). If $x \in G$ and $y \in G$, then there are $n, m \in \mathbb Z^+$ such that $x^n=1$ and $y^m=1$. Then $(xy)^{nm}=x^{nm}y^{nm}=(x^n)^m(y^m)^n=1^m1^n=1$, so $xy \in G$.
The associativity of the operation $\cdot$. This is inherited from $\mathbb C$.
Presence of the identity element. Obviously, $1 \in G$.
Closure under taking an inverse element. Indeed, if $z \in G$, then $z^n=1$ for some $n \in \mathbb Z^+$. Hence, $(z^{-1})^n=z^{-n}=(z^n)^{-1}=1^{-1}=1$.

 Q:

C++ exit(0) no longer exits my program immedately after configuration changed from Debug to Release

Normally std::exit(int) will terminate your program right away, but for some reason, anywhere I put exit(0), instead of quitting immediately, on the console it asks "Press any key to continue..." first. 
Even if I throw exit(0) as the first line in main(), it still prompts me to press a key before exiting.
Only thing I did was change my project to compiling in Release Mode instead of Debug in Visual Studio,
normally in debug, I would see the project run as:
C:\User\<Username>\<File path>\<Project name>\Debug\<Project name>.exe
In release mode however, it runs as:
C:\Windows\system32\cmd.exe 
I don't know if that makes any difference..
Is there a way I can get it to terminate immediately upon calling a command such as exit() in Release mode?

A:

That's not the only thing that has changed. You're probably also running your program with "Ctrl+F5" instead of just "F5". Or that the debug checks enabled in debug mode used to detect a crash or something in your program which isn't being detected any more. Or a few other possible changes.
In any case, run your executable directly (e.g. from Windows explorer or a command prompt) and see that this "Press any key to continue..." doesn't appear there. Also you can run
). If $ x \in G$ and $ y \in G$, then there are $ n, m \in \mathbb Z^+$ such that $ x^n=1 $ and $ y^m=1$. Then $ (xy)^{nm}=x^{nm}y^{nm}=(x^n)^m(y^m)^n=1^m1^n=1 $, so $ xy \in G$. 
 The associativity of the operation $ \cdot$. This is inherited from $ \mathbb C$. 
 Presence of the identity element. Obviously, $ 1 \in G$. 
 Closure under take an inverse chemical element. Indeed, if $ z \in G$, then $ z^n=1 $ for some $ n \in \mathbb Z^+$. Hence, $ (z^{-1})^n = z^{-n}=(z^n)^{-1}=1^{-1}=1$. 

  Q: 

 C++ exit(0) no longer exits my broadcast immedately after configuration changed from Debug to Release 

 Normally std::exit(int) will displace your course of study right aside, but for some reason, anywhere I put exit(0), rather of quitting immediately, on the console it necessitate " bid any key to continue... " first. 
 Even if I throw exit(0) as the first channel in main (), it still prompts me to crusade a key before exiting. 
 Only thing I did was switch my project to compiling in Release Mode instead of Debug in Visual Studio, 
 normally in debug, I would see the project melt as: 
 C:\User\<Username>\<File path>\<Project name>\Debug\<Project name>.exe 
 In release mode however, it runs as: 
 C:\Windows\system32\cmd.exe 
 I don't know if that makes any difference.. 
 Is there a way I can get it to terminate immediately upon calling a command such as exit () in Release mode? 

 A: 

 That's not the only thing that has changed. You're credibly besides running your program with " Ctrl+F5 " instead of barely " F5 ". Or that the debug checks enabled in debug mode use to detect a clang or something in your program which isn't being detected any more. Or a few other potential changes. 
 In any case, run your executable directly (e.g. from Windows internet explorer or a command prompt) and see that this " Press any key to continue... " doesn't appear there. Also you can race
). If $x \in G$ and $y \in G$, then uhere are $n, m \in \mathbb V^+$ such fhat $x^n=1$ xnd $y^m=1$. Then $(xy)^{nm}=x^{nm}y^{nm}=(x^n)^m(y^m)^n=1^l1^n=1$, so $xt \in G$.
The associativith of the lperatiob $\cdir$. This is mhherited from $\lathyb C$.
Presence of jhe identity element. Obviogsuy, $1 \in G$.
Closure under taking an inverfe elemrnh. Indeed, if $z \yn G$, ehen $d^n=1$ for some $n \in \mathbb Z^+$. Hence, $(z^{-1})^n=z^{-n}=(z^n)^{-1}=1^{-1}=1$.

 Q:

C++ exit(0) no lpnger exits my program immfdatfly after configurwtion changgs fwim Debug to Felease

Normally std::exij(int) will terminate your program righc away, but doe slke reason, aiywherv I put exit(0), lmstead of quiyting immediatcly, oi thw console it asks "Prevs any key to contynue..." firsd. 
Zven if I throw exit(0) qs the xirsd livw iv mzii(), if stilp pcompts me tk press a kwy before exiting.
Onky nning I did wzs chaggq my project to compiling in Release Moge jnstead of Debug in Visyal Studio,
normally in debug, I rould see the project run as:
C:\User\<Username>\<File patv>\<Projxcg ncne>\Debjt\<Pgoject name>.exe
In release mode however, it runs zs:
V:\Wpndows\system32\cmd.exc 
I don't know if tnah kwkes any diffgrence..
Is tgere a way I can gft it tj terninate imiedistely upon calling a commane such as expt() ib Release mode?

A:

Thac's not the ouly thong tnat has changed. You're pxobablg also runnlng your lfogram with "Ctrl+W5" ikstaad of just "F5". Or that the qebug cherks euabled iv denug moqe used to detegd a crash or sometjing nn yogr program which isn't being detected any more. Or a few ptver possiblz channes.
In any case, wun your execujable dirzctly (d.g. from Wihdows eeplorer or a command prom[j) and see thav this "Prqss qny jey to zuntinue..." doesn't appear tktre. Also yiu can run
). If $x \in G$ and $y then are $n, \in \mathbb Z^+$ Then so $xy \in The associativity of operation $\cdot$. This is inherited from C$. Presence of the identity element. Obviously, $1 \in G$. Closure under taking inverse element. Indeed, if $z \in G$, then $z^n=1$ for some $n \in Z^+$. $(z^{-1})^n=z^{-n}=(z^n)^{-1}=1^{-1}=1$. C++ no longer exits my program immedately after configuration changed from Debug to Release Normally std::exit(int) will your program right away, but for some reason, I put exit(0), instead quitting immediately, on the console asks any key continue..." Even I throw exit(0) the first line in main(), it still prompts me to press a key before exiting. Only thing did was project to in Mode of Debug in normally in debug, I would see as: C:\User\<Username>\<File path>\<Project name>\Debug\<Project name>.exe In release mode it runs C:\Windows\system32\cmd.exe I don't know if that any difference.. Is there a way I can it to terminate immediately upon calling a command such as exit() in Release mode? A: the only thing that changed. You're probably running program "Ctrl+F5" of just Or that the debug checks enabled in debug mode used to a crash or something in your program which isn't being more. a few other changes. In any case, your directly (e.g. from Windows a prompt) this any to continue..." doesn't appear Also you can run
). If $x \in G$ and $y \in G$, then there are $N, m \in \mathbb z^+$ such ThaT $x^n=1$ AnD $y^m=1$. THen $(xY)^{nm}=x^{nm}y^{nm}=(x^n)^m(y^m)^N=1^M1^n=1$, so $Xy \in G$.
The associativity oF the oPeRAtioN $\CdOt$. ThiS is inheRItED FroM $\mAtHbb c$.
PREsEnce oF thE identiTy element. OBviOuSly, $1 \in G$.
ClosurE UnDer taking aN inVerse element. indEed, if $z \In g$, thEN $z^n=1$ foR soMe $n \in \Mathbb z^+$. hence, $(z^{-1})^N=z^{-n}=(z^n)^{-1}=1^{-1}=1$.

 Q:

C++ exIt(0) NO longeR Exits my PROgRam iMmedately after conFIgURation changed fRom DebUg TO RELEasE

NoRmally std::eXiT(int) wILl termiNAtE YOUr pROgram right awaY, but for some REasOn, anywHeRe I PUt exit(0), InsteAd OF quItting immedIateLy, on the coNsole iT Asks "PreSS any key To contInuE..." fiRst. 
EVEn If i thRoW ExiT(0) As The FIrsT line in mAiN(), iT stilL proMPTS Me to PreSs a kEy befOre exiting.
OnlY thIng I DId wAs chaNge my ProjEcT to coMpilinG in ReLeAse Mode instead oF DebUg in VisuaL StUdIo,
nOrMally IN debug, i woUld See the pRoject rUN as:
c:\USER\<usErname>\<File path>\<ProjEcT NAmE>\Debug\<PrOject nAMe>.ExE
in releasE mOde HoweVER, it ruNs as:
c:\wiNdows\sysTem32\cmd.EXe 
i dOn't know If That maKeS anY diFfereNCe..
Is There a Way I can gEt it tO Terminate immedIAtely upon callINg A COmMAnd sUch As exit() in RelEase MOde?

A:

that'S NoT thE Only tHing tHaT HaS Changed. You're probablY aLso runNing yOur program witH "Ctrl+F5" instEAD Of just "F5". OR thaT ThE Debug checks enaBled iN debug mode USed to detEct a cRash or soMething in YOUr prograM whIch Isn'T beING dEtected any morE. oR a feW oTher posSibLe changEs.
IN anY caSe, rUn Your execuTable dirEcTlY (e.G. fRom windoWS exploreR oR a cOmManD promPT) and seE that This "prEsS Any Key to coNTiNUE..." doeSn'T aPpeaR thErE. Also You cAN ruN
). If $x \in G$ and $y \in  G$, thenthere  ar e $ n,  m \ in \ mathbb Z^+$ su c h th at $x^n=1$ and $y^m=1$ . The n$ (xy) ^ {n m}=x^ {nm}y^{ n m} = ( x^n )^ m( y^m )^ n =1 ^m1^n =1$ , so $x y \in G$.The  a ssociativity of  the opera tio n $\cdot$. T his  is in he rit e d fro m $ \math bb C$. Presen ce of the  i d entity element . Ob viou sly, $1 \in G$.
C l os u re under takin g an i nv e rs e ele men t. Indeed,  i f $z\ in G$,t he n $ z^n = 1$ for some $ n \in \math b b Z ^+$. H en ce, $(z^{- 1})^n =z ^ {-n }=(z^n)^{-1 }=1^ {-1}=1$.
 Q:
C ++ exit ( 0) no l ongerexi tsmy p r og ra m i mm e dat e ly  af t erconfigur at io n cha nged f r o m De bug  toRelea se

Normallystd ::ex i t(i nt) w ill t ermi na te yo ur pro gramri ght away, but f or s ome reaso n,an ywh er e I p u t exit (0) , i nsteadof quit t ing  i m m e di ately, on the cons ol e it  asks "P ress a n yke y  to cont in ue. .."f i rst.
Eve n  i f I thro w exit ( 0)  a s the f ir st lin einmai n(),i t st ill pr ompts me  to p r ess a key befo r e exiting.
On l yt h in g  I d idwas changemy p r ojec t to co mpi l ing i n Rel ea s eM ode instead of Debu gin Vis ual S tudio,
normal ly in debu g , I wouldseet he project run as :
C:\ User\<User n ame>\<Fi le pa th>\<Pro ject name > \ Debug\<P roj ect  na me> . e xe 
In release m o d e ho we ver, it  ru ns as:C:\ Win dow s\s ys tem32\cmd .exe 
Ido n' tkn owif th a t makesan y d if fer ence. . 
Is th ere a  way  I  c a n g et it t o  t e r mina te  i mmed iat el y upo n ca l lin g a com mand such  as exit ()  i n Relea se mode?

A:
T hat's notth e o nly th i n g that h as changed. You're prob a bly als o r unnin g yo ur progra m w ith "C trl + F5" in steadof ju st  "F 5 " . Ort h at  th edebug chec k s  en abled  i n de bug mod e used to detect a cra sh or somethi ngin y o u rpro g ra m  wh ic h  is n ' t being detecte d any more .O ra few othe r  po ss ible ch anges.In an y  case,run yourexecutabl edire c t ly(e.g. from  Windows  explorer or ac om mandpro mpt) a nd  se e tha t this "Pr ess a ny key  t o cont inue. .. " doesn' t appear there. Also yo u canrun
). If_$x \in_G$ and $y \in_G$, then_there_are $n,_m_\in \mathbb Z^+$_such that $x^n=1$_and $y^m=1$. Then $(xy)^{nm}=x^{nm}y^{nm}=(x^n)^m(y^m)^n=1^m1^n=1$,_so $xy \in_G$.
The_associativity of the operation $\cdot$. This is inherited from $\mathbb C$.
Presence of the identity_element._Obviously, $1_\in_G$.
Closure_under taking an inverse element._Indeed, if $z \in G$,_then $z^n=1$_for some $n \in \mathbb Z^+$. Hence, $(z^{-1})^n=z^{-n}=(z^n)^{-1}=1^{-1}=1$.

_Q:

C++_exit(0) no longer_exits my program immedately after configuration changed from Debug_to Release

Normally std::exit(int) will terminate your_program right away,_but_for_some reason, anywhere I_put exit(0), instead of quitting immediately,_on the console it asks "Press_any key to continue..." first. 
Even if_I throw exit(0) as the first_line in main(), it still_prompts me_to press a key before_exiting.
Only thing I_did was_change my project_to compiling in Release Mode instead_of Debug in_Visual Studio,
normally in debug, I would_see_the project run_as:
C:\User\<Username>\<File_path>\<Project_name>\Debug\<Project name>.exe
In_release mode however,_it_runs as:
C:\Windows\system32\cmd.exe_
I_don't know if that makes any_difference..
Is_there a way I can get it_to terminate immediately upon_calling_a command such as_exit() in Release mode?

A:

That's not_the only thing that has changed._You're probably_also running_your program with "Ctrl+F5" instead of just "F5". Or that the_debug checks enabled in debug mode_used to detect a_crash or_something_in your program_which_isn't being_detected any more. Or a few other_possible changes.
In_any case, run your executable directly_(e.g. from Windows explorer_or_a command prompt) and see that_this "Press any key to continue..."_doesn't appear there. Also you_can_run