Can the semidirect product of two groups be abelian group?

while I was working through the examples of semidirect products of Dummit and Foote, I thought that it's possible to show that any semdirect product of two groups can't be abelian if the this semidirect product is not the direct product.

Here is my simple idea:

Suppose $H,K$ are two groups and $H\rtimes K$ be the semidirect product of $H$ and $K$. Let $f:K \rightarrow Aut(H)$ be our homomorphism from $K$ into $Aut(H)$, now we know that $H \unlhd H\rtimes K$ but not necessarily $K$.

If $H\rtimes K$ is abelian, then every subgroup of it is normal, so $K$ must be normal but if $K$ is normal then $f$ is the trivial homomorphism and so the semidirect product turns into the direct product, so the semdidirect product in this case is the direct product, which is a contradiction since we supposed that this semidirect product of $H,K$ is not their direct product.

Is this true? Or have I made a mistake and there exists a counterexample?

• Nice work! (+1) Indeed, and the converse holds as well. Jun 20, 2013 at 1:55
• Converse of your claim (which is true) If H and K are abelian, and their semidirect product is explicitly isomorphic to their direct product, then their semidirect product is (obviously) abelian. You've shown the hard part of the biconditional statement! Jun 20, 2013 at 2:26
• @MathsLover: :D)))) Jun 22, 2013 at 5:17
• @BabakS. , by the way , there is another thing you can say during solving mathematical problems which is , اللهم لا سهل الا ما جعلته سهلا وانت تجعل الحزن اذا شئت سهلا , which is translated as " allah , there is no easy thing except those things which you have made them easy , and you are able to make the sadness as a happiness " this is hadis :D
– FNH
Jun 22, 2013 at 5:21
• @MathsLover: Thanks so much. That is a great another one. I'll keep it in my mind. سبحان ربک رب العزه عما یصفون Jun 22, 2013 at 5:25

You are correct. A semi-direct product of $H$ and $K$ is abelian iff $H$ is abelian, $K$ is abelian, and the semi-direct product is explicitly direct (the action is trivial).

When $H$ and $K$ are not abelian you can have bizarre things like $H \rtimes K$ is (isomorphic to) a direct product of $H$ and $K$, even though the action is not trivial. Take $H=K$ for example (to be non-abelian of order 6 to make it nice and clear). I learned this for $A_5 \rtimes A_5$, and was recently reminded it was true for all non-abelian groups $H=K$.

• Is any kind of necessary condition known for those weird examples (where $H\times H\cong H\rtimes_\theta H$ for $1\ne \theta$)? Jun 20, 2013 at 2:04
• I haven't looked closely if all $\theta$ work. Two that work are the "zero" (trivial) action, and the "identity" action as inner automorphisms. For S3 the "other" action works too. Maybe all "inner" actions work for centerless groups (I imagine if $\phi:H \to H$ is any endomorphism, then set $\Delta = \{ (\phi(h),h) : h \in H \}$. Then $H \times H$ is a semi-direct product $H \rtimes \Delta$ with action... I haven't checked, but hopefully $\phi$ once you view $H \leq Aut(H)$. Jun 20, 2013 at 2:12
• @JackSchmidt , let me emphasize my understanding , if $H$ is a non-abelian group ,and , $f:H\rightarrow Aut(H)$ is a homomorphism , then $H\times H \cong H\rtimes_f H$ , is this what you try to say ?
– FNH
Jun 20, 2013 at 2:26
• Almost: If $f:H \to \operatorname{Inn}(H)$ is a homomorphism, and $Z(H)=1$, then I've verified $H \times H \cong H \rtimes_f H$. Jun 20, 2013 at 15:27
• And now I've verified that $Z(H)=1$ is not superfluous ($H=D_8$ is a counterexample, since not all homomorphisms from $H$ to $H/Z(H)$ are induced by homomorphisms from $H$ to $H$ followed by the natural surjection). Jun 20, 2013 at 15:31

Sure. $H\rtimes_\theta K$ is abelian if and only if $H$ and $K$ are abelian and $\theta$ is the trivial homomorphism.

This is pretty easy to prove, so you should try to. Below is the solution if you get stuck.

Sufficiency is obvious. For necessity, if $\theta$ is nontrivial, then some $k\in K$ maps to some $\operatorname{id}\ne \theta_k\in \operatorname{Aut}(H)$, which means that $\theta_k(h)\ne h$ for some $h\in H$. Thus $hk=k\theta_k(h)\ne kh$, so we have two elements that don't commute.

• How do you get $hk = k \theta_k(h)$ in the last line?
– user568976
Apr 10, 2020 at 17:32