Example of $2$ non-isomorphic groups that have the same quotients 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
 A: 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
A: 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.
A: There are finitely presented infinite groups that have no nontrivial finite quotients.
You could take two of these.
There may be easier examples, though.
A: 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].
A: 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 $Z_2$.
