Can $\mathbb{Z}/6\mathbb{Z}$ act freely and properly discontinuously on $\Sigma_4$? Let $\Sigma_m$ denote the closed connected orientable surface of genus $m$. Let $N_m$ denote the closed connected non-orientable surface of genus $m$.
I was wondering which cyclic groups could act freely and properly discontinuously on $\Sigma_4$. 
It is clear that such a group must be finite. So the only possibilities are $\mathbb{Z}/n\mathbb{Z}$. By using Euler characteristics, one can show that the only possibilities are $n=1,2,3$ or 6. 
It is not hard to show that $\Sigma_4$ is the orientable double cover of $N_5$. The deck group of this cover is $\mathbb{Z}/2\mathbb{Z}$.
By considering rotational symmetry, it is also easy enough to see that $\mathbb{Z}/3\mathbb{Z}$ can act on $\Sigma_4$. The quotient of this action is $\Sigma_2$.
So the remaining question is: could $\mathbb{Z}/6\mathbb{Z}$ act freely and properly discontinuously on $\Sigma_4$?
By considering Euler characteristics, one can show that if such an action were to exist, then the quotient would have to be $N_3$.
One thing that I considered was composing some of the covering maps which I'm already aware of. For example, we could compose the covering map $\Sigma_4 \rightarrow \Sigma_2$ (which comes from rotational symmetry) with the orientable double covering map $\Sigma_2 \rightarrow N_3$. This would give a six-sheeted cover $\Sigma_4 \rightarrow N_3$. However, it is not clear to me that the deck group is $\mathbb{Z}/6\mathbb{Z}$.
Another thing that I considered was looking for epimorphisms $\varphi :\pi_1(N_3) \rightarrow \mathbb{Z}/6\mathbb{Z}$. For each such $\varphi$, there exists a regular cover of $N_3$ with deck group $\mathbb{Z}/6\mathbb{Z}$. However, it is not clear to me that any of these covering spaces is orientable.
Any insight would be appreciated.
 A: As a connected sum of $3$ real projective planes, the fundamental group $\pi_1(N_3)$ has the presentation
$$
\pi_1(N_3)=\langle a,b,c\mid a^2b^2c^2\rangle.
$$
Each one of the generators $a,b,c$ is a simple closed loop through one of the $3$ cross-caps of $N_3$, thinking of $N_3$ as $3$ cross-caps glued onto the complement of the $3$ open discs in $S^2$. 
Let
$$
f\colon \pi_1(N_3)\rightarrow\mathbf Z/6
$$
be defined by
$$
f(a)=f(b)=f(c)=1.
$$
Let $\Sigma/N_3$ be the covering associated to $f$. We have to show that $\Sigma$ is orientable.
Let $2\mathbf Z/6$ be the subgroup of $\mathbf Z/6$ generated by $2$. It's index is equal to $2$. The subgroup $f^{-1}(2\mathbf Z/6)$ of
$\pi_1(N_3)$ is the normal subgroup of all the words in $a,b,c,a^{-1},b^{-1},c^{-1}$ that are of even length. The associated covering of $N_3$ is orientation covering $O/N_3$. Therefore, the map $\Sigma\rightarrow N_3$ factorizes through $O\rightarrow N_3$. In particular, $\Sigma$ is orientable. It follows that $\Sigma=\Sigma_4$.
