 Solution

For the closed orientable surface ${\Sigma }_g$, use the fundamental group presentationSince $g\ge 1$, define a homomorphism to $\mathbb{Z}/n$ by sending $a_1$ to $1$ and every other displayed generator to zero. All commutators map to zero, so the relation is respected, and the map is surjective. Its kernel has index $n$. The classification of connected covering spaces supplies a connected $n$-sheeted covering space, including the identity cover when $n=1$.

Every connected finite covering space of ${\Sigma }_g$ is a closed orientable surface: compactness follows from finite degree, and the base orientation lifts through local covering charts. Lift a finite cell decomposition of the base. Every cell has $n$ lifts, so the Euler characteristic under a finite covering gives $\chi (\tilde{\Sigma })=n\chi ({\Sigma }_g)$. If the covering surface has genus $h$, this reads $2-2h=n(2-2g)$. ThereforeThis genus of a finite cover of a closed orientable surface depends only on the sheet number, not on the particular subgroup used to construct the cover.