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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.

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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 $p^3$.

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  • $\begingroup$ That is better. $\endgroup$ – the_fox Jul 11 at 15:07
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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" ]
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    $\begingroup$ I have a suspicion that the OP wants to know whether there are proper nonabelian subgroups for all groups of the given form rather than for some. But of course the question is unclear! $\endgroup$ – Derek Holt Jul 11 at 14:49
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    $\begingroup$ Ah, maybe. I understood the question as asking for one triple $(p,m,k)$ with these properties. $\endgroup$ – the_fox Jul 11 at 14:54

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