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.

Let $h\in H$ and $n\in N$, then we have $n^{-1}hn\in H$ and $hnh^{-1}\in N$. So we get $n^{-1}hnh^{-1}$ is in both $N$ and $H$. As $H\cap N=\{e\}$ we have $n^{-1}hnh^{-1}=e$ so $hn=nh$.
