Timeline for Finiteness theorems for profinite groups
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 11, 2017 at 20:38 | history | edited | Hugo Chapdelaine | CC BY-SA 3.0 |
added 2 characters in body
|
| Jun 4, 2012 at 10:01 | vote | accept | Hugo Chapdelaine | ||
| Jun 3, 2012 at 17:59 | answer | added | Yiftach Barnea | timeline score: 3 | |
| Jun 3, 2012 at 15:04 | comment | added | Hugo Chapdelaine | @Yves, yes indeed you are right. So may be you (or Yiftach) could write the proof. Is there a natural generalization of this result? It seems to me that the notion of profinite $p$-group is really peculiar and does not seem to generalize easily... | |
| Jun 3, 2012 at 14:26 | comment | added | YCor | PS: by normal here I mean normal in $G$ | |
| Jun 3, 2012 at 14:25 | comment | added | YCor | @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). | |
| Jun 3, 2012 at 14:18 | comment | added | Hugo Chapdelaine | @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. | |
| Jun 3, 2012 at 9:06 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
edited title
|
| Jun 3, 2012 at 7:59 | comment | added | YCor | @Yiftach "finite type" means finitely generated (by a word-by-word translation from French). So you can put your comment as an answer. | |
| Jun 3, 2012 at 7:32 | comment | added | Yiftach Barnea | 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. | |
| Jun 3, 2012 at 3:04 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |