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Let $A$ be an abelian group and $G$ a group and let $\alpha:G\rightarrow{\rm Aut}(A)$. I want to show that the semidirect product $A\rtimes _{\alpha }G$ is isomorphic to the direct product $A\times G$ iff $\alpha(g)=id$ for all $g\in G$.

This is true if $G$ is abelian. If $G$ is nonabelian, I think that the only if direction is not true in general. So I will be thankful if someone provides us a counterexample.

Thank you in advance

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    $\begingroup$ Doesn't the requirement that $f=f(\alpha(g))$ for all $g$ imply that $\alpha(g)$ is the trivial automorphism for all $g$? $\endgroup$
    – jonan
    Jan 29 at 2:02
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    $\begingroup$ Since $f \in \text{Aut}(A)$ I would know what $f(a)$ meant if $a \in A$. However, $\alpha(g) \not\in A$, instead $\alpha(g) \in \text{Aut}(A)$, so I don't know what $f(\alpha(g))$ means... Perhaps you intended $f \circ (\alpha(g))$? $\endgroup$
    – Lee Mosher
    Jan 29 at 2:09
  • $\begingroup$ The question is edited. $\endgroup$ Jan 29 at 4:09

3 Answers 3

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For finite groups we can argue that $|(A\times G)'|=|G'|<|(A\rtimes_\alpha G)'|$ whenever $\alpha$ is non-trivial (the latter group has a non-trivial commutator in $A$).

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I think we can concoct a counterexample.

Let $A$ be the (restricted) direct product of countably many groups of order $2$ (i.e. $A$ is countably infinite elementary abelian), and let $G$ be the direct product of $A$ and countably many copies of the dihedral group of order $8$.

Then the direct product $A \times G$ is isomorphic to $G$.

We can define a nontrivial action of $G$ on $A$ by letting one of the direct factors of $G$ of order $2$ interchange two of the factors of $A$, and fix the rest of $A$, and all other direct factors of $G$ act trivially on $A$.

Then in the resulting semidirect product, we lose three factors of order $2$ and gain an extra dihedral factor, but the result is still isomorphic to $G$.

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  • $\begingroup$ Thank you very much for your answer, but how we gain a dihedral factor. I'm confusing with the facts that $Z_{4}\rtimes Z_{2}\cong D_{8}$ and $Z_{4}\ncong Z_{2}\times Z_{2}$. $\endgroup$ Feb 2 at 21:30
  • $\begingroup$ But $D_8$ is also isomorphic to $(Z_2 \times Z_2) \rtimes Z_2$ (where the action is to interchange the two direct factors), and that is where the dihedral factor is coming from in this example. $\endgroup$
    – Derek Holt
    Feb 3 at 8:53
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As a generalisation to Derek's Example, consider $A=E^{\mathbb N}$ with $E$ abelian and $G=E^{\mathbb N}\times F^{\mathbb N}\times (E\rtimes_\alpha F)^{\mathbb N}$. Then $\alpha$ induces a nontrivial action $\phi$ of $G$ on $A$ such that $A\rtimes_\phi G\cong A\times G$.

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