$\DeclareMathOperator{\Aut}{Aut}$Consider a group $G$ with a normal subgroup $N\triangleleft G$ and suppose that $G$ has a subgroup $H$ so that $G = HN$. Then $H$ acts on $N$; in other words, we have a homomorphism $\rho : H\to\Aut(N)$.

If $H\cap N = 1$ then it is well-known that $G$ can be written as a semidirect product $G\cong N\rtimes_\rho H$.

I am wondering about the case when $H\cap N\ne 1$. We can still consider the semidirect product $N\rtimes_\rho H$, and we get a map $N\rtimes_\rho H\to G$ given by $(n,h)\mapsto nh$. That this is a group homomorphism follows directly from the definition of $\rho$, since the commutation relation is the same both in $G$ and in $N\rtimes_\rho H$. This map is also surjective because $HN=G$. Thus $G$ is isomorphic to some quotient of $N\rtimes_\rho H$—in particular, the quotient of $N\rtimes_\rho H$ by $\{(n,n^{-1})\mid n\in N\cap H\}$, which is of course congruent to $N\cap H$.

If $G$ is finite and $H$ and $N$ are proper subgroups of $G$, then this seems pretty nice because it lets us understand $G$ in term of smaller groups. But for some reason I haven't been able to find a reference to this construction, except for in the specific case where $G = Q_8$. (Maybe I'm just looking in the wrong books.) So I'm wondering

  1. Have I made some error in the construction?
  2. Is this a special case of some more general theory?
  3. What is the extent of the usefulness of this sort of construction?

Any references would be very helpful.

  • $\begingroup$ Knowing that something is a quotient of a semidirect product is just a lot less information than knowing it's a semidirect product; in the most extreme case, we could take $H = G$ regardless of $N$. $\endgroup$ – Qiaochu Yuan Aug 3 '15 at 18:47
  • $\begingroup$ Indeed - but the amount of information scales inversely with the size of the intersection, no? $\endgroup$ – ajd Aug 3 '15 at 18:51
  • $\begingroup$ I don't think that there is a semi-product of $N$ and $H$ when $N \cap H \neq 1$ because, by definition, the homomorphism $G \rightarrow G/N$ doesn't induce an isomorphism $H \rightarrow G/N$. $\endgroup$ – GAVD Aug 3 '15 at 19:07

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