On the number of subgroups of semidirect product

Question

Let $G=H\rtimes K$ be the semidirect product of the normal subgroup $H$ and the subgroup $K$. Let $D$ be a subgroup of $G$ such that $K\le D$. Does $D= (D\cap H)\rtimes K$?.

I believe that is true, so under this resut, does the number of subgroups of $G$ containing $K$ is the number of normal subgroups of $H$.

Thank you all.

Answer 1

Suppose that $G = HK$ for some subgroups $H$ and $K$, where $H \cap K = 1$ and $H \trianglelefteq G$.

Here are steps for the solution of the exercise:

• If $D$ is a subgroup of $G$ such that $K \leq D$, show that $D = (H \cap D)K$.
• Conclude that any subgroup of $G$ containing $K$ is of the form $D = H_0K$ for some $H_0 \leq H$.
• Let $H_0 \leq H$. Show that $H_0K$ is a subgroup of $G$ if and only if $kH_0k^{-1} = H_0$ for all $k \in K$.

So the number of subgroups of $G$ containing $K$ is precisely the number of subgroups of $H$ that are normalized by $K$.

In general this might not have that much to do with the normal subgroups of $H$.

Just to give an example, consider $G = H \rtimes K$ for $H = S_n$ and $K = \langle \phi \rangle$, where $\phi$ is an inner automorphism of $H$ acting by conjugation with a transposition. There are not too many normal subgroups of $H$, but there are a quite a few subgroups of $H$ normalized by $\phi$.