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Let $H$ and $N$ be two groups, $\phi$ and $\psi$ be two homomorphisms $H \to \operatorname{Aut}(N)$, and $G_1 = N \rtimes_\phi H$ and $G_2 = N \rtimes_\psi H$ corresponding semidirect products.

What are conditions on $\phi$ and $\psi$ for $G_1 \cong G_2$?

If for some automorphisms $h \in \operatorname{Aut}(H)$ and $n \in \operatorname{Inn}(\operatorname{Aut}(N))$ we have $$\phi = n \circ \psi \circ h\tag1$$ then $G_1 \cong G_2$.

At least for infinite groups, this condition is not necessary: let $X^\omega$ be direct sum of countable number of copies of $X$, let $N = S_3^\omega \times \mathbb Z_3^\omega \times \mathbb Z_3^\omega$ and $H = \mathbb Z_2^\omega \times \mathbb Z_2^\omega$, then trivial homomorphism and homomorphism that sends $i$-th copy of $\mathbb Z_2$ of the first component of $H$ to non-trivial automorphism of $i$-th copy of $\mathbb Z_3$ of the second component of $N$, and sends all elements from the second component of $H$ to trivial automorphism, both give group $\mathbb S_3^\omega \times \mathbb Z_6^\omega$, but there are no automorphisms on $\operatorname{Aut}(N)$ and $H$ that transforms one of this homomorphisms to the other.

Is condition $(1)$ necessary for finite groups? Is there some "good" general condition?

mihaild
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    Does this answer your question? When are semidirect products isomorphic? – suckling pig Jul 19 '22 at 21:59
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    @Cpc I don't think so, in terms of my question the question you linked is to prove that $(1)$ is sufficient when $n$ is trivial ad $n$ is inner. – mihaild Jul 19 '22 at 22:46
  • In your final question, are you interested primarily in a "good" necessary condition? Or would "good" sufficient conditions also be of interest? I'll throw out that my gut feeling is that there is very little to say regarding "good" necessary conditions; I say that only because the innumerable distinct constructions (i.e. "sufficient" conditions) of isomorphisms that I know about are of such different characters. – Lee Mosher Jul 21 '22 at 15:09
  • Let me also add that it might be worthwhile to surf around this site. On my page the "Related" heading contains several hits. – Lee Mosher Jul 21 '22 at 15:15
  • Sure, if there is any sufficient condition less trivial than independent isomorphisms of two groups, I will be interested in it. – mihaild Jul 22 '22 at 08:17
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    @mihaild See this paper. – Nourddine Snanou Mar 15 '23 at 00:44

1 Answers1

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Your criterion for equivalence of semidirect products corresponds to group isomorphisms that preserve the distinguished subgroup $N$ of the semidirect product. You would expect to find examples in which the two semidirect products were isomorphic under an isomorphism that mapped $N$ to a different normal subgroup.

I did a computer calculation (in Magma) and found that the group $\mathtt{SmallGroup}(128,753)$ of order $128$ has, up to equivalence under its automorphism group, two types of normal subgroup $N$ isomorphic to $C_2^4$ (i.e. elementary abelian of order $16$) with complements isomorphic to $D_8$ (dihedral of order $8$).

One of them has $2$ conjugacy classes of complements and the other has $8$, and they are not equivalent as semidirect products according to your criterion.

Derek Holt
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