Characterization of the isomorphic semidirect products

Let $A$ and $G$ be two finite abelian groups and let $\alpha$, $\beta:G\rightarrow{\rm Aut}(A)$. Suppose that $\alpha (G)$ and $\beta (G)$ are conjugate subgroups of ${\rm Aut}(A)$. Are the semidirect products $A\rtimes _{\alpha }G$ and $A\rtimes _{\beta }G$ isomorphic?

I know that this is true for a finite cyclic group $G$ but I don't what to do if $G$ is a finite non-cyclic Abelian group. I think the answer is usually no, so I will be thankful if someone provides me a counterexample.

Thank you in advance.

Solution 1:

If we assume only that the images of $\alpha$ and $\beta$ are conjugate in $\mathrm{Aut}(A)$, then the answer is no. Let $A = C_3$ and $G = C_4 \times C_2$, and consider two surjective homomorphisms $\alpha, \beta: G \to \mathrm{Aut}(A) \cong C_2$ with kernels isomorphic to $C_4$ and $C_2 \times C_2$ respectively. Then each of the two resulting semidirect products has a unique Sylow 3-subgroup $A$, whose centralizer is $A \times \ker \alpha$ or $A \times \ker \beta$ respectively. These subgroups are non-isomorphic, so the semidirect products are as well. Concretely: $$A \rtimes_{\alpha} G \cong C_4 \times (C_3 \rtimes C_2) \cong C_4 \times S_3, \text{ while} $$ $$A \rtimes_{\beta} G \cong C_2 \times (C_3 \rtimes C_4) \cong C_2 \times \mathrm{Dic}_{12},$$ where both semidirect products above are given by the unique nontrivial actions.

Solution 2:

That is not the answer to the question. Here is a condition under which the named groups are isomorphic.

If $f\in\operatorname{Aut} A$ is chosen such that for any $x\in G$ and $a\in A$ holds $$ f^{-1}\alpha(x)f(a)=\beta(x)(a) $$ and multiplications in the group $A\rtimes_\alpha G$ are given by the rule $$ (a,x)(b,y)=(a+\alpha(x)(b),xy) $$ (we use the additive notation of the operation in $A$), then the mapping $(a,x)\to (f(a),y)$ is an isomorphism of the group $A\rtimes_\alpha G$ to the group $A\rtimes_\beta G$.