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QUESTION: Let $g \geq 4$, $S(g)$ be the fundamental group of the genus $g$ surface, and $G$ be finitely generated (the number of generators $\leq 3$) group with abelianization of rank less than equal $2$. Assume that there exist a surjection $\phi: S(g) \rightarrow G$. Is it true that the kernel of $\phi$ contains at least one non separating loop of the surface? If it is any helpful, you can assume $G$ is a perfect group.

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The Simple Loop Conjecture is as follows.

Every non-injective map from a surface group to a 3-manifold group kills a simple closed curve.

As there are all sorts of 3-manifolds with abelianisation of rank two, I think the answer to your question is unknown.

UPDATE: Sorry, I wrote the above too hastily. I should have said 'I think that the kernel is not known to contain such a loop.' On the other hand, there may well be examples of such maps with no simple loops in the kernel. You could try looking at Louder's recent preprint on the Simple loop conjecture for limit groups, for instance.

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Thanks! Are there cases of $3$-manifolds (with infinite $\pi_1)$) for which we know Simple Loop Conjecture holds? Would it be helpful if we assume that the group $G$ has three (or less) generators and abelianization of rank 0. – Thom Aug 22 '11 at 14:36
@Thom - The simple loop conjecture holds for three-manifolds of the form surface cross interval, by work of Gabai. – Sam Nead Aug 22 '11 at 16:06
A quick search of MathSciNet shows that the Simple Loop Conjecture holds for Seifert-fibred 3-manifolds and, more generally, graph manifolds. As there are very complicated hyperbolic 3-manifolds with trivial H_1 and two generators, I would be surprised if those assumptions help much. – HJRW Aug 22 '11 at 19:03
Thanks Henry! I am convinced that the answer to my question is unknown. – Thom Aug 23 '11 at 15:02
@Henry: The question in fact has nothing to do with 3-manifolds. What one needs is to kill long and complicated self-intersecting loops and at the same time kill the first homology by imposing relations of the form $x=$ long commutator where $x$ is a standard generator of the surface group. The goal is to produce a quotient with no simple loops in the kernel. I think it is quite doable and the example should exist. I am not sure now about the number of generators ($\le 3$) of the quotient, but that also should be possible to achieve with some effort. – Mark Sapir Jul 24 '12 at 15:00

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