 Product of commuting elements of coprime order

Let commuting elements $a,b$ have finite coprime orders $m,n$. Their cyclic subgroups have trivial intersection by Lagrange's theorem. If $(ab{)}^k=1$, then $a^k=b^{-k}$ belongs to this intersection, so $m\mid k$ and $n\mid k$. Thus $mn\mid k$. Conversely, $(ab{)}^{mn}=1$ because the elements commute. This proves the displayed order formula.