1. [Solved] Another Group Theory problem...

This time, it's about proving 2 elements have the same order:-

Let G be a group and x be an element of G. The order of x is the least positive number such that $x^n$ = e.. That is, n > 1 satisfies

$x^n$ = e and $x^m$ =/= e for all 1<m<n

For x in G prove that x and x-1 have the same order
.

So, I've had a fair whack at the question, but I'm not sure if I'm going in the write direction.

$x^{n/2}$. $x^{n/2}$ = $x^n$= e

$x.x^{-1}$ = e
$x.x^{-1}$ = $x.x^-1$
= $x^{1-1}$
= $x^0$
=1 ---> e (identity is 1)

$x^n$ = e
$(x^{n})^{-1} = e^{-1}$
$x^{-n} = e^{-1}$

Note: $e^{-1}$ = $e$

Therefore,

$x^{-n} = e = x^{n}$
$x^{-n} = x^{n}$
$x^{n}/x^{-n} = e$
$x^{n}.x^{n} = e$
$x^{2n} = 1 = e$
$e = 1 = x^{0}= x^{-n}= x^{n}= x^{2n}$

THUS. If
$x^{n}$ = 1 = e
$x^{n}.x^{-n}$ = 1 = e
$x^{-n}$ = 1 = e
$x^{n}$ = $x^{-n}$

Therefore they have the same order.

That is huge im sorry, but yeh. I dunno if it's correct. Any pointers?

2. Hi!
You formulated the problem well, but very few lines of your solution attempt make sense (what is $x^{n/2}$ ??, why two symbols for an identity element?, what does the symbol "/" mean? ). In the language of groups, there are originally no 'powers'. $x^n$ is just our mere shorthand for $x \cdot x \cdot \ldots \cdot x$ with $n$ occurrences of $x$, $n \in \mathbb{N}$, with agreement that $x^0 = e$.
You included the definition of order of an element, so let's use it to solve our problem.
Let $n \geq 1$ be the order of $x$.

$
(x^{-1})^n \cdot x^n = \underbrace{(x^{-1} \cdot x^{-1} \cdot \ldots \cdot x^{-1})}_{n} \cdot \underbrace{(x \cdot x \cdot \ldots \cdot x)}_{n}=
$

$
= \underbrace{x^{-1} \cdot x^{-1} \cdot \ldots \cdot (x^{-1}}_{n} \cdot \underbrace{x) \cdot x \cdot \ldots \cdot x}_{n}=
$

$
= \underbrace{(x^{-1} \cdot x^{-1} \cdot \ldots \cdot x^{-1})}_{n-1} \cdot \underbrace{(x \cdot x \cdot \ldots \cdot x)}_{n-1}= \ldots = x^{-1} \cdot x = e
$

because group operation is associative. This shows that $(x^{-1})^n = (x^n)^{-1}$ but $(x^n)^{-1} = e^{-1} = e$, so we verified that $(x^{-1})^n = e$ which is the first requirement for $n$ to be the order of $x$.
Now let's verify the second requirement. We argue by contradiction. Assume that there is a number $m$, $1 \leq m < n$, such that $(x^{-1})^m = e$. Then, in the same way as in the first step, we show that $(x^{-1})^m \cdot x^m = e$. But $(x^{-1})^m = e$, thus $x^m = e$. And this contradicts our assumption that $n$ is the order of $x$.

3. Thank you for your help. The "x $n/2$" was to denote 0.5x, and the "/" was to denote fractions (except for when it was used as =/= which was my attempt at "does not equal.).

Thank you VERY much for the clarification.

4. Originally Posted by exphate
Thank you for your help. The "x $n/2$" was to denote 0.5x, and the "/" was to denote fractions (except for when it was used as =/= which was my attempt at "does not equal.).

Thank you VERY much for the clarification.
Well, even if you are using real numbers, $x^{n/2}$ does not mean the same thing as $0.5x$.

But the more serious problem here is that $0.5x$ has no meaning whatsoever in an arbitrary group $G$. As for fractions they could be defined in an appropriate way by defining $x/y$ by $xy^{-1}$, but you should be aware that such a notation is never used in group theory and you would do well to drop it altogether.
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