# Finiteness theorems for profinite groups

Let $G$ be a profinite group which fits in 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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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. –  Yiftach Barnea Jun 3 '12 at 7:32
@Yiftach "finite type" means finitely generated (by a word-by-word translation from French). So you can put your comment as an answer. –  YCor Jun 3 '12 at 7:59
@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. –  Hugo Chapdelaine Jun 3 '12 at 14:18
@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). –  YCor Jun 3 '12 at 14:25
PS: by normal here I mean normal in $G$ –  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$as we wanted. Note that it sufices to ask that$N/([N,N]N^{p})$(EDIT: see comments below) is finitely generatd as a$K$-module. - I had to break the first paragraph in a strange place since the latx got crazy. – Yiftach Barnea Jun 3 '12 at 18:01 at the end you mean$N/([N,N]N^p)$, not$N/([N,K]N^p)$. – YCor Jun 3 '12 at 19:44 Yves, you are right. I was trying to be too clever,$[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]$, but I have to check it to be sure. So I'll leave this to the reader and I will fix my mistake. – Yiftach Barnea Jun 3 '12 at 22:02 @Yiftach: no you can't, even for a split extension. We can have$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$. – YCor Jun 3 '12 at 22:49 @Yves: again you are right. I am more confident that it is true if$G$is pro-$p\$. But it is too late to think about it now. –  Yiftach Barnea Jun 3 '12 at 23:13