Timeline for Finiteness theorems for profinite groups
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 11, 2017 at 21:25 | history | edited | Mikhail Borovoi | CC BY-SA 3.0 |
LaTeX and spelling corrected
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| Jun 4, 2012 at 16:04 | comment | added | Yiftach Barnea | @Yves: I formulated a question mathoverflow.net/questions/98778/… | |
| Jun 4, 2012 at 14:59 | comment | added | Yiftach Barnea | @Yves: it is nor clear that you could pass to a finite quotient. For that $N$ must have an open subgroup which is invariant under the $K$-action. Why would there always be such a subgroup? | |
| Jun 4, 2012 at 12:52 | comment | added | YCor | @Yiftach: it seems that passing to a finite quotient of the pro-$p$ group $N\rtimes K$ you get a contradiction. | |
| Jun 4, 2012 at 11:34 | comment | added | Yiftach Barnea | @Yves: if $N$ is finitely generated, then of cousre the whole question is trivial. So my example above is trivial. However, the following question is worth thinking of. Is there an example of $N$ and $K$ pro-$p$ groups such that $K$ acts on $N$ and $N=[N,K]$. As I said above $N$ cannot be finitely generated. So the first interesteing case is $N$ an infinite dimension vector space over $\mathbb{F}_p$. | |
| Jun 4, 2012 at 11:10 | comment | added | YCor | @Yiftach: I removed my last comment: I thought at $N$ as pro-$p$, not $G$. The initial question is whether $G$ is finitely generated: if $G$ is pro-$p$, solving this is not a problem. | |
| Jun 4, 2012 at 10:01 | vote | accept | Hugo Chapdelaine | ||
| Jun 4, 2012 at 9:24 | comment | added | Yiftach Barnea | @Yves: I should add at least if $N$ is reasonably nice, i.e. it has a base for the topology at the identity made of normal subgroups invariant under the $K$ action, for example $N$ is f.g. Otherewise, I will have to think about it. | |
| Jun 4, 2012 at 9:18 | comment | added | Yiftach Barnea | @Yves, what do you mean by cyclic of 2 elements? You cannot have $N=[N,K]$ if both $N$ and $K$ are pro-$p$, the same as for $p$-groups. | |
| Jun 3, 2012 at 23:13 | comment | added | Yiftach Barnea | @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. | |
| Jun 3, 2012 at 22:49 | comment | added | YCor | @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 $x \mapsto -x$. | |
| Jun 3, 2012 at 22:04 | history | edited | Yiftach Barnea | CC BY-SA 3.0 |
added 27 characters in body
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| Jun 3, 2012 at 22:02 | comment | added | Yiftach Barnea | 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. | |
| Jun 3, 2012 at 19:44 | comment | added | YCor | at the end you mean $N/([N,N]N^p)$, not $N/([N,K]N^p)$. | |
| Jun 3, 2012 at 18:01 | comment | added | Yiftach Barnea | I had to break the first paragraph in a strange place since the latx got crazy. | |
| Jun 3, 2012 at 17:59 | history | answered | Yiftach Barnea | CC BY-SA 3.0 |