Question

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 criterias for isomorphism of semidirect products?

Answer 1

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

$G = \langle x,y \mid x^{29}= y^{29}=z^7=1, xy=yx, x^z=x^7, y^z=y^{16} \rangle,$ 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!

Answer 2

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.