most recent 30 from http://mathoverflow.net 2013-05-22T16:53:25Z http://mathoverflow.net/feeds/question/62758 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62758/distinct-non-homeomorphic-profinite-topologies-on-a-given-abstract-group Stephan F. Kroneck 2011-04-23T16:28:35Z 2011-04-24T13:47:53Z

<p>Just a silly little question which arose in connection with infinite Galois groups and their Krull topology:- can a given abstract group be endowed with distinct, non-homeomorphic, profinite topologies ? (I asked this question several years ago on the Topology Q+A and was told the question is undecidable and has something to do with supercompact cardinals). As I'm not that well-versed as concerns large cardinals etc., could someone verify/elucidate this please ? Thank you in advance ! Stephan.</p>

http://mathoverflow.net/questions/62758/distinct-non-homeomorphic-profinite-topologies-on-a-given-abstract-group/62821#62821 Yiftach Barnea 2011-04-24T09:19:08Z 2011-04-24T11:15:44Z

<p>As Agol said in a finitely generated profinite group every subgroup of finite index is open. Therefore, the topology is unique and detremined by the algebra. This was first proved by Serre for pro-$p$ groups and eventually Nikolov and Segal proved it for any profinite groups.</p> <p>Now, take $\mathbb{F}_p[[t]]$ formal power series over a field of $p$-elements and take their its abelian group. Then it is a metric pro-$p$ group which is the same as being countably based at $1$. On the other hand, take a vector space over $\mathbb{F}_p$ of a countable dimension and take its pro-$p$ complition. I am almost sure (so you might want to check the details) that in both cases you have a vector space of dimesnion $2^{\aleph_0}$ so the groups are isomorphic abstractly. But in the first case the topology is countably based at $1$ (and therefore in any points) while it is not countably based at $1$ in the second case. </p> <p>You can read more about similar situation in Wilson's book on profinite groups on the chapter on free group. </p>

http://mathoverflow.net/questions/62758/distinct-non-homeomorphic-profinite-topologies-on-a-given-abstract-group/62841#62841 Jonathan Kiehlmann 2011-04-24T13:47:53Z 2011-04-24T13:47:53Z

<p>Yes. </p> <p>I have classified some abelian examples: there are uncountably many pairwise non-homeomorphic pro-$p$ topologies that can be placed on the (unrestricted) product of any countable collection of cyclic $p$-groups of unbounded exponent. </p> <p>The results are presented here, but I am in the process of redrafting <a href="http://arxiv.org/abs/1101.3005" rel="nofollow">http://arxiv.org/abs/1101.3005</a></p>