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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 $\varphi_2$?

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    $\begingroup$ Two conditions under which they are isomorphic are the following. (1) If $f \colon H \rightarrow H$ is an automorphism of $H$ then $N \rtimes_{\varphi} H$ and $N \rtimes_{\varphi \circ f} H$ are isomorphic groups. (2) If $f$ is an automorphism of $N$ then conjugation by $f$ is an automorphism of ${\rm Aut}(N)$, and if we denote that by $\gamma_f$ (that is, $\gamma_f \colon {\rm Aut}(N) \rightarrow {\rm Aut}(N)$ by $\gamma_f(h) = f \circ h \circ f^{-1}$), then $N \rtimes_{\varphi} H \cong N \rtimes_{\gamma_f \circ \varphi} H$. $\endgroup$
    – KCd
    Commented Oct 16, 2013 at 0:16
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    $\begingroup$ @KCd: Are these also necessary conditions? $\endgroup$
    – Michael Albanese
    Commented Oct 16, 2013 at 1:09
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    $\begingroup$ No, those conditions are not necessary. If $\varphi$ is a trivial action (that is, $\varphi$ is the trivial homomorphism $H \rightarrow {\rm Aut}(N)$) then the two constructions I described are also trivial actions, so it only produces the direct product $N \times H$. It is possible, however, for two groups to have their direct product be isomorphic to a nontrivial semidirect product, and you can't get to a nontrivial semidirect product by the two methods I described. I'll give an example of such groups in my next comment. $\endgroup$
    – KCd
    Commented Oct 16, 2013 at 14:07
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    $\begingroup$ Let $G = {\rm GL}_n({\mathbf R})$. It has normal subgroup $N = {\rm SL}_n({\mathbf R})$ and the subgroup of diagonal matrices ${\rm diag}(a,1,1,\dots,1)$ with $a \not= 0$ is a complement to $N$ that is isomorphic to ${\mathbf R}^\times$. This leads to a nontrivial semidirect product: $G \cong N \rtimes {\mathbf R}^\times$. The center of $G$ is the scalar diagonal matrices, and this is isomorphic to ${\mathbf R}^\times$. When $n$ is odd the center is complementary to $N$ in $G$, so $G \cong N \times {\mathbf R}^\times$ too. For odd $n > 1$ this shows the two conditions I gave are not necessary. $\endgroup$
    – KCd
    Commented Oct 16, 2013 at 14:13
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    $\begingroup$ A particular case of @KCd's comment is that if $\varphi_1$ and $\varphi_2$ are monomorphisms with the same image, then the corresponding semi-direct products are isomorphic. $\endgroup$
    – Prahlad Vaidyanathan
    Commented Jun 21, 2016 at 9:22

2 Answers 2

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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.

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  • $\begingroup$ You say a map between two semidirect products of $H$ and $K$ either sends $H$ to $H$ or maps $H$ onto $K$. Could you write more carefully what you intended? As written it is false. For instance, if $\phi$ and $\psi$ are both trivial, then you're saying a map (= homomorphism) from $H \times K$ to itself sends $H$ to $H$ or $H$ onto $K$. But if $H$ is any nontrivial group and $K = H \times H$, then the map $H \times K \rightarrow H \times K$ where $(x,(y,z)) \mapsto (y,(x,z))$ is an automorphism not sending $H$ to $H$ (as $H \times \{1\}$) or $H$ onto $K$ (as $\{1\} \times K$). $\endgroup$
    – KCd
    Commented Oct 16, 2013 at 13:54
  • $\begingroup$ Sorry, that was not clear. I have edited it. If $H$ and $K$ are such that there does not exist a map from $H$ onto $K$ then a map $H\rtimes_{\phi} K\rightarrow H\rtimes_{\psi}K$ must send $H$ to $H$. This mirrors what happens if $K\cong\mathbb{Z}$. $\endgroup$
    – user1729
    Commented Oct 16, 2013 at 13:58
  • $\begingroup$ But my counterexample still works: if $H$ is any nontrivial group and we set $K = H \times H$ then there usually isn't a surjective homomorphism from $H$ onto $K$ (e.g., if $H$ is finite) and the map I wrote down from $H \times K$ to itself does not send $H$ to $H$: inside this (trivial) semidirect product the image of $H$ is all $(1,(x,1))$, which is not $H$ inside $H \times K$. $\endgroup$
    – KCd
    Commented Oct 16, 2013 at 13:59
  • $\begingroup$ Hmm, yes, I see. I think it should be "...no non-trivial map from $H$ to a subgroup of $K$" (that was what I had written earlier, but thought it was superflous so it didn't make it to the final post). I will get my pen and paper out and check... $\endgroup$
    – user1729
    Commented Oct 16, 2013 at 14:04
  • $\begingroup$ Yup, no non-trivial maps works. You basically go $H\hookrightarrow H\rtimes_{\phi}K\rightarrow H\rtimes_{\psi}K\twoheadrightarrow K$ and looking at where $H$ is sent we see that in the isomorphism $H$ must be sent to a subgroup of $H$. By considering $H\hookrightarrow H\rtimes_{\psi}K\rightarrow H\rtimes_{\phi}K\twoheadrightarrow K$ using the inverse isomorphism you get that $H$ is mapped to $H$ under the isomorphism. $\endgroup$
    – user1729
    Commented Oct 16, 2013 at 14:09
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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.

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