# When are two semidirect products isomorphic?

Let $N$, $H$ be groups, and $\varphi : H \to \operatorname{Aut}(N)$ a group homomorphism. Then we can form $N \rtimes_{\varphi} H$, the semidirect product of $N$ and $H$ with respect to $\varphi$.

Is there a way to determine if $\varphi_1, \varphi_2 : H \to \operatorname{Aut}(N)$ lead to isomorphic semidirect products by just comparing $\varphi_1$ and $\varphi_2$?

#### Solutions Collecting From Web of "When are two semidirect products isomorphic?"

No, even if $H\cong\mathbb{Z}$. There is a very interesting paper of Arzhantseva, Lafont and Minasyan, Isomorphism versus commensurability for a class of finitely presented groups, where they discuss the isomorphism of semidirect products with the infinite cyclic group. Writing $\widehat{\phi}$ for the outer automorphism of $H$ corresponding to $\phi\in\operatorname{Aut}(H)$, they prove the following result.

Theorem: Suppose $K\cong\mathbb{Z}$ and $H$ does not surject onto $\mathbb{Z}$. Then two semidirect products $H\rtimes_{\phi} K$ and $H\rtimes_{\psi}K$ are isomorphic if and only if $\widehat{\phi}$ is conjugate to $\widehat{\psi}$ or $\widehat{\psi}^{-1}$ in $\operatorname{Out}(H)$, the outer automorphism group of $H$.

This is especially nice as it allows them to construct groups with insoluble isomorphism problem in an especially elementary way. The proof is to note that there are finitely presented groups $H$ which don’t map onto $\mathbb{Z}$ and whose outer automorphism group has insoluble word problem. Then taking such a group $H$, we see that $H\rtimes_{\phi}\mathbb{Z}\cong H\times\mathbb{Z}$ if and only if $\phi$ is inner, which is undecidable as $\operatorname{Out}(H)$ has insoluble word problem.

The question now is: What if we replace $\mathbb{Z}$ with an arbitrary group $K$? Well, one can prove that a map $H\rtimes_{\phi} K\rightarrow H\rtimes_{\psi}K$ must send $H$ to $H$ if every homomorphism from $H$ to $K$ has trivial image. It seems that the rest of their proof needs $K\cong\mathbb{Z}$, but I am not entirely sure. (The relevant proof is Proposition $2.1$ of their paper, if anyone else wants to try and make it work?) So basically, I don’t know about the general case, but I find the case when $K\cong\mathbb{Z}$ very interesting. Such a group is called a mapping torus, and they are much-studied.

There is a theorem by Taunt (in “Remarks on the Isomorphism Problem in the Theories of Construction of Finite Groups”), which gives a nice characterization if $\gcd(|N|,|H|)=1$. I’m citing from “Construction of Finite Groups” by Besche and Eick.

Let $N,H$ be finite (soluble) groups with $\gcd(|N|,|H|)=1$. Furthermore let $\psi_i:H\rightarrow Aut(N)$ for $i=1,2$ be two homomorphisms. Define $G_i:=N\rtimes_{\psi_i}H$.

Then $G_1\cong G_2$, iff there exists automorphisms $\alpha\in Aut(N)$ and $\beta\in Aut(K)$ such that $(h^\beta)^{\psi_2}=(h^{\psi_1})^\alpha$ for all $h\in H$.

(i.e. $(\psi_2(\beta(h))=\alpha^{-1}\psi_1(h)\alpha\quad \forall h\in H$ )

The solubility is not necessary, therefore you can omit it.