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Let $G , H$ be two finitely generated residually finite groups such that $F(G)=F(H)$. Where $F(G)$ denotes the isomorphism classes of finite quotients of $G$. Can we say that $G\cong H$?

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see this question: mathoverflow.net/questions/39973/… –  Ian Agol Mar 11 '12 at 9:52
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I notice that someone has voted to close this question as an exact duplicate. It seems to me that the question it is not a duplicate of the question that Agol links to, although several of the answers there are relevant. –  HJRW Mar 11 '12 at 11:31
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4 Answers

In fact (see Agol's link), $F(G)=F(H)$ if and only if the profinite completions $\widehat{G}$ and $\widehat{H}$ are isomorphic. I believe that there are non-isomorphic finitely generated virtually abelian (in particular, residually finite) groups with isomorphic profinite completions---see Mark Sapir's answer in Agol's link for some references.

Furthermore, Bridson and Grunewald answered a question of Grothendieck by constructing examples of pairs of finitely presented, residually finite groups $H\subseteq G$ such that $H$ is a proper subgroup of $G$ but the inclusion $H\to G$ induces an isomorphism of profinite completions $\widehat{H}\to\widehat{G}$.

Very recently, Bridson and I have used these kinds of constructions to prove that the isomorphism problem for profinite completions of finitely presented, residually finite groups is undecidable.

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As originally shown by Steve Humphries (J. of Algebra, 1988, I believe) there exist free finitely generated subgroups of $SL(n, \mathbb{Z})$ which surject under every quotient modulo $m.$ In fact it is true (but not yet published) that a random two-generator subgroup of $SL(n, \mathbb{Z})$ has the Humphries properties with probability bounded away from zero, for $n > 2.$ This gives a (large) family of counterexample to the OP's question.

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Is this a theorem of yours, Igor? –  HJRW Mar 12 '12 at 8:03
    
@HW: Yes, with Inna Capdeboscq... –  Igor Rivin Mar 12 '12 at 13:38
    
Nice! (And some more characters...) –  HJRW Mar 19 '12 at 9:15
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There are infinitely many metabelian groups with the same finite quotients, see Pickel, P. F. Metabelian groups with the same finite quotients. Bull. Austral. Math. Soc. 11 (1974), 115–120.

On the other hand, for many relatively free groups, including free metabelian groups, the genus (i.e. the number of groups with the same finite quotients) is finite, see Gupta; Noskov, G. A. On the genus of certain metabelian groups. Algebra Colloq. 5 (1998), no. 1, 49–66.

See also Grunewald, Fritz; Zalesskii, Pavel; Genus for groups. J. Algebra 326 (2011), 130–168.

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If $G,H$ are arithmetic groups, then Aka studies when $F(G)=F(H)$, see http://arxiv.org/abs/1107.4147.

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