Semidirect product, normal subgroup exercise

Question

Let $G$ be a group and let $H,K$ be subgroups of $G$ such that $G=H \rtimes K$.

(i) Show that if $K \lhd G$, then $kh=hk$ for all $h \in H, k \in K$.

(ii) Deduce that $G$ is abelian if and only if $H,K$ are abelian and $K \lhd G$

I am completely lost with the exercise. First of all, the hypothesis is that $G$ is isomorphic to the external semidirect product of $H$ and $K$, but it doesn't say anything about the morphism $\rho:K \to Aut(H)$, I know that the external semidirect product multiplication is $(h_1,k_1).(h_2,k_2)=(h_1\rho(k_1)(h_2),k_1k_2)$. I don't know how to use the fact that $K$ is normal and the structure of the semidirect product in order to show (a) or (b), I would appreciate hints and suggestions to do both parts of the problem.

Answer 1

Let $K \lhd G$. Then, $h^{-1}(k^{-1}hk)$ lies in $H$ as $H$ is normal, and $(h^{-1}k^{-1}h)k$ lies in $K$ as $K$ is normal. As their intersection is $1$, we have that $h^{-1}k^{-1}hk=1$, i.e. $hk=kh$.

Now, If $H$, $K$ are abelian, and $K \lhd G$, then the above computation shows that all elements of $G$ commute, and so $G$ is abelian.

The converse is trivial (why?)

Note that we do not require the actual automorphism $\rho$ for any of the above computations.

Answer 2

Hint: For (i), show that the commutator $[h,k] = h^{-1}k^{-1}hk\in H\cap K$, for $h\in H$ and $k\in K$. You need that $H$ and $K$ are both normal for this.