Let $G$ be a commutative group, and let $g$ be an element of $G$ be an element of maximal finite order, then $|h|\leq |g|$. Prove that in fact if $h$ is finite order in $G$, then $|h|$ divides $|g|$.
This is what I have:
Proof by contradiction If $|h|$ is finite but doesn't divide $|g|$, then there is a prime integer $p$ such that $|g|=rp^m$, $|h|=sp^n$, with $r$ and $s$ relatively prime to $p$ and $m < n$.
(This is where I do not know where to go.)