Free Actions of Finite Groups on Surface of Genus $2$

Question

Let $\Sigma_2$ be the orientable surface of genus $2$. I am trying to show that if $G$ is a finite group that acts freely on $\Sigma_2$, then $G$ has order $1$ or $2$. My first thought is that, we would then get a finite cover $\Sigma_2 \to \Sigma_2/G$. If I can show that $\Sigma_2/G$ must also be a surface of genus $g$, then I get that $g = 1$ or $g = 2$, and I'm done. But, I don't have any result that says that $\Sigma_2$ doesn't cover anything that isn't $\Sigma_1$ or $\Sigma_2$. I am answering this question in the confines of a first class in algebraic topology that includes fundamental group, covering spaces, and homology (chapters $0$, $1$, and $2$ of Hatcher). Any help is appreciated.

Answer 1

The following relation is verified: $e(\Sigma_2)=|G|e(\Sigma_2/G)$ where $e(\Sigma_2)$ is the Euler characteristic. Since $e(\Sigma_2)=-2$, we deduce that $|G|$ divides $2$ and $G=1$ or $\mathbb{Z}/2$.

Euler characteristic of covering space of CW complex

Answer 2

The trouble is that it is false that only order 1 and 2 groups act on $\Sigma_2$.

Perhaps you intended your question to have a stronger hypothesis, namely by requiring the action to be free, and then the Euler characteristic arguement given by @TsemoAristide gives the answer.

But if you intended your question to be as stated, without a hypothesis that the action is free, then there are several other finite groups that can act. For example, gluing opposite sides of an octagon gives $\Sigma_2$, and the symmetry group of the octagon therefore acts, which is a dihedral group of order 16. Also, gluing opposite sides of a decagon gives $\Sigma_2$, with an action of a dihedral group of order 20. There are still more, arising from orbifold quotients of $\Sigma_2$.