Commutator of two normal subgroups with trivial intersections

So I tried looking everywhere but I'm really lost with this proof. I need to to show that:

Let $H$, $K$ be normal subgroups of a group $G$, satisfying $H\cap K = \{e\}$. Let $h\in H$, $k \in K$.

(a) Prove that the commutator $w = hkh^{-1}k^{−1}$ satisfies $w \in H$ and $w \in K$. Then conclude that $w = e$

(b) Prove that $hk = kh$, for all $h \in H$, $k \in K$.

Solution 1:

(a) Since $K$ is normal $hkh^{-1}\in K$ and therefore $hkh^{-1}k^{-1}\in K$. The same argument proves that $hkh^{-1}k^{-1}\in H$.

(b) $hkh^{-1}k^{-1}=e\implies hkh^{-1}=k\implies hk=kh$.

Solution 2:

For (a), note that $hkh^{-1} \in K$ since $K$ is normal. As $K$ is a subgroup, you now see that $hkh^{-1}k^{-1} \in K$. The claim on $H$ membership is proved the same way. Since $w \in H \cap K = \{e\}$ there is no choice but $w=e$.