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I'm looking for an example of $2$ non isomorphic groups $G_1,G_2$ that are finitely generated and presented that have the same finite quotients (up to isomorphism) .

Thanks

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  • $\begingroup$ nice question. I posted an answer to take a finite simple group cross a free group, but this is wrong. $\endgroup$
    – hunter
    Commented Mar 4, 2017 at 9:26
  • $\begingroup$ Almost the same question here. $\endgroup$ Commented Mar 4, 2017 at 12:04

5 Answers 5

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In my answer to this question, I wrote down finite presentations from a paper by Baumslag and Solitar of two non-isomorphic groups that are each isomorphic to a quotient group of the other. So these two groups have exactly the same quotient groups, not just the same finite quotient groups.

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There are finitely presented infinite groups that have no nontrivial finite quotients.

You could take two of these.

There may be easier examples, though.

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There is a famous example of a non-residually finite one-relator group, $$\langle a, b \mid (ab)^{(ab)^a} = (ab)^2\rangle,$$ due to Baumslag, Miller and Troeger, which has the same finite quotients (the finite cyclic groups) as the infinite cyclic group. It's a particular case of a more general result. This appears in the following paper (which I just happened to have handy).

Gilbert Baumslag, Charles F. Miller III and Douglas Troeger, Reflections on the residual finiteness of one-relator groups, Groups Geom. Dyn. 1 (2007), 209-219. PDF

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The Thompson groups $T < V$ are nonisomorphic, finitely presented, infinite simple groups---so the only finite quotient they have in common is the trivial group.

Cannon-Floyd-Parry's survey article gives a presentation for $T$ with 3 generators and 2 relations, but maybe someone else has knocked it down to 2 generators since then. The presentation they gave for $V$ was less friendly (4G, 14R). Bleak and Quick give a 2-generator, 7-relator presentation for $V$ in arXiv:1511.02123 [math.GR].

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Let $Z_k$ be the cyclic group of order $k$.

Consider, $A = Z_2 \times Z_2$ and $B = Z_4$.

These are not isomorphic to each other and the only non-trivial quotient they have is $Z_2$.

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    $\begingroup$ But $A$ does not have $B$ as a finite quotient and $B$ does not have $A$ so this does answer the question. $\endgroup$
    – Derek Holt
    Commented Mar 5, 2017 at 3:42

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