Let $G$ be a finite group of order $n$ and $m$ elements of $G$ of order exactly $2$, prove that $n-m$ is odd.

Question

I am self-studying Algebra from Hungerford. I got stuck on the following question. I am stuck on this question.

Let $G$ be a finite group, of order $n$, and let $m$ be the number of elements $g \in G$ of order exactly $2$. Prove that $n−m$ is odd. Deduce that if $n$ is even then $G$ necessarily contains elements of order $2$.

Answer 1

Elements of order two are their own inverses. Otherwise $\{x,x^{-1}\} $ come in pairs. Then there's the identity, another singleton. The result follows.

(I have tacitly used uniqueness of inverses.)