To prove a statement about finite groups of even order.

Question

If $G$ is a group of even order then prove that there is an element $a$ in $G$ such that $a$ does not equal the identity element $e$ and $a^2=e$. I just know the fact that the order of element divides the order of group and $a^2=e$ for all $a$ belonging to $G$ is an abelian group. How can I prove the statement using these facts? Even if theres another simple way to do this then it will be appreciated.

Answer 1

I think a good strategy would be to consider that every element must have an inverse in $G,*$, where $*$ is the binary operation of the group. If you know the size of the group is $2n: n\in \mathbb{Z}^+$, what can you say about the "pairing" of elements and their inverses?

(Hint: there are $2n-1$ elements not including the identity).

Answer 2

Hint: Take a look at the set $$ A=\{g\in G| g^2\neq e\} $$ Note that $e\not\in A$.