Hi. I'm trying to prove the following:

Let $\displaystyle G$ be a finite group, and let $\displaystyle S=\{x\in{G}|x\neq{x}^{-1}\}$. Show that the set $\displaystyle S$ has an even number of elements.

So, Here's my thoughts.

Let the number of all elements$\displaystyle x$ such that $\displaystyle x\neq{x}^{-1}$ be $\displaystyle m$ so that there are $\displaystyle m$ elements in $\displaystyle G$ not equal to their own inverses. But,$\displaystyle G$ is a group and so there exists for every element $\displaystyle x\in{G}$ a $\displaystyle x^{-1}\in{G}$ such that $\displaystyle x*x^{-1}=e$ the identity element of $\displaystyle G$. Thus, there are also $\displaystyle m$ inverses corresponding to the $\displaystyle m$ elements belonging to $\displaystyle S$, but these elements also belong to $\displaystyle S$ since $\displaystyle (x^{-1})^{-1}=x$, Therefore, there are $\displaystyle 2m$ elements in $\displaystyle S$. Therefore, there are an even number of elements in $\displaystyle S$

Is this Right. The thing that concerns me is the fact that from this argument $\displaystyle m=2m$. Can someone help me phrase this better? I'm new to this subject and I've only recently became mature enough to construct proofs like this.