prove a property for the stabiliser of 0 by the map apply action of a discrete subgroup of Aut(D) It’s a note from Donaldson’s Riemann Surface,section 3.2.3.
Aut(D) is the holomorphic automorphism group of the unit disk in the complex plane. Explicitly, all möbius holomorphic maps:
$$
z \mapsto \frac{\alpha z+\beta}{\bar{\beta} z+\bar{\alpha}}
$$
where $|\alpha|^{2}-|\beta|^{2}=1$. Now we have the following proposition. It is isomorphic to PSL(2, R). And with the metric defined by  $\left|\alpha-\alpha_{1}\right|^{2}+\left|\beta-\beta_{1}\right|^{2}$  it forms a topological group.
Let G be a discrete group in Aut(D) and Z be the stabilizer of 0 in G. Hence Z is a finite cycle group with $\beta=0$. But how to prove that:
there is some
$k<1$ such that if $\gamma \in G$ is $\gamma(z)=(\alpha z+\beta) /(\bar{\beta} z+\bar{\alpha})$, where $|\alpha|^{2}-|\beta|^{2}=1$, and if $|\beta| \leq k|\alpha|$, then $\gamma$ is in $Z$.
How to prove this property? Donaldson says it’s easy but I’m really stuck here for a long time. I tried to prove it by contradiction. I figured it out that if no such k exists then there is some element in G getting closer and closer to the circle in Aut(D). But I canot prove that it conflicts with the discreetness of G. Thanks!
 A: With some hints from Free discrete action vs discrete subgroup. Maybe I found a proof, though lossing many details, and it's not totally rigorous:
theorem
Let $\Gamma$ be a discrete subgroup of holomorphic automorphisms group Aut($D$), which is the set of all Mobius maps of the form
$$
 z \mapsto \frac{\alpha z+\beta}{\bar{\beta} z+\bar{\alpha}}
 $$
where $|\alpha|^{2}-|\beta|^{2}=1$, and modulo $\{-1, 1\}$. Let $Z$ be the stabilizer of $0$ in $\Gamma$. Then:

*

*$Z$ is a finite cyclic group.

*There is some  $k<1$ such that if $\gamma \in \Gamma$ is $\gamma(z)=(\alpha z+\beta) /(\bar{\beta} z+\bar{\alpha})$, where $|\alpha|^{2}-|\beta|^{2}=1$, and if $|\beta| \leq k|\alpha|$, then $\gamma$ is in $Z$.

Proof
Elements in the stabilizer of $0$ in $\Gamma$ has the form $z \mapsto e^{i\theta}z$. Since $\Gamma$ is discrete, and the intersection of a discrete subspace in another subspace is still discrete, and the intersection of two subgroup is still a subgroup, $Z$ is a discrete subgroup of the stabilizer of $0$ in Aut($0$), which is a circle. Hence $Z$ is a finite cyclic group. Otherwise it would be dence (Subgroup of the unit circle under complex multiplication). Hence property 1 holds.
Suppose property 2 does not hold. Then for any $n \in \Bbb{N}^{+}$, there is $\gamma_{n} = (\alpha_n z+\beta_n) /(\bar{\beta}_n z+\bar{\alpha}_n) \in \Gamma $,  where $|\alpha_n|^{2}-|\beta_n|^{2}=1$, and $|\beta_n| \leq |\alpha_n|/n$, such that $\gamma_n$ is not in $Z$. Then $|\alpha_n| \rightarrow 1, |\beta_n| \rightarrow 0 $. Hence there is some subsequence $\{n_m\}$, such that $\{\alpha_{n_m}\}$ converges to some complex number $e^{i\theta}$. Without loss of generality, suppose $\{(\alpha_n, \beta_n)\} \rightarrow (e^{i\theta}, 0)$.
Since $(1, 0)$ is in $Z \subset \Gamma$, by the discreteness of $\Gamma$, there is some neighborhood $N$ of it such that $N \cap \Gamma = \varnothing$. If $\theta$ defined above satisfying that $\theta/\pi$ is rational, there is some positive $M$ such that $e^{iM\theta} = 1$ and hence $(e^{iM\theta}, 0) = (1, 0) \in N$. Otherwise since every subgroup of the circle is either finite cyclic or dense, there is some positive integer $M$, such that we still have $(e^{iM\theta}, 0) \in N$.
Define $F: \text{Aut}(D) \rightarrow \text{Aut}(D) $ by $\gamma \mapsto \gamma^{M}$ (applying itself $M$ times). Then $F$ is a continuous function from $\text{Aut}(D)$ to itself. Hence by $\{(\alpha_n, \beta_n)\} \rightarrow (e^{i\theta}, 0)$, apply it to the image of $F$, the limit $F(\alpha_n, \beta_n) \rightarrow (e^{iM\theta}, 0)$ holds. Since $F(\alpha_n, \beta_n)$ is self composition of element in $\Gamma$, it's in $\Gamma $ too. And since $(e^{iM\theta}, 0) \in N$, there has to be some $F(\alpha_n, \beta_n) \in \Gamma$. This conflicts with the discreteness of $\Gamma$.
Hence property 2 holds.
