# Semidirect product with subgroup

Let $G,H$ be groups with an action of $H$ on $G$, meaning a group homomorphism $H\to Aut(G)$, and let $S\le H$ be a subgroup, not necessarily normal.

Consider the semidirect products $G \rtimes H$ and $G \rtimes S$, formed respectively with the action above, and its restriction on $S$. Now the latter is a subgroup of the former. What can we say in general about the relationship between the two products, are there known results?

For example, consider the cosets of the inclusion $S\to H$ and the cosets of the inclusion $G\rtimes S\to G\rtimes H$. Are they isomorphic?

• When you write "an action" you mean, I presume, "an action by automorphisms", right? And... what "cosets" ? – DonAntonio Mar 27 '17 at 11:13
• "are the cosets isomorphic" does not mean anything. Please clarify. – Derek Holt Mar 27 '17 at 11:33
• Edited, I hope it's clearer now. – geodude Mar 27 '17 at 11:53
• You are still asking whether cosets are isomorphic, which doesn't mean anything. If $T$ is a set of coset representatives of $S$ in $H$, then $\{(1_G,t) : t \in T\}$ is a set of coset representatives of $G \rtimes S$ in $G \rtimes H$. – Derek Holt Mar 27 '17 at 12:24

As written in the comments, what "cosets isomorphic" means is not clear. However, there is a natural map $f$ from $H/S$ to $G\rtimes H/G\rtimes S$ defined by sending $h\cdot S$ to $(1,h)\cdot G\rtimes S$.
What you can check is that $f$ is one-to-one and onto. Furthermore for any $h,h'$ in $H$ :
$$f(h\cdot(h'\cdot S))=h\cdot ( h'\cdot G\rtimes S)\text{.}$$
In other words $f$ is an isomorphism if $H/S$ and $G\rtimes H/G\rtimes S$ are given the structure of $H$-set (i.e. sets with an action of $H$) and if you don't assume $S$ normal in $H$ this is the only isomorphism you can have.
• Yes, I meant exactly isomorphic as $H$-sets! – geodude Apr 11 '17 at 8:58