Start by proving that, in an abelian group, if $g$ has order $a$, and $h$ has order $b$, and $\gcd(a,b)=1$, then $gh$ has order $ab$. Clearly, $(gh)^{ab}=1$, so $gh$ has order dividing $ab$. Now show that if $(gh)^s=1$ for some $s\lt ab$ then you can find some $r$ such that $(gh)^{rs}$ is either a power of $g$ or of $h$ and not the identity. Details left to the reader.