Q:

Normal Subgroups that Intersect Trivially

Let $H$ and $K$ be normal subgroups of a group $G$ such that they intersect trivially.
Why is it then the case that 
$$hk=kh;\;\;\;\; \forall h\in H,\;\forall k\in K?$$

A:

Note that $hkh^{-1}k^{-1} = (hkh^{-1})k^{-1} \in K$, because $hkh^{-1}\in K$ (by normality of $K$) and $k^{-1}\in K$.
Also, $hkh^{-1}k^{-1} = h(kh^{-1}k^{-1})\in H$, because $kh^{-1}k^{-1}\in H$ (by normality of $H$), and $h\in H$.
So $hkh^{-1}k^{-1} \in H\cap K=\{1\}$. So $hkh^{-1}k^{-1}=1$.
Note that you don't even need $H$ and $K$ to be normal. You just need $H$ and $K$ to normalize each other, that is, $H\subseteq N_G(K)$ and $K\subseteq N_G(H)$.

A:

Because $hkh^{-1}k^{-1}$ is both in $H$ and $K$.

