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$H$ is a group and $\alpha: H \rightarrow \operatorname{Aut}(H)$ is a nontrivial homomorphism. Does $H \rtimes_{\alpha} H \ncong H \times H$ necessarily?

This is a follow-up to this thread.

Hagen gives there a nice counterexample for the case $N \rtimes_{\alpha} H \cong N \times H$, where $H \neq N$, so it cannot be applied here.

I'm trying to find a counterexample or prove that $H \rtimes_{\alpha} H \cong H \times H$ for quite some time now, but hasn't made any significant progress.

Any ideas?

Community
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Brassican
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1 Answers1

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Let $N=H$ be any nonabelian group. Then the normal subgroup $N \times 1$ of $G:=N \times H$ has a complement $H_1 = \{(n,n) : n \in N \}$, and then $G$ is the internal semidirect product $N \rtimes H_1$, and $H_1$ acts nontrivially on $N$.

The map $\alpha$ that defines this semidirect product maps each $h \in H$ to the inner automorphism of $N$ that it induces.

Derek Holt
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