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Assume that we have two residually finite groups $G$ and $H$. Which properties of $G$ and $H$ could be used to show that their pro-finite (or pro-p) completions are different?

I asked a while ago in the group-pub mailing list whether finite presentability is such a property but Lubotzky pointed out that it is not the case. A finitely presented and an infinitely presented group can have isomorphic pro-finite completions.

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Interestingly, property T is now known NOT to be such a property: see this preprint of Aka. – JSE Sep 27 '10 at 3:51
up vote 5 down vote accepted

There's a theorem that two finitely generated residually finite groups have the same profinite completions if and only if they have the same finite quotients. A reference for the statement of this is Theorem 2 of this paper, but they cite Ribes and Zalesskii for the proof.

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@Ian: Right. Somehow I forgot about it. – Mark Sapir Sep 26 '10 at 4:29

It is known that in a topologically finitely-generated profinite group, every subgroup of finite index is open. (See this paper.)

If $G$ is a finitely-generated residually finite group, then the profinite completion $\hat{G}$ contains $G$ as a dense subgroup, and is therefore topologically finitely generated. If $F$ is a finite-index subgroup of $\hat{G}$, it follows that $F$ is open, so every coset of $F$ contains elements from $G$, and therefore $F \cap G$ is a finite-index subgroup of $G$ with the same index. This defines an isomorphism between the the lattice of finite-index subgroups of $\hat{G}$ and the lattice of finite-index subgroups of $G$.

As long as $G$ and $H$ are finitely generated, this gives a rather strong invariant that can be used to distinguish $\hat{G}$ from $\hat{H}$. Specifically, $\hat{G}$ and $\hat{H}$ can only be isomorphic if the lattices of finite-index subgroups of $G$ and $H$ are isomorphic. Moreover, the isomorphism between these lattices must preserve the permutation action on the cosets of each finitely-generated subgroup. In particular, the lattices of finite-index normal subgroups of $G$ and $H$ must also be isomorphic, in a way that preserves the isomorphism types of the finite quotient groups.

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Does this narrow things down sufficiently that for each finitely generated residually finite group $G$, there are only countably many finitely generated residually finite groups $H$ with an isomorphic profinite completion? Or is this obviously false? – Simon Thomas Sep 25 '10 at 22:47
@Simon: See Pyber, László Groups of intermediate subgroup growth and a problem of Grothendieck. Duke Math. J. 121 (2004), no. 1, 169--188. – Mark Sapir Sep 26 '10 at 0:02
@Mark: Many thanks ... that was the answer that I was hoping for! – Simon Thomas Sep 26 '10 at 1:51

For polycyclic groups, a result of Grunewald, Pickel and Segal (Polycyclic groups with isomorphic finite quotients. Ann. of Math. (2) 111 (1980), no. 1, 155--195.) says that the class of f.g. polycyclic groups with the same profinite completion (and even with the same sets of finite homomorphic images) is finite. Thus at least for polycyclic groups one can say whether the profinite completions are different just by looking at the groups. Already for metabelian groups the situation is quite different (see Pickel, P. F. Metabelian groups with the same finite quotients. Bull. Austral. Math. Soc. 11 (1974), 115--120. ). There are also some results about free solvable groups. In general, I do not think this (very interesting) question has been studied enough.

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I studied this question for arithmetic group is high-rank. The first two articles in are related. – Menny Apr 29 '12 at 12:13

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