# Semidirect product, normal subgroup exercise

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.

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Note that the hypothesis is that the group $G$ is equal to the internal semi-direct product of $H$ and $K$. It is not necessary, for this problem, to introduce an explicit action of $K$ on $H$; it is implicit in the group operation in $G$, and that is enough. –  James Sep 2 '14 at 2:27
The hypothesis is not that $G$ is isomorphic to the external semidirect product, but rather that it is the internal semidirect product of those. –  tomasz Sep 2 '14 at 2:28
@user16924. Well, if $G$ is equal to an internal semi-direct product, then it is isomorphic to the corresponding external semi-direct product (for some choice of the action $\rho$). –  James Sep 2 '14 at 2:41
@user16924: Have a look at Dummit and Foote's algebra book. –  voldemort Sep 2 '14 at 2:46
@voldemort Thanks, I'll think about the other implication and write it down. –  user16924 Sep 2 '14 at 2:47

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.

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The converse is trivial because if $G$ is abelian and $h,h' \in H$, then $hh'=h'h$, since $h,h'$ are also in $G$, the same goes for $G$. Now, let $g \in G$, take an element $gkg^{-1}$ from $gKg^{-1}$, then $gkg^{-1}=(gg^{-1})k=ek=k \in K$, so $gKg^{-1}=K$, from here it follows $K$ is normal. –  user16924 Sep 2 '14 at 3:52
Sorry to bother again but I don't quite see why from $(i)$ one deduces that $H,K$ abelian and $K$ normal implies $G$ abelian. Take $x,y \in G$, then $xy=hk$, I want to show that $xy=yx$ or, equivalently $xyx^{-1}y^{-1}=e$, I don't know how to finish from there. –  user16924 Sep 2 '14 at 3:54
@user16924: Every $x \in G$ can be written as $hk$ for some $h \in H$ and $k \in K$. So, $xy=h_1k_1h_2k_2$. Now, $k_1h_2=h_2k_1$ as $K$ is normal (from above computation). So, $xy=h_1h_2k_1k_2$. Also, $yx=h_2k_2h_1k_1=h_2h_1k_2k_1$. Since $H$ is abelian, $h_1h_2=h_2h_1$. Since $K$ is abelian, $k_2k_1=k_1k_2$. Thus $G$ is abelian. –  voldemort Sep 2 '14 at 4:04

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.

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