Distinct, non-homeomorphic, profinite topologies on a given abstract group ? - MathOverflow 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 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 Answer by Yiftach Barnea for Distinct, non-homeomorphic, profinite topologies on a given abstract group ? 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 Answer by Jonathan Kiehlmann for Distinct, non-homeomorphic, profinite topologies on a given abstract group ? 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>