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Let $Q$ and $N$ be two groups. If there are two homomorphisms $\alpha,\beta : Q \rightarrow \operatorname{Aut}(N)$ then we can construct the semidirect products $G_a = N \rtimes_{\alpha} Q $ and $G_b = N \rtimes_{\beta} Q $.

I'm interested to know for which $\alpha$ and $\beta$ these two groups are isomorphic.

We can assume that both groups have the same underlying set $K = N \times Q$. Let $\psi$ be an automorphism in $\operatorname{Aut}(N)$, and let's write $n^{\psi}$ for $\psi(n)$ where $n \in N$. Let's extend $\psi$ to the whole of $K$ by $\psi(n,q) = (n^{\psi},q)$.

Note that in both groups we can write $n$ for $(n,1)$ and $q$ for $(1,q)$. We will see what it gives when we express the product $qn$ in both groups.

Let us denote by $\circ_a$ and $\circ_b$ the group operations in $G_a$ and $G_b$ respectively. In $G_a$ we have $q \circ_a n = (1,q) \circ_a (n,1) = (n^{\alpha(q)},q)$, by definition of semidirect product.

For $\psi$ to be a homomorphism we have to have $\psi(q \circ_a n ) = q \circ_b \psi(n)$. This equation reduces to $\psi(n^{\alpha(q)}) = n^{\psi \beta(q)}$ or $n^{\alpha(q)} = n^{\psi \beta(q) \psi {-1}}$.

We can conclude that if $\exists \psi \in \operatorname{Aut}(N)$ such that $\alpha(q) = \psi \beta(q) \psi^{-1} \forall q \in Q$ then the two semidirect products are isomorphic.

This does not mean that if $\alpha(q)$ and $\beta(q)$ belong to different automorphism classes they give rise to non isomorphic semidirect products. Examples of this are given here, where cases are given where $\alpha(q)$ is trivial and $\beta(h)$ is not and here where similar and additional conditions are discussed (without proof) for finite groups.

Shaun
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Marc Bogaerts
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  • The phrase "automorphism of ${\rm Aut}(N)$" is a bit confusing. You just showed the condition that $\alpha,\beta$ are conjugate is sufficient - do you mean to ask if it's necessary? I really doubt there is any good further characterization of when this happens, because isomorphisms have no requirement to play nice with internal semidirect product structure. – anon Feb 05 '17 at 17:52
  • Oops! I meant necessary. – Marc Bogaerts Feb 05 '17 at 17:56
  • Possible duplicate of Is there a nontrivial semidirect product of two groups isomorphic to their direct product? – Moishe Kohan Feb 05 '17 at 19:31
  • Consider $S_n$, with $n\geq 7$. It is a semidirect product $A_n \rtimes C_2$ in multiple ways. We can take the generator of $C_2$ to induce conjugation by $(12)$, or by $(12)(34)(56)$, etc... These are not conjugate under $Aut(N)\cong S_n$. – verret Feb 05 '17 at 20:19
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    Also possible duplicate of http://math.stackexchange.com/questions/527800 – Derek Holt Feb 05 '17 at 20:55
  • As for your edit: Do you realize that the direct product is the semidirect product $\rtimes_\alpha$ where the homomorphism $\alpha$ is trivial? Can the trivial homomorphism be conjugate to a nontrivial one? – Moishe Kohan Feb 06 '17 at 09:10
  • @MoisheCohen Indeed, this is the case. This is exaclty the reason why I'm interested in the non trivial cases where different automorphisms give rise to isomorphic semiproducts. Do you trealize that if two automorphisms are different this does not necessarily mean that one of them is trivial? – Marc Bogaerts Feb 06 '17 at 09:33
  • @MarcBogaerts: Your question was: If $A_\ltimes_\alpha B\cong A\ltimes_\beta B$, does it follow that $\alpha$ is conjugate to $\beta$ in the group of automorphisms. The examples in the link provide counter examples with $\beta=1$. Are you now asking for a classification of counter examples? Or for counter examples with neither one of the homomorphisms being trivial? For the latter, Derek's answer to the linked question gives you such. – Moishe Kohan Feb 06 '17 at 09:39
  • @MoisheCohen I understand the confusion now. I'll try to clarify. In the examples I studied I found that for every conjugacy class the groups were isomorphic, but I also found out that automorphisms from different classes gave rise tot the same groups. I proved that the set of automorphism giving rise to a group contain conjugacy classes completely or are disjoint from them. My question was how to characterize these sets. – Marc Bogaerts Feb 06 '17 at 14:24
  • @MarcBogaerts: I see, that would be another story. Please, edit your question accordingly and add links to the answers mentioned by me and by Derek. – Moishe Kohan Feb 06 '17 at 14:27
  • @MoisheCohen: Should I remove my rebuttal or reedit it? – Marc Bogaerts Feb 06 '17 at 14:30
  • @MarcBogaerts: I think you should rewrite the entire question so that other people do not get confused either. I will then try to remove my "close" vote (I am not sure how it works and if it is possible). – Moishe Kohan Feb 06 '17 at 14:32

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