Semidirect product and direct product

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.

• Doesn't the requirement that $f=f(\alpha(g))$ for all $g$ imply that $\alpha(g)$ is the trivial automorphism for all $g$? Jan 29 at 2:02
• 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 $f \circ (\alpha(g))$? Jan 29 at 2:09
• The question is edited. Jan 29 at 4:09

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 $$A$$).

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 $$G$$.

• 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 $Z_{4}\ncong Z_{2}\times Z_{2}$. Feb 2 at 21:30
• 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. Feb 3 at 8:53

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 $$A\rtimes_\phi G\cong A\times G$$.