Show it is "abelian"

Question

Let $G$ be a group or order $m*n$, and $H$,$K$ be two normal subgroups of order each $m$ and $n$. Assume the intersection of $H$ and $K$ is the identity element of $G$. Show that $hk$=$kh$ for all $k$ $\in$ $K$ and $h$ $\in$ $H$.

Answer 1

The orders of $G$, $H$, and $K$ don't matter here. The important assumptions are that $H$ and $K$ are normal in $G$ and that $H\cap K=\{e\}$. Here's a hint: to show that $hk=kh$ for any $h\in H$ and $k\in K$, consider the element $hkh^{-1}k^{-1}$. Show that it lies both in $H$ and in $K$, and then the desired result follows (because there's only one element that they have in common).