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

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.)

• See Cauchy's Theorem. Jun 17 '20 at 20:59
• 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 Jun 17 '20 at 21:07
• Cauchy is not needed here. In fact, I think, the result was known long before Cauchy. Jun 17 '20 at 21:07
• I don't think $n\mid m$ follows from $g^m=e$ and $|G|=n$ Jun 17 '20 at 21:09

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.