Let $N,H$ be groups, $\phi \colon H\rightarrow Aut(N)$ be a homomorphism, $f\in Aut(N)$ and $\hat{f}$ be an inner automorphism of $Aut(N)$ induced by $f$. Then $$N\rtimes_{\phi} H \cong N\rtimes_{\hat{f}\circ\phi}H.$$ Also, if $\psi \in Aut(H)$, then $$N\rtimes_{\phi}H\cong N\rtimes_{\psi\circ\phi}H.$$

What are sufficient conditions for isomorphism of semidirect products? Are there any other criteria for isomorphism of semidirect products?

  • $\begingroup$ This post is relevant: mathoverflow.net/questions/52812/semidirect-products $\endgroup$ – Zev Chonoles Feb 5 '11 at 4:18
  • $\begingroup$ uo to some extent! But, I would like to ask necessary and sufficient conditions for isomorphism of semidirect products. Obviously, every homomorphism from H to Aut(N) will give a semidirect product, but using these two facts, we can consider less homomorphimsms to construct all possible semidirect products. Even then, some semidirect products become isomorphic. It happens in construction of small p-groups. $\endgroup$ – Marshall Feb 5 '11 at 4:35
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    $\begingroup$ The word criterias does not exist. The plural of criterion is criteria. $\endgroup$ – Chandan Singh Dalawat Feb 5 '11 at 6:29
  • $\begingroup$ What is the inner automorphism induced by an arbitrary automorphism? $\endgroup$ – Alex B. Feb 5 '11 at 9:03
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    $\begingroup$ Presumably $\hat{f} : \mathrm{Aut}(N) \rightarrow \mathrm{Aut}(N)$ sends $g$ to $f^{-1} \circ g \circ f$. $\endgroup$ – Siksek Feb 5 '11 at 9:29

This is not an answer to your question, but there is another way that an isomorphism can arise. It is possible for two semidirect products $N_1 \rtimes H_1$ and $N_2 \rtimes H_2$ (with $N_1 \cong N_2$, $H_1 \cong H_2$) to be isomorphic as groups, but for there to be no isomorphism that maps $N_1$ to $N_2$. An example of this is the group

$G = \langle x,y,z \mid x^{29}= y^{29}=z^7=1, xy=yx, x^z=x^7, y^z=y^{16} \rangle,$ which is an extension of $C_{29} \times C_{29}$ by $C_7$. Let $N_1$ and $N_2$ be the normal subgroups of order 29 generated by $x$ and $y$. Then $G/N_1 \cong G/N_2$ is the unique nonabelian group of order $29 \times 7$, but there is no automorphism of $G$ that maps $N_1$ to $N_2$, so this group can be expressed as a semidirect product of $C_{29}$ by the nonabelian group of order $29 \times 7$ in two different ways.

You might prefer ot restrict your attention to isomorphisms between $N_1 \cong N_2$ and $H_1 \cong H_2$ that map $N_1$ to $N_2$. If you do that, then I don't know the answer to your question in general, but I believe that in the special case when $N$ is an elementary abelian $p$-group, all isomorphisms arise in the ways you have described in your post. I don't feel like trying to write down a proof right now!

  • $\begingroup$ In your presentation of the group $G$, shouldn't $z$ be one of the generators? $\endgroup$ – Andrew James Kelley Oct 5 '19 at 0:33
  • $\begingroup$ Yes thanks, I have corrected it. $\endgroup$ – Derek Holt Oct 5 '19 at 7:14

Aside from the two ways you've mentioned, I think there are just two other ways of getting an isomorphism (i.e. any isomorphism of semidirect products with same $N$ and $H$ is some combination of all four):

  • Change $H$ to a different complement. This corresponds to multiplying $\phi$ pointwise by inner automorphisms of $N$. These are classified by (non-abelian) 1-cocycles $\mathrm{Z}^1(H,N)$, or since you've already incorporated global conjugations in $N$, you can use the non-abelian cohomology $\mathrm{H}^1(H,N)$. This coincides with the usual cohomology if $N$ is abelian. Note that this doesn't affect the induced map $\tilde{\phi}: H \to \mathrm{Out}(N)$.

  • Change $N$ to a different normal subgroup of the total group which happens to be isomorphic (and have isomorphic complement.) Edit: This case is hard to deal with in general, but one situation it can be avoided is if you only consider semidirect products in which $N$ plays some special role, like being the derived subgroup. Derek elaborates more on this case in his answer.


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