20A05-AGroupOfEvenOrderContainsAnElementOfOrder2.tex

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\pmtitle{a group of even order contains an element of order 2}
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\pmtype{Theorem}
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%\pmkeywords{group order}
%\pmkeywords{order of an element}
%\pmkeywords{even order}
%\pmkeywords{finite group}

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\begin{document}
\begin{proposition*}
Every group of even order contains an element of order $2$.
\end{proposition*}
\begin{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 \PMlinkescapetext{pairing} it with its inverse, from which it follows that $\vert S\vert$ is a \PMlinkname{multiple}{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 $\vert S\vert+\vert G\setminus S\vert=\vert G\vert$, 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 $G$.
\end{proof}
Notice that the above \PMlinkescapetext{proposition} is logically equivalent to the assertion that a group of even order has a non-identity element that is its own inverse. 
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