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Need some help with this, i'm a bit stuck:

Show that if $G$ is a finite group of even order, then $G$ has an odd number of elements of order $2$. Note that $e$ is the only element of order $1$.

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marked as duplicate by YACP, azimut, Andrey Rekalo, Henry T. Horton, Ayman Hourieh Sep 4 '13 at 20:49

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Hint: the elements of $G$ of order greater than $2$ can be paired off with their inverses. –  Geoff Robinson Jan 1 '13 at 23:25
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Every element $x \in G$ has an unique element $x^{-1} \in G$ such that $xx^{-1} = e$. Further $ord(x) = 2 \iff x=x^{-1}$. Then you may split $G$ into elements which are their own inverse and which have a different inverse and $e$. –  André Jan 1 '13 at 23:25
    
As for the converse, see math.stackexchange.com/questions/20066/… –  amWhy Jan 1 '13 at 23:31
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More generally, if $G$ is a finite group and $p^n$ divides the order of $G$, the number of subgroups of $G$ with order $p^n$ is $\equiv 1 \mod p$ (In this case $p=2$ and $n=1$; counting elements of order $2$ is the same as counting subgroups of order $2$) –  Brett Frankel Jan 2 '13 at 1:11
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2 Answers 2

up vote 8 down vote accepted

Let $A$ be the set of all elements of order greater than $2$, and recall that $x$ and $x^{-1}$ have the same order. So convince yourself $$ A=\bigcup_{x\in A}\{x,x^{-1}\}. $$

Conclude that $|A|$ is even. Now why does that imply that there are an odd number of elements of order $2$?

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Oh okay. That makes sense, thanks a lot. –  randomafk Jan 1 '13 at 23:47
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now in the group identity element have order one .thus we are left with odd number of elements.also for each element of the group has a unique inverse. so only even number of elements will covered by the elements which have different inverses from themselves.and we know odd minus even is odd thus we are left with odd number of elements which are the inverses of themselvesi.e which have order two.

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