Kernel of an epimorphism form a surface group to a free group.

Question

Let $S_g = \langle a_1,b_1,\ldots,a_g,b_g \mid [a_1,b_1]\cdots [a_g,b_g] = 1 \rangle$ (i.e. the fundamental group of an orientable closed surface of genus $g$). Let $F_g$ be a free group of genus $g$. Given an epimorphism $\phi : S_g \to H_g$ what can we say about the kernel? Is it free?

EDIT: OK so since subgroups of surface groups are always either finite index and isomorphic to surface groups, or infinite index and isomorphic to free groups, the kernel must be free. For the map $S_g \to F_g$ where the $a_i$ map to the generators of $F_g$ and the $b_i$ map to the identity the kernel is the subgroup generate by the $b_i$ which is free of rank $g$. Will the kernel always be a free group of rank $g$?

Answer 1

If you just want a reference, here is one:

Martin R. Bridson and James Howie, Normalizers in Limit Groups, Math. Ann. 337 (2007), no. 2, 385–394.

They prove that if $G$ is a limit group and $H$ is a nontrivial finitely generated subgroup then either the normalizer of $H$ in $G$ is abelian or $H$ has finite index in its normalizer. Thus, if $H$ is normal finitely generated subgroup in $G$ then either $H$ is of finite index in $G$ or $H$ is trivial or $G$ is abelian.

I will not give the definition of limit groups, but, according to the paper cited above: Examples of limit groups include all finitely generated free or free abelian groups, and the fundamental groups of all closed surfaces of Euler characteristic at most $−2$. The free product of finitely many limit groups is again a limit group, which leads to further examples.

Corollary. If $G$ is isomorphic to the fundamental group of a surface of genus $\ge 2$ then every finitely generated normal subgroup of $G$ is either trivial or has finite index in $G$.

Here is a hyperbolic geometry proof of this corollary, my main reference is

S. Katok, Fuchsian Groups, Chicago Lectures in Mathematics, University of Chicago Press, 1992.

First of all, if $S$ is a closed surface of genus $\ge 2$ then $S$ is homeomorphic to the quotient of the hyperbolic plane ${\mathbb H}^2$ by a discrete torsion-free subgroup $G< PSL(2, {\mathbb R})$ acting on the upper half plane ${\mathbb H}^2$ isometrically with respect to the hyperbolic metric. Groups $G$ as above are called Fuchsian groups (no requirement on compactness of the quotient space). (In general Fuchsian group are allowed to contain elements of finite order, but I will limit myself to torsion free groups only.) Fuchsian groups (rather unimaginatively) are divided in two classes: Groups of the 1st kind and groups of the 2nd kind.

Here are some facts about Fuchsian groups $G$:

1. If $Area({\mathbb H}^2/G)$ is finite then $G$ of the 1st kind.

2. If $Area({\mathbb H}^2/G)<\infty$ and $H$ is a subgroup of $G$ then $[G:H]$ equals the ratio of areas: $$ Area({\mathbb H}^2/H)/ Area({\mathbb H}^2/G). $$ This follows from the fact that area of a surface is multiplicative with respect to degree $d$ coverings ($d$ could be infinite).

3. A nontrivial normal subgroup in a Fuchsian group of the 1st kind is again of the 1st kind, see this MSE question.

4. Each finitely generated Fuchsian group is geometrically finite, i.e. has a finitely sided fundamental polygon.

5. If $H$ is a geometrically finite Fuchsian group of the 1st kind then $Area({\mathbb H}^2/H)<\infty$.

Using these 5 properties, here is a proof of the corollary:

Suppose that $G$ is a Fuchsian group such that ${\mathbb H}^2/G$ is compact. Hence, the area of the quotient is finite, hence, $G$ is of the 1st kind (Property 1). Suppose that $H< G$ is a nontrivial normal subgroup of infinite index. Hence, $Area({\mathbb H}^2/H)=\infty$ (Property 2). Moreover, $H$ is of the 1st kind (Property 3). If $H$ were, in addition, finitely generated, it would be a geometrically finite subgroup of the 1st kind (Property 4), which implies that $Area({\mathbb H}^2/H)< \infty)$ (Property 5). A contradiction.