# Subgroups of semi-direct products of two elementary abelian subgroups. [closed]

First question: Let $$H'$$ be a subgroup of $$H$$ and $$K'$$ a subgroup of $$K$$. Is it true that $$H'\rtimes K'$$ and $$H'\times K'$$ are subgroup of $$H\rtimes K$$?

Second question: Let $$G=(\mathbb{Z}/p\mathbb{Z})^{n}\rtimes (\mathbb{Z}/p\mathbb{Z})^{m}$$. What is the number of subgroups of $$G$$ isomorphic to $$(\mathbb{Z}/p\mathbb{Z})^{2}$$?

Third question: Is it true that if $$\gcd (|H|,|K|)=1$$ then all subgroups of $$H\rtimes K$$ are of the form $$H^{\prime }\rtimes K^{\prime }$$.

## closed as off-topic by Derek Holt, Scientifica, Namaste, José Carlos Santos, Arnaud D.Oct 26 '18 at 13:57

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• For the first, I think you need that $K'$ is invariant under the action of $H'$. But I might be wrong on that. – Cameron Williams Oct 24 '18 at 23:55
• Edited my answer to answer your new third question also. – C Monsour Oct 25 '18 at 12:59

## 1 Answer

First question: If it exists, it's a subgroup, but since conjugation by elements of $$K^{\prime}$$ may take elements of $$H^{\prime}$$ to elements of $$H\setminus H^{\prime}$$, it need not exist.

Second question: That depends on the action involved in the semidirect product. For example, consider $$2^2\rtimes 2$$. If the action is trivial, we get $$2^3$$ which has seven subgroups isomorphic to $$2^2$$. If the action is non-trivial, we get $$D_4$$, which has only two such subgroups.

(Note that $$p^n$$ is traditional shorthand for $$(\mathbb{Z}/p\mathbb{Z})^{n}$$.)

Third question: No, since at the very least, if the action is non-trivial, $$K$$ has conjugates that intersect $$H$$ trivially and are not contained in $$K$$.

• OK thank you very much. But what we can say about this answer math.math.stackexchange.com/questions/1330088/…. Is it false. – Nourddine Snanou Oct 28 '18 at 17:43
• It's fine, but it doesn't say that subgroups are of the form $H'\rtimes K'$ for $H'\le H$ and $K'\le K$. It says there is some $g\in G$ (in fact we can take $g\in H$) such that a subgroup is of the form $H'\rtimes K'$ for $H\in H$ and $K'\in K^g$. – C Monsour Oct 28 '18 at 21:43