Prove if $G$ is abelian, $g\in G$ is of maximal order, and $h \in G$ has finite order, then $|h|$ divides $|g|$.

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I've been playing around with the problem for a while now but haven't managed to make any progress. I would really appreciate a nudge in the right direction -- but please no full solutions.

Attempt:

Assume $|h|$ is finite but does not divide $|g|$. Then there is a prime integer $p$ such that $|g|=p^m r$ and $|h|=p^n s$, with $\gcd(r,s)=1$ and $m<n$. Note that $$\left| g^{p^m} \right| = \frac{|g|}{\gcd\left( p^m, |g| \right)}=r$$ $$\left| h^{s} \right| = \frac{|h|}{\gcd\left( s, |h| \right)}=p^n$$ thus $\gcd(\left| g^{p^m} \right|, \left| h^{s} \right|)=1$, so by a previous exercise we have $$\left| g^{p^m} h^{s} \right|=\left| g^{p^m} \right|\left| h^{s} \right| = p^n r .$$

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Compare $|g^{p^m}h^s|$ to $|g|$.