A Karrass-Solitar theorem for surface groups Let $\Gamma_g$ be a surface group of genus $g \geq 2$. That is, there is a presentation $$\Gamma_g = \langle x_1, y_1, \dots, x_g, y_g \vert \prod_{i = 1}^{g}[x_i,y_i] = 1\rangle$$
Is there a nontrivial, finitely generated $N \lhd \Gamma_g$ of infinite index?
More generally, Can there be a finitely generated $K \leq \Gamma_g$ of infinite index, containing a nontrivial $N \lhd \Gamma_g$?
 A: The answer to both questions is 'no'. This was proved by Greenberg for Fuchsian groups. One outline of the proof is as follows.


*

*Any finitely generated subgroup $H$ of a surface group $\Gamma$ is quasiconvex.  So the map $H\to\Gamma$ induces an injection of Gromov boundaries $\partial H\to\partial\Gamma\cong S^1$.

*If $K\lhd \Gamma$ is non-trivial then the only closed, $K$-invariant subset of $\partial\Gamma$ is the whole of $\partial\Gamma$.

*Therefore $\partial H=\partial\Gamma$.

*Therefore $H$ is of finite index in $\Gamma$.
As you can see, this works for any quasiconvex subgroup of any hyperbolic group. This was proved by Gersten--Short.
A: If $N$ has infinite index in a surface group $S$ then it is free.
(This can be seen topologically, or by invoking a theorem of Strebel on subgroups of Poincare duality groups.)
Now $H^2(S;Z[S])\cong H^1(S/N;H^1(N;Z[S])$, by the LHS spectral sequence,
since $H^p(N;Z[S])=0$ for $p\not=1$.
If $N$ is finitely generated then this is in turn 
$H^1(S/N;Z[S/N)\otimes{H^1(N;Z[N])}$.
But $H^2S;Z[S])$ is infinite cyclic, since $S$ is a surface group.
Therefore $S/N$ and $N$ each have one end, i.e.,
are virtually infinite cyclic, and so $S$ is virtually $Z^2$.
This contradicts genus $\geq2$.
