Subgroup of the semidirect product of two elementary abelian subgroups

Question

Let $H$ and $K$ be two abelian groups. It is well known that a subgroup of the semidirect product $H\rtimes K$ is not in general semidirect product of two subgroups $H'\le H$ and $K'\le K$. But if $G'$ is a subgroup of $H\rtimes K$ such that $K\le G'$, then $G'\simeq (G'\cap H)\rtimes K$ (see Subgroup). How can one use this for the following problem:

Let $G$ be a nonabelian $p$-group of order $p^{m+1}>p^{3}$ such that $G\simeq(\mathbb{Z}/p\mathbb{Z})^{m}\rtimes(\mathbb{Z}/p\mathbb{Z})$. Note that every subgroup of $G$ contains a subgroup of oder $p$.

• Is there a nonabelian subgroups of $G$ of the form $(\mathbb{Z}/p\mathbb{Z})^{k}\rtimes(\mathbb{Z}/p\mathbb{Z})$ with $1<k<m$ ?.

  • If the answer is yes, what is the number of nonabelian subgroups of index $p^{n}$ in $(\mathbb{Z}/p\mathbb{Z})^{m}\rtimes(\mathbb{Z}/p\mathbb{Z})$ (we can consider that $(\mathbb{Z}/p\mathbb{Z})^{m}\rtimes (\mathbb{Z}/p\mathbb{Z})$ is of maximal class).

Any help would be appreciated so much. Thank you all.

Answer 1

Yes, there are such examples.

gap> grps:=AllSmallGroups(3^4);;
gap> for g in grps do
> Print(StructureDescription(g),"\n");
> od;
C81
C9 x C9
(C9 x C3) : C3
C9 : C9
C27 x C3
C27 : C3
(C3 x C3 x C3) : C3
(C9 x C3) : C3
(C9 x C3) : C3
C3 . ((C3 x C3) : C3) = (C3 x C3) . (C3 x C3)
C9 x C3 x C3
C3 x ((C3 x C3) : C3)
C3 x (C9 : C3)
(C9 x C3) : C3
C3 x C3 x C3 x C3
gap> g:=grps[7];
(C3 x C3 x C3) : C3
gap> norm:=List(NormalSubgroups(g),x->StructureDescription(x));
[ "(C3 x C3 x C3) : C3", "C3 x C3 x C3", "C9 : C3", "C9 : C3", 
"(C3 x C3) : C3", "C3 x C3", "C3", "1" ]

Answer 2

Yes, there always exists such a subgroup. Let $g$ be a generator of the complement $K$ of order $p$ in the semidirect product, and let $h$ be an element in the normal subgroup $H$ that lies in $Z_2(G) \setminus Z(G)$. Then $\langle g,h \rangle = \langle [h,g],h \rangle \rtimes \langle g \rangle$ is nonabelian of order $p^3$.