# 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

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

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.

You could take two of these.

There may be easier examples, though.

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

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].

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 $Z_2$.

• But $A$ does not have $B$ as a finite quotient and $B$ does not have $A$ so this does answer the question. Commented Mar 5, 2017 at 3:42