Let $A$ and $G$ be two groups. Let $\alpha:G\rightarrow\operatorname{Bij}(A)$ be a group homomorphism and $\beta: A\rightarrow\operatorname{Bij}(G)$ an anti-homomorphism satisfying some conditions given in Wikipedia. The actions $\alpha$ and $\beta$ induce a group $A\mathbin{_{\alpha}{\bowtie}_{\beta}} G$ called the Zappa-Szép product. This group is a natural generalization of the semidirect product of two groups in which neither of the factors is required to be normal. I know some examples of semidirect products induced by nontrivial actions and isomorphic to direct products. However, I couldn't find a group $H=A\mathbin{_{\alpha}{\bowtie}_{\beta}} G$ isomorphic to the direct product $A\times G$ where the actions $\alpha$ and $\beta$ are both nontrivial. So I will be thankful if someone provides me an example. Notice that I have asked a similar question in math.stackexchange.
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$\begingroup$ Also asked in math.stackexchange. $\endgroup$– Arturo MagidinCommented Aug 13, 2023 at 23:17
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1$\begingroup$ How does an anti-homomorphism from an Abelian group differ from a homomorphism? $\endgroup$– LSpiceCommented Aug 14, 2023 at 0:55
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$\begingroup$ @LSpice Thank you for the remark. An anti-homomorphism from an Abelian group is in fact a homomorphism. So, I must take A an arbitrary group or taking $\beta$ a homomorphism. $\endgroup$– N. SNANOUCommented Aug 14, 2023 at 10:47
2 Answers
Let $S$ be a nonabelian group with a fixed-point-free automorphism $\alpha$. (Such groups for which $\alpha$ has prime order are necessarily nilpotent. I think the smallest example is the nonabelian group of order $7^3$ and exponent $7$, which has a fixed-point-free automorphism of order $3$.)
Now let $G = S \times S$ and define the (diagonal) subgroups $H$ and $K$ of $G$ by $$ H = \{ (s,s) : s \in S \},\quad K = \{(s,\alpha(s)) : s \in S \}.$$ Then $H \cong K \cong S$, $H \cap K = \{1\}$ and $G = HK$, so $G$ is the internal Zappa-Szép product of $H$ and $K$, but neither $H$ nor $K$ is normal in $G$, so this is not equal to a semidirect product.
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$\begingroup$ Thank you very much. Could you reformulate this example for the "external" Zappa-Szép product of H and K. $\endgroup$ Commented Aug 14, 2023 at 11:31
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1$\begingroup$ @N.SNANOU Isn’t it quite straightforward to do that reformulation? Why not do it yourself? $\endgroup$ Commented Aug 14, 2023 at 13:57
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$\begingroup$ @JeremyRickard Yes, that's exactly what I was thinking! $\endgroup$ Commented Aug 14, 2023 at 15:00
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1$\begingroup$ @DerekHolt : It is not really relevant to the main point of your answer, but in general, it is true that a finite group which admits a fixed-point free automorphism of prime order is nilpotent. However, in general, this is not the case if the fpf automorphism does not have prime order. See, for example, the paper by Gorenstein and Herstein on finite groups which admit a fixed point free automorphism of order $4$. $\endgroup$ Commented Aug 14, 2023 at 20:14
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$\begingroup$ @Geoff Robertson Yes, good point. And in fact the automorphism $\alpha$ in my answer does not need to have prime order. $\endgroup$ Commented Aug 14, 2023 at 20:21
I can propose a boring, but pretty general example.
One can check that if a group $G$ acts on itself by conjugation, then the semidirect product $G \rtimes G$ is isomorphic to the direct product $(G \times \{e\}) \times \Delta (G)$, where $\Delta: G \to G \rtimes G$ is the diagonal: $g \mapsto (g, g^{-1})$.
To get the desired example, you can let both groups $L, R$ in the knit product $L \bowtie R$ be of the form $G \times G$, where $G$ is your favourite non-abelian group.
Let's say that $L:=G_{11} \times G_{12}$, and $R := G_{21} \times G_{22}$. Left action $\alpha: L \times R \to R$ is given by $$((l_1, l_2), (r_1, r_2)) \mapsto (l_1 r_1 l^{−1}_1, r_2);$$ right action of $\beta: L \times R \to L$ is given similarly by $$((l_1, l_2), (r_1, r_2)) \mapsto (l_1,r^{−1}_2 l_2 r_2).$$
The resulting knit product $L\mathbin{_{\alpha}{\bowtie}_{\beta}} R$ would be isomorphic to $$(G_{11} \ltimes G_{21}) \times (G_{21} \rtimes G_{22}),$$ which splits (componentwise) by observation in the first paragraph.
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$\begingroup$ Just to be sure I'm understanding, is the action of the left factor given by $\alpha((g_1, g_2), (h_1, h_2)) = (g_1 h_1 g_1^{-1}, g_1 h_2 g_1^{-1}$, and analogously for the action of the right factor? $\endgroup$– LSpiceCommented Aug 16, 2023 at 3:07
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2$\begingroup$ What I mean is: let's say that $L := G_{11} \times G_{12}$ and $R := G_{21} \times G_{22}$. Left action of $L$ on $R$ is given by $((l_1, l_2), (r_1, r_2)) \mapsto (l_1 r_1 l_1^{-1}, r_2)$; right action of $R$ on $L$ is given by $((l_1, l_2), (r_1, r_2)) \mapsto (l_1, r_2^{-1} l_2 r_2)$. $\endgroup$– Denis TCommented Aug 16, 2023 at 4:19
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1$\begingroup$ To make it absolutely clear, resulting knit product is the direct product of $G_{11} \ltimes G_{21}$ and $G_{12} \rtimes G_{22}$. $\endgroup$– Denis TCommented Aug 16, 2023 at 4:23