Surface groups and subgroups of fundamental groups The fundamental group of any closed surface is a surface group.  Let $S_3$ be the orientable surface of genus 3.
Is $\pi_1(S_3)$ isomorphic to an index-3 subgroup of any surface group?
We have 1 2-cell, 6 1-cells, and 1 0-cell.  Thus $\chi(S_3) = -4$.  
If we could show that there was necessarily of covering space $S_3 \to X$ with $\pi_1(X) = G$, $p_\ast(\pi_1(S_3))$ in index-3 subgroup of $G$, then we would know that this is not possible since $4$ is not divisible by $3$.
 A: Such a covering map always exists.  Indeed, suppose more generally that you have two closed surfaces $S$ and $T$ and a monomorphism $\pi_1(S)\to \pi_1(T)$ whose image $H$ is a subgroup of finite index.  Then there exists a finite-sheeted covering $T'\to T$ such that $\pi_1(T')=H$.  Furthermore, $T'$ is also a closed surface, since it is a finite-sheeted cover of a closed surface.  Since closed surfaces are determined up to homeomorphism by their fundamental group, we have $T'\cong S$.  Composing the cover $T'\to T$ with a homeomorphism $S\to T'$ now gives a covering map $p:S\to T$ such that $p_*(\pi_1(S))=H$.
A: In general (for closed orientable surfaces), we have
the short exact sequence of groups
$$1\to\pi_1(S_{k h+1})\to\pi_1(S_{h+1})\to K\to 1,$$
where $|K|=k$.
If you like, you can remember the shape from the more general formula
$$\pi_1(F)\to\pi_1(E)\to\pi_1(B)\to\pi_0(F)\to 1,$$
which is true for any fibration $F\to E\to B$
where $E$ and $B$ are connected.
This, in turn, is the last part of a long exact sequence.
The bit about $|K|=k$ can be derived from the Euler characteristic,
as you did.
So the answer to your question is yes: we're in the case $k=3$ and $h=2$
with the fibration $3\to S_7\to S_3$.
Your argument, however, can be used to show that there is no fibration $3\to S_3\to ???$.
It's very easy to get these things backwards, especially since the $\pi_1(B)\to\pi_0(F)$ is a boundary map.
