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Let $A$ be an abelian group and $G$ a group. Let $\alpha:G\rightarrow{\rm Bij}(A)$ and $\beta: A\rightarrow{\rm Bij}(G)$ be two group homomorphisms. I have tried to prove that the Zappa–Szép product $A{}_{\alpha}{\bowtie}_{\beta} G$ is isomorphic to the direct product $A\times G$ iff $\alpha(g)=id$ and $\beta(a)=id$ for all $g\in G$ and $a\in A$.

The if direction is clear. For the converse, if $G$ is abelian then $A{}_{\alpha}{\bowtie}_{\beta} G$ is abelian and hence $\alpha(g)=id$ and $\beta(a)=id$. If $G$ is nonabelian, I think it is possible for a direct product $A\times G$ to be isomorphic to $A{}_{\alpha}{\bowtie}_{\beta} G$ with $\alpha $ and $\beta$ are both nontrivial, but I couldn't find a counter-example. So I will be thankful if someone provides me a counterexample. Notice that I have asked a similar question in mathoverflow.

Thank you very much.

N. SNANOU
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    If you let $A_5$ act on itself via conjugation and form the group $A_5\rtimes A_5$, you get a group that is isomorphic to $A_5\times A_5$. See, e.g., here. – Arturo Magidin Aug 12 '23 at 03:13
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    @Arturo Magidin Thank you very much. I see you restricted to semidirect products, but I want a counterexample on Zappa–Szép products in which the actions $\alpha $ and $\beta$ are both nontrivial. – N. SNANOU Aug 12 '23 at 09:48
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    And the example did not have one group abelian. It does, however, suggest that there is reason for doubt. – Arturo Magidin Aug 12 '23 at 16:23
  • It's okay to post also in MO after a while, but when you do you should note in both posts that the question is now posted on the other site. It's polite, and it prevents duplication of effort. – Arturo Magidin Aug 13 '23 at 23:17
  • @ArturoMagidin Yes of course you are right. I add a notification to the questions. Thank you. – N. SNANOU Aug 14 '23 at 00:23
  • Why is $A\bowtie G$ abelian if $A$ and $G$ are both abelian? For example, the Klein bottle group $\mathbb{Z}\rtimes\mathbb{Z}$ is non-abelian. – user1729 Aug 14 '23 at 07:58
  • Also, it would help if you defined an antihomomorphism, and said what your motivation for this problem is. – user1729 Aug 14 '23 at 08:01
  • @user1729 I used the assumption $A{}{\alpha}{\bowtie}{\beta} G\cong A\times G$ of the only if direction. Note that, since A is abelian, an anti-homomorphism is in fact a homomorphism. – N. SNANOU Aug 14 '23 at 10:27
  • Ah, ok. Also, you didn't address motivation. (This question is not particularly hard but with a minor twist becomes more complex, which is why I'm wondering.) – user1729 Aug 14 '23 at 10:32
  • @user1729 In fact, I have seen similar questions concerning "the isomorphism between a semidirect product and a direct product" which give some interesting examples proving in some cases the isomorphism between semidirect products and direct product ( see for example this question). Here, I hope finding similar examples for the Zappa-szép products. – N. SNANOU Aug 14 '23 at 11:15
  • @N.SNANOU Sure, but then Derek Holt's answer there gives you a counter-example to the question you asked here. You could also take semi-direct products $\mathbf{Z}\ltimes_{\phi}G$ for $\phi$ inner, etc. – user1729 Aug 14 '23 at 11:31
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    (Here $\alpha$ is trivial and $\beta$ is not, so we obviously have a semidirect product. Your comment above and MO question implies you actually want $\alpha$ to be non-trivial?) – user1729 Aug 14 '23 at 11:32

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