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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.

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  • $\begingroup$ 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. $\endgroup$ – arctic tern Feb 5 '17 at 17:52
  • $\begingroup$ Oops! I meant necessary. $\endgroup$ – Marc Bogaerts Feb 5 '17 at 17:56
  • $\begingroup$ Possible duplicate of Is there a nontrivial semidirect product of two groups isomorphic to their direct product? $\endgroup$ – Moishe Kohan Feb 5 '17 at 19:31
  • $\begingroup$ 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 $Aut(N)\cong S_n$. $\endgroup$ – verret Feb 5 '17 at 20:19
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    $\begingroup$ Also possible duplicate of math.stackexchange.com/questions/527800 $\endgroup$ – Derek Holt Feb 5 '17 at 20:55

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