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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 $\vert S\vert$ 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 pairing it with its inverse, from which it follows that $\vert S\vert$ is a multiple 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 $\vert S\vert+\vert G\setminus S\vert=\vert G\vert$ , which, because $\vert G\vert$ is even, implies that $\vert G\setminus S\vert$ 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 $G$ .

a group of even order contains an element of order 2 is owned by Keenan Kidwell.