# How similar/different are dense subgroups of a compact group.

Let $\Gamma, \Lambda\subset G$ be countably infinite subgroups of a common compact subgroup G. I am interested in properties that one would have to inherit from the other (ie if $\Gamma$ has this property then so does $\Lambda$.)

One initial thing that I thought was that you have a left-right action of $\Gamma\times\Lambda$ on $G$ with the Haar measure. Thus from this you might be able to induce unitary representation from $\Gamma$ to $\Lambda$ much like you would do with measure equivalence and hopefully from this follows some things that we know to be measure equivalence invariants (amenability, property T, etc.) Though I haven't actually verified that this works. Beyond this I can only think of abelian. I would be particularly interested in the case that G is the profinite completion of both $\Gamma$ and $\Lambda$.

Alternatively, are there examples where $\Gamma$ and $\Lambda$ can be quite different.

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There are groups with the same profinite completion, one of which has property (T) and the other does not. arxiv.org/abs/1005.4566 –  Ian Agol Jan 26 '13 at 4:37
Also one can be amenable and another one free non-Abelian. –  Mark Sapir Jan 26 '13 at 5:16
mmm...seems like my initial thoughts are totally wrong –  Owen Sizemore Jan 26 '13 at 5:56
A very simple example of two completely different dense subgroups of $\mathbb{T}$, the circle in $\mathbb{C}$ can be considered by letting $H_1$ be the subgroup of all roots of unity and $H_2$ be the cyclic infinite subgroup generated by $e^{2\pi i \lambda}$ where $\lambda$ is an irrational number. –  Vahid Shirbisheh Jan 26 '13 at 9:29
@Owen: You are confusing lattices in a common locally compact group (the setting where you can induct) and subgroups of a compact group (where you cannot induct). –  Misha Jan 26 '13 at 11:06

Corollary 7.6: Let $p$ be a prime number. Let $\Gamma$ and $\Lambda$ be finitely presented and residually p-finite groups. If $\Gamma$ and $\Lambda$ have isomorphic pro-p-completions, then there first $\ell^2$-Betti number of $\Gamma$ and $\Lambda$ coincides.
There is no assumption in the OP's question that when the compact group is pro-$p$, it is the pro-$p$ completion of the dense subgroup. –  YCor Jan 28 '13 at 18:13