Timeline for A profinite group which is not its own profinite completion?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 30, 2011 at 12:26 | comment | added | Jonathan Kiehlmann | @Benjamin, it is standard when talking about profinite groups. However this is not obvious, and so it is worth mentioning for the benefit of the readers who have not read a textbook on them. | |
| Nov 30, 2011 at 3:33 | comment | added | Benjamin Steinberg | @Jonathan, I thought it was standard that a free profinite group of countable rank means a countable set of generators converging to 1. At least this is the convention used by Ribes and Zalesskii. | |
| Nov 30, 2011 at 3:22 | comment | added | Jonathan Kiehlmann | Benjamin - you need to make sure that your generators tend to the trivial element, otherwise a free profinite group on countable generators isn't free or countably-based. One of those breaks. | |
| Nov 29, 2011 at 23:24 | comment | added | Benjamin Steinberg | Also a free profinite group of countable rank is an example since it maps onto this one and so has uncountably many finite image but it is still countably based. | |
| Nov 29, 2011 at 20:52 | comment | added | Ricky | This is exactly what I mean. | |
| Nov 29, 2011 at 18:22 | comment | added | Colin Reid | More generally, every pro-$p$ group $G$ that is not topologically finitely generated maps onto the one Jonny describes, so $G$ has a non-open subgroup of index $p$. | |
| Nov 29, 2011 at 18:13 | comment | added | Jonathan Kiehlmann | It does mean uncountably many. The easiest way to see this is to note that it maps onto the product of countably many copies of $C_{p}$, which is of course isomorphic to a $2^{\aleph_{0}}$-dimensional $F_{p}$ space. (This is itself a pro-$p$ group with the properties you want to see.) I have a paper on the arxiv all about these sorts of groups. The new version of it should be up tomorrow morning. I will provide the link then. | |
| Nov 29, 2011 at 15:03 | comment | added | Giuseppe | Thank you for your answer. But why does $G$ have many more subgroups of finite index than open subgroups? Does 'many more' here mean uncountably many? | |
| Nov 29, 2011 at 14:14 | history | answered | Ricky | CC BY-SA 3.0 |