Product of commuting elements of coprime order
= Product of commuting elements of coprime order
{title2=$\operatorname{ord}(ab)=\operatorname{ord}(a)\operatorname{ord}(b)$}
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.