Q: Prove that the order of $gh$ is at most of the least common multiple Suppose $G$ is an abelian, and $g,h∈G$. If $g$ and $h$ both have finite order $m$ and $n$, show that the order of $gh$ is at most of the least common multiple of $m$ and $n$.
I am wondering if this can be proved by constructing the least multiple of $m$ and $n$, namely $l$, and going back to show that $(gh)^l=1$, so that $l $ is the order of $gh$. However, I am not sure if this way of proof is valid or not. And how to deal with that "at most"? 

A: Yes, $(gh)^l=g^lh^l=1$.
But no, $(gh)^l=1$ does not mean the order of $gh$ is $l$. Rather it means
that the order of $gh$ is a factor of $l$.
