Let $H$ and $N$ be normal subgroups of a group $G$ with $H \cap N = \{e\}$. Prove that $hn = nh$ for all $h \in H$ and $n \in N$. What does $H \cap N = \{e\}$ imply here? Is it related to the divisibility of the orders of $H$ and $N$? If so, how do I relate that to show commutativity?
Thanks in advance.

Hint: look at the element $(h^{-1}n^{-1}h)n$ and (the same) $h^{-1}(n^{-1}hn)$, where $h \in H$ and $n \in N$. To which subgroup(s) does it belong?
