Let $G$ be a profinite group which fits into the following short exact sequence: $$ 1\rightarrow N\rightarrow G \rightarrow K\rightarrow 1 $$ Assume that $N$ is a pro-$p$ group and that $K$ is topologically finitely generated. Note that $K$ acts naturally by conjugation on $N^{ab}$ and thus we way view $N^{ab}$ as a $\mathbf{Z}_p[[K]]$-module.

Q: If $N^{ab}$ is a finite type $\mathbf{Z}_p[[K]]$-module, does it follow that $G$ is topologically finitely generated?

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    $\begingroup$ Could you please explain what finite type means? Generally, $N$ is generated as a normal subgroup of $G$ by a set $X$ if and only if $N/([N,G]N^p)$ is generated by the image of $X$. So if $N^{ab}$ is a finitely generated $\mathbb{Z}_p[[K]]$-module, then the answer to your question should be true. $\endgroup$ – Yiftach Barnea Jun 3 '12 at 7:32
  • $\begingroup$ @Yiftach "finite type" means finitely generated (by a word-by-word translation from French). So you can put your comment as an answer. $\endgroup$ – YCor Jun 3 '12 at 7:59
  • $\begingroup$ @Yiftach, I think I understand how to do it when the groups are finite. So let $K=<x_1,\ldots,x_r>$ and let $\{n_1,\ldots,n_s\}$ be the generators of the $\mathbf{Z}_p[[K]]$-module. Then by what you said one has that the normal closure of $<x_i*n_j: i,j>$ which I denote by $NC<x_i*n_j: i,j>$ generated $N$. Now if $<x_i*n_j: i,j>$ was contained in a maximal subgroup $M$ of $N$ then because $N$ is a $p$-group we know that $M$ is normal and thus we would have $NC<x_i*n_j: i,j>\subseteq M$ which is a contradiction. $\endgroup$ – Hugo Chapdelaine Jun 3 '12 at 14:18
  • $\begingroup$ @Hugo: the profinite case follows from the finite case (a subgroup is dense iff all its images in finite quotients are surjective; apply this to the normal subgroup generated by a finite subset topologically generating $N$ modulo the commutators). $\endgroup$ – YCor Jun 3 '12 at 14:25
  • $\begingroup$ PS: by normal here I mean normal in $G$ $\endgroup$ – YCor Jun 3 '12 at 14:26

So here is my comment again with slightly more details. Let $Y$ be a finite subset of $G$ such that its image generates $K$. As I was told by Yves, finite type means finitely generated. Thus, let $X$ be a finite subset of $N$ such that $N^{ab}$ is generated by its image as a $\mathbb{Z}_p[[K]]$-module. Then we claim that $H$ the subgroup generated (topologically) by $X \cup Y$ is equal $G$. Since $Y \subseteq H$ we have that $HN/N \cong K$.

Therefore, $HN=G$. So it suffices to show that $N \leq H$. By definition, the action of $K$ on $N^{ab}$ is via the action of $G$. Since $N$ acts trivially on $N^{ab}$ we have that $H$ action on $N^{ab}$ is the same as $HN=G$ action. Hence, the image of $Y^{H}$ generates $N^{ab}$ as a $\mathbb{Z_p}$-module. So $Y^{H}$ generates $N/\Phi(N)$ as a group, where $\Phi(N)=[N,N]N^{p}$ is the Frattini subgroup of $N$. We deduce that $Y^{H}$ generates $N$ (topologically) and $N \leq H$ as we wanted.

Note that it suffices to ask that $N/([N,N]N^{p})$ (EDIT: see comments below) is finitely generated as a $K$-module.

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  • $\begingroup$ I had to break the first paragraph in a strange place since the latx got crazy. $\endgroup$ – Yiftach Barnea Jun 3 '12 at 18:01
  • $\begingroup$ at the end you mean $N/([N,N]N^p)$, not $N/([N,K]N^p)$. $\endgroup$ – YCor Jun 3 '12 at 19:44
  • $\begingroup$ Yves, you are right. I was trying to be too clever, $[N,K]$ does not necessarily has a meaning as $K$ does not necessarily act on $N$. I think it is possible to replace it by $[N,G]$, but I have to check it to be sure. So I'll leave this to the reader and I will fix my mistake. $\endgroup$ – Yiftach Barnea Jun 3 '12 at 22:02
  • $\begingroup$ @Yiftach: no you can't, even for a split extension. We can have $N=[N,K]$ then, but $N$ can be huge. For instance, pick $K$ to be cyclic on 2 elements, and $N$ an infinite direct power of a fixed cyclic group of odd order, $K$ acting by $x \mapsto -x$. $\endgroup$ – YCor Jun 3 '12 at 22:49
  • $\begingroup$ @Yves: again you are right. I am more confident that it is true if $G$ is pro-$p$. But it is too late to think about it now. $\endgroup$ – Yiftach Barnea Jun 3 '12 at 23:13

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