Distinct, non-homeomorphic, profinite topologies on a given abstract group ? 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.
 A: Yes. 
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. 
The results are presented here, but I am in the process of redrafting http://arxiv.org/abs/1101.3005
A: 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. 
