Let $G$ be a finite group of even order. Show that $\exists a\neq e, a=a^{-1}.$

Question

Let $G$ be a finite group of even order. Show that $\exists a\neq e, a=a^{-1}.$

Suppose $\forall a\neq e,a\neq a^{-1}$ and define an equivalence reation on $G$ by $a\sim b$ if and only if $a=b$ or $a=b^{-1}$. Let $S=\{[a]:a\in G\}$ be the set of all equivalence classes. If $a=e, [e]=\{e\}$, and if $a\neq e, [a]=\{a,a^{-1}\}$ with $a\neq a^{-1}$. Then $G=\{e\}\cup(\bigcup\limits_{[a]\in S,[a]\neq [e]}\{a,a^{-1}\})$ and $|G|=2n+1$, a contradiction.

I am wondering if this proof is correct.

Answer 1

Yes, it is correct. But you should say that the union is disjoint because it corresponds to the equivalence classes of your equivalence relation.

Answer 2

Your proof is fine.

Here's an alternative:

Since $\lvert G\rvert$ is even, it is divisible by two. Two is prime. Therefore, by Cauchy's theorem, there exists an $H\le G$ such that $\lvert H\rvert=2.$ Let $h\in H$ such that $h\neq e$. Then, by closure, $h^{-1}=h\in H\subseteq G$.