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, nonhomeomorphic, 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 wellversed as concerns large cardinals etc., could someone verify/elucidate this please ? Thank you in advance ! Stephan.

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


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. 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. You can read more about similar situation in Wilson's book on profinite groups on the chapter on free group. 

