# Conditions on $\alpha, \beta$ under which $A \rtimes_{\alpha} B$ and $A \rtimes_{\beta} B$ are isomorphic

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.

• 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. – arctic tern Feb 5 '17 at 17:52
• Oops! I meant necessary. – Marc Bogaerts Feb 5 '17 at 17:56
• – Moishe Kohan Feb 5 '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 5 '17 at 20:19
• Also possible duplicate of math.stackexchange.com/questions/527800 – Derek Holt Feb 5 '17 at 20:55