If $a$ is a group element, prove that $a$ and $a^{-1}$ have the same order.
I tried doing this by contradiction.
Assume $|a|\neq|a^{-1}|$.
Let $a^n=e$ for some $n\in \mathbb{Z}$ and $(a^{-1})^m=e$ for some $m\in \mathbb{Z}$, and we can assume that $m < n$.
Then $e= e*e = (a^n)((a^{-1})^m) = a^{n-m}$. However, $a^{n-m}=e$ implies that $n$ is not the order of $a$, which is a contradiction and $n=m$.
But I realized this doesn’t satisfy the condition if $a$ has infinite order. How do I prove that piece?