For any surface $\sum_g$ and any cyclic group $C_n$ we can build a surjection $\phi:\pi_1(\sum_g,x_0)\to C_n$ by building a corresponding surface $\sum_{n(g-1)+1}$ and build $p:\sum_{n(g-1)+1}\to\sum_g$ as in the picture below. (We can cut along a "the" standard non-separating curve $\gamma$, copy the result $n$ times, then glue them together cyclically along the copies of $\gamma^+$ and $\gamma^-$)
My question is this: Given a surjection $\phi:\pi_1(\sum_g,x_0)\to C_n$, is there a curve $\gamma\subset\sum_g$ (L in the picture) such that $\phi$ can be realized as above? I.e., such that $\phi$ is the projection $\pi_1(\sum_g,x_0)\to \pi_1(\sum_g,x_0)/p_*(\pi_1(\sum_{n(g-1)+1},\hat{x_0}))$ so that $\sum_{n(g-1)+1}$ is obtained by cutting $\sum_g$ along $\gamma$, copying, and gluing together.
Thanks.
The surjection $\phi:\pi_1(\Sigma_g,x_0)\to C_n$, since $C_n$ is an abelian group, factors through the abelianization $\pi_1(\Sigma_g,x_0)\to H_1(\Sigma_g)$, giving a homomorphism $\phi':H_1(\Sigma_g)\to C_n$.
Consider the image of each standard generator $a_1,b_1,\dots,a_g,b_g\in H_1(\Sigma_g)$ in $C_n$. With $t$ the generator of $C_n$, we have $a_i\mapsto t^{\alpha_i}$ and $b_i\mapsto t^{\beta_i}$ for all $i$, for some $\alpha_i,\beta_i\in\mathbb{Z}$. The GCD of $\alpha_1,\beta_1,\dots,\alpha_g,\beta_g,n$ must be $1$ for $\phi$ to be surjective. Already we have a homomorphism $f:H_1(\Sigma_g)\to \mathbb{Z}$ defined by $a_i\mapsto \alpha_i$ and $b_i\mapsto \beta_i$ such that for the quotient $q:\mathbb{Z}\to C_n$ with $q(1)=t$ we have $q\circ f=\phi'$.
By the Universal Coefficient Theorem as well as Poincaré duality for $\Sigma_g$, $f$ can be considered to be an element of $H_1(\Sigma_g)$. Generically, $f$ is a collection of disjoint oriented closed curves in $\Sigma_g$, and these curves indicate how to evaluate $f$ on a closed loop: calculate the signed intersection number with $f$. By cutting $\Sigma_g$ along $f$, taking $n$ copies of the result, and then gluing the pieces together in the correct way (i.e., using the orientation of a boundary to know whether to glue piece $i$ to piece $i+1$ or to piece $i-1$), one obtains the covering space associated with $\phi$. However, at this point there is no reason for there to be a single curve associated to $f$. It should be noted that $f$ is given by $\sum_i(\beta_ia_i+\alpha_ib_i)$.
Since $\delta=\gcd(\alpha_1,\beta_1,\dots,\alpha_g,\beta_g)$ is coprime to $n$, "dividing by $\delta$" is an automorphism of $C_n$. That is, $t^k\mapsto t^{k/\delta}$, using the suitable multiplicative inverse modulo $n$. Hence, we may assume $\delta=1$. Since the coefficients of $\sum_i(\beta_i a_i+\alpha_i b_i)$ have a GCD of $1$, that means the homology class is primitive, and so it can be represented by a connected simple closed curve.