I want to solve the following question,
Let $G$ be an abelian group of deck transformations acting biholomorphically on the upper half plane. Show that $G$ is infinite cyclic.
My attempt:
We know the automorphism group of upper plane is modular group and the group in the problem is therefore its abelian subgroup.
Consider two elements in this abelian subgroup, we can easily show that this two modular forms share the same two fixed points.
And we know $f$ and $f^n$ share the same fixed points. But is the converse right? Could anyone give me some hint?
At this point it sounds like you have that these elements $f\in G $ must have the same fixed points in $\mathbb{R}\cup\{\infty\} $, and must have the same number of fixed points(either one or two). Breaking it up into cases we have
Case one: each element of $G $ has a unique fixed point $x$. The conjugation by either $z-x $ or $\frac{-1}{z}$ if $x=\infty $ lets us assume without loss of generality that $x=0$. Hence $f\in G$ implies that $f(z)=\frac{az+b}{cz+d}$ for $ad-bc=1$, and $f(0)=0$ so $b=0$. Hence $ad=1$, so we have after rescaling $c $ that $f (z)=\frac{a^2z}{cz+1} $. Since zero is a unique fixed point, $a^2z=cz^2+z$ must have a double zero at zero, and hence $\frac{a^2-1}{c}=0$ which implies that $a^2=1$.
Thus after all that we have that $f (z)=\frac{z}{cz+1}$, and hence $G $ is a subgroup of a group isomorphic to the reals. Since $G $ must also act properly discontinuously, $G $ must be a discrete subgroup of $\mathbb{R}$, and hence infinite cyclic.
The next case is the case when we have two fixed points $x_1$ and $x_2$. Again by conjugation by $\frac{z-x_1}{z-x_2}$(rescaled to have the right determinant) lets us assume without loss of generality that $x_1=0$ and $x_2=\infty $.
Similar to the first case, take $f \in G$, so that $f (z)=\frac{az+b}{cz+d} $. Hence we have that $f (0)=0$ gives $b=0$, and $f (\infty)=\infty$ gives $c=0$, so $f(z)=\frac{a}{d}z$, and $G $ is a subgroup of a group isomorphic to $\mathbb{R}^+$, this time with the multiplicative group structure, which in turn is isomorphic to $\mathbb{R} $ with addition by $\ln$, so as a discrete subgroup $G $ must be infinite cyclic.