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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 $H^{\prime }\rtimes K^{\prime }$.

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

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

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  • $\begingroup$ OK thank you very much. But what we can say about this answer math.math.stackexchange.com/questions/1330088/…. Is it false. $\endgroup$ – Nourddine Snanou Oct 28 '18 at 17:43
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    $\begingroup$ 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 $K'\in K^g$. $\endgroup$ – C Monsour Oct 28 '18 at 21:43

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