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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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  • $\begingroup$ There are groups with the same profinite completion, one of which has property (T) and the other does not. arxiv.org/abs/1005.4566 $\endgroup$
    – Ian Agol
    Commented Jan 26, 2013 at 4:37
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    $\begingroup$ Also one can be amenable and another one free non-Abelian. $\endgroup$
    – user6976
    Commented Jan 26, 2013 at 5:16
  • $\begingroup$ mmm...seems like my initial thoughts are totally wrong $\endgroup$
    – Owen Sizemore
    Commented Jan 26, 2013 at 5:56
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    $\begingroup$ 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. $\endgroup$
    – user23860
    Commented Jan 26, 2013 at 9:29
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    $\begingroup$ @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). $\endgroup$
    – Misha
    Commented Jan 26, 2013 at 11:06

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Since you are interested in positive results (rather than counterexamples) in the case when the pro-finite completions of two groups agree, let me mention the following result from

Martin R. Bridson and Alan W. Reid, Nilpotent completions of groups, Grothendieck pairs, and four problems of Baumslag, http://people.maths.ox.ac.uk/bridson/papers/BReid12/

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.

The techniques in the proof (and hence the result) are more general.

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  • $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Jan 28, 2013 at 18:13

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