Finite group of even order has an element $g \neq e$ such that $g^ 2 = e$

Question

I am trying to prove the following result.

Let $G$ be a finite group of even order. Prove that there exists $g \in G$ where $g^2 = e$ and $g \neq e$.

Here is my attempt.

Since $G$ has even order, $|G| \geq 2$. Hence, there exists some $g \neq e$. Since $G$ is of a finite order, there must exist some power, possibly not minimal, such that $g^m = e$. (Otherwise, the order is infinite.) Let $n$ be the order of $G$. Then $n \mid m$, so $m = nk$ for some $k \in \mathbb{N}$. But $G$ is of even order, so $n = 2j$ for some natural number $j$, so $m=nk=(2j)k = 2(jk)$. We have $$g^m = e \iff g^{2jk} = (g^{jk})^2.$$

The one remaining thing to show is that $g^{jk} \neq e$, but I'm having trouble accomplishing this. (I worry, actually, that we may have $jk = n$, in which case this wouldn't work.)

This approach won’t work—since you took an arbitrary nonidentity element of $G,$ you are essentially trying to prove every nonidentity element of $G$ has order two, which is far from the case, in general — mindfields, Jun 17 '20 at 21:07
Cauchy is not needed here. In fact, I think, the result was known long before Cauchy. — markvs, Jun 17 '20 at 21:07

score 8 · Accepted Answer · answered Jun 17 '20 at 21:06

8

If $G$ does not have $x\ne 1, x^2=1$ then every non-1 element $x$ has the property that $x^{-1}\ne x$. Then we can represent $G$ as a union of $\{1\}$ and several 2-element subsets $\{x,x^{-1}\}$. Hence $|G|$ is odd.

answered Jun 17 '20 at 21:06

markvs

19,653

The most simple solution I have seen in a while. – iam_agf Jun 17 '20 at 22:04

Finite group of even order has an element $g \neq e$ such that $g^ 2 = e$

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