# If $H$ and $K$ are non-isomorphic but of same order, can there be $\varphi_1, \varphi_2$ such that $N\rtimes_{\varphi_1}H\cong N\rtimes_{\varphi_2}K$?

Let $$N$$, $$H$$ and $$K$$ be finite groups, such that $$|H| = |K|$$ and $$H$$ and $$K$$ being non-isomorphic. Can there exist $$\varphi_1$$ and $$\varphi_2$$ such that $$N \rtimes_{\varphi_1} H \cong N \rtimes_{\varphi_2} K$$?

Naming $$G_1 := N \rtimes_{\varphi_1} H$$ and $$G_2 := N \rtimes_{\varphi_1} K$$, I know that $$G_1$$ and $$G_2$$ respectively contain copies of $$H$$ and $$K$$. I think in some special cases it is possible that say if $$K \leqslant G_1$$, then $$G_1$$ would have had a greater order because of containing both $$H$$ and $$K$$. Also I see that whatever $$\varphi_1$$ and $$\varphi_2$$ could be, they cannot both be trivial homomorphisms since then the semidirect product is a direct product and it's clear that $$N \times H \not\cong N \times K$$.

But I wonder if in general we can prove that no such pair $$\varphi_1$$ and $$\varphi_2$$ exists for any arbitrary choice of $$N$$, $$H$$ and $$K$$.

• Well, you can easily get an example with infinite $N$: take $N$ to be a direct sum of countably many copies of $C_4$ and countably many copies of $C_2$; take $N=C_4$, and take $K=C_2\times C_2$, and take the actions to be trivial. So presumably you want to require $N$ to be finite? Commented Feb 7, 2022 at 16:34
• @ArturoMagidin Oh I see, the non-isomorphism for direct products fails. Didn't think of that, as I was working with finite groups. So yes let me require the groups to be finite, I'll edit the question. Commented Feb 7, 2022 at 16:46
• (I meant "take $H=C_4$ and $K=C_2\times C_2$", [$H$ rather than $N$] but you got the example...) Commented Feb 7, 2022 at 16:50

Let $$N = \mathbf{Z}/3 \mathbf{Z}$$, let $$A = \mathbf{Z}/2 \mathbf{Z}$$, and let $$B = \mathbf{Z}/3 \mathbf{Z}$$. If you take the non-trivial action of $$A$$ on $$N$$, you get

$$N \rtimes A = S_3.$$

Now take $$H = A \times B$$, where $$A$$ acts as above and $$B$$ acts trivially. Now clearly

$$N \rtimes H = S_3 \times \mathbf{Z}/3 \mathbf{Z}.$$

On the other hand, if $$K = S_3$$ acts trivially on $$N$$, then clearly

$$N \rtimes K = N \times K = \mathbf{Z}/3 \mathbf{Z} \times S_3 = N \rtimes H,$$

and $$K$$ is not isomorphic to $$H$$.

This is the simplest example of a more general construction. Let $$A$$ be a group, and let $$V$$ and $$W$$ be two representations of $$A$$ as vector spaces over $$\mathbf{F}_p$$ of the same dimension. That means that as abelian groups $$V \simeq W \simeq N$$ for some $$N$$. But now

$$V \rtimes (W \rtimes A) = (V \oplus W) \rtimes A = W \rtimes (V \rtimes A),$$

(here the action of $$W \rtimes A$$ on $$V$$ is the one factoring through the quotient to $$A$$ and similarly elsewhere) and now with $$N=V=W$$ and $$H = W \rtimes A$$ and $$K = V \rtimes A$$ one is done if $$H$$ and $$K$$ are not isomorphic. This will follow if $$W$$ and $$V$$ are "sufficiently different" representations of $$A$$. For example, if $$p$$ has order prime to $$|A|$$, then as representations they are not the same under the action of the automorphism group of $$A$$ (which could include outer automorphisms). The example above corresponds to the group $$A = \mathbf{Z}/2 \mathbf{Z}$$ and its two different representations over $$\mathbf{F}_3$$.