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.
The Simple Loop Conjecture is as follows.
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.
|show 1 more comment|