# A finite group of even order has an odd number of elements of order 2 [duplicate]

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 HouriehSep 4 '13 at 20:49

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
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 $e$. – Epsilon Jan 1 '13 at 23:25
As for the converse, see math.stackexchange.com/questions/20066/… – amWhy Jan 1 '13 at 23:31
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

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 $2$?