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Let $H$ be a group, $G \triangleleft H$ a normal subgroup and $K \leqslant H$ a subgroup. We say $H$ to be a semidirect product of $G$ and $K$, denoted by $H= G \rtimes K$, if $H=GK$ and $G \cap K=1$. Let us recall that $Z(H)= \{ x \in H \ \vert \ \forall h \in H, hx=xh \}$ is the center of $H$. A group $H$ is centerless if the center is trivial. My question is the following one:

If $H$ is centerless, can $H$ always be decomposed as a non-trivial semidirect product?

I can provide the following partial answers:

  • If $H$ is a finite centerless metaabelian group, it is proven here.
  • If $H=A *B$ is a free product it is proven here

Note that by non-trivial semidirect product I mean avoiding cases such as $H=G \times 1$, $H=1 \times K$, or $H=G \times K$.

Juanandete
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    Why is the first point straightforward? If $G$ is a finite nonabelian simple group, the normal subgroup $H$ must be $1$ or $G$. So it is not possible. Of course, $Z(G)=1$ then, too. – Dietrich Burde Jun 05 '24 at 11:32
  • It is not straightforward as you have pointed out, it was wrong. – Juanandete Jun 05 '24 at 11:56

1 Answers1

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To answer the question stated in the OP: No, you can construct a group $H$ which is centerless (i.e. trivial center) and lacks non-trivial semidirect product decomposition. An easy example to see this with is the alternating subgroup $A_5\triangleleft S_5$...but to be a little more general, take any non-abelian simple group and you should get this contradictory outcome.

Why is this the case? That all non-abelian simple groups contradict the posed statement? Recall that simple groups are defined as the groups in which $0$ and $H$ are the only normal subgroups $\triangleleft\ H$. Since the center of a group will be normal, we know that $Z(H)=0$ because $H$ is non-abelian. Since $H\cong N\rtimes G$ will necessarily have an isomorphic copy of $N$ as a normal subgroup (by construction of semi-direct product), then $N=0$ or $H$. We are therefore left with the trivial case.

J.G.131
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  • As you have mentioned simple and non-abelian groups satisfy the condition, what can we say about the non-simple case? – Juanandete Jun 06 '24 at 05:52
  • @Juanandete Secondly, I have outsourced your new question via my post: https://math.stackexchange.com/questions/4928588/non-simple-centerless-group-with-exactly-one-non-trivial-normal-subgroup. I will update you when I hear back from the community. This is a very tricky question, the uses of which I struggle to see for the moment. But hopefully we can get something soon. – J.G.131 Jun 06 '24 at 17:18
  • @Juanandete Update: Derek Holt has given us a counter-example. Therefore, we should be able to conclude that even when excluding simple groups, your revised statement will not hold true. – J.G.131 Jun 06 '24 at 18:20
  • Thank you very much for your time and effort! I've just accepted you answer. :) – Juanandete Jun 07 '24 at 06:44