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Let $X$ be a Riemann Surface of genus $g(X)$ and $G$ a group acting holomorphically and effectively over $X$. I'm reading Miranda and he used twice the fact that the genus $g(X/G) \le g(X)$. He used this fact at least when $g(X)=0,1$.

I don't know if this is a general result or works for this 2 cases. I tried to demonstrate it using Hurwitz but I could not. Please someone help me.

Eren
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If you've read that far into Miranda, you know the quotient map $F:X \to X/G$ is holomorphic. Applying the Riemann-Hurwitz formula to this map you get

$$2g(X)-2=\deg(F)(2g(X/G)-2)+\sum_{p\in X}[\text{mult}_p(F)-1]$$

The sum on the right is non-negative, so $2g(X)-2$ is greater than or equal to $\deg(F)(2g(X/G)-2)$ and $\deg F$ is positive, so $g(X)\geq g(X/G)$.

Arkady
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    You are right, Thanks! And the same applies to any non constant holomorphic image of a Riemann Surface =D. – Eren Mar 11 '16 at 16:25