Let $S$ be a closed orientable surface. For any positive integer $n$, is there a connected $n$-sheeted covering space of $S$? This is certainly not true if $S=S^2$ because $S^2$ is simply-connected. The result is true for $S=T^2$, since $\pi_1(T^2)=\Bbb Z^2$ has an index $n$ subgroup for each $n$. However I cannot handle the other cases. I know that this is equivalent to finding an index $n$ subgroup of $\pi_1(S)$ but the $\pi_1$ for the other surfaces are quite complicated, I think. Any hints?
2 Answers
Since the first homology is the abelianization of the fundamental group, and $H_1(S)=\mathbb{Z}^{2g}$, where $g$ is the genus, we see that we have a surjective composition $\pi_1(S) \rightarrow H_1(S) \rightarrow \mathbb{Z}$, where this last map is just projection onto the first coordinate.
Since taking the preimage of a subgroup $H \leq G$ under a surjective map preserves index, we deduce that $\pi_1(S)$ contains an index n subgroup given by the preimage of $n\mathbb{Z}$ under the above composite. As you mention, covering space theory implies we have the covering you ask for.