q( 3 2) is isomorphic √ 3 / R. is a subfield of R, but ω 2 ∈\n\n12.3 Field Extensions\n\nExercises 12.21. Find the minimal polynomial of 12.22. Find the minimal polynomial of\n\n221\n\n√ √ 3\n\n5+\n\n7 over Q. √ 3 + 3 9 over Q.\n\n12.23. Let K be a finite extension field of F. Show that every element of K is algebraic over F. 12.24. Let K be an extension field of F and L an extension field of K . If a ∈ L is algebraic over F, show that [K (a) : K ] ≤ [F(a) : F]. 12.25. Suppose that we have subfields Fn of K with F1 ⊆ F2 ⊆ F3 ⊆ · · · . Show that ∞ n=1 Fn is a field. √ n 12.26. For each positive integer n, let an = 2 2. If K = ∞ n=1 Q(an ), show that K is an infinite field extension of Q, but every element of K is algebraic over Q. 12.27. Let K be a field extension of F. Show that for every a ∈ K , F(a 2 ) ⊆ F(a). Also, give an explicit example illustrating that we do not have F(a 2 ) = F(a) in general. 12.28. Let K be a field extension of F and a ∈ K . Show that a is algebraic over F if and only if a 2 is algebraic over F. 12.29. Let K be an extension field of C. If a ∈ K is algebraic over C, show that a ∈ C. 12.30. Let K be a finite extension of F. If R is a subring of K containing F, show that R is a field.\n\n12.4 Splitting Fields Let us now take a slightly different perspective from the preceding section. Given a field F, instead of looking at elements of extension fields and finding their minimal polynomials, let us instead take a nonconstant polynomial f (x) ∈ F[x] and see if we can find a field containing F and a root of f (x). For instance, suppose that we only knew about the rational numbers, and we wanted to construct a field having a root of x 2 − 2. Definition 12.13. Let F be a field and let f (x) ∈ F[x] be a nonconstant polynomial. If K is an extension field of F, then we say that f (x) splits over K if there exist a, a1 , . . . , an ∈ K such that f (x) = a(x − a1 ) · · · (x − an ). In particular, K is a splitting field for f (x) if f (x) splits over K , and if L is any subfield of K with F ⊆ L \u0002 K , then f (x) does not split over L.\n\n222\n\n12 Vector Spaces and Field Extensions\n\nTo put this another way, if K is an extension field of F and b1 , . . . , bn ∈ K , write F(b1 , . . . , bn ) for the intersection of all subfields of K containing F and all of the bi . If f (x) ∈ F[x] splits over K , and a1 , . . . , an are the roots of f (x) in K , then K is a splitting field of f (x) if and only if K = F(a1 , . . . , an ). But how to construct such a field? The following observation is helpful. Lemma 12.2. Every nonzero prime ideal in a PID is maximal. Proof. Let I be a nonzero prime ideal in a PID R. Then I = (a), for some a ∈ I . By Lemma 10.2, a is prime. In particular, by Theorem 10.10, a is irreducible. Since I is prime, I \u0007= R. Suppose that J is an ideal of R with I \u0002 J \u0002 R. Let J = (b). Then a ∈ (b), so b|a. As a is irreducible, b is a unit or an associate of a. In the former case, J = R. In the latter, a|b, and hence J = I . Either way, we have a contradiction. \u0002 The next lemma is the key to our construction. Lemma 12.3. Let F be a field and f (x) an irreducible polynomial in F[x]. Let K = F[x]/( f (x)). Then K is a field containing (an isomorphic copy of) F and a root a of f (x). In fact, K = F(a). Proof. We know that F[x] is a Euclidean domain and hence, by Theorem 10.8, a PID. By Theorem 10.11, f (x) is prime. Thus, by Lemma 10.2, ( f (x)) is a prime ideal. The preceding lemma tells us that ( f (x)) is maximal. By Theorem 9.20, K is indeed a field. Define α : F → K via α(b) = b + ( f (x)). It is immediate that α is a homomorphism. If α(b) = 0, then b ∈ ( f (x)), which means that f (x)|b. As b is a constant, b = 0, and hence α is one-to-one. Thus, K contains an isomorphic copy of F, namely α(F). Finally, let us show that K contains a root of f (x). But this root is a = x + ( f (x)). Indeed, f (a) = f (x) + ( f (x)) = 0 + ( f (x)), as required. Clearly, \u0002 F(a) would have to contain x n + ( f (x)) for all n ≥ 0. Thus, K = F(a). Let us combine the preceding lemma with Theorem 12.11. We see that if f (x) ∈ F[x] is irreducible of degree n, then the field K has, as a basis over F, the terms x i + ( f (x)), with 0 ≤ i < n. This allows us for the first time to create finite fields other than Z p , where p is a prime. Example 12.32. Suppose we wish to construct a field of order 125. In view of Theorem 12.7, we would need an extension of degree 3 of Z5 . Consider f (x) = x 3 + 3x 2 + x + 2 ∈ Z5 [x]. By Corollary 11.2, it is irreducible over Z5 if it has no roots in Z5 . There are only five possible roots, and none of them work. Therefore, f (x) is irreducible and F[x]/( f (x)) is a field of order 125. The elements are a0 + a1 x + a2 x 2 + ( f (x)), with ai ∈ Z5 . Addition works in the obvious way. As an example of multiplication, we have (letting I = ( f (x))) (2 + 4x + 3x 2 + I )(1 + 4x + I ) = 2 + 2x + 4x 2 + 2x 3 + I = 2 + 2x + 4x 2 + 2(−3x 2 − x − 2) + I = 3 + 3x 2 + I.\n\n12.4 Splitting Fields\n\n223\n\nWe can now construct splitting fields. Theorem 12.12. Let F be a field and f (x) ∈ F[x] a nonconstant polynomial. Then there is a splitting field of f (x) over F. Proof. First, let us prove the existence of a field extension in which f (x) splits. We proceed by induction on n = deg( f (x)). If n = 1, then F will suffice. Assume that n ≥ 2 and the n − 1 case holds. We know that F[x] is a UFD. Thus, write f (x) = g1 (x) · · · gk (x), where the gi (x) are irreducible in F[x]. By Lemma 12.3, there is an extension field K of F in which g1 (x) has a root, a. Then by Theorem 11.2, g1 (x) = (x − a)h 1 (x), for some h 1 (x) ∈ K [x]. Thus, in K [x], we have f (x) = (x − a)h 1 (x)g2 (x) · · · gk (x). Now, h 1 (x)g2 (x) · · · gk (x) has degree n − 1. Thus, by our inductive hypothesis, it splits in some extension field L of K . Hence, L is an extension field of F, and f (x) splits over L. Let us write f (x) = b(x − b1 ) · · · (x − bn ), with b, b1 , . . . , bn ∈ L. Then \u0002 F(b1 , . . . , bn ) is a splitting field for f (x) over F. But we can go one step further. We want to show that splitting fields are unique up to isomorphism. (The proof is a bit technical, but the result will pay dividends when we classify the finite fields.) To this end, we need to sharpen Corollary 12.1 a bit. If α : R → S is a ring homomorphism, and f (x) = c0 + · · · + cn x n ∈ R[x], then we write α( f (x)) = α(c0 ) + · · · + α(cn )x n ∈ S[x]. Lemma 12.4. Let α : F → K be an isomorphism of fields. Let f (x) ∈ F[x] be an irreducible polynomial. Suppose that a is a root of f (x) in some extension field of F and b is a root of α( f (x)) in some extension field of K . Then there exists an isomorphism β : F(a) → K (b) such that β(c) = α(c) for all c ∈ F and β(a) = b. Proof. Define γ : F[x] → F(a) via γ (g(x)) = g(a). By Lemma 11.1, γ is a homomorphism. By Theorem 12.10, ker(γ ) = ( f (x)). (We assumed that f (x) was monic in that theorem, but that is immaterial here.) In view of Theorem 12.11, γ is onto. Thus, the proof of the First Isomorphism Theorem shows us that the map ρ : F[x]/( f (x)) → F(a) given by ρ(g(x) + ( f (x))) = g(a) is an isomorphism. We also note that if c ∈ F, then ρ(c + ( f (x))) = c and ρ(x + ( f (x))) = a. In precisely the same manner, the map τ : K [x]/(α( f (x))) → K (b) given by τ (h(x) + (α( f (x)))) = h(b) is an isomorphism, τ (d + (α( f (x)))) = d for all d ∈ K and τ (x + (α( f (x)))) = b. Now, the function from F[x] to K [x] mapping each u(x) to α(u(x)) is easily seen to be an isomorphism. Composing that with the obvious homomorphism from K [x] to K [x]/(α( f (x))), we obtain a homomorphism from F[x] onto K [x]/(α( f (x))) with kernel ( f (x)). In view of the First Isomorphism Theorem, we have an isomorphism σ : F[x]/( f (x)) → K [x]/(α( f (x))) given by σ (u(x) + ( f (x))) = α(u(x)) + (α( f (x))). Notice that σ (c + ( f (x))) = α(c) + (α( f (x))) for all c ∈ F, and σ (x + ( f (x))) = x + (α( f (x))).\n\n224\n\n12 Vector Spaces and Field Extensions\n\nFrom Theorem 9.12, we learn that τ σρ −1 : F(a) → K (b) is an isomorphism. Furthermore, if c ∈ F, then τ σρ −1 (c) = τ σ (c + ( f (x))) = τ (α(c) + (α( f (x)))) = α(c), and\n\nτ σρ −1 (a) = τ σ (x + ( f (x))) = τ (x + (α( f (x)))) = b.\n\nLetting β = τ σρ −1 , we are done.\n\n\u0002\n\nThis allows us to prove the uniqueness of splitting fields. Theorem 12.13. Let α : F → K be a field isomorphism, and let f (x) ∈ F[x] be a nonconstant polynomial. If L is a splitting field of f (x) over F, and M is a splitting field of α( f (x)) over K , then there is an isomorphism β : L → M such that β and α agree on F. Proof. We proceed by induction on n = deg( f (x)). If n = 1, then we can only have L = F and M = K . Thus, letting β = α will suffice. Assume that the result is true for polynomials of degree n − 1. As f (x) is a product of irreducibles in F[x], let us say that f (x) = g(x)h(x), where g(x) is irreducible and h(x) ∈ F[x]. Let a be a root of g(x) in L and b a root of α(g(x)) in M. By the preceding lemma, there is an isomorphism γ : F(a) → K (b) such that γ agrees with α on F and γ (a) = b. We have f (x) = (x − a)u(x), for some u(x) ∈ F(a)[x], by Theorem 11.2. Also, γ ( f (x)) = (x − γ (a))γ (u(x)) = (x − b)γ (u(x)) in K (b)[x]. Now, L is a splitting field for u(x) over F(a) and M is a splitting field for γ (u(x)) over K (b). Since deg(u(x)) = n − 1, our inductive hypothesis completes the proof. \u0002 Corollary 12.2. Let F be a field and f (x) ∈ F[x] a nonconstant polynomial. Then any two splitting fields of f (x) over F are isomorphic. Proof. In the preceding theorem, let α : F → F be the identity automorphism. \u0002 Exercises 12.31. Construct an extension field F of Z7 having order 73 . In particular, if F = Z7 (a), what do all of the elements of F look like? To which of these elements is (a 2 + 5a + 4)(3a 2 + 6) equal? 12.32. Construct an extension field F of Z3 having order 81. In particular, if F = Z3 (a), what do all of the elements of F look like? To which of these elements is (a 3 + 2a 2 + 2)(2a 2 + a + 1) equal? √ 12.33. Show that Q( 3 2, ω) is a splitting field of x 3 − 2 over Q, where ω ∈ C, ω3 = 1, but ω \u0007= 1. 12.34. Let F be a field and f (x) ∈ F[x] a nonconstant polynomial. If K is a splitting field of f (x) over F and L is any extension field of F, suppose that α : K → L is a homomorphism satisfying α(c) = c for all c ∈ F. If a ∈ K is a root of f (x), show that α(a) is also a root of f (x).\n\n12.4 Splitting Fields\n\n225\n\n√ 12.35. Find every automorphism of Q( 2). 12.36. Construct a splitting field for x 3 + 2x + 1 over Z3 . Show that it has degree 3 over Z3 . 12.37. Let F be any field and f (x) ∈ F[x] a nonconstant polynomial. If we let g(x) = f (x + 1), show that f (x) and g(x) have the same splitting fields over F. 12.38. Let F be a field and f (x) ∈ F[x] a polynomial with deg( f (x)) = n ∈ N. Show that f (x) has a splitting field K over F with [K : F] ≤ n!.\n\n12.5 Applications to Finite Fields Let us see what we can deduce about finite fields. If F is a finite field, we know that its prime subfield must be isomorphic to Z p , for some prime p. By Theorem 12.7, F must have order p n , for some positive integer n. We will construct a field of order p n and show that, up to isomorphism, there is only one such field. The following concept looks suspiciously like calculus, but is not. Definition 12.14. Let F be a field and f (x) = a0 +a1 x +a2 x 2 +· · ·+an x n ∈ F[x]. Then the formal derivative of f (x) is f (x) = a1 + 2a2 x + · · · + nan x n−1 . Note that this has nothing whatsoever to do with limits, as limits do not necessarily make sense in an arbitrary field. The formula happens to agree with the one used for the derivative of real polynomials. We will also not be disturbed by the fact that the following lemma extends the similarity to calculus. Lemma 12.5. Let F be a field, f (x), g(x) ∈ F[x] and a ∈ F. Then 1. (a f (x)) = a f (x); 2. ( f (x) + g(x)) = f (x) + g (x); and 3. ( f (x)g(x)) = f (x)g(x) + f (x)g (x). Proof. The first two parts follow immediately from the definition. The third is left as Exercise 12.40. \u0002 Definition 12.15. Let F be a field, f (x) ∈ F[x] and a ∈ F. We say that a is a multiple root of f (x) if (x − a)2 | f (x). Example 12.33. In Q[x], 2 is a multiple root of x 5 − 4x 4 + 7x 3 − 7x 2 − 8x + 20, since the polynomial factors as (x − 2)2 (x 3 + 3x + 5). Theorem 12.14. Let F be a field, f (x) ∈ F[x] and let a ∈ F be a root of f (x). Then a is a multiple root of f (x) if and only if f (a) = 0.\n\n226\n\n12 Vector Spaces and Field Extensions\n\nProof. Suppose that a is a multiple root of f (x), say f (x) = (x − a)2 g(x), with g(x) ∈ F[x]. Then by Lemma 12.5, f (x) = 2(x − a)g(x) + (x − a)2 g (x). Thus, f (a) = 0. Conversely, suppose that f (a) = 0. By Theorem 11.2, f (x) = (x−a)h(x), for some h(x) ∈ F[x]. Thus, f (x) = h(x)+(x−a)h (x). As f (a) = 0, we have 0 = h(a) + (a − a)h (a) = h(a). By Theorem 11.2, (x − a)|h(x), and \u0002 hence (x − a)2 | f (x). Corollary 12.3. Let F be a field and let f (x) ∈ F[x] be irreducible. Let K be a splitting field of f (x) over F. If f (x) has a multiple root in K , then f (x) is the zero polynomial. Proof. Let a be the multiple root. Then (multiplying f (x) by a suitable element of F to make it monic), we see that f (x) is the minimal polynomial of a over F. By Theorem 12.10, f (x)| f (x). But if f (x) \u0007= 0, then deg( f (x)) < deg( f (x)), which \u0002 is impossible. Therefore, f (x) is the zero polynomial. Definition 12.16. A field F is said to be perfect if no irreducible f (x) ∈ F[x] has multiple roots in any splitting field of f (x) over F. We digress from our discussion of finite fields to mention the following. Theorem 12.15. Every field of characteristic zero is perfect. Proof. If f (x) = a0 + · · · + an x n , with an \u0007= 0 and n ≥ 1, then f (x) = a1 + · · · + nan x n−1 has leading coefficient nan \u0007= 0. Thus, f (x) is not the zero polynomial. Apply Corollary 12.3. \u0002 Actually, finite fields are perfect too! Let us see why. Lemma 12.6. Let F be a finite field of characteristic p. Then the function α : F → F given by α(a) = a p is an automorphism. Proof. Since F is commutative, α(ab) = (ab) p = a p b p = α(a)α(b), for all a, b ∈ F. By Theorem 8.14, α(a + b) = (a + b) p = a p + b p = α(a) + α(b). If a p = 0, then since F is a field, a = 0. Thus, α is one-to-one. Since F is finite, α is onto as well. \u0002 Theorem 12.16. Every finite field is perfect. Proof. Suppose that F has characteristic p. Let f (x) ∈ F[x] be irreducible. Suppose that f (x) = a0 + a1 x + · · · + an x n . If f (x) has multiple roots in a splitting field, then by Corollary 12.3, f (x) = 0. Thus, kak = 0, for 1 ≤ k ≤ n. If p \u0003 k, then as (k, p) = 1, we may write ku + pv = 1, for some u, v ∈ Z. Therefore, ak = ukak + pvak = 0 + 0 = 0. Thus, f (x) = a0 + a p x p + a2 p x 2 p + · · · + amp x mp . p\n\nIn view of the preceding lemma, there exist bi ∈ F such that bi = ai p . But now Theorem 8.14 tells us that\n\n12.5 Applications to Finite Fields\n\n227 p\n\np\n\n(b0 + b1 x + b2 x 2 + · · · + bm x m ) p = b0 + b1 x p + · · · + bmp x mp = a0 + a p x p + · · · + amp x mp = f (x). That is, f (x) is reducible. This contradiction completes the proof.\n\n\u0002\n\nWhat would an imperfect field look like? Clearly, it would have to be an infinite field of prime characteristic. Exercise 12.44 shows how to construct an imperfect field. Back to the finite fields! Lemma 12.7. Let F be a field of prime characteristic p and let n be a positive n integer. If K = {a ∈ F : a p = a}, then K is a subfield of F. Proof. See Exercise 8.40. Theorem 12.17. Let p be a prime and n a positive integer. Then a field F has order n p n if and only if it is a splitting field of x p −x over the prime subfield, (an isomorphic copy of) Z p . Proof. Let F have order p n . Then U (F) has order p n − 1. Thus, if 0 \u0007= a ∈ F, then n n n a p −1 = 1, and hence a p − a = 0. Clearly, 0 p − 0 = 0 as well. Thus, every element n pn of F is a root of x − x. By Corollary 11.3, x p − x can only have p n roots. Thus, pn x − x splits over F, and surely it cannot split over any smaller field, as all of the n roots must be present. Therefore, F is a splitting field of x p − x over Z p . pn Conversely, let F be a splitting field of x − x over Z p . By Lemma 12.7, the n n roots of x p − x form a subfield K of F. Since x p − x splits over K , we must have pn F = K . Furthermore, the formal derivative of x − x is −1, which has no roots. n Therefore, by Theorem 12.14, x p − x has no multiple roots. In particular, |F| = p n , as required. \u0002 Theorem 12.18. If k is a positive integer, then there is a field of order k if and only if k = p n for some prime p and positive integer n. All fields of order p n are isomorphic. Proof. By Theorem 12.7, a finite field must have order p n . Theorem 12.12 tells us n that x p − x has a splitting field over Z p . By Theorem 12.17, this splitting field has order p n . But Theorem 12.17 also says that every field of order p n is such a splitting field. By Corollary 12.2, these splitting fields are isomorphic. \u0002 The unique (up to isomorphism) field of order p n is called the Galois field of order p n . We can also determine the subfields of a finite field. In order to do so, we will need the following theorem, which is of interest on its own. Theorem 12.19. Let F be a field. Then any finite subgroup G of U (F) is cyclic.\n\n228\n\n12 Vector Spaces and Field Extensions\n\nProof. Since G is a finite abelian group, Theorem 5.3 tells us that it is a direct product of cyclic groups. If all of these cyclic groups have relatively prime orders, then by Theorem 5.4, G is cyclic, and we are done. Otherwise, we may assume that G has a subgroup a\u000e × b\u000e, and there exists a prime p dividing the orders of a and b. By Cauchy’s theorem, a\u000e and b\u000e each contain an element of order p. Thus, G has a subgroup isomorphic to Z p × Z p . But every element of Z p × Z p has order 1 or p. That is, we have at least p 2 roots for the polynomial x p −1 ∈ F[x], which has degree p, giving us a contradiction and completing the proof. \u0002 Theorem 12.20. Let F be a field of order p n , for some prime p and positive integer n. Then every subfield of F has order p m , for some positive divisor m of n. Furthermore, for each positive divisor m of n, F has exactly one subfield of order p m , namely m {a ∈ F : a p = a}. Proof. Let K be a subfield of F. Then K and F have the same prime subfield, (an isomorphic copy of) Z p . By Theorems 12.7 and 12.8, n = [F : Z p ] = [F : K ][K : Z p ]. In particular, if [K : Z p ] = m, then |K | = p m and m|n. m Let m be a divisor of n and let K = {a ∈ F : a p = a}. By Lemma 12.7, K is a subfield of F. Furthermore, the preceding theorem tells us that U (F) is cyclic of order p n − 1. In addition, p n − 1 = ( p m − 1)(1 + p m + p 2m + p 3m + · · · + p n−m ). Thus, ( p m − 1)|( p n − 1). By Corollary 3.3, U (F) has a subgroup G of order p m − 1. m m But every element a of G satisfies a p −1 = 1, and hence a p = a. That is G ⊆ K . Also, 0 ∈ K , and therefore K has at least p m elements. But every element of K is a m root of x p − x, and therefore K can have at most p m elements. To prove the uniqueness of this subfield, suppose that L is another subfield of F with p m elements. Then U (K ) and U (L) are both subgroups of order p m − 1 in U (F). However, Corollary 3.3 tells us that U (F) has only one such subgroup. Therefore, U (K ) = U (L). As the unit group of a field consists of everything except 0, we have K = L, as required. \u0002 Exercises 12.39. Find the smallest field containing exactly 3 proper subfields. 12.40. Let F be a field and f (x), g(x) ∈ F[x]. Show that ( f (x)g(x)) = f (x)g(x) + f (x)g (x). 12.41. Let f (x) ∈ Z5 [x] be an irreducible polynomial of degree 3. If K is a splitting field of f (x) over Z5 , show that |K | = 53 or 56 . 12.42. Let K be a field of order p n for some prime p and positive integer n, having subfields F and L of orders p m and pr , respectively. Find the order of F ∩ L.\n\n12.5 Applications to Finite Fields\n\n229\n\n12.43. Let F be a field and f (x) ∈ F[x] an irreducible polynomial having a multiple root in some extension field of F. Show that char F = p for some prime p, f (x) = a0 + a p x p + a2 p x 2 p + · · · + amp x mp for some ai ∈ F, and that at least one of the ai is transcendental over the prime subfield of F. 12.44. Let Z2 [t] be a polynomial ring over Z2 and F = Z2 (t) its field of fractions. Show that the polynomial x 2 − t ∈ F[x] is irreducible over F, but that it has a multiple root in some extension field of F. In particular, conclude that F is not a perfect field. 12.45. Theorem 12.19 tells us that the unit group of a finite field is cyclic. If char F \u0007= 2, show that the unit group of an infinite field is not cyclic. 12.46. Suppose char F = 2. Let us prove that the preceding exercise still holds. Suppose, to the contrary, that U (F) is cyclic. Let U (F) = a\u000e. 1. Show that F = Z2 (a). 2. If a is algebraic over Z2 , show that F is finite, and we are done. 3. If a is transcendental over Z2 , show that there exists an integer n such that a n = a + 1, and obtain a contradiction. 12.47. Suppose we wrote x 125 − x as a product of irreducibles over Z5 . Show that each of these irreducible polynomials has degree 1 or 3. (Please do not actually write the polynomials!) 12.48. Show that for every prime p and positive integer n, there exists an irreducible polynomial of degree n in Z p [x].\n\nReference 1. Baker, A.: Transcendental Number Theory, 2nd edn. Cambridge University Press, Cambridge (1990)\n\nPart V\n\nApplications\n\nChapter 13\n\nPublic Key Cryptography\n\nIn this short chapter, we talk a bit about cryptography. First, we discuss some classical sorts of private key methods, and their limitations in the modern world. We then look at the first public key cryptographic method.\n\n13.1 Private Key Cryptography Countless methods of encrypting messages have been invented over the centuries, and we will not attempt to give an exhaustive list here. Let us discuss a few well-known codes. One ancient method is known as the Caesar cipher. It could not be much simpler! Each letter in the alphabet is shifted forward three letters. Thus, A becomes D, B becomes E, and Z becomes C. If we wish to send the message HOWDY1 then our encrypted message is KRZGB. Decrypting the message is equally simple; the recipient shifts each letter back three positions. This is not a particularly good code. An opponent who knew that we were using a Caesar cipher could read any intercepted message instantly. We could complicate things a bit by selecting a positive integer k as a key. Instead of shifting letters 3 positions ahead, we would shift them k positions ahead. Decryption would then be a matter of shifting back k positions. We call this an additive cipher. This is better, but not much. There are only 26 possible keys (really 25, as one of them will just leave the message unencrypted). It would not take an opponent long to try all of the possible keys on an intercepted encrypted message, and see which one gives sensible text. But we can be more sophisticated than that. Let ρ be any permutation of the set of letters of the alphabet. We can then encrypt text by replacing each letter with its\n\n1 The\n\nauthor acknowledges that the circumstances under which this message would need to be sent secretly are few and far between. © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_13\n\n233\n\n234\n\n13 Public Key Cryptography\n\nimage under ρ. This is called a simple substitution cipher. For instance, suppose we use Table 13.1. Table 13.1 Encryption table for a simple substitution cipher original text A B C D E F G H I J K L M N O P Q R S T U V W X Y Z encrypted text R V C X N O A Y W B K U J E T D I L Q M Z F S P G H\n\nEncrypting HOWDY, we obtain YTSXG. To decrypt, we apply the inverse of ρ. To put this another way, we flip the rows of the table and, for the sake of convenience, sort by the encrypted letter rather than the original letter, as in Table 13.2. Table 13.2 Decryption table for the simple substitution cipher from Table 13.1 encrypted text A B C D E F G H I J K L M N O P Q R S T U V W X Y Z original text G J C P N V Y Z Q M K R T E F X S A W O L B I D H U\n\nThus, YTSXG decrypts to HOWDY. This is a vast improvement over the additive cipher in terms of security, because the number of possible keys is 26!. Even for a computer, that is a huge number of permutations to consider. It does come at the cost of having a larger key to exchange. Also, if a substantial amount of text is intercepted, this cipher is vulnerable to frequency analysis. That is, in English text, some letters occur much more frequently than others. For instance, E is by far the most common letter, T is the second most common, and so forth. An opponent could look for the most common letters in our text and make an educated guess that those represent T and E. If another moderately common letter occurs between them frequently, it might just be THE. Proceeding in this way, the code could be cracked. Can anything be done about this? There is always the one time pad. This is, in fact, an unbreakable cipher. It is also quite simple. The key is a string of random letters, at least as long as our message to be encrypted. We then assign a numerical value to each letter. We let A be 0, B be 1, and so on, letting Z be 25. To encrypt, we add the value of the first letter of our message to the value of the first letter of our key in Z26 . We then do the same with the second letter of our message and the second letter of our key, until we reach the end of our message. Each of our sums is then converted back to a letter. For instance, say we wanted to encrypt HOWDY, and our randomly selected key was NCVBT. Now, H is 7 and N is 13, so the sum is 20, which is U. Similarly, O and C are 14 and 2 respectively, giving a sum of 16, which is Q. Next, W is 22 and V is 21, giving a sum of 17, which is R. Now, D and B give 3 + 1 = 4, and hence E, and Y and T give 24 + 19 = 17, which is R. Thus, our encrypted text is UQRER. To decrypt, we subtract the value of the key letter from the corresponding encrypted\n\n13.1 Private Key Cryptography\n\n235\n\nletter value. For this message, 20 − 13 = 7, 16 − 2 = 14, 17 − 21 = 22, 4 − 1 = 3 and 17 − 19 = 24, and we obtain HOWDY. Assuming that the key is truly random, is only used once and is kept secret, an opponent who intercepts an encrypted message will be unable to determine anything more than the length of the message. The difficulty with this cipher is that the participants must have the ability to exchange a very large key secretly. In general, anyone who can do that may not need a code! For certain purposes, though, it is ideal. For example, if two people are able to meet once, exchange a briefcase full of random letters, and then leave for distant cities, they will be able to exchange messages while they are apart. In the internet age, the problem is that most encrypted messages are sent between two distant computers, and the computers can never meet secretly to exchange information. None of the schemes we have discussed are suitable. These are all private key methods. That is, the key used must be kept secret. Any opponent who discovered it could easily decrypt an intercepted message, since the ability to encrypt implies the ability to decrypt. Modern codes use public key schemes. The key can be released to an opponent without fear, because in these methods, it is quite possible to be able to encrypt and not be able to decrypt. The next section is devoted to a discussion of the first such scheme. Exercises Spaces and punctuation have been deleted from all messages to be encrypted or decrypted. 13.1. Encrypt the following message using a Caesar cipher: THETREASUREISBURIEDTWENTYPACESNORTHOFTHEPALMTREE 13.2. A message in English has been encrypted using an additive cipher. Decrypt it. BPMBQUMPIAKWUMBPMEITZCAAIQLBWBITSWNUIVGBPQVOA 13.3. Let us define a multiplicative cipher as follows. Assign the letters of the alphabet numerical values as usual (A is 0, B is 1, Z is 25), and choose a positive integer k as a key. Then if a letter with value v appears in the text, encrypt it as kv, with the multiplication taking place in Z26 . Which values of k will produce a valid cipher? 13.4. A message was encrypted using a multiplicative cipher, as in the preceding problem, with k = 7. Decrypt it. WUGCDEGCWERCHCZECRCVAWGANMAWWE FEGBUWWEHZCDXENQWHCJUPCHPCASJAWD 13.5. We establish a simple substitution cipher using the following table. original text A B C D E F G H I J K L M N O P Q R S T U V W X Y Z encrypted text V Y Z X E N A W R I O P C S B D F G H J K L M Q T U Encrypt the following message: TRANSFERTENMILLIONDOLLARSONTUESDAY\n\n236\n\n13 Public Key Cryptography\n\n13.6. Using the same simple substitution cipher as in the preceding problem, a message is encrypted. Decrypt it. YEJJEGJWGEEWBKGHJBBHBBSJWVSVCRSKJEJBBPVJE 13.7. Making use of the key RQPKDFOCMWODKFJDKSKDFVKQUYCHTISOXETX, encrypt the following message with a one time pad. THEDOCUMENTSAREHIDDENBEHINDTHERENOIR 13.8. Making use of the key ICOWLDIFNSXZIEEOWPAMWRUSDMFJEJFJBAWUQH, the following message was encrypted with a one time pad. Decrypt it. EJSJLQOWLULTVXXCBDUDSYYFYQWHEWLAZSSYQY\n\n13.2 The RSA Scheme The RSA Scheme is a public key cipher first described by Ronald L. Rivest, Adi Shamir and Leonard N. Adleman in 1977. In fact, an equivalent system was created by Clifford C. Cocks in 1973, but his work was classified and not made public for more than two decades. For convenience, let us say that Bob will be sending messages to June. In this case, it is June who creates the cipher. She selects two large distinct primes p and q, and lets n = pq. By Theorem 3.19, ϕ(n) = ( p − 1)(q − 1). June chooses a number e, with 1 < e < ϕ(n), such that (e, ϕ(n)) = 1. The public key consists of the numbers e and n. She sends these to Bob, without worrying about whether they are intercepted. She does not, however, tell anyone what p and q are! Bob prepares to send a message m which must be an integer with 0 ≤ m < n. (We will discuss how to convert text to this format shortly.) Bob calculates me ≡ a\n\n(mod n),\n\nwhere 0 ≤ a < n. He then sends the encrypted message a to June. How does June decrypt? The number e was chosen so that e ∈ U (ϕ(n)). Let d be the inverse of e in this group. To put that another way, de ≡ 1\n\n(mod ϕ(n)).\n\nBut now we have the following theorem. Theorem 13.1. Let p and q be distinct primes and n = pq. If k ≡ 1 (mod ϕ(n)), then for any integer b, we have bk ≡ b\n\n(mod n).\n\n13.2 The RSA Scheme\n\n237\n\nProof. First suppose that (b, n) = 1. Then b ∈ U (n). But by Theorem 3.17, |U (n)| = ϕ(n). Thus, by Corollary 3.5, bϕ(n) ≡ 1 (mod n). Now, ϕ(n)|(k − 1), and hence bk−1 ≡ 1 (mod n). Thus, bk ≡ b (mod n). Now, suppose that exactly one of { p, q} divides b. Without loss of generality, say p|b but q \u0002 b. Then b ∈ U (q). As |U (q)| = ϕ(q) = q − 1, we see that bq−1 ≡ 1 (mod q). Thus, b( p−1)(q−1) ≡ 1 (mod q) and hence, as above, bk ≡ b (mod q). Also, b ≡ bk ≡ 0 (mod p). Thus, p and q both divide bk − b. Since p and q are relatively prime, Corollary 2.3 tells us that n = pq|(bk − b), as required. \u0002 Finally, if both p and q divide b, then b ≡ bk ≡ 0 (mod n). Therefore, to decrypt, June calculates a d ≡ (m e )d ≡ m\n\n(mod n).\n\nThe scheme works because June is the only one who knows d. In order to calculate d, an opponent would need to find ϕ(n). But knowing that means being able to calculate p and q (see Exercise 13.10). And it is precisely upon the difficulty of this problem that the security of the system rests. To be sure, if Bob and June were foolish enough to use n = 143, an opponent would be able to find p and q instantly. But what if n had 300 digits? Factoring that into two primes of roughly 150 digits each is certainly beyond human abilities, and even for a computer, it is going to take a very long time. In theory, the cipher would be breakable. But any system that will take a fast computer a trillion years to crack is good enough for most purposes. How do we create our messages? Suppose that n has d digits. Then we will create a message m that is at most d − 1 digits long, so that m < n. If d − 1 is even, we letters; if it is odd, then the will do this by grouping our message into blocks of d−1 2 . Use the same values for letters introduced in the previous blocks will have size d−2 2 section, but make each letter have two digits. Thus, A is 00, B is 01, and Z is 25. Put the numbers from one block together to form a message m. (We have to use the two-digit method. If we dropped the leading zeroes, we would not know if 123 meant 1, 2 and 3 (BCD), 1 and 23 (BX) or 12 and 3 (MD).) If the length of our text message does not split evenly into blocks of the appropriate length, then pad out the last block with random letters. Example 13.1. June decides to create an RSA scheme using the primes p = 113 and q = 137. (Yes, these are much too small to produce a secure system. However, the author is far too lazy to perform calculations using 300-digit numbers, and these will suffice for an illustration.) Then n = pq = 15481 and ϕ(n) = ( p − 1)(q − 1) = 15232. How can June find a suitable e? A prime larger than both p and q will certainly be relatively prime to ϕ(n). June selects e = 151. She then sends the values of n and e to Bob. As n has five digits, Bob knows that he must break his message into two-letter blocks. Suppose he wishes to send the message HOWDY. As the length is not a multiple of 2, he pads it out by adding a Q to the end. Now, HO is 0714, WD is 2203 and YQ is 2416. To encrypt the first message, Bob calculates\n\n238\n\n13 Public Key Cryptography\n\n714151 ≡ 14628\n\n(mod 15481).\n\n2203151 ≡ 2494\n\n(mod 15481)\n\n2416151 ≡ 8498\n\n(mod 15481).\n\nNext, he calculates\n\nand\n\nBob sends June the messages 14628, 2494 and 8498. June must calculate d. Since (e, ϕ(n)) = 1, we know that there exist u, v ∈ Z such that eu + ϕ(n)v = 1. Then d will be u (modulo ϕ(n)). The Euclidean algorithm shows us how to calculate d, and we find that d = 807. Thus, June calculates 14628807 ≡ 714\n\n(mod 15481),\n\n2494807 ≡ 2203\n\n(mod 15481)\n\n8498807 ≡ 2416\n\n(mod 15481).\n\nand\n\nThe original message was, therefore, 071422032416, which converts to HOWDYQ. We should mention a couple of practical points. First, as any power of 0 is 0 and any power of 1 is 1, the messages 0 and 1 will not be encrypted using any RSA scheme. For that matter, since e will always be odd, n − 1 (which is −1 modulo n), will not change either. Given our method of encrypting English text, n − 1 will not arise, but 0 and 1 might. Can we do anything about it? Keep in mind that in the preceding example, we had n = 15481 but the possible messages would have been in the range of 0 to 2525. Would there be any harm in pushing them into the range of 2 to 2527? Surely not! Thus, we can agree to add 2 to every message before encrypting, and then subtract 2 after decrypting. (Do not do this in the exercises!) Another point worth mentioning is that e should be reasonably large. To see why, note that in the example above, we could have used e = 3. This would be a problem if we sent a relatively small message. For example, if we had m = 6, the encrypted message would be 63 = 216. No reduction modulo n takes place! An opponent who intercepted the message could simply take the cube root of 216 and recover the original message, without knowing anything about p and q. If e is large, we can ensure that this is avoided. While modern ciphers are more complex than the RSA scheme, their security invariably rests upon the fact that it is very difficult to factor large numbers. Exercises Spaces have been deleted from all messages to be encrypted or decrypted. Where a letter is needed to pad out a block, use Q.\n\n13.2 The RSA Scheme\n\n239\n\n13.9. Suppose that someone foolishly used numbers as small as n = 1961 and e = 43 to create an RSA scheme. Crack the code by determining d. 13.10. If n is a product of two distinct primes, and both n and ϕ(n) are known, show how to determine the two primes quickly. Illustrate the method using n = 10961, ϕ(n) = 10752. 13.11. Encrypt the message ALGEBRA using an RSA scheme with n = 17399 and e = 149. 13.12. Encrypt the message ABELIANGROUP using an RSA scheme with n = 18203 and e = 191. 13.13. Having set up an RSA scheme using p = 103, q = 179 and e = 151, we receive the following message: 2469, 7093, 14773, 10900, 143. Decrypt it. 13.14. Having set up an RSA scheme using p = 89, q = 167 and e = 181, we receive the following message: 13962, 8768, 7864, 4297, 12341. Decrypt it.\n\nChapter 14\n\nStraightedge and Compass Constructions\n\nWe now apply our knowledge of field extensions in order to answer three questions posed by the ancient Greeks.\n\n14.1 Three Ancient Problems More than 2000 years ago, the ancient Greeks performed many geometric constructions using a straightedge and compass. For our purposes, a straightedge is an infinitely long ruler having no markings on it. If we have constructed two points, then we can use the straightedge to construct the line passing through those points. Furthermore, if we have constructed two points A and B, then for any point C we have constructed (which may or may not be distinct from A and B), we can use the compass to draw a circle centred at C with radius equal to the distance between A and B. Next, we can take any two lines, any two circles, or one of each, that we have constructed, and construct their points of intersection. Then we repeat! The general question is, what can we construct in finitely many steps? Let us discuss a few simple examples that will be of use. Example 14.1. If we have constructed points A and B, let us construct a perpendicular bisector to the line segment AB. To this end, construct a circle centred at A with radius AB, and a circle centred at B with radius AB. Call the intersection points of these circles C and D. Then construct the line through C and D. It is a perpendicular bisector of AB, as illustrated in Figure 14.1. Example 14.2. Suppose that we have constructed points A and B, and the line passing through them. Let us say that we have constructed point C as well, although we do not insist that C ∈ / {A, B}. We claim that we can construct a line through C that is perpendicular to the line through A and B. Without loss of generality, we may assume that C and A are distinct points. Construct the circle centred at C with radius © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1_14\n\n241\n\n242\n\n14 Straightedge and Compass Constructions\n\nFig. 14.1 Construction of a perpendicular bisector of AB C\n\nA\n\nB\n\nD\n\nAC. If it intersects the line through A and B at a single point (which must necessarily be A), then the line through A and C will suffice, as illustrated in Figure 14.2. Otherwise, suppose that the circle meets the line at points A and D. Then the line we are looking for is the perpendicular bisector of AD, which the preceding example allows us to construct. See Figure 14.3. Example 14.3. Suppose that we have constructed three points A, B and C, that are not collinear. The three points must lie on a circle. Let us construct the centre of the circle, and hence the circle itself. Using Example 14.1, construct the perpendicular bisector of the chord AB. It must pass through the centre of the circle. Similarly, we can construct the perpendicular bisector of BC, and it too passes through the centre of the circle. Therefore, the point of intersection D of the two lines we have just constructed is the centre of the circle, and we can construct the circle itself, as it is centred at D and has radius AD. See Figure 14.4. For all the remarkable geometric constructions that were performed in antiquity, some problems could not be solved at the time. Question 14.1. (Squaring the Circle). Given an arbitrary circle, can we construct a square having the same area? As we shall see, if we are given a square, we can construct another square whose area is twice that of the first square. If we extend our constructions into three dimensions, we have the following. Question 14.2. (Doubling the Cube). Given an arbitrary cube, can we construct another cube having twice the volume of the first cube? If we are given three distinct points A, B and C, then we can construct a bisector of the angle ∠ABC. That is, we can construct a point D such that ∠D BC = 21 ∠ABC. See Exercise 14.5. This naturally led to the following question.\n\n14.1 Three Ancient Problems\n\n243\n\nFig. 14.2 Construction of a perpendicular to AB passing through C (first case)\n\nC\n\nA\n\nB\n\nFig. 14.3 Construction of a perpendicular to AB passing through C (second case)\n\nC\n\nA\n\nD\n\nB\n\nQuestion 14.3. (Trisecting the Angle). Given any three distinct points A, B and C, can we construct a point D such that ∠D BC = 13 ∠ABC? In fact, all three questions have a negative answer, but it was not until the nineteenth century that the tools of modern algebra allowed a proof to be given. Exercises 14.1. Suppose that we have points A and B, and the distance from A to B is 1. Construct points C and D such that the distance from C to D is 1.5. 14.2. Suppose that we have points A and B, and the distance √ from A to B is 1. Construct points C and D such that the distance from C to D is 2.\n\n244\n\n14 Straightedge and Compass Constructions\n\nFig. 14.4 Construction of the circle passing through A, B and C\n\nB C A\n\nD\n\n14.3. Given two points A and B, construct a point C such that ABC is an equilateral triangle. 14.4. Given two points A and B, construct points C, D and E all lying on the circle centred at A and passing through B, such that BC D E is a square. 14.5. Given three points A, B and C, construct a point D such that ∠D BC = 1 ∠ABC (where ∠ABC is assumed to be at most 180◦ ). 2 14.6. Given three points A, B and C, construct a point D such that ∠D BC = 2∠ABC (where ∠ABC is assumed to be at most 90◦ ). 14.7. Given two points A and B, construct points C and D on the circle centred at A and passing through B, such that BC D is an equilateral triangle. 14.8. Suppose we are given three points A, B and C, not collinear. Construct the inscribed circle for the triangle ABC; that is, construct a circle that lies inside the triangle but intersects each side at a single point.\n\n14.2 The Connection to Field Extensions In order to tackle these problems, we need to be able to discuss them in algebraic terms. Let us formalize our procedure. We will begin with two points. (Without those, we cannot construct any lines or circles, and so we get nowhere.) Let us identify these with the points (0, 0) and (1, 0) in the plane. We let P1 = {(0, 0), (1, 0)}. Then we proceed as follows. For every positive integer i, take all pairs of distinct points A and B in Pi , and draw the line through A and B. Also, for every pair of distinct points A and B in Pi , and for every point C in Pi (where C may or may not be in {A, B}), draw the circle centred at C with radius equal to the distance between A and B. Let Q i be the set of all lines and circles obtained in this way. Then let Pi+1 be the set of all points of intersection of any two distinct lines, any two distinct circles, or any line and any circle in Q i . We note that each Pi and Q i is a finite set, with Pi ⊆ Pi+1 and Q i ⊆ Q i+1 for all i.\n\n14.2 The Connection to Field Extensions\n\n245\n\nDefinition 14.1. A line or circle in the plane is constructible if it is in some Q i . A point in the plane is constructible if it is in some Pi . A real number r is constructible if the point (r, 0) is constructible. Let us start by proving what numbers we can construct, and then see what limits there are upon constructibility. Lemma 14.1. Let r ∈ R. Then the following are equivalent: 1. 2. 3. 4.\n\nr is constructible; −r is constructible; the point (0, r ) is constructible; and the point (0, −r ) is constructible.\n\nProof. If r = 0, there is nothing to do, so assume that r \u0006= 0. Suppose that (1) holds. Then let A = (0, 0) and B = (r, 0). We can construct the circle centred at A with radius AB and the line through A and B (namely, the x-axis). They intersect at C = (−r, 0), giving (2). As in Example 14.1, construct the perpendicular bisector of BC, which is the y-axis. The circle we constructed above intersects it at (0, r ) and (0, −r ), giving (3) and (4). By symmetry, (2) implies (1) as well. If we assume (3), then again, we can construct a circle centred at (0, 0) with radius |r |. As we are given (0, 0) and (1, 0), we can construct the x-axis, which intersects the circle at (r, 0), giving (1). By symmetry, (4) implies (1) as well. \u0003 Note from the proof that since (0, 0) and (1, 0) are constructible, so are the xand y-axes. Lemma 14.2. Let a, b ∈ R. Then the point (a, b) is constructible if and only if the numbers a and b are constructible. Proof. Suppose that (a, b) is constructible. As in Example 14.2, construct a line through (a, b) perpendicular to the x-axis. It intersects the x-axis at (a, 0), and so a is constructible. Then construct the line through (a, b) perpendicular to the y-axis. It intersects the y-axis at (0, b). Hence, by the preceding lemma, b is constructible. Conversely, let a and b be constructible. Then the points (a, 0) and (0, b) are constructible. Construct the line perpendicular to the x-axis through (a, 0). Similarly, construct the line perpendicular to the y-axis through (0, b). These two lines meet at (a, b). \u0003 Theorem 14.1. The constructible numbers form a subfield of R. Proof. By definition, 1 is constructible. Suppose that a and b are constructible. We would like to show that a + b and a − b are constructible. If b = 0, there is nothing to do. Otherwise, construct the circle centred at (a, 0), the radius of which is the distance from (0, 0) to (b, 0). It intersects the x-axis at the points (a + b, 0) and (a − b, 0). See Figure 14.5.\n\n246 Fig. 14.5 Construction of a + b and a − b\n\n14 Straightedge and Compass Constructions y\n\na–b\n\nb\n\na\n\na+b\n\nx\n\nIf b \u0006= 0, we also need to construct ab−1 . As it is sufficient to construct −ab−1 , we may assume that a ≥ 0 and b > 0. But in view of the preceding lemma, we can construct the points (−b, 0), (0, a) and (0, −1). As in Example 14.3, we can construct the circle passing through these points. Either geometrically or through algebraic manipulation (see Exercise 14.10), we can prove that this circle intersects the x-axis at (ab−1 , 0). See Figure 14.6. \u0003 Thus, ab−1 is constructible. Theorem 8.12 completes the proof. As Q is the prime subfield of R, we now know that every rational number is constructible. But we can say more. √ Theorem 14.2. If a is a positive constructible number, then so is a. Proof. As a and 1 are constructible, Lemma 14.1 tells us that we can construct the points A = (0, a) and B = (0, −1). By Example 14.1, we can construct the perpendicular bisector of AB and, hence, its midpoint C = (0, a−1 ). Now construct 2 the circle with centre C and radius AC. This circle intersects the x-axis at√ (d, 0) and a. Thus, (−d, 0) for some d > 0. Again using Exercise 14.10, we see that d = √ a is constructible. \u0003 Corollary 14.1. Suppose there exist fields Q = F0 ⊆ F1 ⊆ · · · ⊆ Fk , where each Fi+1 is a quadratic extension of Fi and Fk ⊆ R. Then every element of Fk is constructible. Proof. We noted above that every element of F0 is constructible. Thus, by induction, it suffices to show that if every element of a field F is constructible, and [K : F] = 2, then every element of K is constructible. Now, if a ∈ K , but a ∈ / F, then {1, a} is linearly independent over F and hence, in view of Theorem 12.6, a basis for K . In\n\n14.2 The Connection to Field Extensions\n\n247\n\nFig. 14.6 Construction of ab−1\n\ny\n\na\n\n–b\n\nab–1\n\nx\n\n–1\n\nparticular, K = F(a). By Theorem 12.9, a is algebraic over F and, in particular, Theorem 12.11 tells us that the minimal polynomial has degree 2. Say that it is √ −b± b2 −4c 2 . By the preceding theorem, x + bx + c, with b, c ∈ F. But then a = 2 √ b2 − 4c is constructible. But now Theorem 14.1 tells us that a is constructible, and hence so is every element of F(a) = K , as required. \u0003 \u0002 √ √ 4 Example 14.4. The number 3 + 2 + 5 15 is constructible.\u0003Let F0 = Q, F1 = \u0002 \u0002 √ √ √ √ F0 ( 3), F2 = F1 ( 15), F3 = F2 ( 2 + 5 15) and F4 = F3 ( 2 + 5√ 15). It is clear that each extension is of degree at most 2, since if a ∈ F, then\u0002either a ∈ F or √ √ 4 the minimal polynomial of a over F is x 2 − a. Furthermore, 3 + 2 + 5 15 ∈ F4 . Now, let us try to restrict the sorts of numbers that can be constructed. Lemma 14.3. Let F be a subfield of R. Suppose that we have two distinct points A and B such that the coordinates of both points lie in F. Then the line through A and B has an equation of the form ax + by = c, for some a, b, c ∈ F. If C is any point with coordinates in F, then the circle centred at C with radius equal to the distance between A and B has an equation of the form (x − d)2 + (y − e)2 = f , for some d, e, f ∈ F. Proof. Let us say that A = (a1 , a2 ), B = (b1 , b2 ) and C = (c1 , c2 ). Then the equation of the line is (b2 − a2 )x + (a1 − b1 )y = a1 b2 − a2 b1 , and we can see that the\n\n248\n\n14 Straightedge and Compass Constructions\n\ncoefficients are in F. Similarly, the equation of the circle is (x − c1 )2 + (y − c2 )2 = \u0003 (a1 − b1 )2 + (a2 − b2 )2 , which is of the correct form. Readers familiar with linear algebra will not find the next lemma surprising, as the solution to a system of linear equations can be found using only addition, subtraction, multiplication and division. Lemma 14.4. Let F be a subfield of R. Suppose that we have two lines, ax +by = c and d x + ey = f , where a, b, c, d, e, f ∈ F, and that the two lines intersect at a single point. Then that point has coordinates in F. Proof. If ae = bd, then the lines are parallel (or identical), which is \u0005not permit\u0004 ce−b f a f −cd , ae−bd , and these ted. Assume otherwise. Then the point of intersection is ae−bd coordinates lie in F. \u0003 Lemma 14.5. Let F be a subfield of R. Suppose that we have a line ax + by = c and a circle (x − d)2 + (y − e)2 = f , with a, b, c, d, e, f ∈ F. If the line and circle intersect, then there is a nonnegative number g ∈ F such that the coordinates of the √ intersection point(s) lie in F( g). Proof. As a and b cannot both be 0, without loss of generality, say a \u0006= 0. Then . Substituting into the equation of the circle, we obtain x = c−by a \u0006\n\nc − by −d a\n\n\u00072 + (y − e)2 = f.\n\nSimplifying, we obtain an equation of the form uy 2 +√vy + w = 0, for some 2 v 2 −4uw u, v, w ∈ F. Furthermore, u = ab2 +1 > 0. Then y = −v± 2u . If v 2 −4uw < 0, then the line and circle do not intersect, contradicting our assumption. Thus, let √ √ , we see that x ∈ F( g) as g = v 2 − 4uw ≥ 0. Then y ∈ F( g), and as x = c−by a well. \u0003 Lemma 14.6. Let F be a subfield of R. Suppose that we have two distinct circles (x − a)2 + (y − b)2 = c and (x − d)2 + (y − e)2 = f , with a, b, c, d, e, f ∈ F. If these circles intersect, then there is a nonnegative g ∈ F such that the coordinates √ of the intersection point(s) lie in F( g). Proof. Subtracting one equation from the other, we obtain (2a − 2d)x + (2b − 2e)y = f − c + a 2 + b2 − d 2 − e2 . This is the equation of a line unless 2a − 2d = 2b − 2e = 0. But in the latter case, the circles have the same centre, meaning that they are identical or do not intersect, so we may assume otherwise. Thus, we are now looking at the intersection of a circle and a line (with coefficients in F), and Lemma 14.5 applies. \u0003 Time to put it all together!\n\n14.2 The Connection to Field Extensions\n\n249\n\nTheorem 14.3. A real number a is constructible if and only if there exist subfields Fi of R such that Q = F0 ⊆ F1 ⊆ · · · ⊆ Fk , where each Fi+1 is a quadratic extension of Fi and a ∈ Fk . Proof. One direction of the theorem is given by Corollary 14.1. Let us prove the other. Suppose that a is constructible. Referring to the sets Pi and Q i from the definition of constructibility, let K i be the intersection of all subfields of R containing all of the coordinates of the points in Pi . Then each K i+1 is an extension field of K i . We claim that for each i, there exist fields F j with Q = F0 ⊆ F1 ⊆ · · · ⊆ Fm = K i , where each F j+1 is a quadratic extension of F j . Our proof is by induction on i. If i = 1, then Pi = {(0, 0), (1, 0)} and hence K i = Q and there is nothing to do. Assume that our claim holds for i. Then we have Q = F0 ⊆ F1 ⊆ · · · ⊆ Fm = K i , where each F j+1 is a quadratic extension of F j . Now, by Lemma 14.3, every line and circle in Q i has coefficients in K i . Furthermore, Lemmas 14.4, 14.5 and 14.6 tell us that every possible intersection point of two distinct lines,√ two distinct circles or a line and a circle in Pi has coordinates either in K i or in K i ( b), for some nonnegative being b ∈ K i . But Pi is a finite set, so there are only finitely many values of b √ , . . . , b . For each k, 1 ≤ k ≤ n, let F = F ( bk ). used. Let them be b 1 n m+k m+k−1 √ If bk ∈ Fm+k−1 , then Fm+k = Fm+k−1 , and we can discard Fm+k . Otherwise, x 2 − bk is irreducible over Fm+k−1 , and so Fm+k is a quadratic extension of Fm+k−1 , by Theorem 12.11. But the last of the Fm+k is, by definition, K i+1 , establishing the claim. Since a is constructible, (a, 0) ∈ Pi , for some i, hence a ∈ K i and the proof is complete. \u0003 Corollary 14.2. Let a ∈ R. If a is constructible, then a is algebraic over Q and, in fact, its minimal polynomial over Q has degree 2m for some nonnegative integer m. Proof. Using Fi as in the statement of the preceding theorem, we have a ∈ Fk , and by Theorem 12.8, [Fk : Q] = [Fk : Fk−1 ][Fk−1 : Fk−2 ] · · · [F1 : Q] = 2k . k\n\nThus, by Lemma 12.1, the numbers 1, a, a 2 , . . . , a 2 are linearly dependent over Q. That is, a satisfies a nonzero polynomial over Q. By Theorem 12.11, [Q(a) : Q] < ∞ and therefore, by Theorem 12.8, 2k = [Fk : Q] = [Fk : Q(a)][Q(a) : Q]. That is, [Q(a) : Q] divides 2k , and so it is 2m , for some m. By Theorem 12.11, the \u0003 degree of the minimal polynomial of a over Q is 2m . Please note that while the condition given in Theorem 14.3 is necessary and sufficient for a to be constructible, the condition given in the corollary is not. It is possible to find a real number a whose minimal polynomial over Q has degree 4, but such that a is not constructible.\n\n250\n\n14 Straightedge and Compass Constructions\n\nExercises 14.9. Let a and b be nonzero real numbers. If a is constructible and b is not, show that neither a + b nor ab is constructible. If c is not constructible, show by example that b + c may or may not be constructible. 14.10. Let a, b, c and d be positive real numbers and suppose that a circle in the plane passes through the points (a, 0), (−b, 0), (0, c) and (0, −d). Show that ab = cd. 14.11. Are the following numbers constructible? \u0002 √ √ 4 1. √2 + √5 − 3 3 2. 3 + 3 √ √ 14.12. Is 3 2 + 3 4 constructible? 14.13. Let a be a real root of the polynomial x 6 − 15x 5 + 27x 4 − 12x 3 + 30x 2 − 21x + 87. Is a constructible? 14.14. Let a be a real root of the polynomial x 6 −6x 4 +12x 2 −8. Is a constructible?\n\n14.3 Proof of the Impossibility of the Problems We now have the machinery necessary to answer the three questions from Section 14.1. First, let us look at squaring the circle. We may as well assume that our two initial points are the centre of the circle, (0, 0), and a point on the circle, (1, 0). Thus, we can construct the unit circle immediately. Its area is π . If we √ were to construct a square with area π , we would need to construct an edge of length π . The following theorem tells us that we cannot. √ Theorem 14.4. The number π is not constructible. √ Proof. If π were constructible, then by Theorem 14.1, π would be constructible as well. But as we mentioned in Section 12.3, π is transcendental over Q. This contradicts Corollary 14.2. \u0003 As we discussed in Section 14.1, doubling a square is possible. Indeed, if we √ can construct a side with length s, then by Theorems 14.1 and 14.2, the number s 2 is also constructible, and this will be the side length of a square with twice the area. To deal with the problem of doubling the cube, without worrying about going into the third dimension, we may simply suppose that one edge extends between our two initial points, and thus has length 1. This would lead to a cube with √volume 1. To obtain a cube with volume 2, we would need an edge with length 3 2. This is not going to happen. √ Theorem 14.5. The number 3 2 is not constructible.\n\n14.3 Proof of the Impossibility of the Problems\n\nProof. By Example 12.27, the √ minimal polynomial of Corollary 14.2 tells us that 3 2 is not constructible.\n\n251\n\n√ 3 2 over Q is x 3 − 2. But then \u0003\n\nFinally, what about trisecting an angle? Some angles can be trisected (see Exercise 14.15). But not all. Indeed, we will show that an angle of 60◦ (or π3 , as we will be doing a bit of trigonometry) cannot be trisected. In view of Theorems 14.1 and 14.2, √ 1 3 the numbers 0, 1, 2 and 2 are all constructible. Thus, by Lemma 14.2, we can \u0004 √ \u0005 construct the points A = 21 , 23 , B = (0, 0) and C = (1, 0). Then ∠ABC = π3 . Thus, we do not need to assume anything extra to obtain the angle. If we could find a point D such that ∠D BC = π9 , then we could draw the line through D and B, and point would be \b with the unit circle centred at B. An intersection \b \b then \b intersect cos π9 , sin π9 . By Lemma 14.2, this would require cos π9 to be constructible. However, the following theorem dashes any hopes of that. \b Theorem 14.6. The number cos π9 is not constructible. Proof. For any θ ∈ R, note that cos(3θ ) = cos(2θ ) cos(θ ) − sin(2θ ) sin(θ ) = (cos2 (θ ) − sin2 (θ )) cos(θ ) − 2 sin2 (θ ) cos(θ ) = cos3 (θ ) − 3 sin2 (θ ) cos(θ ) = cos3 (θ ) − 3(1 − cos2 (θ )) cos(θ ) = 4 cos3 (θ ) − 3 cos(θ ). Let θ =\n\nπ . 9\n\nThen cos(3θ ) = 21 , and so 1 = 4 cos3 (θ ) − 3 cos(θ ). 2\n\n\b That is, cos π9 satisfies the polynomial 8x 3 − 6x − 1. If this polynomial were reducible over Q, then by Corollary 11.2, it would have a rational root. By Theorem 11.3, the only possible roots are ±1, ± 12 , ± 14 , ± 18 . But none of these work. Thus, \b 8x 3 − 6x − 1 is irreducible over Q. Therefore, the minimal polynomial of cos π9 over Q is x 3 − 43 x − 18 . Corollary 14.2 completes the proof. \u0003 Exercises All angles are expressed in radians. 14.15. Show that it is possible to trisect a right angle using straightedge and compass. 14.16. Show that the angles π/6 and 2π/3 cannot be trisected using straightedge and compass.\n\n252\n\n14 Straightedge and Compass Constructions\n\n14.17. In the next two problems, we will show that it is impossible to construct an angle of θ = 2π/7. If it were possible, then as we are given the points (0, 0) and (1, 0), we would also be able to construct (cos(θ ), sin(θ )). In particular, the number cos(θ ) would be constructible. Let us show that it is not. To this end, for each n, 1 ≤ n ≤ 6, express cos(nθ ) as a linear combination over Q of cosk (θ ), 0 ≤ k ≤ 3. 14.18. Let θ = 2π/7. 1. Show that cos(θ ) + sin(θ )i is a complex root of 1 + x + x 2 + x 3 + x 4 + x 5 + x 6 . 2. Show that 1 + cos(θ ) + cos(2θ ) + cos(3θ ) + cos(4θ ) + cos(5θ ) + cos(6θ ) = 0. 3. Use the answer to the preceding exercise to show that cos(θ ) is a root of 8x 3 + 4x 2 − 4x − 1. 4. Conclude that cos(θ ) is not constructible. 14.19. Suppose we are given the vertices A = (0, 0), B = (1, 0) and C = (c1 , c2 ) of a triangle. Show that we can construct points D, E and F such that the triangles ABC and D E F are similar, but D E F has twice the area of ABC. 14.20. Suppose that we are forced to perform our constructions using a straightedge and a collapsing compass. That is, any time we lift the compass, it collapses. In particular, we cannot directly use it to construct a circle centred at A with radius equal to the distance between B and C. All we can do is take two points A and B that we have constructed, and construct a circle centred at A and passing through B. Show that this does not change the set of constructible numbers in any way. That is, show that any number that was constructible before is still constructible using a straightedge and collapsing compass.\n\nAppendix A\n\nThe Complex Numbers\n\nThe complex numbers are an extension of the real numbers. Definition A.1. A complex number is a formal expression a + bi, with a, b ∈ R. The set of all complex numbers is denoted C. We define addition and multiplication on C via (a + bi) + (c + di) = (a + c) + (b + d)i and (a + bi)(c + di) = (ac − bd) + (ad + bc)i for all a, b, c, d ∈ R. Example A.1. Observe that (2+3i)+(5−9i) = 7−6i and (2+3i)(5−9i) = 37−3i. We identify the real number a with the complex number a+0i. A complex number 0 + bi, with b ∈ R, is said to be purely imaginary. We simply write bi for such a number. In particular, note that i 2 = −1. Also, if u = a + bi, write −u = −a − bi. Let us summarize a few properties concerning complex addition. Theorem A.1. Let u, v, w ∈ C. Then 1. 2. 3. 4. 5.\n\nu + v ∈ C; u + v = v + u; (u + v) + w = u + (v + w); u + 0 = u; and u + (−u) = 0.\n\nProof. The calculations are all straightforward. For instance, to show (2), we note that (a + bi) + (c + di) = (a + c) + (b + d)i = (c + a) + (d + b)i = (c + di) + (a + bi). The remaining parts are left to the reader. © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1\n\n\u0002 253\n\n254\n\nAppendix A: The Complex Numbers\n\nSimilarly, we can list some properties of complex multiplication. Theorem A.2. Let u, v, w ∈ C. Then 1. 2. 3. 4. 5. 6.\n\nuv ∈ C; uv = vu; (uv)w = u(vw); u(v + w) = uv + uw; 1u = u; and if u \u0003= 0, then there exists a z ∈ C such that uz = 1.\n\nProof. Again, all of the calculations in (2) through (5) are straightforward. For instance, to prove (3), let u = a + bi, v = c + di and w = e + f i, with a, b, c, d, e, f ∈ R. Then (uv)w = ((ac − bd) + (ad + bc)i)(e + f i) = (ace − bde − ad f − bc f ) + (ac f − bd f + ade + bce)i = (a + bi)((ce − d f ) + (c f + de)i) = u(vw). (6) If u = a + bi, then let z =\n\na a 2 +b2\n\nb i. a 2 +b2\n\n\u0002\n\n(Readers who have finished Chapter 3 will realize that Theorem A.1 shows that C is an abelian group under addition. Those who have completed Chapter 8 will understand that the two theorems combined show that C is a field.) Let us discuss a simple example of a way in which the complex numbers differ from the real numbers. Definition A.2. If z ∈ C and n is a positive integer, then we say that z is a primitive nth root of unity if z n = 1 but z m \u0003= 1 for any positive integer m < n. In R, the only roots of unity are 1 and −1. But we see immediately that in C, we have a primitive fourth root of unity, i. We can say more, however. We will need this well-known theorem due to Abraham de Moivre. Theorem A.3 (De Moivre’s Theorem). Let θ ∈ R. Then (cos(θ ) + sin(θ )i)n = cos(nθ ) + sin(nθ )i, for any positive integer n. Proof. We proceed by induction on n. If n = 1, there is nothing to do. Assume that the theorem holds for n. Then (cos(θ ) + sin(θ )i)n+1 = (cos(nθ ) + sin(nθ )i)(cos(θ ) + sin(θ )i) = (cos(nθ ) cos(θ ) − sin(nθ ) sin(θ )) + (cos(nθ ) sin(θ ) + sin(nθ ) cos(θ ))i = cos((n + 1)θ ) + sin((n + 1)θ )i,\n\nAppendix A: The Complex Numbers\n\n255\n\n\u0002\n\nas required. Corollary A.1. Let n be a positive integer. Then cos nth root of unity in C.\n\n\u0002 2π \u0003 n\n\n+ sin\n\n\u0002 2π \u0003 n\n\ni is a primitive\n\nProof. By Theorem A.3, \u0004 \u0004 \u0005 \u0004 \u0004 \u0005 \u0005m \u0005 \u0004 \u0005 2π 2mπ 2π 2mπ cos = cos + sin i + sin i, n n n n for any positive integer m. If m = n, then we obtain cos(2π )\u0002+ sin(2π )i = 1. On \u0003 \u0003 = 1. \u0002 the other hand, if 1 ≤ m < n, then 0 < mn < 1, and hence cos 2mπ n Example A.2. Letting n = 3, we obtain a primitive cube root of unity, namely, √ −1 3 + i. 2 2\n\nAppendix B\n\nMatrix Algebra\n\nLet us discuss a few definitions and basic properties of matrices. The entries in the matrices will come from rings and fields. Readers who are not yet familiar with these terms can simply assume that the entries are real numbers. Definition B.1. Let R be a ring and m and n positive integers. Then an m ×n matrix over R is an array of elements of R with m rows and n columns. If our matrix is A, then we write ai j for the (i, j)-entry of A; that is, ⎛\n\na11 a12 ⎜ a21 a22 ⎜ A=⎜ . . ⎝ .. .. am1 am2 Example B.1. If we let A=\n\n··· ··· .. .\n\n⎞ a1n a2n ⎟ ⎟ .. ⎟ . . ⎠\n\n· · · amn\n\n\u0004 \u0005 467 , 385\n\nthen A is a 2 × 3 matrix over R. Furthermore, a12 = 6 and a21 = 3. Definition B.2. Let A and B be m × n matrices over a ring R. Then their sum A + B is the m × n matrix C such that ci j = ai j + bi j for all i and j. Example B.2. Working with 2 × 2 matrices over R, we have \u0004\n\n\u0005 \u0004 \u0005 \u0004 \u0005 36 46 7 12 + = . 25 12 3 7\n\nFor any m ×n matrix A, we also let −A be the m ×n matrix B such that bi j = −ai j for all i and j. Furthermore, the m × n zero matrix has every entry 0. We denote this matrix by 0. Let us list a few properties of matrix addition. © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1\n\n257\n\n258\n\nAppendix B: Matrix Algebra\n\nTheorem B.1. Let R be a ring, and m and n positive integers. If A, B and C are m × n matrices over R, then 1. 2. 3. 4. 5.\n\nA + B is an m × n matrix over R; A + B = B + A; (A + B) + C = A + (B + C); A + 0 = A; and A + (−A) = 0.\n\nProof. The first part is contained in the definition. The other parts are all obtained by calculating the (i, j)-entry of each side. For instance, to prove (3), we note that the (i, j)-entry of (A+ B)+C is (ai j +bi j )+ci j , whereas the (i, j)-entry of A+(B +C) is ai j + (bi j + ci j ), and these are equal. The rest of the proof is left to the reader. \u0002 Anyone who has read Chapter 3 will note that Theorem B.1 implies that the m × n matrices over a ring form an abelian group under addition. Definition B.3. Let A be an m × n matrix over a ring R. If r ∈ R, then the scalar multiple r A is the m × n matrix B such that bi j = rai j for all i and j. Example B.3. Working with 3 × 2 matrices over R, we have ⎛\n\n⎞ ⎛ ⎞ 1 3 5 15 5 ⎝12 6⎠ = ⎝60 30⎠ . 1 2 5 10 Here are a few properties of scalar multiplication. Theorem B.2. Let R be a ring and m, n ∈ N. If A and B are m × n matrices over R and r, s ∈ R, then 1. 2. 3. 4.\n\nr A is an m × n matrix over R; r (A + B) = r A + r B; (r + s)A = r A + s A; and r (s A) = (r s)A.\n\nProof. (1) is clear from the definition. Each of the other parts is proved by calculating the (i, j)-entry of both sides of the equation. For instance, the (i, j)-entry of r (A+ B) is r (ai j + bi j ), whereas that of r A + r B is rai j + r bi j , but these are the same, establishing (2). The rest of the proof is left to the reader. \u0002 If F is a field, then Theorems B.1 and B.2, when combined with the obvious fact that 1A = A, show us that the m × n matrices over F form a vector space over F, as discussed in Chapter 12. Matrix multiplication is a bit different.\n\nAppendix B: Matrix Algebra\n\n259\n\nDefinition B.4. Let R be a ring, and let A be a k × m matrix over R, and B an m × n matrix over R. Then the product AB is the k × n matrix C such that ci j = ai1 b1 j + ai2 b2 j + · · · + aim bm j , for all i and j. Example B.4. Let\n\n\u0004\n\n132 A= 203 be matrices over R. Then\n\n\u0005\n\n⎞ 3415 and B = ⎝2 1 2 0⎠ 1261\n\n\u0004 \u0005 11 11 19 7 AB = . 9 14 20 13\n\nIf R is a ring with identity, then we also have the n × n identity matrix In , which is the n × n matrix A such that aii = 1 for all i and ai j = 0 if i \u0003= j. For instance, ⎛\n\n⎞ 100 I 3 = ⎝0 1 0 ⎠ . 001 For any positive integer n, write Mn (R) for the set of n × n matrices over a ring R. Theorem B.3. Let R be a ring, n a positive integer, and A, B, C ∈ Mn (R). Then 1. 2. 3. 4. 5.\n\nAB ∈ Mn (R); (A + B)C = AC + BC; A(B + C) = AB + AC; (AB)C = A(BC); and if R is a ring with identity, then In A = AIn = A.\n\nProof. (1) follows from the definition. (2) The (i, j)-entry of (A + B)C is (ai1 + bi1 )c1 j + (ai2 + bi2 )c2 j + · · · + (ain + bin )cn j , whereas the (i, j)-entry of AB + AC is (ai1 c1 j + ai2 c2 j + · · · + ain cn j ) + (bi1 c1 j + bi2 c2 j + · · · + bin cn j ), and these are equal. (3) is similar to (2). (4) Through repeated applications of (2) and (3), we can reduce (4) to the case where each of A, B and C has at most one nonzero entry. But then it is trivial.\n\n260\n\nAppendix B: Matrix Algebra\n\n(5) Let D = In . Then the (i, j)-entry of D A is di1 a1 j + di2 a2 j + · · · + din an j = ai j . Thus, In A = A. The proof that AIn = A is similar.\n\n\u0002\n\nAs discussed in Chapter 8, we have now proved that if R is a ring, then so is Mn (R), for any positive integer n. Furthermore, if R is a ring with identity, then so is Mn (R). It is, however, worth mentioning, that Mn (R) need not be commutative, even if R is. For instance, in M2 (R), \u0004\n\n11 01\n\n\u0005\u0004 \u0005 \u0004 \u0005 \u0004 \u0005 \u0004 \u0005\u0004 \u0005 10 21 11 10 11 = \u0003= = . 11 11 12 11 01\n\nDefinition B.5. Let F be a field and n a positive integer. Then a matrix A ∈ Mn (F) is said to be invertible if there exists a B ∈ Mn (F) such that AB = B A = In . In this case, we call B the inverse of A and write B = A−1 . Example B.5. In M2 (R), the matrix \u0004 \u0005 32 A= 75 is invertible, as A−1 =\n\n\u0004\n\n\u0005 5 −2 . −7 3\n\nIn most linear algebra courses, a couple of different methods of finding the inverse of a matrix are presented (often just in Mn (R), but the same methods work in Mn (F), for any field F). There is, however, a shortcut for determining if a matrix is invertible. Definition B.6. Let F be a field and n a positive integer. If A ∈ Mn (F), then the determinant of A, det(A), is an element of F defined recursively as follows. If n = 1, then det((a11 )) = a11 . If n > 1, then for any 1 ≤ i, j ≤ n, let Ai j ∈ Mn−1 (F) be the matrix obtained by discarding row i and column j of A. Then det(A) = a11 det(A11 ) − a12 det(A12 ) + a13 det(A13 ) − · · · + (−1)n+1 a1n det(A1n ). Example B.6. In M2 (F), we have det\n\n\u0005\u0005 \u0004\u0004 a11 a12 = a11 a22 − a12 a21 . a21 a22\n\nExample B.7. In M3 (R), let\n\nAppendix B: Matrix Algebra\n\n261\n\n⎞ 253 A = ⎝ 1 4 6⎠ . 897 Then det(A) = 2 det\n\n\u0004\u0004 \u0005\u0005 \u0004\u0004 \u0005\u0005 \u0004\u0004 \u0005\u0005 46 16 14 − 5 det + 3 det 97 87 89\n\n= 2(−26) − 5(−41) + 3(−23) = 84. We conclude with the following result. Theorem B.4. Let F be a field and n a positive integer. If A, B ∈ Mn (F), then 1. det(AB) = det(A) det(B); and 2. A is invertible if and only if det(A) \u0003= 0. Proof. We will prove the n = 2 case. The general case can be found in standard introductory linear algebra textbooks. (1) Observe that \u0004 \u0005 a11 b11 + a12 b21 a11 b12 + a12 b22 AB = . a21 b11 + a22 b21 a21 b12 + a22 b22 Thus, det(AB) = (a11 b11 + a12 b21 )(a21 b12 + a22 b22 ) − (a11 b12 + a12 b22 )(a21 b11 + a22 b21 ). On the other hand det(A) det(B) = (a11 a22 − a12 a21 )(b11 b22 − b12 b21 ), and these are equal. (2) If det(A) \u0003= 0, then let B = (det(A))−1\n\n\u0004\n\n\u0005 a22 −a12 . −a21 a11\n\nIt is easy to verify that AB = B A = I2 ; thus, B = A−1 . Suppose, on the other hand, that det(A) = 0. If AB = I2 , then by (1), det(A) det(B) = det(I2 ) = 1, which is impossible. \u0002\n\nSolutions\n\nSolutions to the odd-numbered problems.\n\nProblems of Chapter 1 1.1 S ∩ T = {3}, S ∪ T = {1, 2, 3, 4}, S\\T = {1, 2}, T \\S = {4} and S × T = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}. 1.3 Let a ∈ R ∪ T . Then a ∈ R or a ∈ T . If a ∈ R, then as R ⊆ S, we have a ∈ S, and hence a ∈ S ∪ T . If a ∈ T , then a ∈ S ∪ T . 1.5 Take a ∈ R ∪ (S ∩ T ). Then a ∈ R or a ∈ S ∩ T . If a ∈ R, then a ∈ R ∪ S and a ∈ R ∪ T , so a ∈ (R ∪ S) ∩ (R ∪ T ). If a ∈ S ∩ T , then a ∈ S and a ∈ T . Therefore, a ∈ R ∪ S and a ∈ R ∪ T , so a ∈ (R ∪ S) ∩ (R ∪ T ). Thus, R ∪ (S ∩ T ) ⊆ (R ∪ S) ∩ (R ∪ T ). Conversely, suppose that a ∈ (R ∪ S) ∩ (R ∪ T ). If a ∈ R, then a ∈ R ∪ (S ∩ T ). If a ∈ / R, then as a ∈ R ∪ S, we must have a ∈ S and, similarly, a ∈ T . Thus, a ∈ S ∩ T , and hence a ∈ R ∪ (S ∩ T ). That is, (R ∪ S) ∩ (R ∪ T ) ⊆ R ∪ (S ∩ T ). 1.7 (2, 3), (2, 4), (2, 5), (3, 8). 1.9 Reflexive? Yes. If a ∈ R, then a − a = 0 ∈ Q, so aρa. Symmetric? Yes. If aρb, then a − b ∈ Q, so b − a = −(a − b) ∈ Q, and hence bρa. Transitive? Yes. If aρb and bρc, then a − b, b − c ∈ Q. But then a − c = (a − b) + (b − c) ∈ Q, so aρc. 1.11 (1) A relation is a subset of {1, 2, 3} × {1, 2, 3}. This Cartesian product has 9 elements, and therefore 29 = 512 subsets. (See Exercise 1.4.) (2) A relation ρ is symmetric provided 1ρ2 if and only if 2ρ1, 1ρ3 if and only if 3ρ1 and 2ρ3 if and only if 3ρ2. In short, we do not get to decide if 2ρ1, 3ρ1 or 3ρ2, once all the other possibilities are decided. Thus, only 6 of the 9 possible pairs remain to be determined, so the total number is 26 = 64. © Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1\n\n263\n\n264\n\nSolutions\n\n1.13 Reflexivity: As a − a = 3 · 0, we have a ∼ a for all a ∈ N. Symmetry: If a ∼ b, then a − b = 3k, and hence b − a = 3(−k); thus, b ∼ a. Transitivity: Suppose a ∼ b and b ∼ c. Then a − b = 3k, b − c = 3l, for some k, l ∈ Z. Thus, a − c = (a − b) + (b − c) = 3(k + l); that is, a ∼ c. It is an equivalence relation. As for the classes, = {1, 4, 7, . . .}, = {2, 5, 8, . . .} and = {3, 6, 9, . . .}. 1.15 Reflexivity: As |a| = |a|, we have a ∼ a for all a ∈ Z. Symmetry: If a ∼ b, then |a| = |b|. Therefore, |b| = |a| and hence b ∼ a. Transitivity: If a ∼ b and b ∼ c then |a| = |b| = |c|, and hence a ∼ c. It is an equivalence relation. The classes are = {0}, = {1, −1}, = {2, −2}, and so on. 1.17 Note that {1} is a subset of {1, 2}, but {1, 2} is not a subset of {1}. Therefore, ∼ is not symmetric, and hence not an equivalence relation. 1.19 Reflexivity: If (a, b) ∈ R2 , then 3a−b = 3a−b, so (a, b) ∼ (a, b). Symmetry: If (a, b) ∼ (c, d), then 3a−b = 3c−d, so 3c−d = 3a−b and hence (c, d) ∼ (a, b). Transitivity: If (a, b) ∼ (c, d) and (c, d) ∼ (e, f ), then 3a − b = 3c − d = 3e − f , and hence (a, b) ∼ (e, f ). Also, (a, b) ∈ [(4, 2)] if and only if 3a−b = 3·4−2 = 10; that is, if and only if b = 3a − 10. Thus, [(4, 2)] = {(a, 3a − 10) : a ∈ R}. 1.21 Let a ∼ b if and only if either both or neither of a and b lie in {1, 2, 3}. Reflexivity and symmetry are clear. Suppose a ∼ b and b ∼ c. If a ∈ {1, 2, 3}, then b ∈ {1, 2, 3} and hence c ∈ {1, 2, 3}. Similarly if a ∈ / {1, 2, 3}. Thus, ∼ is an equivalence relation. The classes are = {1, 2, 3} and = {4, 5, 6, . . .}. 1.23 If α(a) = α(b), then 2a − 1 = 2b − 1, and hence a = b. Thus, α is one-to-one. But there is no a ∈ {1, 2, 3, 4} such that α(a) = 2, so α is not onto. 1.25 If α(a) = α(b), then 23a−5 = 23b−5 . Taking the base 2 logarithm, we have 3a −5 = 3b−5, and hence a = b. Thus, α is one-to-one. If c ∈ T , then α(((log2 c)+ 5)/3) = c. Therefore, α is onto as well. In fact, we will use β : T → S given by β(c) = ((log2 c) + 5)/3. We have β(α(a)) = β(23a−5 ) = (log2 (23a−5 ) + 5)/3 = a, for all a ∈ S. 1.27 (1) and (3) are binary operations, as ab ∈ N for all a, b ∈ N, and 3 ∈ N. But (2) is not, as 1 ∗ 2 = −1 ∈ / N. 1.29 Surely β is onto. If t ∈ T , then there exists an r ∈ R such that (βα)(r ) = t. But then β(α(r )) = t. However, α need not be. To see this, let R and S be the set of real numbers and let T be the set of nonnegative real numbers. Let α(r ) = r 2 and is not onto,√as there is no r ∈ R such that α(r ) = −1. However, β(s) = s 2 . Then α √ if t ∈ T , then β(α( 4 t)) = β( t) = t; thus, βα is onto. 1.31 (1) For each of the m elements a of S, there are n possible choices for α(a), so n m . (2) If n < m, the answer is 0, as the m elements of S need to map to m different places. Suppose n ≥ m and let S = {a1 , . . . , am }. Then there are n choices for α(a1 ), leaving n − 1 choices for α(a2 ), and so on. The answer is n(n − 1) · · · (n − m + 1).\n\nSolutions\n\n265\n\nProblems of Chapter 2 2.1 Apply induction. When n = 1, both sides are 1. Assume the result for n, then prove it for n + 1: 1 + · · · + n + (n + 1) = n(n + 1)/2 + (n + 1) = (n + 1)(n + 2)/2 = (n + 1)((n + 1) + 1)/2, as required. 2.3 (1) This is the Binomial Theorem with a = b = 1. (2) This is the Binomial Theorem with a = 1, b = −1. 2.5 (1) By induction on n. We have nothing to prove for n = 1, so we begin with n = 2. Here, (1 + a)2 = 1 + 2a + a 2 > 1 + 2a, as a is positive. Assume the result for n, and prove it for n + 1. But (1 + a)n+1 = (1 + a)n (1 + a) > (1 + na)(1 + a) = 1 + (n + 1)a + na 2 > 1 + (n + 1)a, as a > 0. (2) Apply (1) with a = (n − 1)/n, then take nth roots. 2.7 By strong induction on n. If n = 1 or 2, the result is obvious, so assume that n > 2 and the result is true for smaller n. Then f n = f n−1 + f n−2 ≤ (7/4)n−2 + (7/4)n−3 = (7/4)n−3 (7/4 + 1) < (7/4)n−1 , since 11/4 < (7/4)2 . 2.9 By strong induction on the area a = r c of the bar. If the area is 1, then r = c = 1 and no actions are necessary. Suppose the area is a > 1 and the result is true for bars of smaller area. Then a break turns the bar into two bars with areas b and c, both less than a. By our inductive hypothesis, it will take b − 1 and c − 1 actions, respectively, to break down these two bars. We have already used 1 action, so the total is 1 + (b − 1) + (c − 1) = (b + c) − 1 = a − 1, as required. Alternative solution: We must turn 1 bar into r c bars. Each action adds one bar. So we need r c − 1 actions. 2.11 (1) 57 = 20(2) + 17; 20 = 17(1) + 3; 17 = 3(5) + 2; 3 = 2(1) + 1; 2 = 1(2) + 0. Thus, (57, 20) = 1. (2) 117 = 51(2) + 15; 51 = 15(3) + 6; 15 = 6(2) + 3; 6 = 3(2) + 0. Thus, (117, 51) = 3. 2.13 Let us write b = ac and a = bd, with c, d ∈ Z. Then a = acd; that is, a(1 − cd) = 0. If a = 0, then as b = ac, we have b = 0 as well. Otherwise, 1 − cd = 0, so cd = 1. Thus, d ∈ {1, −1}, so a ∈ {b, −b}. 2.15 Let d = (b, c). Then d|c and c|a, so d|a. But also d|b. As (a, b) = 1, we must have d = 1. 2.17 If a and n are relatively prime, write au + nv = 1. Then n(−v) = au − 1. On the other hand, if (a, n) = d > 1 and au − 1 = nb, for some b ∈ Z, then 1 = au − nb. Now, d|a and d|n, so d|1, which is impossible. 2.19 By strong induction on n. It is clear if n ≤ 4. So let n > 4 and suppose that it is true for smaller n. Then f n = f n−1 + f n−2 = ( f n−2 + f n−3 ) + f n−2 = 2 f n−2 + f n−3 = 2( f n−3 + f n−4 ) + f n−3 = 3 f n−3 + 2 f n−4 . If 4|n, then 4|(n − 4), so 3| f n−4 , and hence 3| f n . Suppose that 4 \u0002 n. Then 4 \u0002 (n − 4), so 3 \u0002 f n−4 . If 3| f n , then 3|( f n − 3 f n−3 ) = 2 f n−4 . As (3, 2) = 1, we see that 3| f n−4 , giving us a contradiction.\n\n266\n\nSolutions\n\n2.21 3528 = 23 · 32 · 72 , 30030 = 2 · 3 · 5 · 7 · 11 · 13 and 220000 = 25 · 54 · 11. 2.23 Let d = ( p, n). As d is a positive integer and d| p, we can only have d = 1 or p. If d = 1, we are done. If d = p, then as d|n, we are done. 2.25 If pi |( p1 · · · pk + 1), then as pi | p1 · · · pk , we have pi |( p1 · · · pk + 1 − p1 · · · pk ) = 1, which is impossible. 2.27 By Corollary 2.4, p|a. Let us say a = pb, with b ∈ Z. Then a n = p n bn , so p n |a n . 2.29 (1) and (2) are clearly commutative, but (3) is not, since 1∗2 = 1 but 2∗1 = 2. 2.31 (1) No. There is no e ∈ Z such that 2e + 1 = 2. (2) Yes, let e = 0. Then a ∗ e = a = e ∗ a for all a ∈ Z. 2.33 (1) 4. (2) (4 · 5)25 = (−1)25 = −1 = 6. 2.35 (2) We have [a] + ([b] + [c]) = [a] + [b + c] = [a + (b + c)] = [(a + b) + c] = [a + b] + [c] = ([a] + [b]) + [c]. (4) Note that [a] + = [a + 0] = [a]. (5) Observe that [a] + [−a] = [a + (−a)] = . 2.37 2 · 10 = 4 · 5 = 6 · 10 = 8 · 15 = 12 · 5 = 14 · 10 = 16 · 5 = 18 · 10 = 0. If a ∈ {1, 3, 7, 9, 11, 13, 17, 19}, there is no such b. 2.39 If a 2 = 1 in Z p , then a 2 ≡ 1 (mod p); that is, p|(a 2 −1) = (a −1)(a +1). By Euclid’s lemma, p|(a − 1) or p|(a + 1). That is, a ≡ 1 (mod p) or a ≡ −1 ≡ p − 1 (mod p). If p = 8, then 1, 3, 5 and 7 are solutions. 2.41 Proceeding as in the proof of Theorem 2.13, we have d1 = 70, d2 = 30 and d3 = 21. Solving 3u 1 + 70v1 = 1 using the Euclidean algorithm, we get u 1 = −23 and v1 = 1. Solving 7u 2 + 30v2 = 1, we get u 2 = 13 and v2 = −3. Solving 10u 3 + 21v3 = 1, we get u 3 = −2 and v3 = 1. Thus, our answer is a = 70 · 1 · 2 + 30(−3)(4) + 21 · 1 · 3 = −157. (As our answer is only unique modulo 3 · 7 · 10 = 210, we can also add 210 and get 53.)\n\nProblems of Chapter 3 \u0004\n\n\u0005 1234 . 4231 \u0004 \u0005 1234 (2) . 1324 \u0004 \u0005 1234 (3) . 2413\n\n3.1 (1)\n\nSolutions\n\n267\n\n3.3 There are n places to map 1, then n − 1 to map 2, and so on. So there are n! permutations. If we fix 2, then the other four numbers can be arranged at will, so there are 4! = 24 possibilities. 3.5 Closure: Yes, the composition of two functions is a function. Associativity: Yes, the composition of functions is associative. Identity: Yes, we have the identity function sending each element of {1, 2, 3, 4, 5} to itself. Inverses: No. Define α : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} via α(i) = 1 for all i. There is no possible function β : {1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} such that αβ is the identity function. 3.7 (1) 012 0012 1120 2201 (2) (0, 0) (1, 0) (2, 0) (0, 1) (1, 1) (2, 1)\n\n(0, 0) (0, 0) (1, 0) (2, 0) (0, 1) (1, 1) (2, 1)\n\n(1, 0) (1, 0) (2, 0) (0, 0) (1, 1) (2, 1) (0, 1)\n\n(2, 0) (2, 0) (0, 0) (1, 0) (2, 1) (0, 1) (1, 1)\n\n(0, 1) (0, 1) (1, 1) (2, 1) (0, 0) (1, 0) (2, 0)\n\n(1, 1) (1, 1) (2, 1) (0, 1) (1, 0) (2, 0) (0, 0)\n\n(2, 1) (2, 1) (0, 1) (1, 1) (2, 0) (0, 0) (1, 0)\n\n3.9 Take gi ∈ G, h i ∈ H . Now, (g1 , h 1 )(g2 , h 2 ) = (g2 , h 2 )(g1 , h 1 ) if and only if (g1 g2 , h 1 h 2 ) = (g2 g1 , h 2 h 1 ); that is, if and only if g1 g2 = g2 g1 and h 1 h 2 = h 2 h 1 . 3.11 (1) Division is not associative; for instance, (1/2)/3 \u0003= 1/(2/3). (2) There is no inverse for 2 (or anything else other than 1). 3.13 Yes. Let G be our set. If a + bi, c + di ∈ G, then (a + bi)(c + di) = (ac−bd)+(ad +bc)i. Now, (ac−bd)2 +(ad +bc)2 = a 2 c2 +b2 d 2 +a 2 d 2 +b2 c2 = (a 2 + b2 )(c2 + d 2 ) = 1 · 1 = 1, so (a + bi)(c + di) ∈ G. Complex multiplication is associative. Clearly 1 ∈ G, and it will serve as the identity. If a + bi ∈ G, then (a + bi)(a − bi) = a 2 + b2 = 1. Now, a 2 + (−b)2 = 1, so a − bi ∈ G and (a + bi)−1 = a − bi. 3.15 Yes. To show closure, note that (a/ p n ) + (b/ p m ) = (ap m + bp n )/ p m+n ∈ G. Addition of rational numbers is certainly associative, and 0 = 0/ p is the additive identity. The additive inverse of a/ p n is −a/ p n ∈ G. 3.17 (1) aca −1 b−1 c−1 . (2) a −1 c−1 b−1 a.\n\n268\n\nSolutions\n\n3.19 a b c d\n\na d a b c\n\nb a b c d\n\nc b c d a\n\nd c d a b\n\n3.21 If (1) holds, then for any g, h ∈ G, let a = g, b = hg, c = gh. Then ab = ca = ghg, so by assumption, hg = gh, and G is abelian. If (2) holds, then whenever ab = ca, we have ab = ac, so by cancellation, b = c. 3.23 (1) |Z12 | = 12. Also, |0| = 1, |1| = |5| = |7| = |11| = 12, |2| = |10| = 6, |3| = |9| = 4, |4| = |8| = 3 and |6| = 2. (2) |Z2 × Z4 | = 8. Also, |(0, 0)| = 1, |(1, 0)| = |(1, 2)| = |(0, 2)| = 2 and every other element has order 4. 3.25 |a 3 | = 20/(3, 20) = 20/1 = 20, |a 12 | = 20/(12, 20) = 20/4 = 5 and |a 15 | = 20/(15, 20) = 20/5 = 4. 3.27 We are looking for the smallest positive integer n such that (a, b)n = (e, e); that is, such that a n = e and bn = e. But a n = e if and only if 12|n and bn = e if and only if 18|n. Thus, we want the smallest positive integer n divisible by both 12 and 18. The order is 36. 3.29 (1) Note that a n = e if and only if (a n )−1 = e−1 ; that is, if and only if (a −1 )n = e. (2) Recall that conjugates have the same order. Also, ab = b−1 (ba)b. 3.31 First, note that U (8) has exactly three elements of order 2, namely 3, 5 and 7. Suppose that a and b are distinct elements of order 2. Now, (ab)2 = a 2 b2 = e2 = e, since G is abelian. Furthermore, if ab = e, then a = b−1 = b, since b has order 2. But this is impossible. Thus, |ab| = 2. Furthermore, if ab = a, then b = e and if ab = b, then a = e. Thus, a, b and ab are distinct elements of order 2. Let c be a fourth distinct element of order 2. By the same argument, ac has order 2. If ac = a, then c = e. If ac = b, then c = a −1 b = ab. If ac = ab, then c = b. None of these are true. Thus, ac is a fifth distinct element of order 2. 3.33 (1) Yes. Clearly H contains the identity matrix. If A, B ∈ H , then det(AB) = det(A) det(B) = 1 · 1 = 1, so AB ∈ H . Furthermore, det(A−1 ) = 1/ det(A) = 1, so A−1 ∈ H . (2) No. H does not contain the identity. (3) Yes. First, we see that 0 = 0/1 ∈ H . Next, if a/b, c/d ∈ H , then (a/b) + (c/d) = (ad + bc)/(bd) ∈ H , and −(a/b) = (−a)/b ∈ H . 3.35 Let F be any flip and R any rotation. Drawing out the effects of each operation, we find that F R = R −1 F. This is R F if and only if R = R −1 . Letting R = R360/n , we find that R \u0003= R −1 . Thus, no flip is central. In fact, R = R −1 if and only if\n\nSolutions\n\n269\n\nR = R0 or R = R180 . If n is odd, there is no R180 , so Z (D2n ) = {R0 }. If n is even, we see that R180 commutes with every flip, and surely with every rotation, so Z (D2n ) = {R0 , R180 }. 3.37 Let H and K be subgroups of G. If a, b ∈ H ∩ K , then a, b ∈ H , so ab ∈ H . Similarly, ab ∈ K , and therefore ab ∈ H ∩ K . By the same argument, a −1 ∈ H , a −1 ∈ K , so a −1 ∈ H ∩ K . Finally, as H and K are subgroups, e ∈ H and e ∈ K , so e ∈ H ∩ K . The argument for an arbitrary intersection is similar. 3.39 (1) \u000e0\u000f = {0}, \u000e1\u000f = \u000e3\u000f = \u000e7\u000f = \u000e9\u000f = \u000e11\u000f = \u000e13\u000f = \u000e17\u000f = \u000e19\u000f = Z20 , \u000e2\u000f = \u000e6\u000f = \u000e14\u000f = \u000e18\u000f = {0, 2, 4, 6, 8, 10, 12, 14, 16, 18}, \u000e4\u000f = \u000e8\u000f = \u000e12\u000f = \u000e16\u000f = {0, 4, 8, 12, 16}, \u000e5\u000f = \u000e15\u000f = {0, 5, 10, 15}, \u000e10\u000f = {0, 10}. (2) \u000e1\u000f = {1}, \u000e3\u000f = \u000e11\u000f = {1, 3, 9, 11}, \u000e5\u000f = \u000e13\u000f = {1, 5, 9, 13}, \u000e7\u000f = {1, 7}, \u000e9\u000f = {1, 9}, \u000e15\u000f = {1, 15}. 3.41 Label the vertices of the regular n-gon from 1 to n, counterclockwise. Then notice that a rotation leaves the vertices in counterclockwise order, whereas a flip changes them to clockwise. This makes clear what must happen in each case. 3.43 (1) We have \u000ea 1 \u000f = G, \u000ea 2 \u000f = {e, a 2 , a 4 , a 6 , a 8 , a 10 }, \u000ea 3 \u000f = {e, a 3 , a 6 , a 9 }, \u000ea 4 \u000f = {e, a 4 , a 8 }, \u000ea 6 \u000f = {e, a 6 }, \u000ea 12 \u000f = {e}. (2) As Z12 is cyclic of order 12 with generator 1, we have \u000e1\u000f = Z12 , \u000e2\u000f = {0, 2, 4, 6, 8, 10}, \u000e3\u000f = {0, 3, 6, 9}, \u000e4\u000f = {0, 4, 8}, \u000e6\u000f = {0, 6}, \u000e0\u000f = {0}. 3.45 A positive integer is not relatively prime to pn if and only if it is divisible by p. Thus, we are excluding p, 2 p, 3 p, . . . , p n . There are p n−1 such numbers. 3.47 It does follow, as H ∩ K is a subgroup of H , and every subgroup of a cyclic group is cyclic. 3.49 Let G = \u000ea\u000f be a cyclic group of order n. Now, |a i | = n/(n, i), for every integer i. In particular, each element of G has order dividing n. Now, for every k dividing n, the number of elements of order k is ϕ(k). Thus, the sum of the ϕ(k) is the number of elements in G, namely, n. 3.51 (1) If a ∈ H ∩ K has order n, then \u000ea\u000f is a subgroup of order n in both H and K . As |H | = |K | = n, this means that H = K = \u000ea\u000f, which is impossible. (2) If G has no elements of order n, we are done. Otherwise, take a ∈ G of order n. Then \u000ea\u000f has ϕ(n) elements of order n. If those are all of the elements in G, we are done. Otherwise, find b ∈ / \u000ea\u000f of order n. Then \u000eb\u000f contains ϕ(n) elements of order n, and by (1), \u000ea\u000f and \u000eb\u000f have no elements of order n in common. Thus, we now have 2ϕ(n) elements of order n. Repeat. If the process stops, we have a multiple of ϕ(n). If not, we have infinitely many. 3.53 (1) Left cosets: 0+ H = {. . . , −4, 0, 4, 8, . . .}, 1+ H = {. . . , −3, 1, 5, 9, . . .}, 2 + H = {. . . , −2, 2, 6, 10, . . .}, 3 + H = {. . . , −1, 3, 7, 11, . . .}. As G is abelian, the right cosets are the same. (2) Left cosets: R0 H = {R0 , F2 }, R90 H = {R90 , F4 }, R180 H = {R180 , F1 }, R270 H = {R270 , F3 }. Right cosets: H R0 = {R0 , F2 }, H R90 = {R90 , F3 }, H R180 = {R180 , F1 }, H R270 = {R270 , F4 }.\n\n270\n\nSolutions\n\n3.55 Let G = pq. If H ≤ G, then |H | divides |G|, so |H | ∈ {1, p, q, pq}. As H is a proper subgroup, the order is not pq. But the trivial group is cyclic, as is any group of prime order. 3.57 As H ∩ K is a subgroup of H and K , its order divides both 28 and 65. But (28, 65) = 1, so we can only have H ∩ K = {e}. 3.59 By Exercise 3.30, a1 · · · ak has order 1 or 2. But a group of odd order has no elements of order 2. 3.61 Suppose otherwise, and let h 1 (H ∩ K ), . . . , h n+1 (H ∩ K ) be distinct left cosets / H ∩ K . Since h i−1 h j ∈ H , we have of H ∩ K in H . Then if i \u0003= j, we have h i−1 h j ∈ −1 / K . That is, h 1 K , . . . , h n+1 K are distinct left cosets of K in G, contradicting hi h j ∈ the assumption that [G : K ] = n.\n\nProblems of Chapter 4 4.1 (1) Yes. Clearly H contains the identity matrix. If A, B ∈ H , then det(AB) = det(A) det(B) ∈ Q, and det(A−1 ) = 1/ det(A) ∈ Q; thus, AB, A−1 ∈ H , and H is a subgroup. Also, if C ∈ G L 2 (R), then det(C −1 AC) = det(C −1 ) det(A) det(C) = det(A) det(C −1 ) det(C) = det(AC −1 C) = det(A) ∈ Q; thus, C −1 AC ∈ H , so H is normal. \u0004 \u0005 \u0004 \u0005 \u0004 \u0005 \u0004 \u0005 −1 10 11 11 1 −1 (2) No, = , which is not diagonal. 02 01 01 0 2 4.3 Let N = {e, a}. If b ∈ G, then b−1 ab ∈ N . But if b−1 ab = e, then a = beb−1 = e; impossible. Thus, b−1 ab = a, and a is central; naturally, e is always central. 4.5 Let N and K be normal subgroups of G. By Exercise 3.37, N ∩ K is a subgroup. Let a ∈ N ∩ K and g ∈ G. Since a ∈ N , we have g −1 ag ∈ N . Similarly, g −1 ag ∈ K , so g −1 ag ∈ N ∩ K . The proof of the generalization is similar. 4.7 If a ∈ G, then a −1 H a is a subgroup of order n, so a −1 H a = H . 4.9 If g ∈ G, n ∈ N , then n −1 g −1 ng ∈ N ; thus, g −1 ng = n(n −1 g −1 ng) ∈ N . 4.11 We know that |a| is divisible by 5. Also, (a N )5 = eN , so a 5 ∈ N . By Lagrange’s theorem, (a 5 )14 = e. Thus, the order of a divides 70. So |a| ∈ {5, 10, 35, 70}. To see that these are all possible, let G = Z70 , N = \u000e5\u000f and let a be 14, 7, 2 and 1, respectively. 4.13 For both parts, G = D8 × Z will suffice. We have Z (G) = \u000eR180 \u000f × Z. As G/Z (G) has order 4, it clearly satisfies (2). As for (1), it remains to show that D8 /\u000eR180 \u000f is abelian. But this can be seen by examining the group table from Exercise 4.12.\n\nSolutions\n\n271\n\n4.15 Let |a N | = 42. As G is finite, we know that a has finite order, and so its order is a multiple of 42, say 42n. But then a n has order 42. It need not hold for infinite groups. Indeed, let G = Z, N = \u000e42\u000f and a = 1. We see that |1 + N | = 42, but every nonidentity element of G has infinite order. 4.17 By Exercise 4.16, a −1 b−1 ab ∈ K , for all a, b ∈ G. Similarly, a −1 b−1 ab ∈ N . But K ∩ N = {e}, so a −1 b−1 ab = e, and hence ab = ba. 4.19 Clearly e ∈ N . If a, b ∈ N , say a k = bl = e, for some k, l ∈ N, then (ab)kl = (a k )l (bl )k = el ek = e; thus, ab ∈ N . Also, |a| = |a −1 |, so a −1 ∈ N . Thus, N is a subgroup. As G is abelian, it is normal. Take any c ∈ G. If, for some n ∈ N, we have (cN )n = eN , then cn ∈ N ; that is, cn has finite order, so cnm = e for some m ∈ N. In other words, c ∈ N , so cN = eN . 4.21 (1) It is a homomorphism. Indeed, α(ab) = log10 (ab) = log10 a + log10 b = α(a) + α(b). It is one-to-one, as if α(a) = 0, then log10 a = 0, so a = 1; that is, ker(α) = {1}. It is also onto, as if b ∈ R, then α(10b ) = b. (2) It is not a homomorphism, as β(0 + 0) = 1 \u0003= 2 = β(0) + β(0). 4.23 We have α((a, b)(c, d)) = α((ac, bd)) = ac(bd)−1 = ab−1 cd −1 (since U (16) is abelian), and this is α((a, b))α((c, d)). Thus, α is a homomorphism. Also, \u000e7\u000f = {1, 7}. Now, α((a, b)) = 1 if and only if ab−1 = 1; that is, we have the pairs (1, 1), (3, 3), . . . , (15, 15). Similarly, α((a, b)) = 7 if and only if a = 7b, so we have the pairs (7, 1), (5, 3), (3, 5), (1, 7), (15, 9), (13, 11), (11, 13), (9, 15). 4.25 (1) Not necessarily. For instance, H could be the trivial group. (2) Yes. Let h ∈ H have order n. As α is onto, say α(g) = h. Since G is finite, |g| < ∞, so |h| divides |g|. Let us say that |g| = mn. Then |g m | = n. 4.27 Note that gh = α((g, h)) = α((e, h)(g, e)) = α((e, h))α((g, e)) = hg. 4.29 Let H = G/N . Define α : G → H via α(g) = g N . It is a homomorphism, as α(g1 g2 ) = g1 g2 N = g1 N g2 N = α(g1 )α(g2 ), and g ∈ ker(α) if and only if g N = eN ; that is, if and only if g ∈ N . 4.31 (1) Count the elements of order 2. (2) One is abelian and the other is not. (3) We know that Z is cyclic. Suppose that Z × Z = \u000e(a, b)\u000f. Then there exists an n ∈ Z such that (1, 0) = n(a, b). Since n cannot be 0, we see that b = 0. Similarly, a = 0. But this is impossible. \u0004 \u0005 10 . Note that α(a + b) = 4.33 Define α : Z → G L 2 (R) via α(a) = a1 \u0004 \u0005 \u0004 \u0005\u0004 \u0005 1 0 10 10 = = α(a)α(b). Thus, α is a homomorphism. In para+b 1 a1 b1 ticular, α(Z) = G is a subgroup of G L 2 (R). Furthermore, if α(a) is the identity matrix, then a = 0; thus, α is one-to-one. Therefore, Z is isomorphic to α(Z) = G.\n\n272\n\nSolutions\n\n4.35 Define α : H → a −1 H a via α(h) = a −1 ha. Then for any h, k ∈ H , we have α(hk) = a −1 hka = a −1 haa −1 ka = α(h)α(k); thus, α is a homomorphism. By definition, it is onto. Also, if h ∈ ker(α), then a −1 ha = e; therefore, h = aa −1 = e, and α is one-to-one. 4.37 If n > 1 is a positive integer, then nZ is a proper subgroup which is infinite cyclic, and therefore isomorphic to Z. 4.39 Define α : G → G via α((a1 , a2 , . . .)) = (0, a1 , a2 , . . .). It is a homomorphism; indeed, if a = (a1 , a2 , . . .) and b = (b1 , b2 , . . .), then α(a + b) = α(a) + α(b) = (0, a1 + b1 , a2 + b2 , . . .). Furthermore, it is one-to-one; if α(a) = (0, 0, . . .), then clearly a = (0, 0, . . . ). Thus, G is isomorphic to α(G), which is a proper subgroup of G. 4.41 Define α : G → Z via α((a, b)) = a − b. Now, α is a homomorphism, since α((a, b) + (c, d)) = α((a + c, b + d)) = (a + c) − (b + d) = (a − b) + (c − d) = α((a, b)) + α((c, d)). Also, α is onto, since for any a ∈ Z, α((a, 0)) = a. Finally, the kernel is the set of all (a, b) such that a − b = 0; that is, ker(α) = N . Apply the First Isomorphism Theorem. 4.43 Define α : R → H via α(r ) = cos(2πr ) + sin(2πr )i (where we are working in radians). As cos2 (θ ) + sin2 (θ ) = 1 for any θ ∈ R, we see that α(R) ⊆ H . Furthermore, for any a, b ∈ R such that a 2 + b2 = 1, we can surely find r ∈ R such that cos(2πr ) = a and sin(2πr ) = b; thus, α(R) = H . To show that α is a homomorphism, calculate α(r + s) and α(r )α(s) and use trigonometric identities. Finally, the kernel is the set of all r ∈ R such that cos(2πr ) = 1 and sin(2π s) = 0; that is, ker(α) = Z. Apply the First Isomorphism Theorem. ⎞ ⎛ ⎞ ⎛ 1ab 110 4.45 (1) If ⎝0 1 c ⎠ commutes with ⎝0 1 0⎠, then c = 0. Similarly, a = 0. But 001 001 matrices with a = c = 0 are easily seen to commute with everything in G, so those matrices form the centre. ⎛⎛ ⎞⎞ 1ab (2) Define α : G → Z × Z via α ⎝⎝0 1 c ⎠⎠ = (a, c). We can see that 001 ⎛⎛\n\n⎞⎛ ⎞⎞ ⎛⎛ ⎞⎞ 1ab 1d e 1 a + d e + af + b c + f ⎠⎠ α ⎝⎝0 1 c ⎠ ⎝0 1 f ⎠⎠ = α ⎝⎝0 1 001 00 1 0 0 1 ⎛⎛ ⎞⎞ ⎛⎛ ⎞⎞ 1ab 1d e = (a + d, c + f ) = α ⎝⎝0 1 c ⎠⎠ + α ⎝⎝0 1 f ⎠⎠ , 001 00 1 so α is a homomorphism. It is clearly onto. Furthermore, its kernel is precisely Z (G), as we found in the first part. Now apply the First Isomorphism Theorem.\n\nSolutions\n\n273\n\n4.47 As α(ab) = (ab)m = a m bm = α(a)α(b) (since G is abelian), we know that α is a homomorphism. If α(a) = e, then a m = e, and hence |a| divides m. But by Lagrange’s theorem, |a| divides n as well. Since (m, n) = 1, we can only have |a| = 1. Thus, α is one-to-one. As G is finite, it must be onto as well. 4.49 Let α be an automorphism of Z2 × Z2 . Now, any homomorphism sends the identity to the identity. As α is one-to-one, α((1, 0)) ∈ {(1, 0), (0, 1), (1, 1)}. Furthermore, once α((1, 0)) is chosen, that leaves only two possibilities for α((0, 1)). Once both of these are decided, there is only one option left for α((1, 1)). So there are only 3 · 2 = 6 possible automorphisms. This does not mean that all of them are necessarily automorphisms, but as it happens, they are. To see this, note that every group of prime order is cyclic, and groups of order 4 are abelian. Thus, since Example 4.26 shows us that there are noncommuting automorphisms, no order less than 6 is possible. So all of the functions we have considered are actually automorphisms. Also, every group of order 6 is isomorphic to Z6 or D6 . As the automorphism group is nonabelian, it must be D6 . 4.51 As α(e) = e, we see that e is in our set. If α(a) = a and α(b) = b, then α(ab) = α(a)α(b) = ab, so ab is in our set. Also, α(a −1 ) = (α(a))−1 = a −1 , so a −1 is in our set. 4.53 As α(\u000ea\u000f × {e}) ⊆ \u000ea\u000f × {e}, let us say that α((a, e)) = (a i , e). Similarly, α((e, b)) = (e, b j ) and α((a, b)) = (a, b)k . But then (a i , b j ) = (a k , bk ). That is, α((a, e)) = (a k , e) and α((e, b)) = (e, bk ). Then for any r, s ∈ Z, α((a r , bs )) = α((a, e))r α((e, b))s = (a k , e)r (e, bk )s = (a r , bs )k . 4.55 If m is an integer, then we\u0002 know\u0003 that α(m) \u0002 \u0003 = α(m ·\u0002 1)\u0003 = m · α(1). If n is a nonzero integer, then α(m) = α n · mn = nα mn . Thus, α mn = n1 α(m) = mn α(1).\n\nProblems of Chapter 5 5.1 Let H = \u000e3\u000f and K = \u000e31\u000f. We see that |H | = 8, |K | = 2. As the group is abelian, both subgroups are normal, and H ∩ K = {1}. Thus |H K | = 8 · 2/1 = 16 = |U (32)|, so U (32) = H × K . 5.3 Note that 5(a, b) = (0, 0) if and only if 5a = 0 and 5b = 0. But 5a = 0 for all a, whereas 5b = 0 if and only if b ∈ {0, 5, 10, 15, 20}. Thus, 5 · 5 = 25 elements satisfy 5(a, b) = 0. Now, these elements have order dividing 5, so we need only exclude the identity, which has order 1; thus, there are 24 elements of order 5. As 25(a, b) = (0, 0) for all a and b, we see that every element has order 1, 5 or 25. We have found 25 elements not having order 25, which means that 5 · 25 − 25 = 100 elements have order 25. 5.5 If D8 = H × K , then |H ||K | = 8. As the subgroups are both proper, |H | = 4 and |K | = 2 (or vice versa). By Corollaries 4.2 and 4.3, H and K are abelian, so D8 is abelian, giving us a contradiction.\n\n274\n\nSolutions\n\n5.7 Let G = Z2 × Z2 , N1 = \u000e(1, 0)\u000f, N2 = \u000e(0, 1)\u000f and N3 = \u000e(1, 1)\u000f. As G is abelian, normality is not an issue. We can see that G = N1 N2 , so surely G = N1 N2 N3 . Also, each Ni ∩N j = {(0, 0)}. But we cannot have G = N1 ×N2 ×N3 , since the order is wrong. 5.9 It does not follow. Let G = Z2 × Z2 and H = Z2 . Define α : G → H via α((a, b)) = a+b. We have α((a, b)+(c, d)) = a+b+c+d = α((a, b))+α((c, d)), so α is a homomorphism. As α((0, 0)) = 0 and α((1, 0)) = 1, we see that α is onto. Now, G = \u000e(1, 0)\u000f × \u000e(0, 1)\u000f, but α(\u000e(1, 0)\u000f) = α(\u000e(0, 1)\u000f) = H ; thus, the intersection of the images is not trivial, so we do not have a direct product in H . 5.11 (1) Z3 × Z7 . (2) Z81 , Z27 × Z3 , Z9 × Z9 , Z9 × Z3 × Z3 , Z3 × Z3 × Z3 × Z3 . (3) Z8 ×Z25 ×Z49 , Z4 ×Z2 ×Z25 ×Z49 , Z2 ×Z2 ×Z2 ×Z25 ×Z49 , Z8 ×Z5 ×Z5 ×Z49 , Z4 × Z2 × Z5 × Z5 × Z49 , Z2 × Z2 × Z2 × Z5 × Z5 × Z49 , Z8 × Z25 × Z7 × Z7 , Z4 × Z2 × Z25 × Z7 × Z7 , Z2 × Z2 × Z2 × Z25 × Z7 × Z7 , Z8 × Z5 × Z5 × Z7 × Z7 , Z4 × Z2 × Z5 × Z5 × Z7 × Z7 , Z2 × Z2 × Z2 × Z5 × Z5 × Z7 × Z7 . 5.13 As |U (56)| = ϕ(56) = 24, the possibilities are Z8 × Z3 , Z4 × Z2 × Z3 and Z2 × Z2 × Z2 × Z3 . But running through the elements of U (56), we see that none have order larger than 6. As Z8 × Z3 and Z4 × Z2 × Z3 both have elements of order 12, it must be Z2 × Z2 × Z2 × Z3 . 5.15 We see that G is isomorphic to a direct product of groups of the form Z pni , for various n i ∈ N. But if n i > 1, then such a group has elements of order p 2 . 5.17 Solving 5u + 7v = 1 in Z, one possible solution is u = 3, v = −2. Then a = a 5u+7v = (a 5 )3 (a 7 )−2 . Now |a 15 | = 7 and |a −14 | = |a 21 | = 5. 5.19 We proceed by strong induction on |G|. There is nothing to do if |G| = 1, so we start the induction with |G| = 2. In this case, p = 2 and G has an element of order 2. Let |G| > 2 and assume the result for groups of smaller order. If e \u0003= b ∈ G, then choose some prime q dividing |b|. Let a = b|b|/q . Then |a| = q. If q = p, we are done. Otherwise |G/\u000ea\u000f| = |G|/q, and this is still divisible by p. By our inductive hypothesis, G/\u000ea\u000f has an element c\u000ea\u000f of order p. Thus, c p ∈ \u000ea\u000f, so c pq = e. Hence, |cq | = 1 or p. But if cq = e, then (c\u000ea\u000f)q = e\u000ea\u000f. As |c\u000ea\u000f| = p, this is impossible. 5.21 (1) 8, 2, 3, 3, 25, 7, 7. (2) 2, 2, 2, 3, 9, 27. 5.23 p 3 , q 2 , r ; p 2 , p, q 2 , r ; p, p, p, q 2 , r ; p 3 , q, q, r ; p 2 , p, q, q, r ; p, p, p, q, q, r . 5.25 It is obviously the case for n = 1. For larger n, we claim that it is true if and only if n is a product of distinct primes. If n = p1 · · · pk , where the pi are all distinct, then the only possible list of elementary divisors is p1 , . . . , pk , so the groups are all isomorphic. On the other hand, if p 2 |n for some prime p, then we have the cyclic group of order n and Z p × Zn/ p . Since ( p, n/ p) = p > 1, we see that this group is not cyclic.\n\nSolutions\n\n275\n\n5.27 The list of elementary divisors of G 1 × G 2 is obtained by combining the lists of elementary divisors of G 1 and G 2 . Similarly for G 1 × G 3 . If these lists are the same, then deleting the elementary divisors of G 1 from each list, we see that G 2 and G 3 have the same elementary divisors, and hence are isomorphic. 5.29 We know that G is isomorphic to Z2n1 × · · · × Z2nk . If 2(a1 , . . . , ak ) = (0, . . . , 0), then 2ai = 0 for all i; that is, each ai has order 1 or 2. But a cyclic group of order 2ni has only one element of order 2, so there are only two such ai , for each i. In total, we get 2k elements. But we must exclude the identity, so our number is 2k − 1. 5.31 (1) Remember that q + Z = r + Z if and only if q − r ∈ Z. This is basically the same as Example 1.19, using Q instead of R. (2) We have b(a/b + Z) = a + Z = 0 + Z. Thus, |a/b + Z| ≤ b. But if c ∈ N and c(a/b + Z) = 0 + Z, then ca/b ∈ Z; that is, b|ac. Since (a, b) = 1, this means that b|c. In particular, c ≥ b, so the order is b. 5.33 We have α(a + b) = n(a + b) = na + nb = α(a) + α(b), so α is a homomorphism. If a ∈ G, then since G is divisible, there exists a b ∈ G such that nb = a. Thus, α(b) = a, and α is onto. But it is not necessarily an isomorphism. Let G be the Prüfer p-group and take n = p. Then we see that 1/ p + Z ∈ ker(α). 5.35 If N is a subgroup of G, take a + N ∈ G/N . Then for any n ∈ N, there exists a b ∈ G such that nb = a. Therefore, n(b + N ) = a + N , and G/N is divisible. However, Q is divisible but Z is not, as there is no b ∈ Z such that 2b = 1. 5.37 Let G/N = \u000ea N \u000f. If g N ∈ G/N , then g N = (a N )k , for some k ∈ Z. That is, g = a k n, for some n ∈ N . In other words, G = \u000ea\u000fN . If e \u0003= b ∈ \u000ea\u000f ∩ N , then b = a l ∈ N , for some 0 \u0003= l ∈ Z. If l < 0, then we may replace b with b−1 , so let l > 0. Then (a N )l = eN , which means that a N has finite order in G/N . As G/N = \u000ea N \u000f and G/N is infinite cyclic, this is impossible. Therefore, G = \u000ea\u000f× N . It remains only to show that \u000ea\u000f is infinite cyclic. It is surely cyclic, and if |a| = k, then again, (a N )k = eN gives us a contradiction.\n\nProblems of Chapter 6 6.1 (1) (2 4 7 6)(3 5) (2) (1 2 5)(3 6 4)(7 8) 6.3 (1) (1 4 2)(3 6 7 5). (2) Writing the permutation as a product of disjoint cycles, we get (1 2)(3 5 4), so the inverse is (1 2)(3 4 5). 6.5 An element of order 3 must be a product of one or more disjoint 3-cycles. Let us count the 3-cycles (a b c). There are 9 choices for a, 8 for b and 7 for c. But\n\n276\n\nSolutions\n\n(a b c) = (b c a) = (c a b), so we must divide by 3, giving 9 · 8 · 7/3 = 168. For pairs of disjoint 3-cycles, we get 9 · 8 · 7 · 6 · 5 · 4/(3 · 3 · 2) = 3360, using the same argument and the fact that the order of the two cycles is irrelevant. Finally, to get three disjoint 3-cycles, we have 9!/(3 · 3 · 3 · 3!) = 2240, again noting that the three cycles can be permuted as we please. Our total is 5768. 6.7 If τ exists, then it has order k. Thus, if k is even, then τ 2 has order k/2, and therefore it cannot be a k-cycle. So suppose that k is odd. Let τ = σ (k+1)/2 . Then τ 2 = σ k+1 = σ , as σ has order k. Furthermore, as 2((k + 1)/2) + (−1)k = 1, we know that ((k + 1)/2, k) = 1. The preceding exercise tells us that τ is a k-cycle. 6.9 Let |σ | = 105 = 3 · 5 · 7. We know that |σ | is the least common multiple of the lengths of its cycles in the disjoint cycle decomposition. The product of a 3-cycle, a 5-cycle and a 7-cycle would work, so m = 15 is a possibility. Can there be a smaller value? There must surely be a cycle whose length is a multiple of 7 and a divisor of 105. If it is smaller than 15, it can only be 7. Similarly for 3 and 5. Thus, m = 15 is the smallest possible value. Let |τ | = 125. The only way to make this happen is for the disjoint cycle decomposition for τ to include a 125-cycle. We see that n = 125. 6.11 (1) even (2) odd 6.13 Without loss of generality, the possible products are (1 2)(1 2) = (1), having order 1, (1 2)(1 3) = (1 3 2), having order 3 and (1 2)(3 4), having order 2. 6.15 It is certainly impossible if n is 2 or 3, as the groups are too small. But if n ≥ 4, then An has the subgroup {(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. (It contains the identity, and closure is easily checked.) This subgroup is not cyclic. Indeed, if σ is a permutation of order 4, then its disjoint cycle decomposition is a product of one or more 4-cycles and, possibly, some 2-cycles. But a 4-cycle by itself is odd, so we need n ≥ 6 to get something like (1 2 3 4)(5 6) ∈ An . 6.17 The order of an element in Sn is the least common multiple of the lengths of its disjoint cycles. If this order is odd, then these cycles all have odd length. But a cycle of odd length is even. 6.19 We see by inspection that n = 1 falls into the second category and n = 2 and 3 fall into the third category. Let n ≥ 4. By Exercise 6.17, all elements of odd order lie in An . As An contains half the elements of Sn , we see that there are at least as many elements of even order as of odd order, and they can only be equal if every element of An has odd order. However, (12)(34) ∈ An has order 2. So if n ≥ 4, we are in the first category. 6.21 Such a subgroup would have index 2, and therefore be normal, by Theorem 4.1. But A5 is simple.\n\nSolutions\n\n277\n\n6.23 It can. Note that A6 has an isomorphic copy of A5 as a proper subgroup. (Just use the exact same permutations as in A5 , assuming that each fixes the number 6.) 6.25 Let N be a nontrivial proper normal subgroup of A4 . By the preceding exercise, N contains no 3-cycles, so N is a subgroup of the group exhibited in Example 6.11. If it is not the same group, then it can only have order 2. But by Exercise 4.3, a normal subgroup of order 2 is central. However, the elements of order 2 in A4 are the products of two disjoint transpositions, and these are not central. For instance, (1 2)(3 4)(1 2 3) \u0003= (1 2 3)(1 2)(3 4). 6.27 In view of the preceding exercise, it suffices to show that each (1 i), 2 ≤ i ≤ n, is the product of such transpositions. We proceed by induction on i, beginning with i = 2. There is nothing to do there, so assume the result for i and prove it for i + 1, when 1 < i < n. However, (i (i + 1))(1 i)(i (i + 1)) = (1 (i + 1)), completing the proof.\n\nProblems of Chapter 7 \u0004 \u0005 \u0004 \u0005 ab 11 7.1 Let A = ∈ G L 2 (R). Then A commutes with if and only if cd 01 a = a + c, a + b = b + d and c + d = d; that is, if\u0004and only \u0005 if c = 0 and a = d. ab Thus, the matrices in the centralizer have the form , where a, b ∈ R (and 0a a \u0003= 0, so that the matrix is invertible). \u0004 \u0005 23 7.3 We always have C(H ) ⊆ N (H ). Let A = , and suppose that B ∈ N (H ). 56 Then B −1 AB ∈ H , so B −1 AB = An , for some integer n. However, det(B −1 AB) = det(A) = −3, whereas det(An ) = (det(A))n = (−3)n . We conclude that n = 1, and hence B −1 AB = A. 7.5 As always, C(H ) ⊆ N (H ). Let H = {e, a}. If b ∈ N (H ), then b−1 eb = e and we must have b−1 ab ∈ H . If b−1 ab = e, then a = e, which is impossible. Therefore, b−1 ab = a, and b ∈ C(H ). 7.7 Take b ∈ C(a). As C(a) is a subgroup, b−1 ∈ C(a), so b−1 a = ab−1 . Inverting, we get a −1 b = ba −1 ; thus, b ∈ C(a −1 ). This means that C(a) ⊆ C(a −1 ) ⊆ C((a −1 )−1 ) = C(a). 7.9 (1) As H has prime order, it is abelian, so H ≤ C(H ). In particular, |C(H )| is divisible by 11 and divides 77, so it is 11 or 77. If it is 11, we must have C(H ) = H . Otherwise, H ≤ Z (G). In the same way, we now have |Z (G)| = 11 or 77. Since G is not abelian, Z (G) = H . But this contradicts Corollary 4.1. (2) Suppose otherwise. Combining (1) with Theorem 7.3, and noting that H is normal, we have G/H isomorphic to a subgroup of Aut(H ). By Theorem 4.14, H is\n\n278\n\nSolutions\n\nisomorphic to Z11 , and Theorem 4.22 tells us that Aut(H ) is isomorphic to U (11). But this is a group of order 10 and cannot have a subgroup of order 7. 7.11 {R0 }, {R180 }, {R90 , R270 }, {F1 , F2 }, {F3 , F4 }. 7.13 It does not follow. Let G = S3 , H = \u000e(1 3)\u000f and K = \u000e(1 3 2)\u000f. Now consider the subgroups \u000e(1 2)\u000f and \u000e(2 3)\u000f. As (1 3)−1 (1 2)(1 3) = (2 3) = (1 3 2)−1 (1 2)(1 3 2), it follows immediately that these subgroups are both H - and K -conjugate. However, H ∩ K = {(1)}, so they are not (H ∩ K )-conjugate. 7.15 If each [G : C(a)] in the class equation is divisible by p 2 , then since |G| is also divisible by p 2 , we must have p 2 dividing the order of |Z (G)|, which is not the case. Thus, since each [G : C(a)] divides p n , one of them must be p. It follows that |C(a)| = p n−1 . 7.17 (1) No. Groups of order 25 are abelian, so all conjugacy classes would have just one element. (2) Yes, S3 . (3) No, the identity is always in a conjugacy class by itself. 7.19 Suppose that b−1 ab = a −1 . Then b−2 ab2 = b−1 a −1 b = (b−1 ab)−1 = a. That is, b2 ∈ C(a). If G has odd order, so does b. Thus, write |b| = 2m − 1, for some m ∈ N. Then (b2 )m = b, so b ∈ C(a). But then b−1 ab = a, so a = a −1 . That is, a 2 = e. As a has odd order, a = e, giving us a contradiction. 7.21 Sylow 2-subgroup: \u000e25\u000f×\u000e7\u000f. Sylow 5-subgroup: \u000e(4, 0)\u000f. Sylow 7-subgroup: \u000e(0, 2)\u000f. 7.23 We have |G| = 2 · 3 · 72 . The number of Sylow 7-subgroups is 1 + 7k, for some nonnegative integer k, and divides 6. The only possible solution is k = 0. 7.25 Let H be a Sylow p-subgroup of G. By definition, its order is p m , the largest power of p dividing |G|. Thus, n ≤ m. By Exercise 7.16, H has a subgroup of order pn . 7.27 By the Second Isomorphism Theorem, H N /N is isomorphic to H/(H ∩ N ). In particular, its order divides |H | and is therefore a power of p. Furthermore, H ≤ H N ≤ G, so |G|/|H N | is a divisor of |G|/|H |. In particular, |G|/|H N | is relatively prime to p. However, [G/N : H N /N ] = (|G|/|N |)/(|H N |/|N |) = |G|/|H N |. Thus, H N /N is indeed a Sylow p-subgroup of G/N . 7.29 The number of Sylow 7-subgroups is 1 + 7k and divides 12. Thus, it is 1, and the Sylow 7-subgroup is normal. 7.31 The number of Sylow 17-subgroups is 1 + 17k and divides 256, so it is 1 or 256. If it is 1, the 17-Sylow subgroup is normal. If it is 256, then we note that each Sylow 17-subgroup is cyclic and has 16 elements of order 17. Distinct groups of prime order intersect trivially, so we have 16 · 256 = 4096 elements of order 17. This leaves only 256 other elements. But this is the size of a Sylow 2-subgroup, so there can be only one, and it is normal.\n\nSolutions\n\n279\n\n7.33 The number of Sylow p-subgroups is 1 + kp and divides q. If it is not 1, it is q, so p|(q − 1), giving us a contradiction. Thus, the Sylow p-subgroup is normal. Similarly, the number of Sylow q-subgroups is 1 + lq and divides p 2 . Thus, it is 1, p or p 2 . Suppose it is not 1. If it is p, then q|( p − 1), and since ( p − 1)|( p 2 − 1), we have a contradiction. If it is p 2 , we again obtain a contradiction. Therefore, the Sylow q-subgroup is normal as well. Thus, G is the direct product of its Sylow subgroups. Now, groups of order a prime or the square of a prime are abelian, and we are done. 7.35 The number of Sylow 3-subgroups is 1 + 3k and divides 19, so it is 1 or 19. If it is 1, then there are 2 elements of order 3. If it is 19, then there are 38, since subgroups of prime order intersect trivially. 7.37 Let H be a Sylow 7-subgroup and K a Sylow 17-subgroup. The number of Sylow 7-subgroups is 1 + 7k and divides 85, so it is 1 or 85. If it is 1, then H is normal. By Theorem 4.5, H K is a subgroup, and its order is 7 · 17/1 = 119. So assume that there are 85 Sylow 7-subgroups. Then we have 6 · 85 = 510 elements of order 7. The number of Sylow 17-subgroups is 1 + 17l and divides 35, so it is 1 or 35. If it is 1, then K is normal, and as above, we are done. Otherwise, we get 16 · 35 = 560 elements of order 17. But we now have too many elements. 7.39 It is not abelian, so we can rule out the two abelian groups. It has an element of order 6, namely (R120 , 1), so we can rule out A4 . But it has no elements of order 4, unlike the group H from Example 7.14, which has ((1 2), 1). Thus, it must be D12 . 7.41 It suffices to show that every cyclic subgroup is normal, for then if K ≤ Q 8 and a ∈ K , b ∈ Q 8 , we have b−1 ab ∈ \u000ea\u000f ≤ K . As 1 and −1 are central, we need not worry about them. The remaining cases just involve checking, for instance, that j −1 i j = − ji j = k j = −i = i −1 ∈ \u000ei\u000f. 7.43 (1) We have αa,b αc,d (x) = αa,b (cx +d) = a(cx +d)+b = acx +ad +b. Thus, αa,b αc,d = αac,ad+b ∈ G, since p \u0002 a and p \u0002 c imply that p \u0002 ac; that is, ac is not 0 in Z p . Thus, we have closure. Composition of functions is always associative. The identity is α1,0 . To find the inverse of αa,b , note that we want ac = 1 and ad + b = 0. Now, (a, p) = 1, so write au + pv = 1, for some u, v ∈ Z. Thus, au = 1 in Z p . Let c = u. Similarly, letting d = −ub, we have ad + b = −aub + b = 0 in Z p . To see that the group is not abelian, note that α2,0 and α1,1 do not commute. (2) Let H = {αa,b ∈ G : a ∈ {1, 2, 4}}. Closure checks out as above, since products of 1, 2 and 4 remain in {1, 2, 4} in Z7 . Clearly α1,0 ∈ H , so H ≤ G. There are 3 choices for a and 7 for b, so |H | = 21. Also, H is not abelian for the same reason given in the first part. 7.45 The number of Sylow p-subgroups is 1 + kp and divides q. As p > q, it is 1. The number of Sylow q-subgroups is 1 + lq and divides p, so it is 1 or p. But if it is p, then q|( p − 1), which is not allowed. Therefore, both Sylow subgroups are normal, and G is the direct product of Z p and Zq , and hence isomorphic to Z pq , as ( p, q) = 1.\n\n280\n\nSolutions\n\nProblems of Chapter 8 8.1 The addition table is found in Table 3.1. For multiplication, the table is as follows. 01234 000000 101234 202413 303142 404321 8.3 It is easy to see that R is closed under addition and contains {0}. Thus, since it is a finite set, it is an additive subgroup of Z15 , which is an abelian group. Furthermore, R is closed under multiplication in Z15 , and we know that this multiplication operation is associative and satisfies the distributive laws. Therefore, R is a ring. It is certainly commutative, and we can see that 6 is the identity. 8.5 It is not a ring, as it does not satisfy the distributive laws. Let α(x) = x 2 , β(x) = x and γ (x) = 2x. Then (α ◦ (β + γ ))(x) = 9x 2 , but (α ◦ β)(x) + (α ◦ γ )(x) = 5x 2 . 8.7 It is easy to see that the sum of two matrices in R also lies in R. Also, matrix addition is commutative and associative. The zero matrix is the additive identity, and negatives of matrices in R lie in R. Thus, R is an abelian group under addition. The product of two matrices in R is easily seen to be in R. Furthermore, matrix multiplication is associative and satisfies the distributive laws. Therefore,\u0004R is a\u0005ring. 11 It contains the identity matrix, so it is a ring with identity. However, and 00 \u0004 \u0005 01 do not commute, so it is not a commutative ring. 01 8.9 Not necessarily. Consider the additive group Z p , but define a multiplication operation via ab = 0 for all a and b. Clearly this operation is associative and the distributive laws are satisfied. Thus, we have a ring with p elements, but there is no identity. 8.11 (1) a 2 + ba − ab − b2 . (2) a 3 − a 2 b − aba − ba 2 + ab2 + bab + b2 a − b3 . 8.13 We have b = b1 = bac = 1c = c. 8.15 No, use R = Z2 ⊕ Z2 . 8.17 Note that (a + bi) − (c + di) = (a − c) + (b − d)i ∈ R and (a + bi)(c + di) = (ac − bd) + (ad + bc)i ∈ R, for all a, b, c, d ∈ Z. Also, 0 ∈ R. Thus, R is a subring. In addition, R is a unital subring, as it contains 1 + 0i, the identity of C.\n\nSolutions\n\n281\n\n\u0004 \u0005 \u0004 \u0005 00 00 8.19 Certainly R contains the zero matrix. If a, b ∈ R, then − = 0a 0b \u0004 \u0005 \u0004 \u0005\u0004 \u0005 \u0004 \u0005 0 0 00 00 0 0 ∈ R, and = ∈ R. Thus, R is a subring. It is a ring 0 a−b 0a 0b 0 ab \u0004 \u0005 00 with identity, as serves as the identity. But the identity of M2 (R) is not there, 01 so it is not a unital subring. 8.21 We have (0, 0) ∈ T . If r1 , r2 ∈ R, then (r1 , 0) − (r2 , 0) = (r1 − r2 , 0) ∈ T and (r1 , 0)(r2 , 0) = (r1r2 , 0) ∈ T . 8.23 We have 0 = 0a ∈ S. If r1 , r2 ∈ R, then r1 a − r2 a = (r1 − r2 )a ∈ S and (r1 a)(r2 a) = (r1 ar2 )a ∈ S. 8.25 Not necessarily. Let R = Q, S = Z and a = 2. Then 1/2 ∈ T , but (1/2)2 ∈ / T. 8.27 We have 1 ∈ R. If a + bi, c + di ∈ R, then (a + bi) − (c + di) = (a − c) + (b − d)i ∈ R. Furthermore, (a + bi)(c + di) = (ac − bd) + (ad + bc)i ∈ R. If c + di \u0003= 0, then (c + di)(c − di) = c2 + d 2 , which is a nonzero rational number, c d so the inverse of c + di is c2 +d 2 − c2 +d 2 i ∈ R. 8.29 We have (r, s) ∈ U (R ⊕ S) if and only if there exist r1 ∈ R, s1 ∈ S such that rr1 = r1r = 1 and ss1 = s1 s = 1; that is, if and only if r ∈ U (R) and s ∈ U (S). 8.31 If a 2 = a, then a(a − 1) = 0, so since there are no zero divisors, a = 0 or 1. An integral domain must have these two elements. 8.33 By Exercise 8.20, K ∩ L is a subring. As 1 ∈ K and 1 ∈ L, we have 1 ∈ K ∩ L. Also, if 0 \u0003= a ∈ K ∩ L, then a −1 ∈ K and a −1 ∈ L, so a −1 ∈ K ∩ L. Thus, K ∩ L is a subfield. The proof for an arbitrary collection of subfields is similar. 8.35 We have (a 10 )4 = (b10 )4 and (a 13 )3 = (b13 )3 . That is, a 40 = b40 and a 39 = b39 . So, a 39 a = b39 b = a 39 b. If a = 0, then since b40 = 0 and there are no zero divisors, b = 0. If a \u0003= 0, then cancelling a 39 , we obtain a = b. 8.37 (1) 7. (2) 0. 8.39 As 1 cannot have infinite order in a finite additive group, we know that char R = p, for some prime p. Thus, pa = 0 for all a ∈ R, so every element of R has additive order 1 or p. If |R| is divisible by some prime q \u0003= p, then by Cauchy’s theorem, R has an element of additive order q, which is impossible. Thus, the only prime dividing |R| is p. 8.41 (1) We have (1 + a)(1 − a + a 2 − a 3 + · · · + (−1)n−1 a n−1 ) = 1. k k (2) Let char R = p and choose k such that p k > n. Then (1 + a) p = 1 + a p pk n pk −n pk (using the Freshman’s Dream). But a = a a = 0, so (1 + a) = 1.\n\n282\n\nSolutions\n\nProblems of Chapter 9 9.1 (1) (0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5). (2) (0, 0), (2, 0), (0, 3), (2, 3). 9.3 By Exercise 8.20, I ∩ J is a subring. Take a ∈ I ∩ J and r ∈ R. Then a ∈ I implies ra, ar ∈ I . Similarly, ra, ar ∈ J , so ra, ar ∈ I ∩ J , and I ∩ J is an ideal. The argument for an arbitrary collection of ideals is similar. 9.5 (1) Let R = 2Z, a = 2. \u0004 \u0005 \u0004 \u0005 10 b0 (2) Let R = M2 (R), a = . Every matrix in S is of the form , for 00 c0 \u0004 \u0005 11 some b, c ∈ R. Clearly a ∈ S, but multiplying on the right by , we get a matrix 11 not in S, so S is not an ideal. 9.7 Let R = Z8 , I = (2) and J = (4). 9.9 We know from Exercise 3.42 that G is an additive group. It is clearly abelian. It is also closed under multiplication. Furthermore, (a1 , a2 , . . .)((b1 , b2 , . . .) (c1 , c2 , . . .)) = (a1 b1 c1 , a2 b2 c2 , . . .) = ((a1 , a2 , . . .)(b1 , b2 , . . .))(c1 , c2 , . . .). Thus, we have associativity of multiplication. The distributive law follows similarly, and we have a ring. By Exercise 3.42, H is an additive subgroup of G, so it remains only to check absorption. If (a1 , a2 , . . .) ∈ H , (b1 , b2 , . . .) ∈ G, then only finitely many of the ai are different from 0, so only finitely many of the ai bi are different from 0, and (a1 b1 , a2 b2 , . . .) ∈ H . Similarly for (b1 a1 , b2 a2 , . . .). 9.11 The addition table may be found in Table 4.2 (replacing each instance of “N ” with “I ”). The multiplication table follows. 0+I 1+ I 2+I 3+ I 4+ I\n\n0+I 0+I 0+I 0+I 0+I 0+I\n\n1+ I 0+I 1+ I 2+I 3+ I 4+ I\n\n2+I 0+I 2+I 4+ I 1+ I 3+ I\n\n3+ I 0+I 3+ I 1+ I 4+ I 2+I\n\n4+ I 0+I 4+ I 3+ I 2+I 1+ I\n\n9.13 Expanding, we obtain 8x 4 +2x 3 +7x 2 +5x +2+I . Now, 2(x 3 +6x 2 +2) ∈ I , so 2x 3 + I = −12x 2 −4+ I . Also, 8x(x 3 +6x 2 +2) ∈ I , so 8x 4 + I = −48x 3 −16x + I . Similarly, 48x 3 + I = −288x 2 − 96 + I . Thus, our answer is 288x 2 + 96 − 16x − 12x 2 − 4 + 7x 2 + 5x + 2 + I = 283x 2 − 11x + 94 + I . 9.15 By the preceding exercise, R/(I ∩ J ) is commutative if and only if ab − ba ∈ I ∩ J for all a, b ∈ R. But this happens if and only if ab − ba ∈ I and ab − ba ∈ J for all a, b ∈ R; that is, if and only if R/I and R/J are commutative. 9.17 The only ideals of F are {0} and F, so 81 and 1.\n\nSolutions\n\n283\n\n9.19 If (a + I )n = 0 + I , then a n ∈ I , and hence there exists an m ∈ N such that (a n )m = 0; that is, a nm = 0, which means that a ∈ I , and hence a + I = 0 + I . 9.21 (1) No, as α(1 · 1) = 2 but α(1)α(1) = 4. (2) Yes. If f (x), g(x) ∈ R[x], then α( f (x) + g(x)) = f (2) + g(2) = α( f (x)) + α(g(x)) and α( f (x)g(x)) = f (2)g(2) = α( f (x))α(g(x)). 9.23 For any r1 , r2 ∈ R, we have β(α(r1 + r2 )) = β(α(r1 ) + α(r2 )) = β(α(r1 )) + β(α(r2 )), and similarly for multiplication. 9.25 It is a homomorphism, as α((a, b) + (c, d)) = α((a + c, b + d)) = (a + c, 0) = α((a, b)) + α((c, d)), and similarly for multiplication. The kernel is {0} ⊕ Z. Furthermore, α −1 (2Z ⊕ 3Z) = 2Z ⊕ Z. 9.27 Let S = R/I and define α : R → S via α(a) = a + I . We have α(a + b) = a + b + I = (a + I ) + (b + I ) = α(a) + α(b), and similarly for multiplication, so α is a homomorphism. Also, a ∈ ker(α) if and only if a + I = 0 + I ; that is, if and only if a ∈ I . 9.29 Let α : F → K be a homomorphism. Now, ker(α) is an ideal of F. As F is a field, this means that ker(α) = {0} or F. In the former case, α is one-to-one, which is impossible, as K has fewer elements than F. In the latter case, α(a) = 0 for all a, so this is the only possible homomorphism. 9.31 (1) The additive groups are not isomorphic. (See Exercise 4.31.) (2) One has an identity, the other does not. 9.33 Let α : R → S be an isomorphism. We claim that α(Z (R)) ⊆ Z (S). But if a ∈ Z (R), then for any r ∈ R, we have ar = ra, and hence α(a)α(r ) = α(r )α(a). As α is onto, α(a) commutes with everything in S. Thus, restricting α to Z (R), we have a one-to-one homomorphism into Z (S). But if b ∈ Z (S), then for any r ∈ R, we have α(r )b = bα(r ). Letting b = α(c), this means that α(r )α(c) = α(c)α(r ); that is, α(r c) = α(cr ), and since α is one-to-one, r c = cr . In particular, b = α(c) ∈ α(Z (R)). Therefore, α : Z (R) → Z (S) is onto as well, and hence an isomorphism. 9.35 Let K be the field of fractions, and define α : F → K via α(a) = [a, 1] for all a ∈ F. If a, b ∈ F, then α(a + b) = [a + b, 1] = [a, 1] + [b, 1] = α(a) + α(b) and α(ab) = [ab, 1] = [a, 1][b, 1] = α(a)α(b); thus, α is a homomorphism. If [a, 1] = [0, 1], then a = 0, so α is one-to-one. Furthermore, if a, b ∈ F, with b \u0003= 0, then α(ab−1 ) = [ab−1 , 1] = [a, b]; thus, α is onto. 9.37 No: Z and Q are certainly not isomorphic (Q is a field but Z is not), however, we already know that the field of fractions of Z is isomorphic to Q, and by Exercise 9.35, the field of fractions of Q is isomorphic to Q as well. 9.39 (1) Note that \u0004\u0004 \u0005 \u0004 \u0005\u0005 \u0004 \u0005 \u0004\u0004 \u0005\u0005 \u0004\u0004 \u0005\u0005 ab e f a+e c+g ab e f α + = =α +α . cd g h b+ f d +h cd g h\n\n284\n\nSolutions\n\nFurthermore, \u0004\u0004 \u0005 \u0004 \u0005\u0005 \u0004 \u0005 \u0004\u0004 \u0005\u0005 \u0004\u0004 \u0005\u0005 ab e f ae + bg ce + dg e f ab α = =α α . cd g h a f + bh c f + dh g h cd Also, it is clear that applying α twice returns the original matrix. (2) requires similar computations. 9.41 Define α : R ⊕ S → S via α((r, s)) = s. If ri ∈ R, si ∈ S, then α((r1 , s1 ) + (r2 , s2 )) = α((r1 + r2 , s1 + s2 )) = s1 + s2 = α((r1 , s1 )) + α((r2 , s2 )), and similarly for multiplication. Thus, α is a homomorphism. If s ∈ S, then α((0, s)) = s, and hence α is onto. Finally, (r, s) ∈ ker(α) if and only if s = 0; that is, ker(α) = R ⊕{0}. Apply the First Isomorphism Theorem. 9.43 Define α : Z[x] → Z5 via α( f (x)) = [ f (0)], where the square brackets denote the congruence class in Z5 . If f (x), g(x) ∈ Z[x], then α( f (x) + g(x)) = [ f (0) + g(0)] = [ f (0)] + [g(0)] = α( f (x)) + α(g(x)), and similarly for multiplication. Thus, α is a homomorphism. If [a] ∈ Z5 , then letting f (x) be the constant polynomial a, we see that α( f (x)) = a; thus, α is onto. Furthermore, as f (0) is the constant term of f (x), we see that ker(α) is precisely I . Now apply the First Isomorphism Theorem. 9.45 The first part is the Third Isomorphism Theorem. To see the second part, note that 3Z/12Z = {0 + 12Z, 3 + 12Z, 6 + 12Z, 9 + 12Z} is a commutative ring having identity 9+12Z. Furthermore, its characteristic is 4. Thus, it has a subring isomorphic to Z4 . As the ring only has four elements, the ring is itself isomorphic to Z4 . 9.47 Note that (1 + i)(1 − i) = 2 ∈ (2), and yet neither 1 + i nor 1 − i is a multiple of 2 in R. The ideal is not prime and hence, as R is a commutative ring with identity, not maximal. 9.49 Let I be a prime ideal. Then R/I is an integral domain. But a finite integral domain is a field (see Theorem 8.10), so R/I is a field, and hence I is maximal. 9.51 Let R = 2Z4 = {0, 2} and I = {0}. Now, I is surely maximal, since if it got any larger, it would be R. But it is not prime, as 2 ∈ / I , but 2 · 2 = 0 ∈ I . 9.53 In a field, the only element that is not a unit is 0, and {0} is an ideal. In Z pn , we know (see Exercise 8.30) that the units are precisely the elements a that are relatively prime to p n . In other words, the elements that are not units are those that are divisible by p, so ( p) is the ideal in question. 9.55 Use P = R ⊕ I . As I \u0003= R, we see that P \u0003= R ⊕ R. Also, if (a, b)(c, d) ∈ P, then bd ∈ I . As I is prime, either b or d is in I , and hence (a, b) or (c, d) is in P.\n\nSolutions\n\n285\n\nProblems of Chapter 10 10.1 f (x) − g(x) = 9x 4 + 8x 3 + 4x 2 + 6x + 4, f (x)g(x) = 4x 7 + 7x 6 + 2x 5 + x 4 + 7x 2 + 4x + 5. 10.3 q(x) = 5x 2 + 6x + 3, r (x) = 4x 2 + 1. 10.5 No. According to the preceding exercise, x is not a unit. 10.7 Suppose that char R[x] = n > 0. Then, in particular, for every constant polynomial a, we have na = 0. Thus, 0 < char R ≤ n. On the other hand, if char R = m > 0, then for any f (x) ∈ R[x], we note that for each coefficient ai appearing in f (x), we have mai = 0; thus, m f (x) = 0, and 0 < char R[x] ≤ m. The only remaining case is where char R and char R[x] are both 0. 10.9 Certainly 0 ∈ S[x]. Take f (x), g(x) ∈ S[x]. Then all coefficients of f (x) and g(x) lie in S. The coefficients of f (x) − g(x) are differences of elements of S, and hence lie in S, so f (x) − g(x) ∈ S[x]. Similarly, the coefficients of f (x)g(x) are sums of products of elements of S and thus lie in S. Hence, S[x] is a subring. Let S be an ideal. If f (x) ∈ S[x] and g(x) ∈ R[x], then the coefficients of f (x)g(x) are sums of products, where each term in the sum is an element of S multiplied by an element of R, and therefore lies in S. Thus, f (x)g(x) ∈ S[x]. Similarly, g(x) f (x) ∈ S[x]. 10.11 As a and ab are associates, write ab = au, where u is a unit. If a \u0003= 0, then cancellation gives b = u. 10.13 Let a be a unit. Then for any 0 \u0003= b ∈ R, we have b = a(a −1 b). Thus, ε(a) ≤ ε(b), so ε(a) is indeed the smallest possible value, n. Now suppose that a is not a unit. We can write 1 = aq +r , where q, r ∈ R and either r = 0 or ε(r ) < ε(a). In the former case, a is a unit, which is a contradiction. In the latter case, ε(a) is not the smallest possible value. 10.15 We have \u0005 \u0004 \u0005 \u0004 7 3 3 13 2 19 + − x3 − x − x+ f (x) = g(x) 2 2 2 2 2 \u0005\u0004 \u0005 \u0004 \u0005 \u0004 3 4 46 72 2 144 216 13 2 19 7 x − x+ − x− + x + x+ , g(x) = − x 3 − 2 2 2 2 7 49 49 49 49\n\nand since 72 x 2 + 144 x + 216 divides − 27 x 3 − 13 x 2 − 19 x + 23 , the former is a gcd. We 49 49 49 2 2 must make it monic, so multiplying by 49/72, we get ( f (x), g(x)) = x 2 + 2x + 3. 10.17 Beginning with the second of the two equations in the solution to Exercise 10.15, we see that\n\n286\n\nSolutions\n\n\u0004 \u0004 \u0005\u0004 \u0005\u0005 7 3 4 46 49 13 19 g(x) − − x 3 − x 2 − x + − x− 72 2 2 2 2 7 49 \u0004 \u0004 \u0004 \u0005\u0005 \u0004 \u0005\u0005 3 4 46 49 g(x) − f (x) − g(x) − x− = 72 2 7 49 \u0005 \u0004 \u0005 \u0004 23 7 5 7 x+ + g(x) − x − . = f (x) 18 36 12 18\n\n( f (x), g(x)) =\n\n10.19 We must apply the Euclidean algorithm. Let us use the notation established in Example 10.8, taking u = 5 + 7i and v = 1 + 3i. Now, (1 + 3i)(1 − 3i) = 10, so uv−1 = (5 + 7i)(1 − 3i)/10 = 2.6 − 0.8i. Thus, we have m = 3 and n = −1, so q = 3 − i and r = (5 + 7i) − (1 + 3i)(3 − i) = −1 − i. That is, 5 + 7i = (1 + 3i)(3 − i) + (−1 − i). For the next step, we let u = 1 + 3i and v = −1 − i. But (−1 − i)(−1 + i) = 2, so uv−1 = (1 + 3i)(−1 + i)/2 = −2 − i. Therefore, 1 + 3i = (−1 − i)(−2 − i) + 0. Thus, −1 − i is a gcd of 5 + 7i and 1 + 3i.\n\n√ 10.21 Using the notation as in Example 10.15, we note that N (1 + 2 5i) = 21. √ If 1 + 2 5i = uv, then N (u)N (v) = 21, and assuming without loss of generality that N (u) ≤ N (v), we have N (u) = 1 or 3. As in Example 10.15, N (u) = √ 3 is 5i impossible and N (u) = 1 means u ∈ {1, −1}. In particular, u is a unit and 1 + 2 √ √ √ (1 + 2 5i)|21. is irreducible. However, (1 + 2 5i)(1 − 2 5i) = 21 = 3 · 7. Thus, √ But as N (3) =√9 and N (7) = 49, we cannot possibly have 1 + 2 5i dividing 3 or 7. Thus, 1 + 2 5i is not prime. 10.23 Combine the preceding two exercises with Theorem 10.11.\n\n10.25 Not necessarily. We know that Z is a Euclidean domain, but Z[x] is not a PID, hence not a Euclidean domain. 10.27 This is essentially the same as Exercise 2.24. 10.29 Suppose not, and let 0 \u0003= a ∈ R be a nonunit. Let In = (a n ). As a n |a n+1 , we see that In+1 ⊆ In . Suppose that In = In+1 . Then a n ∈ (a n+1 ); that is, a n = a n+1 b, for some b ∈ R. Cancelling a n , we get 1 = ab. Thus, a is a unit, giving us a contradiction. 10.31 Using the notation in Example 10.8, we have ε(1 + i) = 2. As we noted in that example, if u and v are in our ring, and uv = 1 + i, then ε(u)ε(v) = 2. But ε(u) and ε(v) are nonnegative integers, so without loss of generality, ε(u) = 1. This means u ∈ {±1, ±i}, and so u is a unit. Thus, 1 + i is irreducible. However, we know that R is a Euclidean domain, and hence a PID, so every irreducible is prime, by Theorem 10.11.\n\nSolutions\n\n287\n\n10.33 Not necessarily. We know that Q[x] is a UFD, and yet its subring R discussed in Example 10.18 is not. √ √ 10.35 Let a = 2, b = 5, c = 2 + 6i and d = 2 √ − 6i. Clearly ab = cd = 10. Defining a norm as in Example 10.15 via N (m + n 6i) = m 2 + 6n 2 , we see from the same calculation that N (uv) = N (u)N (v) for all u, v ∈ R. Suppose that uv = 2. Then N (u)N (v) = 4, so either N (u) = N (v) = 2 (which is impossible) or N (u) = 1 and N (u) = 4 (or vice versa). But this means u is 1 or −1. In particular, u is a unit, so 2 is irreducible. Similar calculations show that b, c and d are irreducible. It is immediate that neither a nor b divides c or d. That R is not a UFD follows from the definition. 10.37 Let p be an irreducible of R. Then p is a nonzero nonunit. Suppose that p|ab, for some a, b ∈ R. If a is a unit, then b and ab are associates, so p|b. If a = 0, then p|a. Similarly if b is zero or a unit. So let a and b be nonzero nonunits. We may write a = p1 · · · pk and b = q1 · · · ql , where the pi and q j are irreducible. By the preceding exercise, p divides some pi or some q j . Without loss of generality, say p|q1 . Since q1 |b, we have p|b.\n\nProblems of Chapter 11 11.1 (1) As 5 is a root and the degree is greater than 1, no. (2) Trying each possible root in Z7 , we see that this polynomial has no root there. Thus, since the degree is 3, it is irreducible. (3) No, since it factors as (x 2 + 4)(x 2 + 4). 11.3 The possibilities are x 3 + ax 2 + bx + c, where a, b, c ∈ {0, 1}. If c = 0, then 0 is a root, so c = 1. Also, 1 is a root of x 3 + 1 and x 3 + x 2 + x + 1, so we can rule them out. The remaining polynomials are x 3 + x 2 + 1 and x 3 + x + 1. Both have degree 3, and neither has a root, so they are irreducible. 11.5 Let h(x) = f (x)− g(x). If f (x) \u0003= g(x), then h(x) is not the zero polynomial. Say deg(h(x)) = n. Then h(x) can have at most n roots, but h(a) = f (a)−g(a) = 0 for all a ∈ F, giving us a contradiction. 11.7 No, take a, b ∈ R such that ab \u0003= ba. Let r = a, f (x) = x and g(x) = b. Then α( f (x)g(x)) = α(bx) = ba, whereas α( f (x))α(g(x)) = ab. 11.9 As deg(x 2 + 1) = 2, the polynomial is reducible if and only if it has a root m ∈ {0, 1, . . . , p − 1}. Factoring out x − m, we can only be left with x − n, for some n ∈ {0, 1, . . . , p − 1}. Thus, x 2 + 1 = x 2 − (m + n)x + mn. That is, x 2 + 1 is reducible if and only if there exist m, n ∈ {0, 1, . . . , p − 1} such that p|(m + n) and p|(mn − 1). Given the range of values for m and n, we can only have m + n ∈ {0, p}. But m and n cannot possibly both be 0, so m + n = p.\n\n288\n\nSolutions\n\n11.11 (1) The only possible rational roots are ±1, ±2. But none of these work, so it has no rational roots. (2) The possible rational roots are of the form m/n, where m|(−2) and n|6. Trying all of the possibilities, we see that −1/2 and 2/3 are roots. 11.13 (1) Looking first for rational roots, we know that they must be integers and divide 18. We find that 3 is a root, so we have (x − 3)(x 3 − 7x 2 + 14x − 6). Now, 3 is also a root of x 3 − 7x 2 + 14x − 6, so we have (x − 3)2 (x 2 − 4x + 2). By Eisenstein’s criterion, we are now done. (2) Looking first for rational roots, we see that they can only be ±1, ±2. In fact, −2 is a root, so we have (x + 2)(x 3 + x + 1). Now, the only possible rational roots of x 3 + x + 1 are 1 and −1, and these do not work. A degree 3 polynomial with no roots is irreducible, so we are done. 11.15 Note that f (x) is a constant polynomial if and only if f (x + a) is a constant polynomial. Thus, we may assume that both have degrees larger than 1. Suppose that f (x) = g(x)h(x), where g(x) and h(x) are nonconstant polynomials. Then f (x +a) = g(x +a)h(x +a). As g(x +a) and h(x +a) are nonconstant polynomials, it follows that f (x + a) is reducible. The converse is similar. 11.17 It is irreducible, using the preceding exercise with p = 7. 11.19 (1) We know that 2 − 3i must also be a root of f (x), so f (x) is divisible by (x − (2 + 3i))(x − (2 − 3i)) = x 2 − 4x + 13. Performing the division, we get f (x) = (x 2 − 4x + 13)(x − 7), so the third root is 7. (2) Here, 1 + i is also a root, so f (x) is divisible by (x − (1 − i))(x − (1 + i)) = x 2 − 2x + 2. Performing the division, we get f (x) = (x 2√− 2x + 2)(x 2 + √ 2x + 3). By the quadratic equation, the remaining roots are −1 + 2i and −1 − 2i. 11.21 (1) By Eisenstein’s criterion, the polynomial is irreducible over Q. In R[x], √ √ √ 4 4 2 we can factor it as (x − 10)(x + 10)(x + 10). In C[x], we factor further and √ √ √ √ get (x − 4 10)(x + 4 10)(x − 4 10i)(x + 4 10i). (2) Using the Rational Roots Theorem, we find that 2 is a root. Thus, we can factor it as (x − 2)(x 2 + 3x + 11). But x 2 + 3x + 11 is irreducible over R, hence over Q, so we are done in C, we use the quadratic equation and √those two cases. For √ get (x − 2)(x − (−3 + 35i)/2)(x − (−3 − 35i)/2). 11.23 The roots must also include 2 + 5i and 4 − i, so we can use (x − (2 − 5i))(x − (2 + 5i))(x − (4 + i))(x − (4 − i))(x − 6) = x 5 − 18x 4 + 150x 3 − 768x 2 + 2293x − 2958. 11.25 (1) Reducing modulo 5, we get x 3 + 2x + 1. We see that it has no roots in Z5 , and since it has degree 3, the polynomial is irreducible in Z5 [x], and hence f (x) is irreducible in Q[x]. (2) Reducing modulo 3, we get x 4 + x 2 + 2. This has no roots in Z3 , but we must rule out the possibility of a product of two polynomials of degree 2. We may assume\n\nSolutions\n\n289\n\nthat both such polynomials are monic, and we can only have something of the form (x 2 + ax + 1)(x 2 + bx + 2) = x 4 + x 2 + 2. Comparing coefficients, we find that a + b = 0 and 2a + b = 0. Thus, a = b = 0. But (x 2 + 1)(x 2 + 2) = x 4 + 2 \u0003= x 4 + x 2 + 2. Thus, our polynomial is irreducible in Z3 [x], and f (x) is irreducible in Q[x]. 11.27 The monic polynomials of degree 2 are precisely those of the form x 2 +ax +b, with a, b ∈ F. There are thus n 2 of them. Such a polynomial is reducible if and only if it factors as (x −c)(x −d), with c, d ∈ F. When c = d, there are n choices. If c \u0003= d, there are n choices for c and n − 1 for d. Of course, (x − c)(x − d) = (x − d)(x − c), so we get n(n − 1)/2 possibilities, for a total of n + n(n − 1)/2 = n(n + 1)/2 reducible polynomials. By unique factorization, all of them are distinct. Thus, the number of irreducibles is n 2 − n(n + 1)/2 = n(n − 1)/2. 11.29 (1) x 4 + 1 = (x 2 + a)(x 2 − a). (2) x 4 + 1 = (x 2 + ax − 1)(x 2 − ax − 1). (3) x 4 + 1 = (x 2 + ax + 1)(x 2 − ax + 1).\n\nProblems of Chapter 12 12.1 No, as x n + 1 and x n lie in V but their difference, 1, does not. 12.3 We have 0 ∈ U and 0 ∈ W , so 0 ∈ U ∩ W . If v1 , v2 ∈ U ∩ W , then v1 , v2 ∈ U , so v1 + v2 ∈ U . Similarly, v1 + v2 ∈ W , so v1 + v2 ∈ U ∩ W . If a ∈ F, then av1 ∈ U and av1 ∈ W , so av1 ∈ U ∩ W . The argument for an arbitrary collection of subspaces is similar. 12.5 As 0 ∈ U , we have 0 = α(0) ∈ α(U ). (This follows immediately from the fact that α is, by definition, a homomorphism of additive groups.) Also, if α(u 1 ), α(u 2 ) ∈ α(U ) and a ∈ F, then α(u 1 ) + α(u 2 ) = α(u 1 + u 2 ) ∈ α(U ), since u 1 + u 2 ∈ U , and aα(u 1 ) = α(au 1 ) ∈ α(U ), since au 1 ∈ U . 12.7 It is. As 2 · 0 + 3 · 0 + 7 · 0 = 0, we see that (0, 0, 0) ∈ W . Suppose that (a1 , b1 , c1 ), (a2 , b2 , c2 ) ∈ W and a ∈ F. Then 2(a1 +a2 )+3(b1 +b2 )+7(c1 +c2 ) = (2a1 +3b1 +7c1 )+(2a2 +3b2 +7c2 ) = 0+0 = 0, so (a1 +a2 , b1 +b2 , c1 +c2 ) ∈ W . Also, 2aa1 + 3ab1 + 7ac1 = a(2a1 + 3b1 + 7c1 ) = a · 0 = 0; thus, (aa1 , ab1 , ac1 ) ∈ W. 12.9 We have v + v + v = 1v + 1v + 1v = (1 + 1 + 1)v = 0v = 0. 12.11 (1) As 3(1, 3, 5) + 2(2, 1, 4) − 1(7, 11, 23) = (0, 0, 0), they are linearly dependent. (2) Suppose that a(1, 3, 4)+b(2, 2, 1)+c(3, 6, 3) = (0, 0, 0). Then a+2b+3c = 3a + 2b + 6c = 4a + b + 3c = 0. Thus, 3a − b = a − 2b = 0. We see immediately that a = b = 0, and hence c = 0. Therefore, the vectors are linearly independent.\n\n290\n\nSolutions\n\n12.13 (1) No. If they did, then as (1, 0, 2) + (2, 5, 3) = (3, 5, 5), the vectors are linearly dependent, which means that some proper subset would form a basis for Q3 . But Q3 is 3-dimensional over Q, so this is impossible. (2) Yes. We claim that the vectors are linearly independent. If a(1, 0, 2) + b(2, 3, 5) + c(0, 0, 4) = (0, 0, 0), we see immediately that b = 0, from which it follows that a = 0 and then c = 0. Thus, we can add vectors to this set to find a basis for Q3 . But again, we are in a space with dimension 3, so no more vectors can be added. Therefore, the vectors span the space. 12.15 If the field is C, we can see that every unique and \u0004 matrix \u0005 \u0004 can \u0005 be\u0004written \u0005 \u0004in a \u0005 10 01 00 00 obvious way as a linear combination of , , , , so these 00 00 10 01 matrices dimension is 4. Working over R, we would also need \u0004 \u0005 \u0004form\u0005a basis \u0004 \u0005and\u0004the \u0005 i 0 0i 00 00 , , , , so the dimension is 8. 00 00 i 0 0i 12.17 Let dim V = n. If n = 0, then V = {0} and the only possible subspace is {0}, so there is nothing to do. So assume that n ≥ 1. If W = {0}, then again, there is nothing to do. So assume that there exists 0 \u0003= w1 ∈ W . Then w1 is, by itself, linearly independent. If w1 spans W , then we have a basis for W . If not, then there exists a w2 ∈ W such that w2 is not a scalar multiple of w1 . But now w1 and w2 are linearly independent. If they span W , we have a basis. Otherwise, find w3 ∈ W such that w3 is not a linear combination of w1 and w2 . Repeat this procedure. We cannot possibly go beyond wn , as V cannot have n + 1 linearly independent vectors. Thus, W has a basis consisting of at most n elements, so dim W ≤ dim V . If W \u0003= V , then we can add to the basis for W to obtain a basis for V , which means we must have dim W < dim V . 12.19 Suppose that a1 α(v1 ) + a2 α(v2 ) + · · · + an α(vn ) = 0, for some ai ∈ F. Then α(a1 v1 + · · · + an vn ) = 0. As α is one-to-one, a1 v1 + · · · + an vn = 0. But the vi are linearly independent. Thus, a1 = · · · = an = 0. √ √ √ 12.21 Let a = 5 + 7. Then a 2 = 12 + 2 35, so (a 2 − 12)2 = 140. Thus, a satisfies f (x) = x 4 −24x 2 +4. We must show that f (x) is irreducible over Q. If it has a root in Q, then by the Rational Roots Theorem, the root must lie in {±1, ±2, ±4}. But none of these work. The only other possibility is that f (x) is the product of two polynomials of degree 2. By Theorem 11.4, they may be assumed to be in Z[x]. Up to a factor of −1, and noting that there is no x 3 term in f (x), the factorization must be (x 2 + bx + c)(x 2 − bx + d), for some b, c, d ∈ Z. As there is no x term in f (x), either b = 0 or c = d. If b = 0, we have c + d = −24 and cd = 4. No integers can possibly satisfy these equations. So, assume that c = d. We are left with the cases (x 2 + bx + 2)(x 2 − bx + 2) and (x 2 + bx − 2)(x 2 − bx − 2), for some integer b. These possibilities yield, respectively, 4 − b2 = −24 and −4 − b2 = −24. Neither of these equations has a solution in Z. 12.23 Suppose that [K : F] = n. If a ∈ K , then 1, a, a 2 , . . . , a n are linearly dependent over F, by Lemma 12.1. Thus, there exist bi ∈ F, not all zero, such that b0 + b1 a + b2 a 2 + · · · + bn a n = 0. That is, a is a root of b0 + b1 x + · · · + bn x n .\n\nSolutions\n\n291\n\n12.25 Let L = ∞ n=1 Fn . As 1 ∈ F1 , we have 1 ∈ L. Suppose that r, s ∈ L. Then r ∈ Fm , s ∈ Fn , for some m, n ∈ N. Letting k be the larger of m and n, we have r, s ∈ Fk . Thus, r − s ∈ Fk ⊆ L and, if s \u0003= 0, r s −1 ∈ Fk ⊆ L. 12.27 As a ∈ F(a) and F(a) is a field, we must have a 2 ∈ F(a). Also, F ⊆ F(a). As F(a 2 ) is the intersection of all subfields of K containing F and a 2 , it follows that F(a 2 ) ⊆ F(a). For the second part, let F = Q and a = i. Then Q(a 2 ) = Q(−1) = Q, but Q(a) contains i, so the fields are different. 12.29 The minimal polynomial of a is irreducible over C. By the Fundamental Theorem of Algebra, this minimal polynomial has degree 1, and must therefore be x − a ∈ C[x]; thus, a ∈ C. 12.31 Note that f (x) = x 3 + x + 1 is irreducible over Z7 . (It has degree 3 and no roots in Z7 .) Thus, F = Z7 [x]/( f (x)) will work. Letting a = x + ( f (x)), we know that the elements of F are the linear combinations of 1, a and a 2 over Z7 . Also, a is a root of f (x), so a 3 = −a − 1 = 6a + 6 and a 4 = (6a + 6)a = 6a 2 + 6a. Thus, (a 2 + 5a + 4)(3a 2 + 6) = 3a 4 + a 3 + 4a 2 + 2a + 3 = 3(6a 2 + 6a) + (6a + 6) + 4a 2 + 2a + 3 = a 2 + 5a + 2. 12.33 If a and b are any two roots of x 3 − 2, then (ab−1 )3 = 1, so ab−1 is one of is ω. Also, (ω2 )3√= 1, so the roots of x 3 − 1. One such root in C is 1 and another √ 3 2 3 ω is the third complex root of x − 1. Thus, since Q( 2, ω) must contain 3 2, 1, ω and√ω2 , we see that it contains every root of x 3 − 2; in particular, x 3 − 2 splits over over any subfield,√then that subfield Q( 3 2, ω). On the other hand, if x 3 − 2 splits √ √ 3 2 and ω2 3 2. As it is a field, would have to contain all three roots, namely, 3 2, ω √ this means it must contain ω as well, so it is all of Q( 3 2, ω). √ √ √ 12.35 Note that Q( 2) is a splitting field of x 2 −2 over Q. (Both roots, 2 and − 2 are in the field, and would have to be in any splitting field.) As an automorphism α must map the identity to the identity, we see immediately that α(c) = c for all c ∈ Z. Similarly, if m, n ∈ Z with n > 0, then m = α(m) = α(n(m/n)) = nα(m/n). Thus, √ 2) must be a root of x 2 − 2; α(c) = c for all√ c ∈ Q. By the preceding exercise, α( √ √ in particular, α( 2) ∈ {√ 2, − 2}. In √ the former case, α is the identity function. In the latter case, α(a + b 2) = a − b 2 for all a, b ∈ Q. By Lemma 12.4, this is an automorphism. 12.37 Let K be a splitting field for f (x) over F. Say that in K [x], we have f (x) = a(x −a1 )(x −a2 ) · · · (x −an ). Then g(x) = a(x +1−a1 )(x +1−a2 ) · · · (x +1−an ) = a(x − (a1 − 1))(x − (a2 − 1)) · · · (x − (an − 1)). Since the ai lie in K , so do the ai − 1; thus, g(x) splits over K . Furthermore, for g(x) to split, all of the ai − 1 must be present, and hence so must all of the ai . Thus, we cannot make K any smaller and have g(x) split, so K is a splitting field for g(x). Showing that splitting fields for g(x) must be splitting fields for f (x) involves a similar argument. 12.39 If |F| = p n , for some prime p and positive integer n, then F has one proper subfield for each integer m, 1 ≤ m < n, with m|n. The first value n that works is 6, so the smallest such field has order 26 = 64. Specifically, it is the splitting field of x 64 − x over Z2 .\n\n292\n\nSolutions\n\n12.41 Let a ∈ K be a root of f (x). Then [Z5 (a) : Z5 ] = 3. If all roots of f (x) lie in Z5 (a), then K = Z5 (a), and |K | = 53 . Otherwise, in Z5 (a)[x], we have f (x) = (x − a)g(x), where g(x) is an irreducible polynomial of degree 2. Letting b be a root of g(x) in K , we see that [Z5 (a, b) : Z5 (a)] = 2. Furthermore, in Z5 (a, b), the polynomial f (x) splits into linear factors, so K = Z5 (a, b). Now, [K : Z5 ] = [K : Z5 (a)][Z5 (a) : Z5 ] = 2 · 3 = 6, and |K | = 56 . 12.43 Every field of characteristic 0 is perfect, so char F = p, for some prime p. The fact that f (x) = a0 + a p x p + · · · + amp x mp follows exactly as in the proof of Theorem 12.16. Suppose that all of the ai are algebraic over the prime subfield, (an isomorphic copy of) Z p . Then [Z p (a0 ) : Z p ] < ∞. Also, a p is algebraic over Z p , and hence over Z p (a0 ), so [Z p (a0 , a p ) : Z p (a0 )] < ∞. Thus, [Z p (a0 , a p ) : Z p ] = [Z p (a0 , a p ) : Z p (a0 )][Z p (a0 ) : Z p ] < ∞. In the same way, [Z p (a0 , a p , . . . , amp : Z p ] < ∞, which means that Z p (a0 , a p , . . . , amp ) is a finite field, and hence perfect. If f (x) is irreducible over F, it is surely irreducible over Z p (a0 , . . . , amp ). An irreducible polynomial over a perfect field cannot have multiple roots in any extension field. 12.45 If it were cyclic, it would be infinite cyclic. But note that −1 ∈ U (F), and −1 has order 2. An infinite cyclic group has no such element. 12.47 Let F be the splitting field of x 125 − x over Z5 . We know that it has order 125. Let f (x) ∈ Z5 [x] be an irreducible factor of x 125 − x. If a ∈ F is a root of f (x), then [Z5 (a) : Z5 ] = deg( f (x)). But Z5 (a) is a subfield of F. A subfield of a field of order 53 can only have order 5 or 53 . Thus, deg( f (x)) = 1 or 3.\n\nProblems of Chapter 13 13.1 WKHWUHDVXUHLVEXULHGWZHQWBSDFHVQRUWKRIWKHSDOPWUHH 13.3 We need k to be relatively prime to 26. If it is not, then letting d = (26, k), we see that both 0 and 26/d will be encrypted as 0, so decryption will be impossible. On the other hand, if (k, 26) = 1, then k ∈ U (26), so we can decrypt by multiplying by k −1 . (If k ≡ 1 (mod 26), then multiplying by k does not change the text at all, so it would be reasonable to rule out this key as well.) 13.5 JGVSHNEGJESCRPPRBSXBPPVGHBSJKEHXVT 13.7 KXTNRHIOQJHVKWNKSVNHSWOXCLFAAMJSKSBO √ 13.9 Writing n = pq, the smaller of p and q must certainly be less than n, so we only need to try primes up to 44. We discover that p = 37 and q = 53. Thus, ϕ(n) = 36 · 52 = 1872. To find d, we use the Euclidean algorithm. In particular, 1872 = 43(43) + 23; 43 = 23(1) + 20; 23 = 20(1) + 3; 20 = 3(6) + 2;\n\nSolutions\n\n293\n\n3 = 2(1)+1; 2 = 1(2)+0. Thus, 1 = 3(1)+2(−1) = 3(1)+(20(1)+3(−6))(−1) = 20(−1) + 3(7) = 20(−1) + (23(1) + 20(−1))(7) = 23(7) + 20(−8) = 23(7) + (43(1)+23(−1))(−8) = 43(−8)+23(15) = 43(−8)+(1872(1)+43(−43))(15) = 1872(15) + 43(−653). Therefore, 43(−653) ≡ 1 (mod 1872). As we need d to be positive, adding 1872, we get d = 1219. 13.11 We must break our message into blocks of length 2. As we have an odd number of letters, we add a Q to the end. Then AL is 0011, GE is 0604, BR is 0117 and AQ is 0016. Next, 11149 ≡ 5581 (mod 17399), 604149 ≡ 2315 (mod 17399), 117149 ≡ 4926 (mod 17399) and 16149 ≡ 9527 (mod 17399), so our encrypted message consists of the four numbers 5581, 2315, 4926 and 9527. 13.13 Note that n = 103·179 = 18437 and ϕ(n) = 102·178 = 18156. To find d, we apply the Euclidean algorithm. Namely, 18156 = 151(120) + 36; 151 = 36(4) + 7; 36 = 7(5) + 1; 7 = 1(7) + 0. Thus, 1 = 36(1) + 7(−5) = 36(1) + (151(1) + 36(−4))(−5) = 151(−5) + 36(21) = 151(−5) + (18156(1) + 151(−120))(21) = 18156(21) + 151(−2525). Therefore, −2525e ≡ 1 (mod ϕ(n)). As we need d to be positive, we add 18156 and get d = 15631. We now calculate 246915631 ≡ 1514 (mod 18437), 709315631 ≡ 1124 (mod 18437), 1477315631 ≡ 1314 (mod 18437), 1090015631 ≡ 1208 (mod 18437) and 14315631 ≡ 11 (mod 18437) (which we remember to write as 0011). So, our message is 15141124131412080011, which translates to POLYNOMIAL.\n\nProblems of Chapter 14 14.1 Construct the line through A and B. Next, construct the circle centred at A with radius AB. Say it meets the line at B and C. Construct the circle centred at B with radius AB, and say that it meets the line at A and E. Then the distance from C to E is 3, so if we construct the perpendicular bisector of C E, and it meets the line at D, then the distance from C to D is 1.5. 14.3 Construct the circle centred at A with radius AB and the circle centred at B with radius AB. Let C be either of the intersection points of these circles. By construction, the three sides of ABC have the same length. 14.5 We begin by constructing the line through B and A. Next, construct the circle centred at B with radius BC. It meets the line through B and A at two points; let E be the one of those points on the same side of B as A. Then replacing A with E, we may assume that in our original angle, A and C were equidistant from B. Construct the line through A and C, then construct the perpendicular bisector of AC. These two lines meet at the desired point, D.\n\n294\n\nSolutions\n\n14.7 Proceeding as in the solution to Exercise 14.3, construct a point E such that AB E is an equilateral triangle. It must lie on the circle. Now do the same thing with A and E; that is, construct a point C (the one that is different from B) such that AEC is an equilateral triangle. Again, C must be on the circle. Now do the same with A and C, and construct a new point F on the circle such that AC F is an equilateral triangle. Performing the same construction for A and F, we obtain a new point D on the circle such that AF D is an equilateral triangle. And then in the same way, we can construct a point G on the circle such that ADG is an equilateral triangle. But now B EC F DG is a regular hexagon, so BC D is an equilateral triangle. 14.9 As the constructible numbers form a field, if a + b were constructible, then a+b−a = b would also be constructible, giving a contradiction. Similarly, if ab were constructible, then the field of constructible numbers would include a −1 ab = b. Now, if −b were constructible then b would be constructible as well, so letting c = −b, we see that b + c = 0 is constructible. On the other hand, if we let c = b, then we get b + c = 2b. If this were constructible, then since 1/2 is also constructible, we would find that b would be constructible as well. 14.11 (1) Yes. As all integers are constructible and the field of constructible numbers is closed √ under √ the taking of square roots of its nonnegative elements, we see that 2 + 5 − 3 is constructible, and then we can take the square root twice to obtain this element. √ √ a root of x 3 − 3. (2) No. Once again, we know that 3 is constructible, but 3 3 is √ By Eisenstein’s criterion, this polynomial is irreducible √ over Q, so 3 3 has minimal polynomial x 3 − 3. As the degree is not a power of 2, 3 3 is not constructible. By Exercise 14.9, the sum of a number that is constructible with one that is not is not constructible. 14.13 By Eisenstein’s criterion, this polynomial is irreducible over Q. Thus, it is the minimal polynomial of a over Q. As the degree is not a power of 2, a is not constructible. 14.15 We will prove a stronger statement, that an angle of π/6 can be constructed. We are given the points (0, 0) and (1, 0). To√ obtain such an angle, we only need to construct the point √ (cos(π/6), sin(π/6)) = ( 3/2, 1/2). But by Theorems 14.1 and 14.2, the numbers 3/2 and 1/2 are constructible, so the point is constructible. 14.17 There is nothing to do for n = 1. When n = 2, we have cos(2θ ) = 2 cos2 (θ )− 1. For the n = 3 case, we look to the proof of Theorem 14.6, and see that cos(3θ ) = 4 cos3 (θ ) − 3 cos(θ ). To handle the remaining cases, we simply note that cos(θ ) = cos(−θ ) = cos(2π − θ ). Thus, cos(4θ ) = cos(8π/7) = cos(6π/7) = cos(3θ ) and, similarly, cos(5θ ) = cos(2θ ) and cos(6θ ) = cos(θ ).\n\nSolutions\n\n295\n\n√ 14.19 Let D = A. √ We know that the number 2 is constructible, which means that the point E = ( 2, 0) is constructible. Construct the circle centred at A and passing through C. It will meet the x-axis (which we can construct) at (m, 0), where √ m is the distance from A to C. Thus, the number m is constructible, and so m 2 √ is constructible. In particular, we can construct √ the point (m 2, 0). Now draw the the line through A circle centred at A and passing through (m 2, 0). It intersects √ and C at a point F, where the distance from A to F is m 2 and F is on the same side of A as C. The triangles ABC and D E F are similar. As the side lengths are √ increased by a factor of 2, the area is increased by a factor of 2.\n\nIndex\n\nA Abel, 38 Abelian group, 38, 135 Absorption, 149 Additive cipher, 233 Additive identity, 26, 29, 135 Additive inverse, 26, 29 Additive notation, 42, 45, 135 Adleman, 236 Algebraic, 217 Algebraically closed, 200 Alternating group, 107 Ascending chain condition, 183 Associate, 180 Associative, 26, 29, 36, 38, 135 Automorphism inner, 82 of groups, 81 of rings, 163 power, 84 Automorphism group, 81\n\nB Basis, 212 Bijective, 11 Binary operation, 12 Binomial Theorem, 17\n\nC Caesar cipher, 233 Cancellation law, 43, 143 Cartesian product, 4 Cauchy, 92 Cauchy’s theorem, 125 for abelian groups, 92\n\nCayley, 101 Cayley’s theorem, 101 Centralizer, 115 Centre, 50, 141 Characteristic, 147 Chinese Remainder Theorem, 30 Cipher additive, 233 Caesar, 233 multiplicative, 235 simple substitution, 234 Class equation, 120 Closed, 26, 29, 36, 38, 135 Cocks, 236 Collapsing compass, 252 Commutative, 26, 29 Commutative ring, 136 Compass, 241 collapsing, 252 Complex numbers, 3, 253 Composite, 24 Composition, 11 Congruence class, 28 Congruent, 27 modulo a subgroup, 57 Conjugacy class, 119 Conjugate, 47, 120 Constant polynomial, 172 Constructible circle, 245 Constructible line, 245 Constructible number, 245, 249 Constructible point, 245 Content, 195 Coset left, 58, 152 right, 59 Cycle, 102\n\n© Springer International Publishing AG, part of Springer Nature 2018 G. T. Lee, Abstract Algebra, Springer Undergraduate Mathematics Series, https://doi.org/10.1007/978-3-319-77649-1\n\n297\n\n298 Cycle notation, 102 Cyclic group, 45, 54, 74 Cyclic subgroup, 50\n\nD Decomposable group, 97 Degree of a field extension, 215 of a polynomial, 172 de Moivre, 254 de Moivre’s theorem, 254 Derivative, 225 Descartes, 4 Determinant, 260 Dihedral group, 52 Dimension, 213 Direct product, 40 external, 85 internal, 85 Direct sum, 136 Disjoint cycle decomposition, 102, 103 Disjoint cycles, 102 Distributive law, 26, 29, 135 Divisible, 20, 177, 193 Divisible group, 98 Division algorithm, 19 for polynomials, 174 Doubling the cube, 242, 250\n\nE Eisenstein, 196 Eisenstein’s criterion, 197 Element, 3 Elementary divisors, 93, 94 Empty set, 3 Equivalence class, 7 Equivalence relation, 6 Euclid, 21, 24 Euclidean algorithm, 22 for Euclidean domains, 178 Euclidean domain, 176 Euclidean function, 176 Euclid’s lemma, 24 Euler, 55 Euler phi-function, 55 Evaluation, 191 Even permutation, 106 Extension field, 208 finite, 215 quadratic, 215 simple, 217 External direct product, 85\n\nIndex F Factor group, 65 Factorial, 17 Factor ring, 152 Factor Theorem, 193 Fibonacci sequence, 19 Field, 144 extension, 208 finite, 215 quadratic, 215 simple, 217 finite, 227 Galois, 227 imperfect, 227 of fractions, 162 of quotients, 162 perfect, 226 splitting, 222 Finite-dimensional, 213 Finite extension, 215 Finite field, 227 Finite group, 45 Finite order, 46 First Isomorphism Theorem for Groups, 78 First Isomorphism Theorem for Rings, 165 First Sylow Theorem, 122 Flip, 52 Formal derivative, 225 Freshman’s Dream, 147 Function, 10 bijective, 11 injective, 10 one-to-one, 10 onto, 11 surjective, 11 Fundamental Theorem of Algebra, 200 Fundamental Theorem of Arithmetic, 24 Fundamental Theorem of Finite Abelian Groups, 91\n\nG Galois, 108 Galois field, 227 Gauss, 195 Gaussian integers, 177 Gauss’s lemma, 195 Gcd, 20, 178 General linear group, 40 Generator, 45, 49, 151 Greatest common divisor, 20, 178 Group, 38 abelian, 38, 135 alternating, 107 automorphism, 81\n\nIndex cyclic, 45, 54, 74 decomposable, 97 dihedral, 52 divisible, 98 factor, 65 finite, 44 general linear, 40 indecomposable, 97 infinite, 45 inner automorphism, 82 of complex numbers, 38 of integers, 38 of integers modulo n, 38 of integers relatively prime to n, 39 of order 2 p, 75 of order 4, 75 of order 8, 129 of order 12, 132 of order 15, 128 of order p 2 , 120 of order p n , 120, 125 of order pq, 125 of order pqr , 126 of prime order, 74 of rational numbers, 38 of real numbers, 38 of units, 143 quaternion, 129 quotient, 65 simple, 108 special linear, 63 symmetric, 37, 40, 101 trivial, 39 Group automorphism, 81 Group homomorphism, 69 Group identity, 38 Group isomorphism, 72 Group operation, 38 Group table, 38\n\nH Homomorphism of groups, 69 of rings, 155\n\nI Ideal, 149 maximal, 167 prime, 169 principal, 151 Identity, 36, 38, 136 additive, 26, 29, 135 group, 38\n\n299 multiplicative, 26, 29, 136 Identity matrix, 259 Image, 71 Imperfect field, 227 Indecomposable group, 97 Index, 58 Induction, 16 strong, 18 Infinite-dimensional, 213 Infinite group, 45 Infinite order, 46 Injective, 10 Inner automorphism, 82 Inner automorphism group, 82 Integers, 3 Integers modulo n, 28 Integral domain, 143 Internal direct product, 85 Intersection, 4 Invariant factor, 95 Invariant factor decomposition, 95 Inverse, 36, 38, 143, 260 additive, 26, 29 multiplicative, 26 Inverse image, 71 Invertible matrix, 260 Involution, 165 symplectic, 165 transpose, 165 Irreducible, 184 Irreducible polynomial, 191 Isomorphic groups, 72 Isomorphic rings, 159 Isomorphism of groups, 72 of rings, 159\n\nJ Jordan, 108\n\nK Kernel, 69, 156 Key, 233 private, 235 public, 235\n\nL Lagrange, 58 Lagrange’s theorem, 58 Leading coefficient, 172 Leading term, 172 Least common multiple, 104\n\n300 Left coset, 58, 152 Linear combination, 210 Linear dependence, 211 Linear independence, 211 Linear transformation, 210\n\nM Mathematical induction, 16 strong, 18 Matrix, 257 identity, 259 invertible, 260 zero, 257 Matrix ring, 136 Maximal ideal, 167 Minimal polynomial, 218, 219 Modular arithmetic, 27 Monic polynomial, 172 Motzkin, 183 Multiple, 20 Multiple root, 225 Multiplicative cipher, 235 Multiplicative identity, 26, 29, 136 Multiplicative inverse, 26 Multiplicative notation, 42\n\nN Natural numbers, 3 N /C Theorem, 117 Nonempty set, 3 Norm, 184 Normalizer, 116 Normal subgroup, 61, 63\n\nO Odd permutation, 106 One time pad, 234 One-to-one, 10 One-to-one correspondence, 11 Onto, 11 Order finite, 46 infinite, 46 of a group, 45 of a group element, 46\n\nP Partition, 7 p-element, 88 Perfect field, 226 Permutation, 12, 35\n\nIndex even, 106 odd, 106 p-group, 88, 125 Prüfer, 98 PID, 182 Polynomial, 171 constant, 172 irreducible, 191 minimal, 218, 219 monic, 172 primitive, 195 reducible, 191 zero, 172 Polynomial degree, 172 Polynomial ring, 136, 172 Power, 45 Power automorphism, 84 Preimage, 71 Prime, 24, 183 Prime factorization, 24 Prime ideal, 169 Prime number, 24 Prime subfield, 166 Primitive nth root of unity, 254 Primitive polynomial, 195 Principal ideal, 151 Principal ideal domain, 182 Private key, 235 Proper subgroup, 49 Proper subset, 4 Proposition, 15 Prüfer, 98 Prüfer p-group, 98 Public key, 235 Purely imaginary, 253\n\nQ Quadratic extension, 215 Quaternion group, 129 Quotient, 20 Quotient group, 65 Quotient ring, 152\n\nR Rational numbers, 3 Rational Roots Theorem, 195 Real numbers, 3 Reducible polynomial, 191 Reflexive, 5 Relation, 5 equivalence, 6 reflexive, 5 symmetric, 5\n\nIndex transitive, 6 Relative complement, 4 Relatively prime, 20 Remainder, 20 Remainder Theorem, 192 Right coset, 59 Ring, 135 commutative, 136 factor, 152 of complex numbers, 136 of integers, 136 of integers modulo n, 136 of matrices, 136 of polynomials, 136, 172 of rational numbers, 136 of real numbers, 136 quotient, 152 with identity, 136 Ring automorphism, 163 Ring homomorphism, 155 Ring isomorphism, 159 Rivest, 236 Root, 192 multiple, 225 Rotation, 52 RSA scheme, 236 S Scalar multiplication, 207 Schönemann, 196 Second Isomorphism Theorem for Groups, 79 Second Isomorphism Theorem for Rings, 166 Second Sylow Theorem, 123 Set, 3 empty, 3 nonempty, 3 Set difference, 4 Shamir, 236 Simple extension, 217 Simple group, 108 Simple substitution cipher, 234 Span, 211 Special linear group, 63 Splits, 221 Splitting field, 222 Squaring the circle, 242, 250 Straightedge, 241 Strong induction, 18 Subfield, 145 prime, 166 Subgroup, 48 cyclic, 49\n\n301 normal, 61, 63 of a cyclic group, 54 of index 2, 62 proper, 49 Subring, 140 unital, 141 Subset, 4 proper, 4 Subspace, 209 Surjective, 11 Sylow, 122 Sylow p-subgroup, 122 Sylow theorems, 122 Symmetric, 5 Symmetric group, 37, 40, 101 Symmetry of a polygon, 52 Symplectic involution, 165\n\nT Third Isomorphism Theorem for Groups, 80 Third Isomorphism Theorem for Rings, 167 Third Sylow Theorem, 123 Transcendental, 217 Transitive, 6 Transpose, 165 Transposition, 105 Trisecting the angle, 243, 251 Trivial group, 39\n\nU UFD, 185 Union, 4 Unique factorization domain, 185 Unit, 143 Unital subring, 141 Unit group, 143\n\nV Vector space, 207 finite-dimensional, 213 infinite-dimensional, 213\n\nW Well Ordering Axiom, 15\n\nZ Zero divisor, 142 Zero matrix, 257 Zero polynomial, 172" ]
[ null ]
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https://dispersivewiki.org/DispersiveWiki/index.php?title=Hartree_equation&amp;action=edit
[ "# Hartree equation\n\nThe Hartree equation is of the form\n\n$i\\partial _{t}u+\\Delta u=V(u)u\\,$", null, "where\n\n$V(u)=\\pm |x|^{-n}*|u|^{2}\\,$", null, "and $0", null, ". It can thus be thought of as a non-local cubic Schrodinger equation; the cubic NLS is in some sense a limit of this equation as $n\\rightarrow d,$", null, "(after suitable normalization of the kernel $|x|^{-n}\\,$", null, ", which would otherwise blow up). The analysis divides into the short-range case $n>1\\,$", null, ", the long-range case $0", null, ", and the borderline (or critical) case $n=1\\,.$", null, "Generally speaking, the smaller values of $n\\,$", null, "are the hardest to analyze. The + sign corresponds to defocusing nonlinearity, the - sign corresponds to focusing.\n\nThe $H^{1}\\,$", null, "critical value of $n\\,$", null, "is 4, in particular the equation is always subcritical in four or fewer dimensions. For $n<4\\,$", null, "one has global existence of energy solutions. For $n=4\\,$", null, "this is only known for small energy.\n\nIn the short-range case one has scattering to solutions of the free Schrodinger equations under suitable assumptions on the data. However this is not true in the other two cases HaTs1987. For instance, in the borderline case, at large times t the solution usually resembles a free solution with initial data $y$", null, ", twisted by a Fourier multiplier with symbol $e^{iV({\\hat {y}})\\log(t)}$", null, ". (This can be seen formally by applying the pseudo-conformal transformation, discarding the Laplacian term, and solving the resulting ODE GiOz1993.) This creates modified wave operators instead of ordinary wave operators. A similar thing happens when $1/2n<1\\,$", null, "but $\\log(t)\\,$", null, "must be replaced by $t^{n-1}/(n-1)\\,.$", null, "The existence and mapping properties of these operators is only partly known:\n\n• When $n\\geq 2\\,$", null, "and $n=1\\,,$", null, "the wave operators map ${\\hat {H^{s}}}$", null, "to ${\\hat {H^{s}}}\\,$", null, "for $s>1/2\\,$", null, "and are continuous and open Na-p3 (see also GiOz1993)\n• For $n>1\\,$", null, "and $n\\geq 1\\,$", null, "this is in NwOz1992\n• In the defocusing case, all solutions in suitable spaces have asymptotic states in $L^{2}\\,$", null, ", and one has asymptotic completeness when $n>4/3\\,$", null, "HaTs1987.\n• For $n<1,n\\geq 3\\,,$", null, "and $1-n/2", null, "this is in Na-p4\n• In the Gevrey and real analytic categories there are some large data results in GiVl2000, GiVl2000b, GiVl2001, covering the cases $n\\leq 1\\,$", null, "and $n\\geq 1.\\,$", null, "• For small decaying data one has some invertibility of the wave operators HaNm1998\n\nA variant of the Hartree equations is the Schrodinger-Poisson system." ]
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https://usense.be/38107/calculate-vibrating-feeder-capacity.html
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https://answers.everydaycalculation.com/lcm/315-240
[ "Solutions by everydaycalculation.com\n\n## What is the LCM of 315 and 240?\n\nThe lcm of 315 and 240 is 5040.\n\n#### Steps to find LCM\n\n1. Find the prime factorization of 315\n315 = 3 × 3 × 5 × 7\n2. Find the prime factorization of 240\n240 = 2 × 2 × 2 × 2 × 3 × 5\n3. Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the lcm:\n\nLCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7\n4. LCM = 5040\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to find LCM of upto four numbers in your own time:" ]
[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]
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https://latestart.savingadvice.com/2019/02/
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Array\n(\n => 14.248.77.166\n)\n\n => Array\n(\n => 157.35.224.133\n)\n\n => Array\n(\n => 183.83.38.27\n)\n\n => Array\n(\n => 182.68.4.77\n)\n\n => Array\n(\n => 122.177.130.234\n)\n\n => Array\n(\n => 103.24.99.99\n)\n\n => Array\n(\n => 103.91.127.66\n)\n\n => Array\n(\n => 41.90.34.240\n)\n\n => Array\n(\n => 49.205.77.102\n)\n\n => Array\n(\n => 103.248.94.142\n)\n\n => Array\n(\n => 104.143.92.170\n)\n\n => Array\n(\n => 219.91.157.114\n)\n\n => Array\n(\n => 223.190.88.22\n)\n\n => Array\n(\n => 223.190.86.232\n)\n\n => Array\n(\n => 39.41.172.80\n)\n\n => Array\n(\n => 124.107.206.5\n)\n\n => Array\n(\n => 139.167.180.224\n)\n\n => Array\n(\n => 93.76.64.248\n)\n\n)\n```\nArchive for February, 2019: Latestart's Personal Finance Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Archive: February, 2019", null, "", null, "", null, "# Archive for February, 2019\n\n## Weather\n\nFebruary 17th, 2019 at 08:15 pm\n\nNot much going on. Survived the polar vortex at end of January. My work closed for a couple of days so no need to go out. This weather has been a bit different here this winter. Warmer than normal one day and colder than normal the next." ]
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http://www.jzus.zju.edu.cn/article.php?doi=10.1631/jzus.2006.A1115&comnowpage=0
[ "Full Text:", null, "<1568>\n\nCLC number: TP39\n\nOn-line Access:\n\nRevision Accepted: 2006-04-17\n\nCrosschecked: 0000-00-00\n\nCited: 0\n\nClicked: 3125\n\nCitations:  Bibtex RefMan EndNote GB/T7714\n\n### -   Go to\n\nArticle info.\n Journal of Zhejiang University SCIENCE A 2006 Vol.7 No.7 P.1115~1123 10.1631/jzus.2006.A1115\n\nAn efficient method for tracing planar implicit curves\n\n Author(s):  YU Zheng-sheng, CAI Yao-zhi, OH Min-jae, KIM Tae-wan, PENG Qun-sheng Affiliation(s):  Computer Science School, Hangzhou Dianzi University, Hangzhou 310018, China; more Corresponding email(s):   [email protected], [email protected] Key Words:  Planar implicit curve, Curve tracing, Continuation method, Geometric modeling Share this article to： More <<< Previous Article|Next Article >>>\n\nYU Zheng-sheng, CAI Yao-zhi, OH Min-jae, KIM Tae-wan, PENG Qun-sheng. An efficient method for tracing planar implicit curves[J]. Journal of Zhejiang University Science A, 2006, 7(7): 1115~1123.\n\n@article{title=\"An efficient method for tracing planar implicit curves\",\nauthor=\"YU Zheng-sheng, CAI Yao-zhi, OH Min-jae, KIM Tae-wan, PENG Qun-sheng\",\njournal=\"Journal of Zhejiang University Science A\",\nvolume=\"7\",\nnumber=\"7\",\npages=\"1115~1123\",\nyear=\"2006\",\npublisher=\"Zhejiang University Press & Springer\",\ndoi=\"10.1631/jzus.2006.A1115\"\n}\n\n%0 Journal Article\n%T An efficient method for tracing planar implicit curves\n%A YU Zheng-sheng\n%A CAI Yao-zhi\n%A OH Min-jae\n%A KIM Tae-wan\n%A PENG Qun-sheng\n%J Journal of Zhejiang University SCIENCE A\n%V 7\n%N 7\n%P 1115~1123\n%@ 1673-565X\n%D 2006\n%I Zhejiang University Press & Springer\n%DOI 10.1631/jzus.2006.A1115\n\nTY - JOUR\nT1 - An efficient method for tracing planar implicit curves\nA1 - YU Zheng-sheng\nA1 - CAI Yao-zhi\nA1 - OH Min-jae\nA1 - KIM Tae-wan\nA1 - PENG Qun-sheng\nJ0 - Journal of Zhejiang University Science A\nVL - 7\nIS - 7\nSP - 1115\nEP - 1123\n%@ 1673-565X\nY1 - 2006\nPB - Zhejiang University Press & Springer\nER -\nDOI - 10.1631/jzus.2006.A1115\n\nAbstract:\nThis paper presents a method for tracing a planar implicit curve f(x, y)=0 on a rectangular region based on continuation scheme. First, according to the starting track-point and the starting track-direction of the curve, make a new function F(x, y)=0 where the same curve with f(x, y)=0 is defined. Then we trace the curve between the two domains where F(x, y)>0 and F(x, y)<0 alternately, according to the two rules presented in this paper. Equal step size or adaptive step size can be used, when we trace the curve. An irregular planar implicit curve (such as the curve with large curvatures at some points on the curve), can be plotted if an adaptive step size is used. Moreover, this paper presents a scheme to search for the multiple points on the curve. Our method has the following advantages: (1) it can plot C0 planar implicit curves; (2) it can plot the planar implicit curves with multiple points; (3) by the help of using the two rules, our method does not need to compute the tangent vector at the points on the curve, and directly searches for the direction of the tracing curve; (4) the tracing procedure costs only one of two evaluations of function f(x, y)=0 per moving step, while most existing similar methods cost more evaluations of the function.\n\n### Reference\n\n Barnhill, R.E., Farin, G., Jordan, M., Piper, B.R., 1987. Surface/surface intersection. Computer Aided Geometric Design, 4(1-2):3-16.", null, "Bresenham, J.E., 1965. Algorithm for computer control of a digital plotter. IBM Systems Journal, 4(1):25-30.\n\n Bresenham, J.E., 1977. A linear algorithm for incremental digital display of circular arcs. Comm. ACM, 20(2):100-106.", null, "Cai, Y.Z., 1990. Numerical Control Rendering Using Positive-negative Method. Zhejiang University Press, Hangzhou.\n\n Chandler, R.E., 1988. A tracking algorithm for implicitly defined curves. IEEE Computer Graphics and Applications, 8(2):83-89.", null, "Cohen, E., 1976. A method for plotting curves defined by implicit equation. Computer Graphics (SIGGRAPH’76), 10(2):263-265.", null, "Lennon, W.J., Jordan, B.W., Holm, B.C., 1973. An improved algorithm for the generation of nonparametric curves. IEEE Transactions on Computers, C-22:1052-1060.\n\n Lopes, H., Oliverira, J.B., de Figueiredo, L.H., 2002. Robust adaptive polygonal approximation of implicit curves. Computer & Graphics, 26(6):841-852.", null, "Martin, R., Shou, H., Voiculescu, I., Bowyer, A., Wang, G., 2002. Comparision of interval methods for plotting algebraic curves. Computer Aided Geometric Design, 19(7):553-587.", null, "Shou, H.H., Martin, R.R., Wang, G.J., Bowyer, A., Voiculescu, I., 2005. A Recursive Taylor Method for Algebraic Curves and Surfaces. In: Dokken, T., Jüttler, B. (Eds.), Computational Methods for Algebraic Spline Surface (COMPASS). Springer, p.135-155.\n\n van Aken, J., 1984. An efficient ellipse-drawing algorithm. IEEE Computer Graphics and Applications, 3:24-35.\n\n van Aken, J., Novak, M., 1985. Curve-drawing algorithms for raster displays. ACM Transactions on Graphics, 4(2):147-169.", null, "", null, "" ]
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https://www.ruprict.net/post/2008-10-06-the-bitter-coder-tutorials-binsor-style-injecting-service-arrays/
[ "### The Bitter Coder Tutorials, Binsor Style: Injecting Service Arrays\n\nMon, Oct 6, 2008 3-minute read\n\nPrevious posts in the series:\n\nThis post feeds off off Part 12 and is based on Alex’s post here.  We are going to rework the last post and remove the decorator pattern.  Instead, we’ll create a calculator that manages any number of other calculators, which are injected as an array.\n\nSo, in keeping with my shameless plagirizing of Alex’s code, here is the abstract class for our new calculators:\n\nnamespace BitterCoder.Tutorials.Binsor.Core\n\n{\n\npublic abstract class AbstractCalculator\n\n{\n\npublic abstract decimal Calculate(decimal currentTotal, Order order);\n\n}\n\n}\n\nAnd now our total calculator:\n\nusing System.Linq;\n\nnamespace BitterCoder.Tutorials.Binsor.Core\n\n{\n\npublic class TotalCalculator : AbstractCalculator\n\n{\n\nprivate static decimal CalculateTotal(Order order)\n\n{\n\nreturn order.Items.Sum(item => item.CostPerItem*item.Quantity);\n\n}\n\npublic override decimal Calculate(decimal currentTotal, Order order)\n\n{\n\nreturn currentTotal + CalculateTotal(order);\n\n}\n\n}\n\n}\n\nAnd the GST calculator:\n\nnamespace BitterCoder.Tutorials.Binsor.Core\n\n{\n\npublic class GSTCostCalculatorNoDecorator : AbstractCalculator\n\n{\n\nprivate decimal _gstRate = 1.125m;\n\npublic decimal GstRate\n\n{\n\nget { return _gstRate; }\n\nset { _gstRate = value; }\n\n}\n\nprivate static bool IsNewZealand(Order order)\n\n{\n\nreturn (order.CountryCode == \"NZ\");\n\n}\n\npublic override decimal Calculate(decimal currentTotal, Order order)\n\n{\n\nif (IsNewZealand(order))\n\n{\n\nreturn (currentTotal*_gstRate);\n\n}\n\nreturn currentTotal;\n\n}\n\n}\n\n}\n\nThe shipping calculator:\n\nnamespace BitterCoder.Tutorials.Binsor.Core\n\n{\n\npublic class ShippingCalculatorNoDecorator : AbstractCalculator\n\n{\n\nprivate decimal _shippingCost = 5.0m;\n\npublic decimal ShippingCost\n\n{\n\nget { return _shippingCost; }\n\nset { _shippingCost = value; }\n\n}\n\n{\n\nset { _fragileShippingPremium = value; }\n\n}\n\nprivate decimal GetShippingTotal(Order order)\n\n{\n\ndecimal shippingTotal = 0;\n\nreturn order.Items.Sum(item =>\n\n{\n\ndecimal itemShippingCost = ShippingCost*item.Quantity;\n\nreturn shippingTotal += itemShippingCost;\n\n});\n\n}\n\npublic override decimal Calculate(decimal currentTotal, Order order)\n\n{\n\nreturn currentTotal + GetShippingTotal(order);\n\n}\n\n}\n\n}\n\nLastly, our calculator of calculators, the reworked DefaultCalculator:\n\npublic class DefaultCalculatorNoDecorator : ICostCalculator\n\n{\n\npublic DefaultCalculatorNoDecorator(AbstractCalculator[] _calculators)\n\n{\n\nthis._calculators = _calculators;\n\n}\n\npublic decimal CalculateTotal(Order order)\n\n{\n\ndecimal currentTotal = 0;\n\nreturn _calculators.Sum(calc => calc.Calculate(currentTotal, order));\n\n}\n\n}\n\nSo, we'll inject an array of calculators (total, shipping, and gst) into our default calculator. Binsor, away!\n```component \"default.calculator\", ICostCalculator,DefaultCalculatorNoDecorator:\n_calculators=[@total.calculatornodec,@shipping.calculatornodec,@gst.calculatornodec]\n\ncomponent \"total.calculatornodec\", AbstractCalculator,TotalCalculator\n\ncomponent \"gst.calculatornodec\", AbstractCalculator, GSTCostCalculatorNoDecorator:\nGstRate=Convert.ToDecimal(1.20)\n\ncomponent \"shipping.calculatornodec\", AbstractCalculator, ShippingCalculatorNoDecorator:" ]
[ null ]
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https://gmatclub.com/forum/olive-is-creating-a-five-digit-code-using-the-digits-0-through-9-how-261351.html
[ "GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 20 Aug 2019, 16:22", null, "### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\nYour Progress\n\nevery week, we’ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.", null, "### Request Expert Reply", null, "# Olive is creating a five-digit code using the digits 0 through 9. How\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager", null, "", null, "G\nJoined: 01 May 2016\nPosts: 77\nLocation: United States\nConcentration: Finance, International Business\nGPA: 3.8\nOlive is creating a five-digit code using the digits 0 through 9. How  [#permalink]\n\n### Show Tags\n\n1\n6", null, "00:00\n\nDifficulty:", null, "", null, "", null, "45% (medium)\n\nQuestion Stats:", null, "67% (02:46) correct", null, "33% (02:53) wrong", null, "based on 76 sessions\n\n### HideShow timer Statistics\n\nOlive is creating a five-digit code using the digits 0 through 9. How many different codes can she create with exactly two prime digits if no digits can be repeated?\n\nA. 252\n\nB. 3,120\n\nC. 3,456\n\nD. 14,400\n\nE. 30,240\n\n_________________\nMath Expert", null, "V\nJoined: 02 Sep 2009\nPosts: 57155\nRe: Olive is creating a five-digit code using the digits 0 through 9. How  [#permalink]\n\n### Show Tags\n\n1\n1\nTridhipal wrote:\nOlive is creating a five-digit code using the digits 0 through 9. How many different codes can she create with exactly two prime digits if no digits can be repeated?\n\nA. 252\n\nB. 3,120\n\nC. 3,456\n\nD. 14,400\n\nE. 30,240\n\nThe code will be any combination of the following: {P1} {P2} {X} {Y} {Z}, where P1 and P2 are prime digits, and X, Y and Z are other digits (all distinct).\n\nThere are four single-digit primes: 2, 3, 5, and 7 and six other digits: 0, 1, 4, 6, 8, and 9.\n\nThe number of pairs of primes, P1 and P2, out of four primes is 4C2 = 6;\nThe number of triplets of non-primes, X, Y and Z, our of six is 6C3 = 20.\n\nTherefore, 6*20*5! = 14,400 (multiplying by 5! to account for different arrangements of 5 digits {P1} {P2} {X} {Y} {Z}).\n\nAnswer: D.\n_________________\nTarget Test Prep Representative", null, "D\nStatus: Founder & CEO\nAffiliations: Target Test Prep\nJoined: 14 Oct 2015\nPosts: 7399\nLocation: United States (CA)\nRe: Olive is creating a five-digit code using the digits 0 through 9. How  [#permalink]\n\n### Show Tags\n\nTridhipal wrote:\nOlive is creating a five-digit code using the digits 0 through 9. How many different codes can she create with exactly two prime digits if no digits can be repeated?\n\nA. 252\n\nB. 3,120\n\nC. 3,456\n\nD. 14,400\n\nE. 30,240\n\nFrom 0 through 9, the prime digits are 2, 3, 5, and 7, and thus the non-prime digits are 0, 1, 4, 6, 8, and 9.\n\nThe number of ways to select the 2 prime digits is 4C2 = (4 x 3)/2 = 6.\n\nThe number of ways to select the 3 non-prime digits is 6C3 = (6 x 5 x 4)/(3 x 2) = 20.\n\nTherefore, there are 6 x 20 = 120 ways to create these codes if order doesn’t matter. However, for each code created, the order matters. For example, the code 27014 is different from 70412, 42017, etc. For each of 120 codes, since there are 5 different digits, there are 5! = 120 ways to arrange them. Thus the total number of codes can be created is:\n\n120 x 120 = 14,400\n\nAnswer: D\n_________________\n\n# Scott Woodbury-Stewart\n\nFounder and CEO\n\[email protected]\n\nSee why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews\n\nIf you find one of my posts helpful, please take a moment to click on the \"Kudos\" button.\n\nIntern", null, "", null, "B\nJoined: 28 May 2012\nPosts: 1\nGMAT 1: 680 Q42 V41", null, "Re: Olive is creating a five-digit code using the digits 0 through 9. How  [#permalink]\n\n### Show Tags\n\nWhy do you use the combinations formula here?\n\nI understand that there are 4 primes and 6 non-primes. Therefore, I thought the path would be as follows: 4 (options for primes) * 3 (options for primes - 1) * 6 (options for non-primes) * 5 (options for non-primes -1 ) * 4 (options for non-primes -2). What is wrong with this logic?\nIntern", null, "", null, "Joined: 20 Feb 2019\nPosts: 1\nRe: Olive is creating a five-digit code using the digits 0 through 9. How  [#permalink]\n\n### Show Tags\n\nSynchrony wrote:\nWhy do you use the combinations formula here?\n\nI understand that there are 4 primes and 6 non-primes. Therefore, I thought the path would be as follows: 4 (options for primes) * 3 (options for primes - 1) * 6 (options for non-primes) * 5 (options for non-primes -1 ) * 4 (options for non-primes -2). What is wrong with this logic?\n\nYou didn't take into account the way of arranging the digits!", null, "Re: Olive is creating a five-digit code using the digits 0 through 9. How   [#permalink] 27 Feb 2019, 08:55\nDisplay posts from previous: Sort by\n\n# Olive is creating a five-digit code using the digits 0 through 9. How\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne\n\n#### MBA Resources", null, "", null, "" ]
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https://tex.stackexchange.com/questions/121205/palatino-font-size
[ "# Palatino font size\n\nI'm using Palatino for my dissertation, but 10pt is too small and 11pt is too large (looks childish). I would like to select something in between. Any suggestions on how to do this?\n\n1. (pdflatex, 11pt memoir, fix-cm, renew \\normalsize): this only changes \\normalsize (so not \\small, \\huge, etc.)\n\n\\documentclass[11pt,a4paper,twoside,openright]{memoir}\n\\renewcommand*{\\rmdefault}{ppl}\n\\usepackage{fix-cm}\n\\renewcommand{\\normalsize}{\\fontsize{10.5pt}{12.5pt}\\selectfont}\n\\begin{document}\n{\\tiny This is tiny}\\\\\n{\\scriptsize This is script size}\\\\\n{\\footnotesize This is footnote size}\\\\\n{\\small This is small}\\\\\n{\\normalsize This is normal size}\\\\\n{\\large This is large}\\\\\n{\\Large This is larger}\\\\\n{\\LARGE This is very large}\\\\\n{\\huge This is huge}\\\\\n{\\Huge This is very huge}\n\\end{document}\n\n2. (pdflatex, 11pt memoir, fix-cm, relscale): this doesn't work at all (since you would need to use \\relscale{0.96} whenever changing from \\normalsize to \\small or something)\n\n\\documentclass[11pt,a4paper,twoside,openright]{memoir}\n\\renewcommand*{\\rmdefault}{ppl}\n\\usepackage{fix-cm,relsize}\n\\begin{document}\n\\relscale{0.96}\n{\\tiny This is tiny}\\\\\n{\\scriptsize This is script size}\\\\\n{\\footnotesize This is footnote size}\\\\\n{\\small This is small}\\\\\n{\\normalsize This is normal size}\\\\\n{\\large This is large}\\\\\n{\\Large This is larger}\\\\\n{\\LARGE This is very large}\\\\\n{\\huge This is huge}\\\\\n{\\Huge This is very huge}\n\\end{document}\n\n3. (lualatex, 11pt memoir): this changes all type sizes (\\normalsize, \\small, \\huge, etc.)\n\n\\documentclass[11pt,a4paper,twoside,openright]{memoir}\n\\usepackage{fontspec}\n\\setmainfont[Scale=0.96]{Palatino Linotype}\n\\begin{document}\n{\\tiny This is tiny}\\\\\n{\\scriptsize This is script size}\\\\\n{\\footnotesize This is footnote size}\\\\\n{\\small This is small}\\\\\n{\\normalsize This is normal size}\\\\\n{\\large This is large}\\\\\n{\\Large This is larger}\\\\\n{\\LARGE This is very large}\\\\\n{\\huge This is huge}\\\\\n{\\Huge This is very huge}\n\\end{document}\n\n\nFor example in my case, \\normalsize was 10.91pt and afterwards 10.46pt. Perfect! Note also that in the third option, the system's Palatino font is used, while in the first two, URW Palladio is used. To conclude I would say that lualatex (or perhaps also xelatex etc.) is the way to go. The \\usepackage[scale=0.95]{tgpagella} option mentioned in the comments also works, but this is specific to the font package.\n\nHowever, there is still an issue remaining: how to change the math type size? I'm using \\usepackage{lmodern}.\n\n• Check out this thread: tex.stackexchange.com/questions/83851/… Jun 26, 2013 at 17:43\n• If you don't need math, you can use tgpagella that accepts a \"scale\" option: \\documentclass[11pt]{article} and then \\usepackage[scale=0.95]{tgpagella} will set up the document with the parameters for 11pt size, but the font would be at 10.5; it's common to have a wider interline with Palatino. Jun 26, 2013 at 17:55\n• Remember your supervisor's eyes may not be as good as yours. Using an 11 or 12 point size may be worth your while....\n– user10274\nJun 26, 2013 at 18:16\n• Perhaps the user needs to change their name to Goldi-wox. There's a great book in here somewhere... \"Goldi-wox and the three fonts.\" 10.5 is \"just right.\" Jun 26, 2013 at 18:19\n\nuse also TeX Gyre Pagella Math:\n\n\\documentclass[11pt,a4paper,twoside,openright]{memoir}\n\\usepackage{unicode-math}\n\\setmainfont[Scale=0.96]{TeXGyre Pagella}\n\\setmathfont[Scale=0.96]{TeXGyrePagellaMath-Regular}\n\\begin{document}\n{\\tiny This is tiny}\\\\\n{\\scriptsize This is script size}\\\\\n{\\footnotesize This is footnote size}\\\\\n{\\small This is small}\\\\\n{\\normalsize This is normal size}\\\\\n{\\large This is large}\\\\\n{\\Large This is larger}\\\\\n{\\LARGE This is very large}\\\\\n{\\huge This is huge}\\\\\n{\\Huge This is very huge}\n\n$\\int\\limits_1^\\infty \\frac1{x^2}\\mathrm{d}x=1$\n\n\\end{document}\n\n• This also seems to work: \\makeatletter \\DeclareMathSizes{\\@xipt}{10.5}{8}{6} \\makeatother\n– Wox\nJun 27, 2013 at 15:59" ]
[ null ]
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https://zbmath.org/?q=an:0767.11054
[ "## Divisibility by 2-powers of certain quadratic class numbers.(English)Zbl 0767.11054\n\nThe author looks at divisibility of the strict class number of quadratic fields of discriminant $$8p$$, $$-8p$$ and $$-4p$$ by powers of 2 where $$p\\equiv 1\\pmod 4$$ is prime.\nReviewer: R.Mollin (Calgary)\n\n### MSC:\n\n 11R29 Class numbers, class groups, discriminants 11R11 Quadratic extensions" ]
[ null ]
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https://makeslider.com/slide/poisson-regression-r7xncx
[ "# Poisson Regression", null, "Poisson Regression Caution Flags (Crashes) in NASCAR Winston Cup Races 19751979 L. Winner (2006). NASCAR Winston Cup Race Results for 1975-2003, Journal of Statistics Education, Vol.14,#3, www.amstat.org/publications/jse/v14n3/datasets.winner.html Data Description Units: NASCAR Winston Cup Races (19751979) n=151 Races Dependent Variable: Y=# of Caution Flags/Crashes (CAUTIONS) Independent Variables: X1=# of Drivers in race (DRIVERS) X2=Circumference of Track (TRKLENGTH) X3=# of Laps in Race (LAPS) Generalized Linear Model Random Component: Poisson Distribution for # of Caution Flags Density Function: e X 1 , X 2 , X 3 X 1 , X 2 , X 3 PY y X 1 , X 2 , X 3 y! y y 0,1,2,... Link Function: g(= log( Systematic Component: g ( ) log( ) 0 1 X 1 2 X 2 3 X 3 X 1 , X 2 , X 3 e 0 1 X 1 2 X 2 3 X 3 Testing For Overall Model H0: (# Cautions independent of all predictors) HA: Not all j = 0 (# Cautions associated with at least 1\n\npredictor) Test Statistic: Xobs2 = -2(lnL0-lnL1) Rejection Region: Xobs2 2,3 P-Value: P(23 Xobs2) Where: lnL0 is maximized log likelihood under model H0 lnL1 is maximized log likelihood under model HA NASCAR Caution Flag Example Model : g ( ) 0 Criterion Deviance Scaled Deviance Pearson Chi-Square Scaled Pearson X2 Log Likelihood DF 150 150 150 150 Value 215.4915 215.4915 201.6050 201.6050 410.8784 Value/ DF 1.4366 1.4366 1.3440 1.3440 Model : g ( ) 0 1 X 1 2 X 2 3 X 3 Criterion Deviance Scaled Deviance Pearson Chi-Square\n\nScaled Pearson X2 Log Likelihood DF 147 147 147 147 Value 171.2162 171.2162 158.8281 158.8281 433.0160 Value/ DF 1.1647 1.1647 1.0805 1.0805 2 Test Statistic: X obs 2 ln L0 ln L1 2(410.8784 433.0160) 44.2752 2 2 Rejection Region ( 0.05) : X obs .05,3 7.815 P value: P 32 44.2752 0 Statistical output obtained from SAS PROC GENMOD Testing for Individual (Partial) Regression Coefficients H 0 : j 0 H A : j 0 ^ j Z Test Statistic : zobs ^\n\nSE j ( Z ) P value : 2 P Z zobs 2 2 Test Statistic : X obs P value : P 2 2 1 ^ j ^ SE j 2 X obs 2 ^ 1 - Sided Tests : Confirm sign of j is correct, then \" cut\" P - value in half. NASCAR Caution Flag Example Parameter Intercept Drivers TrkLength Laps DF 1 1 1\n\n1 Estimate -0.7963 0.0365 0.1145 0.0026 Std Error 0.4117 0.0125 0.1684 0.0008 Chi-Square 3.74 8.55 0.46 10.82 Pr>ChiSq 0.0531 0.0035 0.4966 0.0010 Conclude the following: Controlling for Track Length and Laps, as Drivers Cautions Controlling for Drivers and Laps, No association between Cautions and Track Length Controlling for Drivers and Track Length, as Laps Reduced Model: log(Crashes) = -0.6876+0.0428*Drivers+0.0021*Laps Cautions Testing Model Goodness-of-Fit Two Common Measures of Goodness of Fit: Pearsons Chi-Square Deviance\n\nBoth measures have approximate Chi-Square Distributions under the hypothesis that the current model is appropriate for fixed number of combinations of independent variables and large counts ^ y i i n Pearson's Chi - Square : X 2 ^ ^ i 1 V i n y Deviance: G 2 2 yi log ^ i i 1 i 2 ^ ^ ^ where V i i for Poisson Distribution NASCAR Caution Flags Example Null Model Criterion Pearson X2 Deviance\n\nDF 150 150 Value 201.6050 215.4915 Value/ DF 1.3440 1.4366 P-Value 0.0032 0.0004 Full Model Criterion Pearson X2 Deviance DF 147 147 Value 158.8281 171.2162 Value/ DF 1.0805 1.1647 P-Value 0.2386 0.0838 Note that the null model clearly does not fit well, and the full model fails to reject the null hypothesis of the model being appropriate (however, we have many combinations of Laps, Track Length, and Drivers)\n\nSAS Program options ps=54 ls=76; data one; input serrace 6-8 year 13-16 searace 23-24 drivers 31-32 trklength 34-40 laps 46-48 road 56 cautions 63-64 leadchng 71-72; cards; 1 1975 1 35 2.54 191 1 5 13 ... 151 1979 31 37 2.5 200 0 6 35 ; run; /* Data set one contains the data for analysis. Variable names and column specs are given in INPUT statement. I have included ony first and last observations */ /* The following model fits a Generalized Linear model, with poisson random component, and a constant mean: g(mu)=alpha is systematic component, g(mu)=log(mu) is the link function: mu=e**alpha */ proc genmod; model Cautions = / dist=poi link=log; run; /* The following model fits a Generalized Linear model, with poisson random component, g(mu)=alpha + beta1*drivers + beta2*trkength + beta3*laps is systematic component, g(mu)=log(mu) is the link function: mu=e**alpha + beta1*drivers + beta2*trkength + beta3*laps */\n\nproc genmod; model Cautions = drivers trklength laps / dist=poi link=log; run; quit; SPSS Output Goodness-of-Fit Test Used when there are many distinct levels of explanatory variables Based on lumping together cases based on their predicted values into J (often 10 is used) groups Compares observed and expected counts by group based on Deviance and Pearson residuals. For Poisson model (where obs is observed, exp is expected): Pearson: ri = (obsi-expi)/expi X2=ri2 Deviance: di = (obsi* log(obsi/expi)) G2=2 di2 Degrees of Freedom: J- p-1 where p=#Predictor Variables NASCAR Caution Flags Example ^ i e Group 1 2 3 4 5 6 7 8 9 10 Fitted <3.50 3.50-3.80 3.80-4.08 4.08-4.25\n\n4.25-4.42 4.42-5.15 5.15-5.50 5.50-6.25 6.25-6.70 >6.70 0.6876 0.0428 Di 0.0021Li #Races 15 14 18 20 12 17 15 15 14 11 #Crashes 37 60 72 68 51 100 88 91 94 63 Expected 46.05 50.37 71.24 84.03 52.35 81.39 78.19 87.40\n\n90.81 78.46 Pearson -1.33 1.36 0.09 -1.75 -0.19 2.06 1.11 0.38 0.33 -1.75 Pearson X2 P-value 15.5119 0.0300 Note that there is evidence that the Poisson model does not provide a good fit Computational Approach e y Poisson Probability Mass Function : P (Y y ) y 0,1,2,... y! Systematic Component : g ( ) 0 1 X 1 2 X 2 3 X 3 Link Function : g ( ) log( ) e g ( ) e 0 1 X 1 2 X 2 3 X 3 For Subject i : g ( i ) 0 1 X 1i 2 X 2i 3 X 3i x'i where : 1 X x i 1i X 2i X 3i\n\nx1' ' x2 X ' x n 0 1 2 3 y1 y Y 2 yn n Likelihood Function : L y1 ,..., yn e i iyi yi ! i 1 n l ln( L) e i 1 x'i n '\n\ni yi x i 1 1 2 n ' ' exp e xi e x i yi ! i 1 n n ln y ! i i 1 1 X l x'i x i e yi x i yi i x i yi i 1i X 2i\n\nX 3i yi Computational Approach 1 X l x'i x i e yi x i yi i x i yi i 1i X 2i X 3i 1 X l Setting 0 yi i 1i 0 X' (Y ) 0 X 2i X 3i ' ' 2l x i e x i yi x i x i e x i x'i X' WX where W diag ' ' Setting : G X' WX and g X' ( Y ) leads to the the estimate of via Newton - Raphson algorithm :\n\nln y ^ New ^ Old ^0 ^ Old ^ Old 0 G g with a reasonable staring vector of 0 0 with approximate large - sample estimated variance - covariance matrix : 1 ^ ^ ^ V G 1 X' W X 1" ]
[ null, "https://makeslider.com/assets/img/default.png", null ]
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https://percentagecalculator.pro/percent-off/
[ "# How to calculate discount\n\n1 percent-off 1 is 0.99\n\n### Inputs:\n\nPrice before the discount: \\$\nDiscount: %\n\n### Results:\n\nDiscount:\n\nPrice after discount:\n\n## How to calculate a 1% discount on 0.99?\n\nYou can easily calculate the discount or percent-off, as well as, the final price or the price after the discount of a product by following these steps:\n\n1. Multiply the list price by the discount percentage, then divide it by 100:\n\n1.00 times 1.00 divided by 100 = 1.00 × 1.00 / 100 = 1.00 × 0.01 = 0.01 (This is the discount).\n\n2. Subtract the discount from the price before the discount to get the final price:\n3. Final Price = 1.00 minus 0.01 = 1.00 - 0.01 = 0.99.\n\nSo, an item that costs 1.00, when discounted 1.00 percent, will have the final price equal to 0.99.\n\n## Discount Formulas\n\nTo calculate discount, as explained above, is easy by using the following equations:\n\nDiscount = Price Before the Discount × Discount % / 100\n\nFinal Price = Price Before the Discount - Discount" ]
[ null ]
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http://mkweb.bcgsc.ca/infinity/etc/effects.conf
[ "# click name = click action = fill pages = rand(aleph) npages = 1 position = full_screen page_rate = 0.05 bycol = yes timing = instant decay = fill-instant start = 27:1:2m end = 27:4m rgb = red # click name = flutter action = fill pages = rand(setnoletters) npages = 1 position = full_screen page_rate = 0.1 bycol = yes timing = instant decay = fill-fast start = 37:1:2m end = 37:3m rgb = red # click name = flutter action = fill pages = rand(setnoletters) npages = 1 position = full_screen page_rate = 0.2 bycol = yes timing = instant overwrite = yes decay = fill-fast start = 50:1:2m end = 50:3m rgb = red name = blips action = fill pages = rand(setnoletters) npages = 1 position = full_screen page_rate = 0.1 bycol = yes timing = instant decay = fill-fast start = 58:4:2m end = 59:2m rgb = red name = fill before reals action = fill npages = 1 pages = rand(digit) position = full_screen timing = 1m decay = slow-fade-reals page_rate = 0.05 byrand = yes start = 85:3m end = 86:3m rgb = grey name = fill before powersets action = fill npages = 1 pages = rand(set,greek) position = full_screen timing = vwipe-test decay = vwipe2-test bycol = 1 start = 134:4:2m end = 135:1:2m rgb1 = red rgb1_rate = 0.5 name = dots action = fill pages = rand(setnoletters) npages = 9999 position = full_screen page_rate = 0.1 bycol = yes timing = fill decay = fill start = 180m end = 182m sync = kick #rgb1 = red #rgb1_rate = 0.1 #correct = yes name = dots action = fill pages = rand(setnoletters) position = full_screen page_rate = 0.25 bycol = yes timing = fillhold decay = fillhold start = 182:2m end = 190m #sync = kick #rgb1 = hilite1 #rgb_rate = 0.1 #correct = yes start = 1m name = decay effects 1 action = decay name = snare action = midi_trigger position = full_screen midi_sets = snare midi_action = cascade_red_star_subtle start = 52m end = 65m name = snare action = midi_trigger position = full_screen midi_sets = snare midi_action = cascade_red_star_subtle start = 71m end = 75m name = snare action = midi_trigger position = full_screen midi_sets = snare midi_action = cascade_red_braces_strong start = 140m end = 149m name = snare action = midi_trigger position = full_screen midi_sets = snare midi_action = cascade_set_strong start = 153m end = 154:4m name = snare action = midi_trigger position = full_screen midi_sets = snare midi_action = cascade_set_strong start = 159m end = 165m name = aleph action = aleph position = full_screen timing = aleph decay = aleph start = 183:2m end = 192m #sync = kick rgb = red wipe = end block = end #correct = yes name = decay effects 2 start = 1m action = decay skip = yes name = snare action = midi_trigger position = full_screen midi_sets = snare midi_action = decay start = 57m end = 65m skip = yes name = decay effects 3 start = 1m action = decay" ]
[ null ]
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https://hackage.haskell.org/package/profunctors-5.1.1/docs/Data-Profunctor.html
[ "profunctors-5.1.1: Profunctors\n\nData.Profunctor\n\nDescription\n\nFor a good explanation of profunctors in Haskell see Dan Piponi's article:\n\nSynopsis\n\n# Profunctors\n\nclass Profunctor p where Source\n\nFormally, the class `Profunctor` represents a profunctor from `Hask` -> `Hask`.\n\nIntuitively it is a bifunctor where the first argument is contravariant and the second argument is covariant.\n\nYou can define a `Profunctor` by either defining `dimap` or by defining both `lmap` and `rmap`.\n\nIf you supply `dimap`, you should ensure that:\n\n``dimap` `id` `id` ≡ `id``\n\nIf you supply `lmap` and `rmap`, ensure:\n\n````lmap` `id` ≡ `id`\n`rmap` `id` ≡ `id`\n```\n\nIf you supply both, you should also ensure:\n\n``dimap` f g ≡ `lmap` f `.` `rmap` g`\n\nThese ensure by parametricity:\n\n````dimap` (f `.` g) (h `.` i) ≡ `dimap` g h `.` `dimap` f i\n`lmap` (f `.` g) ≡ `lmap` g `.` `lmap` f\n`rmap` (f `.` g) ≡ `rmap` f `.` `rmap` g\n```\n\nMinimal complete definition\n\nMethods\n\ndimap :: (a -> b) -> (c -> d) -> p b c -> p a d Source\n\nMap over both arguments at the same time.\n\n``dimap` f g ≡ `lmap` f `.` `rmap` g`\n\nlmap :: (a -> b) -> p b c -> p a c Source\n\nMap the first argument contravariantly.\n\n``lmap` f ≡ `dimap` f `id``\n\nrmap :: (b -> c) -> p a b -> p a c Source\n\nMap the second argument covariantly.\n\n``rmap` ≡ `dimap` `id``\n\nInstances\n\n Profunctor (->) Monad m => Profunctor (Kleisli m) Functor w => Profunctor (Cokleisli w) Profunctor (Tagged *) Profunctor (Forget r) Arrow p => Profunctor (WrappedArrow p) Functor f => Profunctor (Costar f) Functor f => Profunctor (Star f) Profunctor p => Profunctor (Environment p) Profunctor p => Profunctor (Closure p) Profunctor p => Profunctor (Codensity p) Profunctor p => Profunctor (Copastro p) Profunctor p => Profunctor (Cotambara p) Profunctor p => Profunctor (Pastro p) Profunctor p => Profunctor (Tambara p) (Functor f, Profunctor p) => Profunctor (Cayley f p) (Profunctor p, Profunctor q) => Profunctor (Rift p q) (Profunctor p, Profunctor q) => Profunctor (Procompose p q) (Profunctor p, Profunctor q) => Profunctor (Ran p q)\n\n## Profunctorial Strength\n\nclass Profunctor p => Strong p where Source\n\nGeneralizing `Star` of a strong `Functor`\n\nNote: Every `Functor` in Haskell is strong with respect to `(,)`.\n\nThis describes profunctor strength with respect to the product structure of Hask.\n\nMinimal complete definition\n\nMethods\n\nfirst' :: p a b -> p (a, c) (b, c) Source\n\nsecond' :: p a b -> p (c, a) (c, b) Source\n\nInstances\n\n Strong (->) Monad m => Strong (Kleisli m) Strong (Forget r) Arrow p => Strong (WrappedArrow p) `Arrow` is `Strong` `Category` Functor m => Strong (Star m) Strong p => Strong (Closure p) Profunctor p => Strong (Tambara p) (Functor f, Strong p) => Strong (Cayley f p) (Strong p, Strong q) => Strong (Procompose p q)\n\nclass Profunctor p => Choice p where Source\n\nThe generalization of `Costar` of `Functor` that is strong with respect to `Either`.\n\nNote: This is also a notion of strength, except with regards to another monoidal structure that we can choose to equip Hask with: the cocartesian coproduct.\n\nMinimal complete definition\n\nMethods\n\nleft' :: p a b -> p (Either a c) (Either b c) Source\n\nright' :: p a b -> p (Either c a) (Either c b) Source\n\nInstances\n\n Choice (->) Monad m => Choice (Kleisli m) Comonad w => Choice (Cokleisli w) `extract` approximates `costrength` Choice (Tagged *) Monoid r => Choice (Forget r) ArrowChoice p => Choice (WrappedArrow p) Traversable w => Choice (Costar w) Applicative f => Choice (Star f) Profunctor p => Choice (Cotambara p) Choice p => Choice (Tambara p) (Functor f, Choice p) => Choice (Cayley f p) (Choice p, Choice q) => Choice (Procompose p q)\n\n## Profunctorial Costrength\n\nclass Profunctor p => Costrong p where Source\n\nAnalogous to `ArrowLoop`, `loop` = `unfirst`\n\nMinimal complete definition\n\nMethods\n\nunfirst :: p (a, d) (b, d) -> p a b Source\n\nunsecond :: p (d, a) (d, b) -> p a b Source\n\nInstances\n\n Costrong (->) MonadFix m => Costrong (Kleisli m) Functor f => Costrong (Cokleisli f) Costrong (Tagged *) ArrowLoop p => Costrong (WrappedArrow p) Functor f => Costrong (Costar f) (Corepresentable p, Corepresentable q) => Costrong (Procompose p q)\n\nclass Profunctor p => Cochoice p where Source\n\nMinimal complete definition\n\nMethods\n\nunleft :: p (Either a d) (Either b d) -> p a b Source\n\nunright :: p (Either d a) (Either d b) -> p a b Source\n\nInstances\n\n Cochoice (->) Applicative f => Cochoice (Costar f) Traversable f => Cochoice (Star f)\n\n## Common Profunctors\n\nnewtype Star f d c Source\n\nLift a `Functor` into a `Profunctor` (forwards).\n\nConstructors\n\n Star FieldsrunStar :: d -> f c\n\nInstances\n\n Functor f => Profunctor (Star f) Traversable f => Cochoice (Star f) Applicative f => Choice (Star f) Functor m => Strong (Star m) Distributive f => Closed (Star f) Functor f => Representable (Star f) Functor f => Sieve (Star f) f Alternative f => Alternative (Star f a) Monad f => Monad (Star f a) Functor f => Functor (Star f a) MonadPlus f => MonadPlus (Star f a) Applicative f => Applicative (Star f a) Distributive f => Distributive (Star f a) type Rep (Star f) = f\n\nnewtype Costar f d c Source\n\nLift a `Functor` into a `Profunctor` (backwards).\n\nConstructors\n\n Costar FieldsrunCostar :: f d -> c\n\nInstances\n\n Functor f => Profunctor (Costar f) Applicative f => Cochoice (Costar f) Functor f => Costrong (Costar f) Traversable w => Choice (Costar w) Functor f => Closed (Costar f) Functor f => Corepresentable (Costar f) Functor f => Cosieve (Costar f) f Monad (Costar f a) Functor (Costar f a) Applicative (Costar f a) Distributive (Costar f d) type Corep (Costar f) = f\n\nnewtype WrappedArrow p a b Source\n\nWrap an arrow for use as a `Profunctor`.\n\nConstructors\n\n WrapArrow FieldsunwrapArrow :: p a b\n\nInstances\n\n Category * p => Category * (WrappedArrow p) Arrow p => Arrow (WrappedArrow p) ArrowZero p => ArrowZero (WrappedArrow p) ArrowChoice p => ArrowChoice (WrappedArrow p) ArrowApply p => ArrowApply (WrappedArrow p) ArrowLoop p => ArrowLoop (WrappedArrow p) Arrow p => Profunctor (WrappedArrow p) ArrowLoop p => Costrong (WrappedArrow p) ArrowChoice p => Choice (WrappedArrow p) Arrow p => Strong (WrappedArrow p) `Arrow` is `Strong` `Category`\n\nnewtype Forget r a b Source\n\nConstructors\n\n Forget FieldsrunForget :: a -> r\n\nInstances\n\n Profunctor (Forget r) Monoid r => Choice (Forget r) Strong (Forget r) Representable (Forget r) Sieve (Forget r) (Const r) Functor (Forget r a) Foldable (Forget r a) Traversable (Forget r a) type Rep (Forget r) = Const r\n\ntype (:->) p q = forall a b. p a b -> q a b infixr 0 Source" ]
[ null ]
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https://www.esaim-m2an.org/articles/m2an/ref/2013/02/m2an120037/m2an120037.html
[ "Subscriber Authentication Point\nFree Access\n Issue ESAIM: M2AN Volume 47, Number 2, March-April 2013 555 - 581 https://doi.org/10.1051/m2an/2012037 21 January 2013\n1. A.C. Antoulas, Approximation of large-scale dynamical systems, Advances in Design and Control. Society for Industrial and Applied Mathematics SIAM, Philadelphia, PA (2005). With a foreword by Jan C. Willems. [Google Scholar]\n2. N. Arada, E. Casas and F. Tröltzsch, Error estimates for the numerical approximation of a semilinear elliptic control problem. Comput. Optim. Appl. 23 (2002) 201–229. [CrossRef] [MathSciNet] [Google Scholar]\n3. M. Barrault, Y. Maday, N.C. Nguyen and A.T. Patera, An ‘empirical interpolation’ method : application to efficient reduced-basis discretization of partial differential equations. C. R. Math. Acad. Sci. Paris 339 (2004) 667–672. [Google Scholar]\n4. P. Benner and E.S. Quintana-Ortí, Model reduction based on spectral projection methods, in Reduction of Large-Scale Systems, edited by P. Benner, V. Mehrmann, D.C. Sorensen, Lect. Notes Comput. Sci. Eng. 45 (2005) 5–48. [Google Scholar]\n5. E. Casas, J.C. De los Reyes and F. Tröltzsch, Sufficient second-order optimality conditions for semilinear control problems with pointwise state constraints. SIAM : J. Optim. 19 (2008) 616–643. [Google Scholar]\n6. E. Casas and F. Tröltzsch, First- and second-order optimality conditions for a class of optimal control problems with quasilinear elliptic equations. SIAM : J. Control Optim. 48 (2009) 688–718. [Google Scholar]\n7. S. Chaturantabut and D.C. Sorensen, Nonlinear model reduction via discrete empirical interpolation. SIAM : J. Sci. Comput. 32 (2010) 2737–2764. [Google Scholar]\n8. A.L. Dontchev, W.W. Hager, A.B. Poore and B. Yang, Optimality, stability, and convergence in nonlinear control. Appl. Math. Optim. 31 (1995) 297–326. [CrossRef] [MathSciNet] [Google Scholar]\n9. M.A. Grepl and M. Kärcher, Reduced basis a posteriori error bounds for parametrized linear-quadratic elliptic optimal control problems. C. R. Acad. Sci. Paris, Ser. I 349 (2011) 873–877. [Google Scholar]\n10. M. Hinze, R. Pinnau, M. Ulbrich and S. Ulbrich, Optimization with PDE Constraints. Springer-Verlag, Berlin. Math. Model. Theory Appl. 23 (2009). [Google Scholar]\n11. M. Hinze and S. Volkwein, Error estimates for abstract linear-quadratic optimal control problems using proper orthogonal decomposition. Comput. Optim. Appl. 39 (2008) 319–345. [Google Scholar]\n12. P. Holmes, J.L. Lumley and G. Berkooz, Turbulence, coherent structures, dynamical systems and symmetry. Cambridge Monographs on Mechanics. Cambridge University Press, Cambridge (1996). [Google Scholar]\n13. A.D. Ioffe, Necessary and sufficient conditions for a local minimum 3 : Second order conditions and augmented duality. SIAM : J. Control Optim. 17 (1979) 266–288. [CrossRef] [MathSciNet] [Google Scholar]\n14. K. Ito and K. Kunisch, Lagrange Multiplier Approach to Variational Problems and Applications. SIAM, Philadelphia (2008). [Google Scholar]\n15. M. Kahlbacher and S. Volkwein, POD aposteriori error based inexact SQP method for bilinear elliptic optimal control problems. ESAIM : M2AN 46 (2012) 491–511. [CrossRef] [EDP Sciences] [Google Scholar]\n16. K. Kunisch and S. Volkwein, Galerkin proper orthogonal decomposition methods for parabolic problems. Numer. Math. 90 (2001) 117–148. [CrossRef] [MathSciNet] [Google Scholar]\n17. O. Lass and S. Volkwein, POD Galerkin schemes for nonlinear elliptic-parabolic systems (2011). Submitted. [Google Scholar]\n18. J.L. Lions, Optimal Control of Systems Governed by Partial Differential Equations. Springer-Verlag, Berlin (1971). [Google Scholar]\n19. K. Malanowski, Convergence of approximations vs. regularity of solutions for convex, control–constrained optimal control problems. Appl. Math. Optim. 8 (1981) 69–95. [CrossRef] [Google Scholar]\n20. K. Malanowski, Two-norm approach in stability and sensitivity analysis of optimization and optimal control problems. Adv. Math. Sci. Appl. 2 (1993) 397–443. [MathSciNet] [Google Scholar]\n21. K. Malanowski, Ch. Büskens and H. Maurer, Convergence of approximations to nonlinear optimal control problems, in Mathematical Programming with Data Perturbations, edited by Marcel-Dekker, Inc. Lect. Notes Pure Appl. Math. 195 (1997) 253–284. [Google Scholar]\n22. I. Neitzel and B. Vexler, A priori error estimates for space-time finite element discretization of semilinear parabolic optimal control problems. Numer. Math. 120 (2012) 345–386. [CrossRef] [MathSciNet] [Google Scholar]\n23. A. Rösch and D. Wachsmuth, Numerical verification of optimality conditions. SIAM J. Control Optim. 47 (2008) 2557–2581. [CrossRef] [MathSciNet] [Google Scholar]\n24. A. Rösch and D. Wachsmuth, How to check numerically the sufficient optimality conditions for infinite-dimensional optimization problems, in Optimal control of coupled systems of partial differential equations, Int. Ser. Numer. Math. 158 (2009) 297–317. [Google Scholar]\n25. E.W. Sachs and M. Schu, A priori error estimates for reduced order models in finance. ESAIM : M2AN 47 (2013) 449–469. [CrossRef] [EDP Sciences] [Google Scholar]\n26. K. Schittkowski, Numerical solution of a time–optimal parabolic boundary-value control problem. JOTA 27 (1979) 271–290. [CrossRef] [Google Scholar]\n27. A. Studinger and S. Volkwein, Numerical analysis of POD a posteriori error estimation for optimal control (2012). [Google Scholar]\n28. T. Tonn, K. Urban and S. Volkwein, Comparison of the reduced-basis and POD a posteriori error estimators for an elliptic linear quadratic optimal control problem. Mathematical and Computer Modelling of Dynamical Systems, Math. Comput. Modell. Dyn. Syst. 17 (2011) 355–369. [Google Scholar]\n29. F. Tröltzsch, Optimal Control of Partial Differential Equations. American Math. Society, Providence, Theor. Methods Appl. 112 (2010). [Google Scholar]\n30. F. Tröltzsch and S. Volkwein, POD a posteriori error estimates for linear-quadratic optimal control problems. Comput. Optim. Appl. 44 (2009) 83–115. [CrossRef] [Google Scholar]\n31. S. Volkwein, Optimal control of a phase-field model using proper orthogonal decomposition. ZAMM Z. Angew. Math. Mech. 81 (2001) 83–97. [CrossRef] [MathSciNet] [Google Scholar]\n32. S. Volkwein, Model Reduction using Proper Orthogonal Decomposition. Lecture notes, Institute of Mathematics and Statistics, University of Konstanz (2011). [Google Scholar]\n\nCurrent usage metrics show cumulative count of Article Views (full-text article views including HTML views, PDF and ePub downloads, according to the available data) and Abstracts Views on Vision4Press platform.\n\nData correspond to usage on the plateform after 2015. The current usage metrics is available 48-96 hours after online publication and is updated daily on week days." ]
[ null ]
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http://topcoder.bgcoder.com/print.php?id=1042
[ "### Problem Statement\n\nWe have to make a uniform random decision among k possible outcomes, but all we have available is an unfair coin.\n\nThe coin has a probability pH of giving Heads and pT = 1 - pH of giving Tails in the first throw. Subsequent throws are dependent on the previous throw, and there is a probability pHT of throwing Tails after a Heads throw and pTH of throwing Heads after a Tails throw. Note that either Heads or Tails always comes out, so the probability of throwing Tails after a Tails throw is pTT = 1 - pTH and the probability of throwing Heads after a Heads throw is pHH = 1 - pHT.\n\nIn order to make a fair decision, we throw the coin n times. If there are k coin result sequences that are equally probable (independent of pH, pHT and pTH), we can map each of these sequences to one of the k original decision outcomes. We repeat this with further coin result sequences, mapping as many of them as possible to the k original decision outcomes. At the end, some of the coin result sequences may remain unmapped. After we throw the coin n times, either one of the mapped sequences will come out, in which case we can directly make the fair decision by choosing the mapped outcome, or one of the unmapped sequences will come out, in which case we still cannot make a fair decision (we could then repeat making n further throws, until one of the mapped sequences comes out).\n\nFor example, let's say k = 3, and n = 6 (we have to choose among three outcomes, and throw the coin 6 times). In this case, the sequences \"HTHHHH\", \"HHTHHH\", \"HHHTHH\" and \"HHHHTH\" are equally probable (the given strings show the outcome of the 6 coin throws in order, with H representing Heads and T representing Tails; each of these sequences has a probability of pH * pHH * pHH * pHT * pTH). So we can map three of these sequences to the three outcomes, and one sequence will remain unmapped. Similarly, we can map the sequences \"HTTHHH\", \"HHTTHH\" and \"HHHTTH\" to the three outcomes. We get similarly six further mappings if we exchange \"H\" and \"T\" in these sequences. It turns out that these are the only mappings that we can find in this case, meaning that we have mapped 12 of the 26=64 possible sequences, with 52 sequences remaining unmapped. Note also that sequences that remain unmapped in one mapping step can still be used in further steps - i.e., if in the example above k was 2 instead of 3, and we mapped 2 of the first 4 mentioned sequences above to the 2 outcomes, the remaining 2 would also get mapped, since they still have the same probability. In general, if we have m equally probable outcomes, m % k of them will remain unmapped, while the other m - (m % k) outcomes will get mapped to some of the original decisions.\n\nYou will be given k, the number of outcomes in the decision you have to make, n, the number of coin throws you perform, and pH, pHT and pTH, the initial and subsequent conditional probabilities of the coin outcome (pH, pHT and pTH from the above description, respectively). You are to return the probability that the n throws will be sufficient to make a decision - i.e., the probability that we will get one of the outcomes that we have mapped to one of the original decision outcomes.\n\n### Definition\n\n Class: RandomChoice Method: decisionProbability Parameters: int, int, double, double, double Returns: double Method signature: double decisionProbability(int k, int n, double pH, double pHT, double pTH) (be sure your method is public)\n\n### Notes\n\n-When checking if two coin result sequences are equally probable, we don't take the given probabilities into account - i.e. the person performing the described procedure does not have any information about the probabilities. The probabilities are given in the method input since they are needed to compute the return value - i.e., after having decided which sequences are still unmapped, to compute the probability of one of these sequences coming up.\n-Your return value must have an absolute or relative error less than 1e-9.\n\n### Constraints\n\n-k will be between 2 and 1000, inclusive.\n-n will be between 1 and 1000, inclusive.\n-pH will be between 0.01 and 0.99, inclusive.\n-pHT will be between 0.01 and 0.99, inclusive.\n-pTH will be between 0.01 and 0.99, inclusive.\n\n### Examples\n\n0)\n\n `2` `4` `0.5` `0.5` `0.5`\n`Returns: 0.25`\n With n=4 we can get the following outcomes, each with the given probability: ```HHHH (pH * pHH * pHH * pHH) THHH (pT * pTH * pHH * pHH) HHHT (pH * pHH * pHH * pHT) THHT (pT * pTH * pHH * pHT) HHTH (pH * pHH * pHT * pTH) THTH (pT * pTH * pHT * pTH) HHTT (pH * pHH * pHT * pTT) THTT (pT * pTH * pHT * pTT) HTHH (pH * pHT * pTH * pHH) TTHH (pT * pTT * pTH * pHH) HTHT (pH * pHT * pTH * pHT) TTHT (pT * pTT * pTH * pHT) HTTH (pH * pHT * pTT * pTH) TTTH (pT * pTT * pTT * pTH) HTTT (pH * pHT * pTT * pTT) TTTT (pT * pTT * pTT * pTT) ``` The outcomes HHTH and HTHH are equally probable (independent of how the individual probabilities pH, pHT, pTH are chosen), so they can be mapped to the k=2 original outcomes. The same is true for the outcomes TTHT and THTT. The rest remains unmapped (there is no other pair with equal probabilities). To compute the return value, we now insert the given values for pH, pHT, pTH, and add the probabilities for the mapped outcomes (4 mapped outcomes, each with probability 0.54, which gives the result 0.25).\n1)\n\n `2` `4` `0.1` `0.8` `0.6`\n`Returns: 0.3648`\n The same as the previous examples, but with different probabilities. The final result is now: 2 * pH * pHH * pHT * pTH + 2 * pT * pTT * pTH * pHT = 2 * 0.1 * (1.0 - 0.8) * 0.8 * 0.6 + 2 * (1.0 - 0.1) * (1.0 - 0.6) * 0.6 * 0.8 = 0.3648.\n2)\n\n `2` `6` `0.5` `0.5` `0.5`\n`Returns: 0.625`\n With n=6, we get the following equally probable outcome n-tuples: ``` 1: (HHHHTH, HHHTHH, HHTHHH, HTHHHH) 2: (HHTHTT, HHTTHT, HTHHTT, HTTHHT) 3: (TTTTHT, TTTHTT, TTHTTT, THTTTT) 4: (TTHTHH, TTHHTH, THTTHH, THHTTH) 5: (HHHTHT, HHTHHT, HTHHHT) 6: (HHHTTH, HHTTHH, HTTHHH) 7: (HHTHTH, HTHHTH, HTHTHH) 8: (HTHTTT, HTTHTT, HTTTHT) 9: (TTTHTH, TTHTTH, THTTTH) 10: (TTTHHT, TTHHTT, THHTTT) 11: (TTHTHT, THTTHT, THTHTT) 12: (THTHHH, THHTHH, THHHTH) 13: (HHTTTH, HTTTHH) 14: (HTHTTH, HTTHTH) 15: (TTHHHT, THHHTT) 16: (THTHHT, THHTHT) ``` Since here k=2, we can map all outcomes in all quadruples (1 - 4) and all pairs (13 - 16) and from each triple (5 - 12) we map a pair and leave one element unmapped. So in total we have mapped 40 outcomes. Since we have pH=pHT=pTH=0.5, the probability of each outcome is 0.56, and the final result is 40 * 0.56 = 0.625.\n3)\n\n `3` `6` `0.5` `0.5` `0.5`\n`Returns: 0.5625`\n The same example as the last one, but now with k=3. Now we can map all triples (5 - 12 from the previous example) to the k=3 outcomes and 3 elements from each quadruple (1 - 4), for a total of 36 mapped outcomes, giving 36 * 0.56 = 0.5625.\n4)\n\n `5` `1` `0.5` `0.5` `0.5`\n`Returns: 0.0`\n With just one throw we can never be sure to make a fair decision.\n5)\n\n `1000` `1000` `0.5` `0.01` `0.01`\n`Returns: 0.9972747235292863`\n Be sure not to timeout.\n\n#### Problem url:\n\nhttp://www.topcoder.com/stat?c=problem_statement&pm=4819\n\n#### Problem stats url:\n\nhttp://www.topcoder.com/tc?module=ProblemDetail&rd=8092&pm=4819\n\ngepa\n\n#### Testers:\n\nPabloGilberto , lbackstrom , vorthys , Olexiy\n\nMath" ]
[ null ]
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https://brilliant.org/problems/photoelectric-effect-4/
[ "# Photoelectric Effect\n\nThe photoelectric effect is the observation that many metals emit electrons when light shines upon them. Electrons emitted in this manner can be called photoelectrons.\n\nThe work function of an element is the minimum energy needed to remove an electron from the solid element. The work function for rubidium is $\\SI[per-mode=symbol]{208.4}{\\kilo\\joule\\per\\mole}$.\n\nWhat is the minimum amount of energy needed to eject a single electron from the surface of rubidium metal?\n\n\nDetails and Assumptions:\n\n• $N_A = 6.022 \\times 10^{23} \\si{\\mole}.$\n• $h = 6.63 \\times 10^{-34}\\si{\\joule\\second}.$\n• $c = 3 \\times 10^8\\si[per-mode=symbol]{\\meter\\per\\second}.$", null, "×" ]
[ null, "https://ds055uzetaobb.cloudfront.net/brioche/uploads/oBkzAExMMg-download.png", null ]
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https://physics.stackexchange.com/questions/615424/extension-of-faddeev-jackiw-first-order-lagrangian-formalism-to-fields
[ "# Extension of Faddeev-Jackiw first-order Lagrangian formalism to fields\n\nIn this paper, Toms discusses the method that Faddeev and Jackiw proposed for quantization of constrained theories. In section III.B, he applies this method to a field theory, but I have several doubts about how this transition is done.\n\nA first-order Lagrangian for a mechanical system can be written as\n\n$$\\begin{equation}\\tag{2.1} L=A_\\alpha(\\xi)\\dot{\\xi}^\\alpha+L_v(\\xi), \\end{equation}$$\n\nwhose Euler-Lagrange equations are\n\n$$\\begin{equation}\\tag{2.2} F_{\\alpha\\beta}\\dot{\\xi}^\\beta=-\\frac{\\partial L_v}{\\partial\\xi^\\alpha}, \\end{equation}$$\n\nwith\n\n$$\\begin{equation}\\tag{2.3} F_{\\alpha\\beta}=\\frac{\\partial A_\\beta}{\\partial\\xi^\\alpha}-\\frac{\\partial A_\\alpha}{\\partial\\xi^\\beta} \\end{equation}$$\n\nbeing the symplectic two-form. So far so good.\n\nLater, on section III.B, he says that the traduction of expresion (2.3) to field theory is\n\n$$\\begin{equation}\\tag{3.41} F_{\\alpha\\beta}=\\frac{\\delta A_\\beta}{\\delta\\xi^\\alpha}-\\frac{\\delta A_\\alpha}{\\delta\\xi^\\beta}, \\end{equation}$$\n\nbut I can't see how this araises from Euler-Lagrange equations, i.e.\n\n$$\\begin{equation}\\tag{EL} \\frac{\\delta S}{\\delta\\phi^\\alpha}=\\frac{\\partial\\mathcal{L}}{\\partial\\phi^\\alpha}-\\partial_\\mu\\Bigg(\\frac{\\partial\\mathcal{L}}{\\partial({\\partial_\\mu\\phi^\\alpha})}\\Bigg)=0, \\end{equation}$$\n\nwith\n\n$$\\begin{equation} S[\\phi]=\\int d^4{x} \\mathcal{L}(\\phi,\\partial\\phi). \\end{equation}$$\n\nOn a field theory, I suppose that a first-orden lagrangian density would be written as\n\n$$\\begin{equation}\\tag{LD} \\mathcal{L}=A_\\alpha(\\phi,\\nabla\\phi)\\dot{\\phi}^\\alpha+\\mathcal{L}_v(\\phi,\\nabla\\phi), \\end{equation}$$\n\nwhere $$\\phi^\\alpha$$ are the fields and $$\\nabla\\phi$$ indicates dependence in spatial derivatives. However, if I insert that lagrangian density into (EL) I can't get an expression equivalent to (2.2). Moreover, the \"$$A$$\"s in expresion (3.41) must be functionals. Which is the relation between those \"$$A$$\"s and the ones in (LD)?\n\n## 1 Answer\n\nOP is essentially asking the following.\n\nHow does the point-mechanical Faddeev-Jackiw formulas from chapter II generalizes to the field-theoretic formulas in section III.B?\n\nWell, let's see. The field-theoretic first-order Lagrangian is $$L~=~\\int\\!d^3{\\bf x}~A_{\\alpha}({\\bf x}) \\dot{\\xi}^{\\alpha}({\\bf x})~-~H.\\tag{2.1'}$$ An infinitesimal variation reads \\begin{align} \\delta L ~\\sim~& \\int\\!d^3{\\bf x}\\left( \\delta A_{\\alpha}({\\bf x}) \\dot{\\xi}^{\\alpha}({\\bf x}) -\\frac{dA_{\\alpha}({\\bf x})}{dt} \\delta\\xi^{\\alpha}({\\bf x})\\right) ~-~\\delta H \\cr ~=~&\\int\\!d^3{\\bf x} \\int\\!d^3{\\bf x}^{\\prime} \\left( \\delta\\xi^{\\alpha}({\\bf x}) \\frac{\\delta A_{\\beta}({\\bf x}^{\\prime})}{\\delta \\xi^{\\alpha}({\\bf x})} \\dot{\\xi}^{\\beta}({\\bf x}^{\\prime}) ~-~ \\dot{\\xi}^{\\beta}({\\bf x}^{\\prime}) \\frac{\\delta A_{\\alpha}({\\bf x})}{\\delta \\xi^{\\beta}({\\bf x}^{\\prime})} \\delta\\xi^{\\alpha}({\\bf x}) \\right) \\cr &~-~ \\delta H\\cr ~=~&\\int\\!d^3{\\bf x} \\int\\!d^3{\\bf x}^{\\prime} \\delta\\xi^{\\alpha}({\\bf x}) F_{\\alpha\\beta}({\\bf x},{\\bf x}^{\\prime})\\dot{\\xi}^{\\beta}({\\bf x}^{\\prime}) ~-~ \\int\\!d^3{\\bf x}~ \\delta \\xi^{\\alpha}({\\bf x}) \\frac{\\delta H}{\\delta \\xi^{\\alpha}({\\bf x})}. \\end{align} Here the $$\\sim$$ symbol means equality modulo total time derivative terms. Also we have defined the components of symplectic 2-form $$F_{\\alpha\\beta}({\\bf x},{\\bf x}^{\\prime})~=~\\frac{\\delta A_{\\beta}({\\bf x}^{\\prime})}{\\delta\\xi^{\\alpha}({\\bf x})}-\\frac{\\delta A_{\\alpha}({\\bf x})}{\\delta\\xi^{\\beta}({\\bf x}^{\\prime})}.\\tag{2.3'/3.41'}$$ (Judging from eq. (3.42) it becomes clear that eq. (3.41) is bi-local.)\n\nThe Euler-Lagrange equations are Hamilton's equations $$\\int\\!d^3{\\bf x}^{\\prime} F_{\\alpha\\beta}({\\bf x},{\\bf x}^{\\prime})\\dot{\\xi}^{\\beta}({\\bf x}^{\\prime}) ~=~ \\frac{\\delta H}{\\delta \\xi^{\\alpha}({\\bf x})}.\\tag{2.2'}$$\n\nReferences:\n\n1. D.J. Toms, Faddeev-Jackiw quantization and the path integral, arXiv:1508.07432.\n• Thank you very much. – AFG Feb 20 at 11:55" ]
[ null ]
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http://mymathforum.com/algebra/342220-help-hard-prove-question.html
[ "User Name Remember Me? Password\n\n Algebra Pre-Algebra and Basic Algebra Math Forum\n\n October 11th, 2017, 06:11 PM #1 Newbie   Joined: Oct 2017 From: Dream Country Posts: 3 Thanks: 1 Help for the hard prove question If n is a positive integer and n is even, prove: (2^(n!)-1) is divisible by (n^2-1). This question confuse me some days. Please help me or give me some hint. Thank you very much. Last edited by skipjack; October 12th, 2017 at 01:57 AM.", null, "October 11th, 2017, 07:27 PM   #2\nMath Team\n\nJoined: May 2013\nFrom: The Astral plane\n\nPosts: 2,304\nThanks: 961\n\nMath Focus: Wibbly wobbly timey-wimey stuff.\nQuote:\n Originally Posted by GGQ", null, "If n is a positive integer and n is even, prove: (2^(n!)-1) is divisible by (n^2-1). This question confuse me some days. Please help me or give me some hint. Thank you very much.\nLet n = 3. Then $\\displaystyle \\frac{2^{3!} - 1}{3^2 - 1} = \\frac{2^6 - 1}{8}$ $\\displaystyle = \\frac{63}{8}$\nSo it doesn't work for n = 3.\n\n-Dan\n\nLast edited by skipjack; October 12th, 2017 at 01:58 AM.", null, "October 11th, 2017, 07:44 PM #3 Newbie   Joined: Oct 2017 From: Dream Country Posts: 3 Thanks: 1 question is \" n is even \" so the example that you take n=3 is illegal. thank you. Thanks from topsquark", null, "October 12th, 2017, 11:35 AM   #4\nMath Team\n\nJoined: May 2013\nFrom: The Astral plane\n\nPosts: 2,304\nThanks: 961\n\nMath Focus: Wibbly wobbly timey-wimey stuff.\nQuote:\n Originally Posted by GGQ", null, "question is \" n is even \" so the example that you take n=3 is illegal. thank you.\nSorry. Yes, I missed that.\n\n-Dan", null, "October 12th, 2017, 12:29 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Since n is an even positive integer, we can take n= 2k for k any positive integer. Since we have a statement that depends on the positive integer k, this pretty much cries out for \"proof by induction\" on k. Base case: k= 1. In that case n= 2, n!= 2 so the statement becomes \"2^2- 1= 4- 1= 3 is divisible by 2^2- 1= 4- 1= 3. 3 is clearly divisible by 3. Now, suppose that 2^((2k)!-1) is divisible by (2k)^2- 1. That is: 2^((2k)!- 1)= p((2k)^2- 1) for some integer, p. Then 2^((2(k+1))!- 1)= 2^(2(k+1)(2k+1)(2k!)- 1)= 2^(2(k+1)(2k+1)(k!-1)+ k+1-1)= 2^k2^((k+1)(k!-1)= 2^k(2^(k!+ 1))^(k+1)= 2^k(p(k^2- 1))^{k+ 1). That last clearly has k^2- 1 as a factor so is divisible by k^2- 1. Thanks from topsquark and GGQ Last edited by Country Boy; October 12th, 2017 at 12:33 PM.", null, "October 12th, 2017, 05:58 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Don't you need it to be divisible by $(2k)^2-1=4k^2-1$? Thanks from topsquark", null, "October 12th, 2017, 10:09 PM   #7\nNewbie\n\nJoined: Oct 2017\nFrom: Dream Country\n\nPosts: 3\nThanks: 1\n\nQuote:\n Originally Posted by Country Boy", null, "Since n is an even positive integer, we can take n= 2k for k any positive integer. Since we have a statement that depends on the positive integer k, this pretty much cries out for \"proof by induction\" on k. Base case: k= 1. In that case n= 2, n!= 2 so the statement becomes \"2^2- 1= 4- 1= 3 is divisible by 2^2- 1= 4- 1= 3. 3 is clearly divisible by 3. Now, suppose that 2^((2k)!-1) is divisible by (2k)^2- 1. That is: 2^((2k)!- 1)= p((2k)^2- 1) for some integer, p. Then 2^((2(k+1))!- 1)= 2^(2(k+1)(2k+1)(2k!)- 1)= 2^(2(k+1)(2k+1)(k!-1)+ k+1-1)= 2^k2^((k+1)(k!-1)= 2^k(2^(k!+ 1))^(k+1)= 2^k(p(k^2- 1))^{k+ 1). That last clearly has k^2- 1 as a factor so is divisible by k^2- 1.\n\nThis part \" 2^((2k)!-1) is divisible by (2k)^2- 1. \" is wrong.\nCorrect is ( 2^(2k)!) -1 is divisible by (2k)^2- 1\nthank you.", null, "Tags hard, integer, prove, question", null, "Thread Tools", null, "Show Printable Version", null, "Email this Page Display Modes", null, "Linear Mode", null, "Switch to Hybrid Mode", null, "Switch to Threaded Mode", null, "Similar Threads Thread Thread Starter Forum Replies Last Post isel Algebra 8 September 22nd, 2013 09:35 AM stuart clark Algebra 1 March 23rd, 2011 11:46 PM riotsandravess Advanced Statistics 4 November 3rd, 2010 06:22 PM riotsandravess Advanced Statistics 0 October 28th, 2010 01:21 PM yehoram Algebra 8 August 9th, 2010 08:37 PM\n\n Contact - Home - Forums - Cryptocurrency Forum - Top", null, "", null, "", null, "", null, "", null, "", null, "" ]
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https://kmmiles.com/2386-6-miles-in-km
[ "# 2386.6 miles in km\n\n## Result\n\n2386.6 miles equals 3840.0394 km\n\n## Conversion formula\n\nMultiply the amount of miles by the conversion factor to get the result in km:\n\n2386.6 mi × 1.609 = 3840.0394 km\n\n## How to convert 2386.6 miles to km?\n\nThe conversion factor from miles to km is 1.609, which means that 1 miles is equal to 1.609 km:\n\n1 mi = 1.609 km\n\nTo convert 2386.6 miles into km we have to multiply 2386.6 by the conversion factor in order to get the amount from miles to km. We can also form a proportion to calculate the result:\n\n1 mi → 1.609 km\n\n2386.6 mi → L(km)\n\nSolve the above proportion to obtain the length L in km:\n\nL(km) = 2386.6 mi × 1.609 km\n\nL(km) = 3840.0394 km\n\nThe final result is:\n\n2386.6 mi → 3840.0394 km\n\nWe conclude that 2386.6 miles is equivalent to 3840.0394 km:\n\n2386.6 miles = 3840.0394 km\n\n## Result approximation\n\nFor practical purposes we can round our final result to an approximate numerical value. In this case two thousand three hundred eighty-six point six miles is approximately three thousand eight hundred forty point zero three nine km:\n\n2386.6 miles ≅ 3840.039 km\n\n## Conversion table\n\nFor quick reference purposes, below is the miles to kilometers conversion table:\n\nmiles (mi) kilometers (km)\n2387.6 miles 3841.6484 km\n2388.6 miles 3843.2574 km\n2389.6 miles 3844.8664 km\n2390.6 miles 3846.4754 km\n2391.6 miles 3848.0844 km\n2392.6 miles 3849.6934 km\n2393.6 miles 3851.3024 km\n2394.6 miles 3852.9114 km\n2395.6 miles 3854.5204 km\n2396.6 miles 3856.1294 km\n\n## Units definitions\n\nThe units involved in this conversion are miles and kilometers. This is how they are defined:\n\n### Miles\n\nA mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.\n\n### Kilometers\n\nThe kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world." ]
[ null ]
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https://tacunasystems.com/knowledge-base/load-cell-minimum-weight/
[ "You are here:\n\n## Preface\n\nThe minimum weight, or even change in weight, that a load cell can sense is important information when selecting the right load cell for a force measurement application. This article explains how to uncover this critical piece of knowledge.\n\n## What Is the Minimum Weight a Load Cell Can Sense?\n\nTo answer this question, one must study the load cell manufacturer’s data sheet giving the product’s specifications. This spec sheet gives a division range which can be used to determine the minimum weight that a load cell can measure. Division ranges vary from load cell to load cell, and usually fall between 500-10,000 divisions.\n\nAnother item of importance on this data sheet is the sensitivity rating. The sensitivity of a system indicates the rate of change of its output as the system input varies. Systems that have low sensitivity ranges or are of lower quality will not be able to detect small changes that occur when a load cell experiences a force from a minimal weight.\n\nThe final section of this paper discusses other important data sheet specifications affecting the choice of load cell for the application. The ability to read the minimum weight required by an application depends on careful consideration of these specifications.\n\nFurthermore, the individual components of a load cell system itself can introduce errors that change the ability of the system to measure small loads accurately. Figure 1 shows a block diagram of a typical load cell system.\n\nThe analog portion of load cell system consists of several strain gauge sensors and a Wheatstone bridge (both built into the load cell), an amplifier and offset stage, an analog-to-digital converter (ADC), and a power supply. Each of these stages can affect the reliability and accuracy of a load cell system.\n\n## How a Load Cell System Works\n\nThe sensing portion of a load cell is a strain gauge. A strain gauge is a passive transducer that changes its resistance proportionally to the stress placed on it. Therefore load cells made with this sensor respond to different weights by a change in resistance. These resistance changes are usually very small, and measured with a circuit known as a Wheatstone Bridge. A Wheatstone bridge has two voltage dividers. Figure 2 shows an example of  a Wheatstone Bridge circuit.\n\n#### Full Scale Output vs Maximum Measurable Output\n\nFigure 2 shows an unbalanced Wheatstone bridge with a measurable voltage difference between the two output terminals. This load cell requires an excitation voltage across the cell power rails, typically in the range of 5-15V. As previously stated, a force applied on the load cell creates a voltage difference between the output terminals from the Wheatstone bridge, usually in the µV range. There is an upper limit to this measurable change, known as the full scale outputThis full scale output of a load cell appears on the manufacturer data sheetIt should be noted however that the maximum voltage difference between the terminals is relative to the excitation voltage. The maximum output equals the full scale output multiplied by the excitation voltage. A full scale output of 3 mV/V and an excitation voltage of 10 V will result in a maximum measurable difference of 30 mV.\n\n#### Divisions\n\nThe smallest measurable difference between the terminals is the maximum output divided by the number of divisions from the load cell. If the load cell in the above example has a division range of 10,000, the smallest detectable voltage change would be 30mv/10,000 = 3 µV. Since this is such a small voltage difference, the system required to measure this change must be very sensitive.\n\n#### Signal Amplification\n\nTypically this output signal undergoes signal amplification. This process boosts the output signal to a detectable level. An instrumentation amplifier is a common type; this amplifier uses two input buffers for high input impedance, and a difference amplifier between the Wheatstone bridge output and the ADC. The ADC takes an analog signal and converts it into a discrete signal usable by a digital display.\n\nFigure 3 shows a standard instrumentation amplifier circuit.\n\nThe instrumentation amplifier stage needs to have high differential gain, high power supply rejection ratio, low drift, low offset and low input bias current . If the instrumentation amplifier has all of these qualifications, the signal-to-noise ratio will be large enough for accurate conversion by the ADC. A direct correlation exists between the number of divisions of an ADC from an indicator and the number of divisions of a load cell. An ADC must have enough high enough resolution, i.e. enough bits, to detect small changes from a high division load cell. Manufacturers of load cell indicators will provide the number of bits of the ADC, and sometimes even provide details on which load cells should be used with the system.\n\n## Load Cell Specifications and Definitions\n\nAs mentioned above, load cells have many important specifications describing their suitability for a particular application and the demands placed on it throughout its lifetime. Figure 4 displays typical specifications found on a load cell data sheet.\n\n#### Maximum Error\n\nAs seen in Figure 4, load cell specifications include error ranges for various causes such as nonlinearity, hysteresis and creep. The sum total of these errors is known as the combined error or maximum possible error. The maximum possible error is important when considering accurate division measurements . If a load cell experiences its maximum possible error, its minimum weight divisions be less reliable. A load cell with a 0.1% combined error has minimum divisions accurate with a maximum error of ±0.1% of rated output. The total error is determined by a number of varying factors, from temperature deviations to magnetic error.  The following definitions are important determinants of the accuracy class of a load cell. Understanding the source of error can help increase accuracy when testing a load cell system. These definitions are provided from the Handbook of Electronic Weighting.\n\n#### Other Definitions\n\nThe following definitions are important determinants of the accuracy class of a load cell. Understanding the source of error can help increase accuracy when testing a load cell system. These definitions are provided from the Handbook of Electronic Weighting.\n\n1. Zero Balance: This is the electrical output signal of the load cell when no weight or load is placed on it.\n2. Non-Linearity: This expresses the maximum deviation of the calibration curve. This curve is obtained by gradually increasing the applied weight from the zero balance level to the rated output of the load cell, and plotting the output against this weight. The smaller the non-linearity, the more accurate measurement we obtain.\n3. Hysteresis: This is the maximum deviation of the output signal for the same applied load. The first set of values are obtained by increasing the applied weight from zero balance to the rated output, while the second readings are obtained by decreasing the rated output to the zero balance level. Assume the x-axis is the applied weight and the y-axis is the output on a plot. The smaller this numerical difference is at the same weight (the smaller the difference in the y values at the same x value), the more accurate measurement we obtain.\n4. Non-Repeatability: This is the maximum difference between the electrical output signal of the load cell for repeated loads under identical environmental and loading conditions. A small value depicts a high system accuracy and reliability.\n5. Creep: This specification becomes very important when a load is constant over a long time, such as for monitoring purposes. Creep is the change in the load cell output signal level with respect to time under a constant load, with all environmental conditions being constant.\n6. Temperature Effect on Output: This is the effect of temperature shifts on the output of the load cell as it tends to introduce errors that affect system accuracy.\n7. Temperature Effects on Zero Balance: Temperature shifts also affect the output signal of the load cell under no-load. To cater for both types of temperature shifts, ensure the load cell design you are using incorporates a temperature compensation technique.\n\n## Regulations and Standards on Load Cell Specifications\n\nMultiple organizations maintain industry-accepted standards for measurement applications. These include organizations such as the International Organization of Legal Metrology (OIML) and the National Type Evaluation Committee (NTEP). These organizations have their own standards and tolerances for different load cell classes. Figure 5 gives an example chart of some tolerances given by OIML.\n\nHighlighted in Figure 5 are the minimum and maximum load cell divisions for class IIIL3 load cells. As can be seen, the minimum and maximum number of divisions for class IIIL3 load cells are 2000 and 10000 respectively. To find the minimum detectable weight of a load cell, simply divide its maximum rated weight by the number of divisions. For example, if a load cell is said to have 10,000 divisions with a capacity of 50,000 lb., the minimum weight that is measurable by the cell will be 5 lb. Load cells are typically given a classification indicating the number of divisions of the load cell. An example of this can be seen in Figure 6.\n\nFigure 6 shows the characteristics of the 102BH class of load cells, manufactured by Anyload. The maximum number of divisions (intervals) for this load cell class is 3000, per the figure. The maximum capacity range for this load cell class is between 11 t and 55 t. From the table we see that its accuracy class is OIML Class C. (See Load Cell Classes: OIML Requirements for further discussion on this point.) Similar data sheets for different load cell classes appear on manufacturers websites.\n\n## Summary\n\nAs this article explains, there are several important specifications that give clues to the minimum weight, and the accuracy of that weight, measurable by any given load cell. The number of divisions and range of weight determine a load cell’s minimum weight detectable . However, the sensitivity ratings and the quality of the system being used to detect the changes from a load cell are also critical factors in displaying accurate weight measurements.\n\n## Sources\n\n1.  Brusamarello, V., Machado de Brito, R., Muller, I., Pereira, C. E., “Load Cells in Force Sensing Analysis – Theory and a Novel Application”, ResearchGate, Jan. 2010.\n2.  Franco, S. “Design with operational amplifiers and analog integrated circuits”. New York: McGraw-Hill. pp. 87, 2015\n3. K. Elis Nordon, “Handbook of Electronic Weighting”, Wiley-VCH, pp. 24-29, Jul. 1998.\n4. “Load Cell Accuracy in Relation to the  Conditions of Use”, Technical Note VPGT-02, Jan. 8 2015. Retrieved from http://www.vishaypg.com/docs/11864/11864.pdf\n5. “Load Cell and Weight Module Handbook”,  Rice Lake Weighing Systems, pp. 9-10, 2010\n6. “OIML Certificate of Conformity”, Number R60/2000-NL1-10.27 , Dec. 2010\n7. “R 60 OIML-CS rev.2”, NIST Handbook 44, pp. 1-4, Jan 5 2018." ]
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http://hackage.haskell.org/package/HaTeX-3.16.0.0/docs/Text-LaTeX-Packages-TikZ-Syntax.html
[ "HaTeX-3.16.0.0: The Haskell LaTeX library.\n\nText.LaTeX.Packages.TikZ.Syntax\n\nDescription\n\nThis module defines the syntax of a TikZ script.\n\nTo generate a TikZ script, first create a `TPath` using data constructors, or alternatively, use a `PathBuilder` from the Text.LaTeX.Packages.TikZ.PathBuilder module.\n\nOnce a `TPath` is created, use `path` to render a picture from it. Use `scope` to apply some parameters to your picture, such line width or color.\n\nSynopsis\n\n# Points\n\ndata TPoint Source\n\nA point in TikZ.\n\nInstances\n\n Show TPoint Render TPoint\n\nPoint using `Measure`s for coordinantes.\n\nPoint using numbers as coordinates.\n\nThree-dimensional point.\n\nMakes a point relative to the previous.\n\n# Paths\n\n## Types\n\ndata TPath Source\n\nType for TikZ paths. Every `TPath` has two fundamental points: the starting point and the last point. The starting point is set using the `Start` constructor. The last point then is modified by the other constructors. Below a explanation of each one of them. Note that both starting point and last point may coincide. You can use the functions `startingPoint` and `lastPoint` to calculate them. After creating a `TPath`, use `path` to do something useful with it.\n\nConstructors\n\n Start TPoint Let `y = Start p`.Operation: Set the starting point of a path.Last point: The last point of `y` is `p`. Cycle TPath Let `y = Cycle x`.Operation: Close a path with a line from the last point of `x` to the starting point of `x`.Last point: The last point of `y` is the starting point of `x`. Line TPath TPoint Let `y = Line x p`.Operation: Extend the current path from the last point of `x` in a straight line to `p`.Last point: The last point of `y` is `p`. Rectangle TPath TPoint Let `y = Rectangle x p`.Operation: Define a rectangle using the last point of `x` as one corner and `p` as the another corner.Last point: The last point of `y` is `p`. Circle TPath Double Let `y = Circle x r`.Operation: Define a circle with center at the last point of x and radius `r`.Last point: The last point of `y` is the same as the last point of `x`. Ellipse TPath Double Double Let `y = Ellipse x r1 r2`.Operation: Define a ellipse with center at the last point of `x`, width the double of `r1` and height the double of `r2`.Last point: The last point of `y` is the same as the last point of `x`. Grid TPath [GridOption] TPoint Node TPath LaTeX Let `y = Node x l`.Operation: Set a text centered at the last point of `x`.Last point: The last point of `y` is the same as the last point of `x`.\n\nInstances\n\n Show TPath Render TPath\n\ndata GridOption Source\n\nConstructors\n\n GridStep Step\n\nInstances\n\n Show GridOption Render GridOption\n\ndata Step Source\n\nConstructors\n\n DimStep Measure XYStep Double PointStep TPoint\n\nInstances\n\n Show Step Render Step\n\n## Critical points\n\nCalculate the starting point of a `TPath`.\n\nCalculate the last point of a `TPath`.\n\n## Functions\n\nAlias of `Line`.\n\n# Parameters\n\ndata Parameter Source\n\nParameters to use in a `scope` to change how things are rendered within that scope.\n\nConstructors\n\n TWidth Measure TColor TikZColor TScale Double TRotate Double Angle is in degrees.\n\nInstances\n\n Show Parameter Render Parameter\n\ndata TikZColor Source\n\nColor models accepted by TikZ.\n\nConstructors\n\n BasicColor Color RGBColor Word8 Word8 Word8\n\nInstances\n\n Show TikZColor Render TikZColor\n\ndata Color Source\n\nBasic colors.\n\nConstructors\n\n Red Green Blue Yellow Cyan Magenta Black White\n\nInstances\n\n Show Color Render Color\n\ndata Word8 :: *\n\n8-bit unsigned integer type\n\n# TikZ\n\ndata TikZ Source\n\nA TikZ script.\n\nInstances\n\n Show TikZ Render TikZ\n\nJust an empty script.\n\npath :: [ActionType] -> TPath -> TikZ Source\n\nA path can be used in different ways.\n\n• `Draw`: Just draw the path.\n• `Fill`: Fill the area inside the path.\n• `Clip`: Clean everything outside the path.\n• `Shade`: Shade the area inside the path.\n\nIt is possible to stack different effects in the list.\n\nExample of usage:\n\n`path [Draw] \\$ Start (pointAtXY 0 0) ->- pointAtXY 1 1`\n\nMost common usages are exported as functions. See `draw`, `fill`, `clip`, `shade`, `filldraw` and `shadedraw`.\n\nscope :: [Parameter] -> TikZ -> TikZ Source\n\nApplies a scope to a TikZ script.\n\ndata ActionType Source\n\nDifferent types of actions that can be performed with a `TPath`. See `path` for more information.\n\nConstructors\n\n Draw Fill Clip Shade\n\nInstances\n\n Show ActionType Render ActionType\n\n(->>) :: TikZ -> TikZ -> TikZ Source\n\nSequence two TikZ scripts.\n\n# Sugar\n\nEquivalent to `path [Draw]`.\n\nEquivalent to `path [Fill]`.\n\nEquivalent to `path [Clip]`.\n\nEquivalent to `path [Shade]`.\n\nEquivalent to `path [Fill,Draw]`.\n\nEquivalent to `path [Shade,Draw]`." ]
[ null ]
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https://www.aplikatif.com/progress-calculation-in-primavera-at-the-wbs-level/
[ "# Understanding Progress Calculation in Primavera at the WBS Level\n\nWhen working with Primavera P6, you may have questioned why the Activity % Complete does not roll up (total) at the Work Breakdown Structure (WBS) level.\n\nThis article will explore the reasons behind this and shed light on how Primavera P6 handles progress calculations.\n\nContents\n\n## Different Units of Measure\n\nOne crucial aspect to consider is that each activity in Primavera P6 can have its unique unit of measure.\n\nFor example, the Excavation activity may use cubic meters (M3) as its unit, while the Form/Pour Concrete Slab activity may employ square meters (M2), and the Install Underground Water Lines activity might adopt meters (M) as its unit.\n\n## The Challenge of Incompatible Units\n\nSince these activities utilize different units of measure, they cannot be directly summed up to calculate the total progress. Primavera P6 does not automatically aggregate the Activity % Complete at the WBS level due to the lack of compatibility for direct summation.\n\n## Progress Calculation in Primavera P6\n\nIn Primavera P6, progress is typically tracked for each activity based on its specific unit of measure. The total progress value for each activity is then rolled up to higher levels in the project hierarchy, such as the WBS or project level, considering the activity’s weighted values or Earned Value Cost calculations.\n\n## Importance of Unit Price and Cost Calculations\n\nSuppose You want to understand progress at the WBS level accurately. In that case, it is crucial to know the unit prices for each activity and calculate the total cost for each activity accordingly.\n\nBy factoring in these unit prices and costs, you can obtain a comprehensive view of the project’s overall progress using the Earned Value Cost analysis.\n\n## Conclusion\n\nIn Primavera P6, the absence of rolled-up Activity % Complete at the WBS level stems from the different units of measure employed by individual activities.\n\nBy understanding this aspect and considering unit prices and costs, project managers can ensure better progress analysis at the WBS or Project level in Primavera P6." ]
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http://connorraylive.com/kindergarten-addition-math-worksheets/13/
[ "# 13. PRINTABLE KINDERGARTEN MATH WORKSHEETS DOMINO ADDITION 3 DOMINO ADDITION SHEET 3", null, "Date : 21 Jan, 2021\n\nCollection : kindergarten addition math worksheets\n\nBy : Johnas Tan\n\nnumber bonds worksheets 2nd grade save kindergarten addition math worksheets printable lovely number bonds\n\n📷", null, "kindergarten math worksheets addition veggie math\n\n📷", null, "📷", null, "kindergarten subtraction worksheets picture addition worksheets\n\n📷", null, "subtraction to 10 sheet 4\n\n📷", null, "printable math worksheets for kindergarten addition and subtraction\n\n📷", null, "📷", null, "📷", null, "kindergarten worksheets kindergarten worksheets free worksheets kids maths worksheets\n\n📷", null, "printable math worksheets for kindergarten addition and\n\n📷", null, "kindergarten addition worksheets 1 and 2\n\n📷", null, "addition homework page from kindergarten addition worksheets 1 6 from www tlsbooks com\n\n📷", null, "📷", null, "📷", null, "kindergarten math worksheets addition garden math\n\n📷", null, "📷", null, "📷", null, "", null, "", null, "" ]
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https://link.springer.com/article/10.1007/s00780-013-0225-4?error=cookies_not_supported&code=f287d0dc-45ee-4869-86f6-07f20169ad20
[ "# Comparative and qualitative robustness for law-invariant risk measures\n\n## Abstract\n\nWhen estimating the risk of a P&L from historical data or Monte Carlo simulation, the robustness of the estimate is important. We argue here that Hampel’s classical notion of qualitative robustness is not suitable for risk measurement, and we propose and analyze a refined notion of robustness that applies to tail-dependent law-invariant convex risk measures on Orlicz spaces. This concept captures the tradeoff between robustness and sensitivity and can be quantified by an index of qualitative robustness. By means of this index, we can compare various risk measures, such as distortion risk measures, in regard to their degree of robustness. Our analysis also yields results of independent interest such as continuity properties and consistency of estimators for risk measures, or a Skorohod representation theorem for ψ-weak convergence.\n\nThis is a preview of subscription content, access via your institution.\n\n## References\n\n1. 1.\n\nAcerbi, C., Tasche, D.: On the coherence of expected shortfall. J. Bank. Finance 26, 1487–1503 (2002)\n\n2. 2.\n\nBelomestny, D., Krätschmer, V.: Central limit theorems for law-invariant coherent risk measures. J. Appl. Probab. 49, 1–21 (2012)\n\n3. 3.\n\nBeutner, E., Zähle, H.: A modified functional delta method and its application to the estimation of risk functionals. J. Multivar. Anal. 101, 2452–2463 (2010)\n\n4. 4.\n\nBiagini, S., Frittelli, M.: A unified framework for utility maximization problems: an Orlicz space approach. Ann. Appl. Probab. 18, 929–966 (2008)\n\n5. 5.\n\nBiagini, S., Frittelli, M.: On the extension of the Namioka–Klee theorem and on the Fatou property for risk measures. In: Delbaen, F., et al. (eds.) Optimality and Risk: Modern Trends in Mathematical Finance. The Kabanov Festschrift, pp. 1–29. Springer, Berlin (2009)\n\n6. 6.\n\nBoussama, F.: Ergodicity, mixing and estimation in GARCH models. Ph.D. thesis, University of Paris 7 (1998)\n\n7. 7.\n\nBreiman, L.: Probability. Classics in Applied Mathematics, vol. 7. SIAM, Philadelphia (1991) (corrected reprint of the 1968 original)\n\n8. 8.\n\nCheridito, P., Li, T.: Risk measures on Orlicz hearts. Math. Finance 19, 189–214 (2009)\n\n9. 9.\n\nCherny, A., Madan, D.: New measures for performance evaluation. Rev. Financ. Stud. 22, 2571–2606 (2009)\n\n10. 10.\n\nCherny, A., Madan, D.: Markets as a counterparty: an introduction to conic finance. Int. J. Theor. Appl. Finance 13, 1149–1177 (2010)\n\n11. 11.\n\nCont, R., Deguest, R., Scandolo, G.: Robustness and sensitivity analysis of risk measurement procedures. Quant. Finance 10, 593–606 (2010)\n\n12. 12.\n\nChung, K.L.: The strong law of large numbers. In: Neyman, J. (ed.) Proceedings of the Second Berkeley Symposium on Mathematical Statistics and Probability, pp. 341–352. University of California Press, Berkeley (1951)\n\n13. 13.\n\nCuevas, A.: Qualitative robustness in abstract inference. J. Stat. Plan. Inference 18, 277–289 (1988)\n\n14. 14.\n\nDelbaen, F.: Coherent risk measures. Cattedra Galileiana, Scuola Normale, Superiore, Pisa (2000). http://www.math.ethz.ch/~delbaen/ftp/preprints/RiskMeasuresGeneralSpaces.pdf\n\n15. 15.\n\nDenneberg, D.: Premium calculation: why standard deviation should be replaced by absolute deviation. ASTIN Bull. 20, 181–190 (1990)\n\n16. 16.\n\nEdgar, G.A., Sucheston, L.: Stopping Times and Directed Processes. Cambridge University Press, Cambridge (1992)\n\n17. 17.\n\nEl Karoui, N., Ravanelli, C.: Cash subadditive risk measures and interest rate ambiguity. Math. Finance 19, 561–590 (2009)\n\n18. 18.\n\nFarkas, W., Koch-Medina, P., Munari, C.-A.: Beyond cash-additive capital requirements: when changing the numéraire fails. Finance Stoch. 18, 145–174 (2014)\n\n19. 19.\n\nFilipović, D., Svindland, G.: The canonical model space for law-invariant convex risk measures is L 1. Math. Finance 22, 585–589 (2012)\n\n20. 20.\n\nFöllmer, H., Schied, A.: Convex measures of risk and trading constraints. Finance Stoch. 6, 429–447 (2002)\n\n21. 21.\n\nFöllmer, H., Schied, A.: Stochastic Finance. An Introduction in Discrete Time, 3rd edn. de Gruyter, Berlin (2011)\n\n22. 22.\n\nGilat, D., Helmers, R.: On strong laws for generalized L-statistics with dependent data. Comment. Math. Univ. Carol. 38, 187–192 (1997)\n\n23. 23.\n\nHampel, F.R.: A general qualitative definition of robustness. Ann. Math. Stat. 42, 1887–1896 (1971)\n\n24. 24.\n\nHuber, P.J., Ronchetti, E.M.: Robust Statistics, 2nd edn. Wiley, New York (2009)\n\n25. 25.\n\nKallenberg, O.: Foundations of Modern Probability. Springer, New York (1997)\n\n26. 26.\n\nKrätschmer, V., Schied, A., Zähle, H.: Qualitative and infinitesimal robustness of tail-dependent statistical functionals. J. Multivar. Anal. 103, 35–47 (2012)\n\n27. 27.\n\nKusuoka, S.: On law invariant coherent risk measures. Adv. Math. Econ. 3, 83–95 (2001)\n\n28. 28.\n\nMizera, I.: Qualitative robustness and weak continuity: the extreme unction. In: Antoch, J., Hušková, M., Sen, P.K. (eds.) Nonparametrics and Robustness in Modern Statistical Inference and Time Series Analysis: A Festschrift in Honor of Professor Jana Jurečková. IMS Collections Festschrift, pp. 169–181. Institute of Mathematical Statistics, Beachwood (2010)\n\n29. 29.\n\nNelson, D.B.: Stationarity and persistence in the GARCH(1, 1) model. Econom. Theory 6, 318–334 (1990)\n\n30. 30.\n\nParthasarathy, K.R.: Probability Measures on Metric Spaces. Probability and Mathematical Statistics, vol. 3. Academic Press, New York (1967)\n\n31. 31.\n\nRuszczynski, A., Shapiro, A.: Optimization of convex risk functions. Math. Oper. Res. 31, 433–451 (2006)\n\n32. 32.\n\nSchied, A.: On the Neyman–Pearson problem for law-invariant risk measures and robust utility functionals. Ann. Appl. Probab. 14, 1398–1423 (2004)\n\n33. 33.\n\nTsukahara, H.: Estimation of distortion risk measures. J. Financ. Econom. (2013). doi:10.1093/jjfinec/nbt005\n\n34. 34.\n\nvan der Vaart, A.W., Wellner, J.A.: Weak Convergence and Empirical Processes. Springer, New York (1996)\n\n35. 35.\n\nvan Zwet, W.R.: A strong law for linear functionals of order statistics. Ann. Probab. 8, 986–990 (1980)\n\n36. 36.\n\nVillani, C.: Topics in Optimal Transportation. American Mathematical Society, Providence (2004)\n\n37. 37.\n\nWang, S.S.: Premium calculation by transforming the layer premium density. ASTIN Bull. 26, 71–92 (1996)\n\n38. 38.\n\nWeber, S.: Distribution-invariant risk measures, information, and dynamic consistency. Math. Finance 16, 419–441 (2006)\n\n39. 39.\n\nYaari, M.: The dual theory of choice under risk. Econometrica 55, 95–115 (1987)\n\n40. 40.\n\nZähle, H.: Marcinkiewicz–Zygmund and ordinary strong laws for empirical distribution functions and plug-in estimators. Statistics (2013). doi:10.1080/02331888.2013.800075\n\nDownload references\n\n## Acknowledgements\n\nThe authors thank Freddy Delbaen, Paul Embrechts, Marco Frittelli and two anonymous referees for comments, which helped to improve a previous draft of the paper.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Volker Krätschmer.\n\n## Appendix: Auxiliary results on the ψ-weak topology\n\n### Appendix: Auxiliary results on the ψ-weak topology\n\nFirst recall from Corollary A.45 in  that the ψ-weak topology on $$\\mathcal{M}_{1}^{\\psi}(\\mathbb{R})$$ is separable and metrizable. The following lemma provides some useful characterizations of ψ-weak convergence; see [26, Lemma 3.4] for a proof.\n\n### Lemma A.1\n\nThe following statements are equivalent:\n\n1. (i)\n\nμ n μ ψ-weakly.\n\n2. (ii)\n\nf n →∫f for every $$f\\in C_{\\psi}(\\mathbb {R})$$.\n\n3. (iii)\n\nf n →∫f for every continuous f with compact support and for f=ψ.\n\n4. (iv)\n\nμ n μ weakly andψ n →∫ψ.\n\nThe following lemma gives a transparent characterization of the ψ-weakly compact subsets of $${\\mathcal{M}}_{1}^{\\psi}$$. Recall that a set $${\\mathcal{N}}\\subset{\\mathcal{M}}_{1}^{\\psi}$$ is called uniformly ψ-integrating if it satisfies (2.12).\n\n### Lemma A.2\n\nA set $${\\mathcal{N}}\\subset{\\mathcal{M}}_{1}^{\\psi}$$ is relatively compact for the ψ-weak topology if and only if there exists a measurable function $$\\phi:\\mathbb{R}\\to[0,\\infty)$$ such that ϕ(x)/ψ(x)→∞ as |x|→∞ and such that\n\n$$\\sup_{\\nu\\in{\\mathcal{N}}}\\int\\phi\\,d\\nu<\\infty.$$\n(A.1)\n\nIn this case, $${\\mathcal{N}}$$ is uniformly ψ-integrating.\n\n### Proof\n\nThe first statement is an immediate consequence of Corollary A.47 in . For bounded ψ, the second statement is trivial. To prove it for unbounded ψ, we assume without loss of generality that ϕ>0. Fix ε>0, and denote by K the left-hand side of (A.1). Choosing M 1>0 so large that ψ(x)/ϕ(x)≤ε/K when |x|≥M 1, and choosing M 0>0 so large that ψ(x)≥M 0 implies |x|≥M 1, we obtain", null, "for all MM 0. That is, (2.12) holds. □\n\n## Rights and permissions\n\nReprints and Permissions\n\n## About this article\n\n### Cite this article\n\nKrätschmer, V., Schied, A. & Zähle, H. Comparative and qualitative robustness for law-invariant risk measures. Finance Stoch 18, 271–295 (2014). https://doi.org/10.1007/s00780-013-0225-4\n\nDownload citation\n\n• Received:\n\n• Accepted:\n\n• Published:\n\n• Issue Date:\n\n### Keywords\n\n• Law-invariant risk measure\n• Convex risk measure\n• Coherent risk measure\n• Orlicz space\n• Qualitative robustness\n• Comparative robustness\n• Index of qualitative robustness\n• Hampel’s theorem\n• ψ-Weak topology\n• Distortion risk measure\n• Skorohod representation\n\n• 62G35\n• 60B10\n• 60F05\n• 91B30\n• 28A33\n\n• D81" ]
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https://au.mathworks.com/matlabcentral/answers/391429-how-can-i-store-the-places-of-two-different-points-at-two-different-time
[ "# How can I store the places of two different points at two different time?\n\n1 view (last 30 days)\nAmina on 29 Mar 2018\nCommented: Elias Gule on 29 Mar 2018\nHello,\nI have to measure a distance between two points. I have to see where the solution across some values quite small, like 0.01. In other words, there should be the first point which is like (x(150), 0.01) at t=150 and the second one should be (x(200), 0.01) at t=200. Distance should be equal (x(200)-x(150)) because y-values are the same.\nMy problem is that how can I store/ calculate the x(150) and x(200) places when the time reaches to t=150 and t=200, respectively?\n\nElias Gule on 29 Mar 2018\nI think this is what you mean: You have a time vector with elements like 150,200,etc,\nand a position vector with elements like 0.01,0.01,etc.\nIf this is true, then I guess the code you that you want is:\nt = [150 200 300 400 500]; % replace with your own time vector\nx = [0.01 0.01 1 2 5]; % replace with your own position/x vector\nt0 = 150; % time for point 1\nt1 = 300; % time for point 2\nx0 = x(ismember(t,t0));\nx1 = x(ismember(t,t1));\ndx= x1 - x0; % delta x\ndy = 0; % since you said the y-values are always the same\ndist = hypot(dx,dy); % the distace between two points\n\nAmina on 29 Mar 2018\nHello Elias,\nIt is partly what I am looking for.\nJust I would like to re-explain what I mean for 0.01. The points are (x(150),y) at t=150 and (x(200)),y) at t=200, so y values are 0.01 everytime. My position vector will change according to time, so I do not have exact values for them.\nElias Gule on 29 Mar 2018\nThat should not be a problem; because as long as you know the time at which you want a corresponding x-value, the code that I supplied simply applies logical indexing to map that particular time instance to a corresponding position.\nlet's say for example, that, at t= 1507, the x-value is 1663, the code will be able to map these two values, as long as the t-vector and the x-vector has the same length.\nThe assumption that I'm making here is that the elements in the t- and x-vectors are mapped position-wise.\nThanks." ]
[ null ]
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https://stats.stackexchange.com/questions/190058/what-lowers-the-p-value-besides-the-sample-size/190134
[ "# What lowers the p-value besides the sample size?\n\nThe title pretty much says it. Sorry, it's so unspecific, I am not asking the question for myself and I am not familiar with stats.\n\nMaybe that is already subsumed in Peter Flom's definition of the effect size, but a higher signal (for me, that is the effect size) to noise ratio: for example, in a regression context, your t-statistic would be $$t=\\frac{\\hat\\beta_j-\\beta_{j0}}{s.e.(\\hat\\beta_j)}$$ where $s.e.(\\hat\\beta_j)$ is the $j$th diagonal element of $s^2(X'X)^{-1}$. Here, $s^2$ is an estimator of $\\sigma^2$, the error variance. A larger $\\sigma^2$ will, all else equal, tend to produce larger $s^2$ and thus translate into a smaller t-statistic and hence a larger $p$-value." ]
[ null ]
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https://tcrhs.buncombeschools.org/staff/hamlin__wayne_-_science/physics_guidebook
[ "# Physics Guidebook\n\nUsing a marble stitched notebook, the Physics Guidebook\nacts as\na Unit/Chapter summary, a cumulative review of\nvocabulary, concepts,\nequations, and graphs.\n\nSection I:  Key Terms/Vocabulary\nStudents independently write the key terms/vocabulary definitions\nas given in the textbook.  This will act as an introduction to the\nterms used in the chapter/unit.\n\nSection II:  Application of Concepts\nAfter having identified the standard text definitions of the important\nkey terms/vocabulary, students will survey the chapter/unit by\nanalyzing the pictures, drawings, diagrams and solved example\nproblems.  Students will now make a statement regarding possible\napplication(s) and the reason/purpose for examining and learning\nthese physical phenomenonIn effect, students will answer for\nthemselves, \"Why do I need to learn this information and when will\nI ever use this information?\"  This section can begin by completing\nthe following statement: \"I believe that understanding and using the\n\nthis information and these concepts can be applied by...\" or by making\na similar statement regarding the possible use and application of the\nexpressed physical phenomenon.  There is no minimum or maximum\nnumber of applications, only the requirement that students examine\nfor themselves, the purpose for learning these specific topics in\nphysics.\n\nSection III:  Concepts\nWorking in groups, students will review their notes and lab activities to\nsummarize the most important concepts of phenomenon.  Groups will\npresent and share their concept information to the class.  Other groups\ncan challenge this information for it's validity and accuracy, as well as\nchange their own summary of the important concepts.\n\nSection IV:  Concepts/Equations\nIndividually, and with the assistance of the teacher, the students will\nlist all the equations used to solve problems.  With each equation,\nthey will list what each variable stands for, what the s.i. unit for each\nquantity is, whether the quantity is a scalar or vector, and any\nrelated derivations that exist to produce other equations.\n\nSection V:  Concepts/Equations/Graphs\nUsing lab activities, notes and textbook examples, students will\nsummarize the basic graphing relationships in reference to slope,\narea and applying linearization of equations to determine constants,\netc...\n\nSection VI:  Summary of Concepts, Key Terms/Vocabulary\nWorking in groups, students will redefine the key terms/ vocabulary\nand basic concepts from section I and II using their own expression\nof words, drawings, sketches, artwork etc... to aid in remembering\nand reviewing information before testing.  Groups will present and\nshare their concept information to the class. Other groups can\nchallenge this information for it's validity and accuracy, as well as\nchange their own summary of the important concepts." ]
[ null ]
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http://www.seio.es/BBEIO/BEIOVol31Num3/files/assets/basic-html/page-11.html
[ " BEIO Volumen 31. Número 3 / Noviembre 2015", null, "", null, "", null, "11 / 132", null, "", null, "", null, "A Review on Functional Data Analysis for Cox processes\n\n217\n\nintensity process characterizes the Cox process and its statistics. If the mean of\n\nthe Cox process is absolutely continuous, it is the integral of the intensity and\n\nthus, is also a stochastic process. Indeed, the Cox process can be defined by\n\nits mean process, as in Serfozo (1972), and a number of statistics can also be\n\ncharacterized by the mean process. In brief, the intensity and the mean are both\n\nstochastic processes which determine the Cox process. Therefore, by inferring\n\none of these processes, the target process and its statistics will be inferred as\n\nwell.\n\nFunctional data analysis permits the estimation of a stochastic process and\n\nensuing forecast by means of principal component prediction models under both\n\nweak and usual conditions from observed sample paths of the process (Valder-\n\nrama et al., 2000). Consequently, no structure or given distribution is assumed,\n\nand thus, this technique is suitable for the development of an alternative infer-\n\nence method for the Cox process. As stated by Biau et al. (2013), despite the\n\nlarge body of research on functional data analysis, relatively few attempts have\n\nbeen made to connect it with stochastic processes. The present paper, then, will\n\nreview the application of functional data analysis to the inference of the intensity\n\nand the mean processes without previous assumptions, employing only the raw\n\ndata observed from the Cox process. In addition, this review will provide the\n\nestimation and prediction of a number of statistics of the Cox process and will\n\npresent a Cox process goodnes-of-fit test.\n\nThe following section contains the definitions of the Cox and compound Cox\n\nprocess and the statistics that will be employed in the subsequent sections. The\n\ninference for the mean process is presented in Section 3, where, additionally, the\n\napplication of this inference to the compound Cox process is derived. Similarly,\n\nSection 4 presents the inference method for the intensity process and its conse-\n\nquences. As a result of this method, a Cox process goodness-of-fit test can be\n\ndeveloped. The hypothesis test is described and is illustrated with an example.\n\nFinally, the conclusions are drawn in the last section, particularly in relation to\n\nthe potential future uses of the Cox process.\n\n2. Cox and compound Cox processes\n\nThis section presents the background necessary for the rest of the paper.\n\nFurther knowledge about other statistics or extensions of the Cox process (CP)\n\ncan be found in the references within the text.\n\n2.1. Cox process\n\nAccording to the classical approach, a CP\n\n{\n\nN\n\n(\n\nt\n\n);\n\nt\n\nt\n\n0\n\n}\n\nwith intensity pro-\n\ncess\n\n{\n\nλ\n\n(\n\nt, x\n\n(\n\nt\n\n));\n\nt\n\nt\n\n0\n\n}\n\nis defined as a conditional Poisson process with inten-\n\nsity process\n\n{\n\nλ\n\n(\n\nt, x\n\n(\n\nt\n\n));\n\nt\n\nt\n\n0\n\n}\n\ngiven the information process\n\n{\n\nx\n\n(\n\nt\n\n);\n\nt\n\nt\n\n0\n\n}\n\n.\n\nIf\n\nthe mean, Λ(\n\nt, x\n\n(\n\nt\n\n)), of the CP is absolutely continuous, then Λ(\n\nt, x\n\n(\n\nt\n\n)) =" ]
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https://reactgo.com/python-get-last-digit-of-number/
[ "Author -  Sai gowtham\n\n# Python - Get the last digit of a number\n\nIn this tutorial, we are going to learn about how to get the last digit of a number in Python with the help of examples.\n\nNote: In Python, strings are the sequence of characters, where the index of the first character is `0`, the second character is `1`, third character is `3` and the last character index `-1`.\n\nConsider, we have a following number:\n\n``pin = 62913``\n\nTo get the last digit of a number:\n\n1. Convert the number to a string using the str() Function.\n\n2. Use the square brackets syntax `[]` to access the last character of a string.\n\n3. Convert the result back to a number to get the last digit of the number.\n\nHere is an example:\n\n``````pin = 67683\nlast_digit_str = str(pin)[-1]\n\nresult = int(last_digit_str) # str to int\nprint(result)``````\n\nOutput:\n\n``3``\n\nIn the example above, we first converted the number to a string, so that we can use the square brackets syntax `[]` on it.\n\nYou can access the characters of a string, by using the square brackets syntax and specifying an index.\n\n`-1` is the index of a last character in the string. At last we have used the int() function to convert the string to a number.\n\nNote: Negative indexes are used to get data from the end of a string.\n\nYou can also shorten the above code in a single line like this:\n\n``````pin = 67683\nlast_digit = int(str(pin)[-1])\n\nprint(last_digit)``````\n\nor You can get the last digit of a number, by dividing the number by 10.\n\nHere is an example:\n\n``````pin = 67683\nlast_digit = pin % 10\n\nprint(last_digit) # 3``````\n\n## How rotate an image continuously in CSS\n\nIn this demo, we are going to learn about how to rotate an image continuously using the css animations.\n\n## How to create a Instagram login Page\n\nIn this demo, i will show you how to create a instagram login page using html and css.\n\n## How to create a pulse animation in CSS\n\nIn this demo, i will show you how to create a pulse animation using css.\n\n## Creating a snowfall animation using css and JavaScript\n\nIn this demo, i will show you how to create a snow fall animation using css and JavaScript.\n\n## Top Udemy Courses", null, "##### JavaScript - The Complete Guide 2023 (Beginner + Advanced)\n116,648 students enrolled\n52 hours of video content\n\\$14.99 FROM UDEMY", null, "##### React - The Complete Guide (incl Hooks, React Router, Redux)\n631,582 students enrolled\n49 hours of video content\n\\$24.99 FROM UDEMY", null, "##### Vue - The Complete Guide (w/ Router, Vuex, Composition API)\n203,937 students enrolled\n31.5 hours of video content\n\\$14.99 FROM UDEMY" ]
[ null, "https://i.udemycdn.com/course/480x270/2508942_11d3.jpg", null, "https://i.udemycdn.com/course/480x270/1362070_b9a1_2.jpg", null, "https://i.udemycdn.com/course/480x270/995016_ebf4.jpg", null ]
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https://edspi31415.blogspot.com/2012/09/cartesian-coordinates-to-pixel-screen.html
[ "## Thursday, September 13, 2012\n\n### Cartesian Coordinates to Pixel (Screen) Coordinates\n\nConverting Cartesian Coordinates to Screen Coordinates\n\nWhen working with computer or calculator graphics, sometimes we have to work with screen coordinates. The screen coordinate system has the following:\n\n1. The upper corner pixel (point) is (0,0).\n2. The lower corner pixel is (A, B).\n3. The x axis increases in the right direction.\n4. The y axis increases in the down direction, the opposite direction of the Cartesian plane.", null, "", null, "In order to to convert a point (x, y) in the Cartesian coordinates to point (xp, yp) in Screen coordinates, first observe the following:\n\n1. Picture the plane you are working with as a screen. If you are working with a calculator or computer, this is fairly easy.\n2. Let Xmin be the least-valued x of the screen (left edge) and Xmax be the most-valued x (right edge). Similarly, let Ymin be the least-valued y of the screen (bottom) and Ymax be the most-valued y of the screen (top).\n\nYou can check with the window settings on a graphing calculator to verify Xmax, Xmin, Ymax, and Ymin.\n\nFor the Hewlett Packard 28 series, 48 series and 50g, the variables XRNG and YRNG list the screen's dimensions, {Xmin, Xmax} for XRNG and {Ymin, Ymax} for YRNG.\n\nTo transform Cartesian coordinates to screen coordinates, we can use transformation matrices for scaling and translation. The general form", null, "show the transformation from coordinates (x, y) to (x', y') where:\n\ns_x = scale of the new x axis. If s_x < 0, the x point is reflected with respective to the x axis.\n\ns_y = scale of the new y axis. If s_y < 0, the y point is reflected with respected to the y axis. In going from Cartesian coordinates to screen coordinates, s_y < 0.\n\nx_s = Is the translation (shift to the new center) in the x-direction. If x_s > 0, the new center is to the right of the original center. Similarly, if x_s < 0, the new center is to the left of the original center.\n\ny_s = Is the translation (shift to the new center) in the y-direction. If y_s > 0, the new center is to the above the original center. Similarly, if y_s < 0, the new center is to the below the original center.\n\nFor the transformation from Cartesian coordinates to screen coordinates:\n\ns_x = A / (Xmax - Xmin)\n\ns_y = -B / (Ymax - Ymin)\n\nSince the new center will be at (Xmin, Ymax):\n\nx_s = -Xmin\n\ny_s = -Ymax\n\nTherefore by matrix multiplication, with x' = xp and y' = yp:\n\nxp = (x - Xmin) * A / (Xmax - Xmin)\n\nyp = (y - Ymax) * -B / (Ymax - Ymin)\n\nOn the HP 48 series and 50g, the translation from Cartesian to screen coordinates can be accomplished with the C→PX function.\n\nExample:\n\nUsing the following window with settings Xmin = -5, Xmax = 5, Ymin = -4, and Ymax = 4, transform the Cartesian coordinate (1, 0) to screen coordinates. The screen has pixel size 130 × 79.\n\nIn this case, A = 130 and B = 79.\n\nThen:\n\nxp = 130 / (5 - (-5)) * (1 - (-5)) = 78\nyp = -79 / (4 - (-4)) * (0 - 4) = 39.5\n\nThe screen coordinate of (1, 0) is {78, 39.5}.\n\nHP 50g Programs\n\nDPIX\n\nThis program figures the distance in pixels between two points stated in Cartesian coordinates.\n\nInput:\n2: coord-y\n1: coord-x\n\nOutput:\n1: Distance\n\nProgram:\n<< C→PX SWAP C→PX - DUP * ΣLIST B→R √ R→B >>\n\nExample:\n\nUsing the following window with settings Xmin = -5, Xmax = 5, Ymin = -4, and Ymax = 4, find the pixel distance between (0,1) and (-1,-2)? (HP 50g pixel dimension 130 × 79)\n\n2: (0,1)\n1: (-1,-2)\nDPIX\n\nAnswer in Decimal mode: #32d. (Approxmiately 32)\n\nARCEZ\n\nThis draws the arc using Cartesian coordinates and the radius in Cartesian scale (the way one would think if drawing on a Cartesian plane). This program does all the necessary transformation for you. You can use it in a program (requiring a PICTURE command later in a program) or alone (press the left arrow button to show the graph screen).\n\nInput:\n4: Cartesian Coordinates (x,y)\n3: Radius in Cartesian form (the way you think)\n2: Beginning Angle\n1: Ending Angle\n\n<< → P R A B\n<< P DUP R +\nC→PX SWAP C→PX - DUP * ΣLIST B→R √ R→B\nP C→PX SWAP A →NUM B →NUM ARC >> >>\n\nExample:\n\nUsing the following window with settings Xmin = -5, Xmax = 5, Ymin = -4, and Ymax = 4, draw a circle with radius 1.5 and center (1,0). (HP 50g pixel dimension 130 × 79)\n\n4: (1,0)\n3: 1.5\n2: 0\n1: 2 * π\nARCEZ\n\nPressing the left arrow button will show the picture below (with no functions selected and additional drawing):", null, "This concludes today's blog entry. As always, thank you for all your comments and support. Have a great day!\n\nThis blog is property of Edward Shore. © 2012\n\n### Casio fx-CG50: Sparse Matrix Builder\n\nCasio fx-CG50: Sparse Matrix Builder Introduction The programs can create a sparse matrix, a matrix where most of the entries have zero valu..." ]
[ null, "https://lh6.googleusercontent.com/-SQTtTp9olfI/UFKVnqCdT7I/AAAAAAAAAcs/luGT8Rvb9ZY/620F830D-4AF0-4226-B7C5-BAE08073B51C.jpg", null, "https://lh4.googleusercontent.com/-2u5_iWA9GLw/UFKVnIM5PZI/AAAAAAAAAck/DOtgBjOIiNc/43229D89-E875-41F5-9A0C-565C8AD5E582.jpg", null, "https://lh5.googleusercontent.com/-1945AHK5MK8/UFKVoINKwOI/AAAAAAAAAc0/Z9KAOpY-YJU/CECC3970-036E-4A12-B378-8BB855C779A1.jpg", null, "https://lh4.googleusercontent.com/-1jY81DxXwhY/UFKVo3BiwII/AAAAAAAAAc8/0Pw24Edg4S0/98AF4810-95E8-4081-B0B1-556731C096FF.jpg", null ]
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https://wiki.libsdl.org/SDL3/SDL_MostSignificantBitIndex32/raw
[ "= SDL_MostSignificantBitIndex32 = Use this function to get the index of the most significant (set) bit in a 32-bit number. This is also the same as the base 2 logarithm of the number. == Syntax == int SDL_MostSignificantBitIndex32(Uint32 x) == Function Parameters == {| |'''x''' |the number to find the MSB of |} == Return Value == Returns the index of the most significant bit of x, or -1 if x is 0. == Code Examples == #include \"SDL_bits.h\" if (bitmask) { int index = SDL_MostSignificantBitIndex32(bitmask); } ---- [[CategoryAPI]], [[CategoryBits]]" ]
[ null ]
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https://askiv.com/voltage-across-a-capacitor-equation-formula-calculation/
[ "# Voltage across a Capacitor Equation, Formula, Calculation\n\n0\n78\n\n### The voltage across a capacitor equation\n\nAs you know the capacitor stores the electricity and releases the same. Here the voltage across a capacitor V(V) in volts is equal to the capacitive reactance Xc in ohms times of the capacitor current in Amps. The capacitor voltage formula will be,\n\nV(V) = I(A)  x Xc\n\nVolts = Amps x capacitive reactance\n\nHere the capacitive reactance Xc is equal to the 0.159 divided by frequency and the capacitance. Hence the voltage across the capacitor equation will be\n\nV(V)  = 0.159 x I(A) / (f(Hz) x C(F))\n\n### Capacitor voltage for the non-sinusoidal wave:\n\nCapacitor voltage Vc is equal to the integration of the current I(A) in amps divided by the capacitance C(F) in Farad. The formula will be,\n\nV = (1/C) x ∫Idt\n\nExample:\n\nLet us calculate the voltage across the 3-farad capacitor and the current flow is 10sin10t Amps.\n\nThe voltage across the capacitor is,\n\nV = (1/3) ∫10*sin10t\n\n= – 3.3 cos10t volts\n\nCapacitor voltage formula with charge:\n\nThe voltage across the capacitor V(V) in volts is equal to the ratio between the charge Q(c) in coulomb to the capacitance C(F) in farad.\n\nThe capacitor voltage V(V) = Q(c) / C(F)" ]
[ null ]
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https://jrodthoughts.medium.com/the-algorithm-that-powers-deep-learning-9037b125a4f4
[ "# The Algorithm that Powers Deep Learning\n\nOk, the title is a bit of an exaggeration but hopefully it caught your attention :)\n\nWhen looking into deep learning models, very frequently you are going to encounter the term stochastic gradient descent(SGD) as an optimization mechanism. SGD is, by far the most common algorithms in deep learning models and is not a stretch to say that nearly all deep learning is, to some extent, powered by SGD., What is the mystical algorithm and why is so relevant to deep learning?\n\nIn its most basic form, SGD is an optimization algorithm. Like many other optimization techniques, SGD focuses on minimizing the cost function of a specific model without drastically impacting the rest of the model. Optimization algorithms are nothing new in machine learning and has been a core part of it since its inception. However, most of the traditional optimization techniques in machine learning needed to be adapted to perform with the large datasets common in deep learning problems. Specifically, SGD is the deep learning adaptation of a common family of optimization methods known as Gradient Descent.\n\nThe general goal of optimization algorithms is to minimize the cost function(also known as error function) of a learning model. If we represent the cost function as c= f(x), the objective of an optimization algorithms would be to minimize c by altering x.\n\nGradient Descent Optimization(GDO) relies on derivatives to minimize the cost function. Derivatives is one of the pillars of calculus and has many applications on different areas of deep learning. GDP relies on a very particular property of derivatives that allows to obtain small changes in the output of a function by scaling its input. Using some mathematical nomenclature, if f’(x) is the first-order derivative of our cost function f(x) and d is a very small number, then we can assert that:\n\nf(x + d)==> f(x) + d*f’(x)\n\nAll those mathematical expressions, simple tell us that we can make small changes in f(x) by modifying x. GDO uses that technique to find different optimization points for a cost function f(x). Even if you are not familiar with GDO, I am sure you have heard of some of its terms such as local minimum or maximum or global minimum or maximum as they are often used indiscriminately in mainstream technical articles.\n\nLocal minimum/maximum refers to a point on which a function f(x) is lower/higher that all its neighboring points. Similarly, global minimum/maximum refers to the absolutely lowest or highest point of the function. In the const of GDO, the goal of the algorithms is to find local minimums that don’t contradict the global minimum.\n\ntraditional GDO techniques often result impractical and prohibitory expensive when dealing with large datasets. Imagine calculating derivatives across billions of data points in a training dataset. SGD improves on classic GDO techniques by uniformly drawing small sets of samples (ranging fro a few dozens to a few hundreds)) from the training datasets and evaluating different optimization functions.\n\nWithout getting into the algorithmic details behind SGD, it its important to highlight that it excels at funding very low values for the cost function very quickly. More importantly, SGD does so while keeping steady computational costs. SGD provides no guarantees that will ever arrives at a local minimum but the tradeoff in terms of speed and resources makes a more viable option than typical GDO algorithms.\n\nOptimization is a very active area of research in the deep learning space. Constantly, new algorithms are being actively tested by researchers and many of them are improvements on SGD. For now, SGD has become a favorite of the deep learning community and is important to understand some of its concepts when applied in deep learning solutions .\n\nWritten by" ]
[ null ]
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https://avesis.gazi.edu.tr/yayin/28db82da-951e-484d-a53b-ab2fc9a4782a/weighted-statistical-relative-invariant-mean-in-modular-function-spaces-with-related-approximation-results
[ "## Weighted Statistical Relative Invariant Mean in Modular Function Spaces with Related approximation Results\n\nNUMERICAL FUNCTIONAL ANALYSIS AND OPTIMIZATION, cilt.39, sa.11, ss.1181-1207, 2018 (SCI İndekslerine Giren Dergi)", null, "", null, "• Yayın Türü: Makale / Tam Makale\n• Cilt numarası: 39 Konu: 11\n• Basım Tarihi: 2018\n• Doi Numarası: 10.1080/01630563.2018.1470096\n• Dergi Adı: NUMERICAL FUNCTIONAL ANALYSIS AND OPTIMIZATION\n• Sayfa Sayıları: ss.1181-1207\n\n#### Özet\n\nThe idea of statistical relative convergence on modular spaces has been introduced by Orhan and Demirci. The notion of sigma-statistical convergence was introduced by Mursaleen and Edely and further extended based on a fractional order difference operator by Kadak. The concern of this paper is to define two new summability methods for double sequences by combining the concepts of statistical relative convergence and sigma-statistical convergence in modular spaces. Furthermore, we give some inclusion relations involving the newly proposed methods and present an illustrative example to show that our methods are nontrivial generalizations of the existing results in the literature. We also prove a Korovkin-type approximation theorem and estimate the rate of convergence by means of the modulus of continuity. Finally, using the bivariate type of Stancu-Schurer-Kantorovich operators, we display an example such that our approximation results are more powerful than the classical, statistical, and relative modular cases of Korovkin-type approximation theorems." ]
[ null, "https://avesis.gazi.edu.tr/Content/images/integrations/small/integrationtype_2.png", null, "https://avesis.gazi.edu.tr/Content/images/integrations/small/integrationtype_1.png", null ]
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https://docs.techsoft3d.com/hps/latest/build/api_ref/cpp/class_h_p_s_1_1_shape_coordinate.html
[ "HPS::ShapeCoordinate Class Reference\n\n`#include <hps.h>`\n\n## Public Member Functions\n\nbool Equals (ShapeCoordinate const &in_that) const\n\nbool operator!= (ShapeCoordinate const &in_that) const\n\nbool operator== (ShapeCoordinate const &in_that) const\n\nShapeCoordinateSetMargins (float in_margin_one, float in_margin_two=0.0f, float in_margin_three=0.0f, float in_margin_four=0.0f)\n\nShapeCoordinate ()\nDefault constructor.\n\nShapeCoordinate (float in_x, float in_y)\n\nShapeCoordinate (float in_x, float in_y, FloatArray const &in_margins)\n\nShapeCoordinate (float in_x, float in_y, size_t in_count, float const in_margins [])\n\nShapeCoordinate (float in_x, float in_y, float in_radius)\n\nShapeCoordinate (float in_x, float in_y, float in_radius, FloatArray const &in_margins)\n\nShapeCoordinate (float in_x, float in_y, float in_radius, size_t in_count, float const in_margins [])\n\nbool ShowMargins (FloatArray &out_margins) const\n\nShapeCoordinateUnsetMargins ()\n\n## Public Attributes\n\nFloatArray margins\n\nfloat x\n\nfloat y\n\n## Detailed Description\n\nThe ShapeCoordinate class is a coordinate used in shape definitions. Two ShapeCoordinates can be used together to define a ShapePoint It is a 2d parametric coordinate.\n\n## ◆ ShapeCoordinate() [1/6]\n\n HPS::ShapeCoordinate::ShapeCoordinate ( float in_x, float in_y )\n\nConstruct a new ShapeCoordinate from an (x, y) pair.\n\nParameters\n in_x A normalized distance over the horizontal text bounds. in_y A normalized distance over the vertical text bounds.\n\n## ◆ ShapeCoordinate() [2/6]\n\n HPS::ShapeCoordinate::ShapeCoordinate ( float in_x, float in_y, FloatArray const & in_margins )\n\nConstruct a new ShapeCoordinate from an (x, y) pair and up to four margins values\n\nParameters\n in_x A normalized distance over the horizontal text bounds. in_y A normalized distance over the vertical text bounds. in_margins A series of up to four distances, normalized over the respective text margins.\n\n## ◆ ShapeCoordinate() [3/6]\n\n HPS::ShapeCoordinate::ShapeCoordinate ( float in_x, float in_y, size_t in_count, float const in_margins[] )\n\nConstruct a new ShapeCoordinate from an (x, y) pair and up to four margins values\n\nParameters\n in_x A normalized distance over the horizontal text bounds. in_y A normalized distance over the vertical text bounds. in_size The size of in_margins. in_margins A series of up to four distances, normalized over the respective text margins.\n\n## ◆ ShapeCoordinate() [4/6]\n\n HPS::ShapeCoordinate::ShapeCoordinate ( float in_x, float in_y, float in_radius )\n\nConstruct a new ShapeCoordinate from an (x, y, radius) tuple.\n\nParameters\n in_x A normalized distance over the horizontal text bounds. in_y A normalized distance over the vertical text bounds. in_radius A normalized distance over the radius of the circle circumscribing the text bounds.\n\n## ◆ ShapeCoordinate() [5/6]\n\n HPS::ShapeCoordinate::ShapeCoordinate ( float in_x, float in_y, float in_radius, FloatArray const & in_margins )\n\nConstruct a new ShapeCoordinate from an (x, y, radius) tuple and up to four margin values.\n\nParameters\n in_x A normalized distance over the horizontal text bounds. in_y A normalized distance over the vertical text bounds. in_radius A normalized distance over the radius of the circle circumscribing the text bounds. in_margins A series of up to four distances, normalized over the respective text margins.\n\n## ◆ ShapeCoordinate() [6/6]\n\n HPS::ShapeCoordinate::ShapeCoordinate ( float in_x, float in_y, float in_radius, size_t in_count, float const in_margins[] )\n\nConstruct a new ShapeCoordinate from an (x, y, radius) tuple and up to four margin values.\n\nParameters\n in_x A normalized distance over the horizontal text bounds. in_y A normalized distance over the vertical text bounds. in_radius A normalized distance over the radius of the circle circumscribing the text bounds. in_count The size of in_margins. in_margins A series of up to four distances, normalized over the respective text margins.\n\n## ◆ Equals()\n\n bool HPS::ShapeCoordinate::Equals ( ShapeCoordinate const & in_that ) const\n\nThis function is used to check an object for equivalence to this.\n\nParameters\n in_that The object to compare to this.\nReturns\ntrue if the objects are equivalent, false otherwise.\n\n## ◆ operator!=()\n\n bool HPS::ShapeCoordinate::operator!= ( ShapeCoordinate const & in_that ) const\ninline\n\nThis function is used to check an object for equivalence to this.\n\nParameters\n in_that The object to compare to this.\nReturns\ntrue if the objects are not equivalent, false otherwise.\n\n## ◆ operator==()\n\n bool HPS::ShapeCoordinate::operator== ( ShapeCoordinate const & in_that ) const\ninline\n\nThis function is used to check an object for equivalence to this.\n\nParameters\n in_that The object to compare to this.\nReturns\ntrue if the objects are equivalent, false otherwise.\n\n## ◆ SetMargins()\n\n ShapeCoordinate& HPS::ShapeCoordinate::SetMargins ( float in_margin_one, float in_margin_two = `0.0f`, float in_margin_three = `0.0f`, float in_margin_four = `0.0f` )\n\nSets the margins for this ShapeCoordinate.\n\nParameters\n in_margin_one The first margin value. in_margin_two The second margin value. in_margin_three The third margin value. in_margin_four The fourth margin value.\nReturns\nA reference to this ShapeCoordinate.\n\n## ◆ ShowMargins()\n\n bool HPS::ShapeCoordinate::ShowMargins ( FloatArray & out_margins ) const\n\nShows the margins for this ShapeCoordinate.\n\nParameters\n out_margins The margins for this ShapeCoordinate.\nReturns\ntrue if margins were set, false otherwise.\n\n## ◆ UnsetMargins()\n\n ShapeCoordinate& HPS::ShapeCoordinate::UnsetMargins ( )\n\nRemoves the margins for this ShapeCoordinate.\n\nReturns\nA reference to this ShapeCoordinate.\n\nThe documentation for this class was generated from the following file:" ]
[ null ]
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https://www.idtech.com/blog/algebra-terms-and-converting-word-problems-into-equations
[ "# How to Translate Word Problems Into Equations", null, "Math is complicated enough when trying to make sense of numbers alone, and even more so when words are thrown into the mix. In this useful Math with Margo tutorial, she’ll show students how to convert word problems into algebra terms and then find an easy solution.\n\nFollow along step-by-step, and your student will understand algebra terms and how they fit into complex math problems.\n\n## Key Vocabulary\n\nBefore we begin, let’s touch on some important vocabulary. What is a variable? An expression? Here’s a quick refresher before we dive into today’s problem.\n\nVariable: In algebra, a variable like x or y symbolizes an unknown value.\n\nExpression: A mathematical expression is a group of numbers, operators, and symbols. Expressions can include addition, multiplication, subtraction, and division.\n\nTerms: A term is a single mathematical expression: one number, operator, or symbol.\n\n## How to Translate Word Problems Into Equations\n\nLet's go through an example problem to get some practice translating word problems into equations.\n\nProblem: A person exchanges 390 pennies for quarters, dimes, and nickels. The number of dimes in the exchange was twice the number of quarters and the number of nickels was twice the number of dimes. How many quarters were in the exchange?", null, "Confused? Never fear! Word problems like this seem complicated at first, but with Margo’s method, soon you’ll understand the key parts.\n\n### Step One: Highlight Important Parts of the Text", null, "### Step Two: Establish your variables\n\n• q represents the # of quarters\n• d represents the # of dimes\n• n represents the # of nickels", null, "### Step Three: Rewrite the Problem as an Equation\n\nSince we have three equations and three variables, we can use substitution to solve for a variable. Which variable is in each equation?\n\nThat’s right, d is.\n\nd = 2q needs to be written as “q =” so we can replace the q in the first equation with something in terms of d. If we divide both sides by 2, we are left with d/2 = q.\n\nn = 2d is already in the form we want. We can substitute n in the first equation with 2d, since they are equal in value.\n\n390 = 25q + 10d + 5n\nd = 2q   ⇒ d/2 = q\nn = 2d", null, "### Step Four: Substitute", null, "Now we have one equation with one variable, and we can simplify it to solve out for d.\n\n### Step Five: Simplify and Combine Terms\n\nWhen a number is pushed up against a variable it implies multiplication, so to get the d variable by itself. We need to do the opposite operation, so divide both sides by 32.5.\n\nRemember, \"d\" represents the number of dimes, so there were 12 dimes in the exchange.", null, "### Step Six: Divide Each Side of the Equation\n\nSince we are ultimately trying to find out how many quarters were in the exchange, we can use the original equation we set up that has a relationship between the number of dimes and the number of quarters.\n\nWe know that d=12, so we can substitute d for 12 and then solve for q.\n\nDivide both sides of the equation by 2 to get q by itself.", null, "### Step Seven: Solve\n\nTherefore, our final answer is that there were 6 quarters in the exchange!\n\nAs the above hopefully shows, math doesn’t need to be a headache; in fact, there are many ways to make math fun! If your child has an aptitude for math, but could use a challenge, a math competition or game could be just the motivator they need.\n\nIf math class isn't not your student’s cup of tea, there are excellent resources to support their learning out there, like this recent post presenting a few fascinating facts about math that are sure to pique the interest of reluctant mathematicians.\n\nPlus, Margo’s math expertise spans other tutorials in solving for x, converting decimals into fractions, and some helpful advice for studying for that next math test or quiz. Our math tips & resources page is a great place to bring the \"M\" in STEM front and center.\n\nFor more personalized support, iD Tech’s top-notch instructors offer one-on-one online math tutoring in subjects ranging from geometry to pre-algebra and statistics to help kids achieve their goals in math.", null, "Virginia started with iD Tech at the University of Denver in 2015 and has loved every minute since then! A former teacher by trade, she has a master's in education and loves working to embolden the next generation through STEM. Outside the office, you can usually find her reading a good book, struggling on a yoga mat, or exploring the Rocky Mountains.\n\nFeatured Posts\n\nCategories\n\nAuthors" ]
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http://jupiqocyxyke.comunidades.net/discrete-mathematics-with-graph-theory-ebook
[ "•", null, "# Discrete Mathematics with Graph Theory ebook\n\nDiscrete Mathematics with Graph Theory ebook\n\n## Discrete Mathematics with Graph Theory. Edgar G Goodaire, Edgar G. Goodaire, Michael M Parmenter, Michael M. Parmenter", null, "Discrete.Mathematics.with.Graph.Theory.pdf\nISBN: 0130920002,9780130920003 | 557 pages | 14 Mb", null, "Discrete Mathematics with Graph Theory Edgar G Goodaire, Edgar G. Goodaire, Michael M Parmenter, Michael M. Parmenter\nPublisher: Prentice Hall\n\nLocation: NIMBioS at the University of Tennessee, Knoxville. Meeting dates: August 16-18, 2010. Graphs are structures consisting of circles and lines connecting them, which are called vertices and edges, respectively. In discrete mathematics, graph theory is the study of graphs. Discrete Mathematics with Graph Theory (2nd Edition), 9780130920003 (0130920002), Prentice Hall, 2001. References Graph theory with application by Narsing Deo Graph Theory in Discrete Mathematics by Rosen. Goodaire, et al., (Prentice Hall, 2002) WW.djvu. Download free pdf ebooks rapidshare, 4shared,uploading,torrent,bittorrent. Download ebook Chromatic Graph Theory (Discrete Mathematics and Its Applications) by Gary Chartrand and Ping Zhang pdf free. Discrete Mathematics with Graph Theory 2nd ed. A connected multigraph has an Euler path but not Euler circuit if and only if it has exactly vertices of odd degree. Tags: 3rd Edition, and, Discrete, Mathematics, solutions manual, test banks, Theory, with Graph. Perhaps even more relevant is what we might call spreadsheet math (customized worksheets, recursive functions, graphs, macro programming). Set theory; Posets; Permutations and combinations; Lattices; Inclusion, Exclusion principle; Graph Theory; Graph coloring; Functions; Discrete Mathematics; Boolean algebra. Disorders of Synaptic Plasticity and Schizophrenia – J. With “theory,” I mean subjects such as discrete mathematics (graph theory and stochastic systems), software engineering (design patterns), and linear algebra. By testbooksolutions in Uncategorized on April 27, 2013 . Topic: Graph Theory and Biological Networks Tutorial. Part 5 - Discrete Mathematics and Graph Theory. Few people ever read a preface, and those who do often just glance at the first few lines.\n\nMore eBooks:" ]
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