# a group of even order contains an element of order 2

###### Proposition.

Every group of even order contains an element of order $2$.

###### Proof.

Let $G$ be a group of even order, and consider the set $S=\{g\in G:g\neq g^{-1}\}$. We claim that $|S|$ is even; to see this, let $a\in S$, so that $a\neq a^{-1}$; since $(a^{-1})^{-1}=a\neq a^{-1}$, we see that $a^{-1}\in S$ as well. Thus the elements of $S$ may be exhausted by repeatedly selecting an element and it with its inverse       , from which it follows that $|S|$ is a multiple (http://planetmath.org/Divisibility) of $2$ (i.e., is even). Now, because $S\cap(G\setminus S)=\emptyset$ and $S\cup(G\setminus S)=G$, it must be that $|S|+|G\setminus S|=|G|$, which, because $|G|$ is even, implies that $|G\setminus S|$ is also even. The identity element  $e$ of $G$ is in $G\setminus S$, being its own inverse, so the set $G\setminus S$ is nonempty, and consequently must contain at least two distinct elements; that is, there must exist some $b\neq e\in G\setminus S$, and because $b\notin S$, we have $b=b^{-1}$, hence $b^{2}=1$. Thus $b$ is an element of order $2$ in $G$. ∎

Notice that the above is logically equivalent to the assertion that a group of even order has a non-identity element that is its own inverse.

Title a group of even order contains an element of order 2 AGroupOfEvenOrderContainsAnElementOfOrder2 2013-03-22 17:11:37 2013-03-22 17:11:37 azdbacks4234 (14155) azdbacks4234 (14155) 20 azdbacks4234 (14155) Theorem msc 20A05