# Maximal finite order of Abelian Groups

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.)

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Your conclusion on $r,s,p,m$ is wrong. – Olivier Bégassat Feb 26 '13 at 22:10
Why would $r$ and $s$ be relatively prime to $m$? – Olivier Bégassat Feb 26 '13 at 22:20
Suppose the $|h|$ doesn't divide $|g|$, then $|h|$ contains a prime factor with multiplicity higher than in $|g|$. This tells us that $gh$ has order greater than that of $g$, contradicting the hypothesis.
Commutativity tells us that $|gh| = \hbox{lcm}(|g|,|h|)$? – orlandpm Jan 4 '14 at 2:18