If $H/\Gamma$ is a compact Riemann surface , the generator of $\Gamma$ is not commutative In the book Compact Riemann Surfaces by Jurgen Jost, the Exercises for 2.4 is asking to prove that: 
Let $H/\Gamma$ be a compact Riemann surface. Show that each nontrivial abelian subgroup of $\Gamma$ is infinite cyclic group. Where H is an upper complex plane equipped with hyperbolic metric $\frac{2}{(z-\bar{z})^2}dzd\bar{z}$
Since is $H/\Gamma$ is compact, $\Gamma$ is a group generated by a finite set $\{g_1,\dots,g_m\}$. So it is suffice to prove that $g_i$ and $g_j$ is not commutative for any $i\neq j$. But I have trouble in finding the contradiction by assuming $g_ig_j=g_jg_i$. Furthermore, each generator maps one side of a fundamental polygon to another side. Different pairs of such sides are carried to each other by different elements of $\Gamma$.
Any help will be appreciated!

Here's a (long) hint. The automorphism group of the upper-half plane is isomorphic with $\operatorname{PSL}(2,\mathbb{R})$, which contains three types of elements: elliptics, parabolics, and hyperbolics, depending on whether the absolute value of the trace is less than, equal to, or greater than $2$. If $\Gamma$ contains an elliptic the quotient will not be a Riemann surface, and if it contains a parabolic there will be a cusp, and hence the quotient won't be compact. Thus $\Gamma$ consists only of hyperbolic elements. Now you need to show that distinct hyperbolic isometries (one not being a power of the other) cannot commute. You can do this by showing that if two hyperbolic isometries commute, then they have the same fixed point set, and hence leave invariant the same axis in $\mathbb{H}$. Then one isometry is a power of the other precisely if their translation distances are integer multiples. And if their translation distances are not integer multiples, the group they generate contains elements of arbitrarily small translation distance, so that the quotient object isn't even Hausdorff.  
