Finite stabiliser subgroup of Möbius transformations for a single point is cylic? Let $G \leq \text{Aut}(\mathbb{P}^1(\mathbb{C}))$ be a finite group. Let $x \in \mathbb{P}^1(\mathbb{C})$, then the lecture notes I am reading assert that $\text{Stab}_G(x)$ is cyclic.   (Dolgachev, McKay Correspondence, p.4)
I have gotten as far as showing that we can assume that $x = 0$, and that some element $g$ of the stabiliser also has another fixed point $\infty$, meaning that $g$ is some transformation $z \mapsto \lambda z$ for $\lambda$ some $n^{th}$ root of unity.
My problem is coming from showing that the whole group is cyclic, if some other element of the stabiliser doesn't fix $\infty$ then surely this can't be true right? 
 A: If $h$ is another element of $G$ that fixes $0$, and if $h$ does not fix $\infty$, then one can use $g$ and $h$ together to produce infinitely many different elements of the group $G$, contradicting that $G$ is finite. 
To see how to do this, it helps to transform the coordinates (conjugating by $z \mapsto 1/z$) so that $g$ and $h$ both fix $\infty$, $g$ fixes $0$, and $h$ does not fix $0$. 
The map $g$ is still a transformation of the same form $z \mapsto \lambda z$ for some root of unity $\lambda$. To say this in geometric language, $g$ is a nontrivial rotation of the Euclidean plane centered on the origin $\mathcal O$ that rotates around $\mathcal O$ by some rational angle $\theta = \pi \, m/n$. 
Similarly, $h$ is a rotation of the Euclidean plane centered on a different point $\mathcal P \ne \mathcal O$ and $h$ rotates around $\mathcal P$ by some rational angle $\phi = \pi \, k/l$.
The rational numbers $m/n$ and $k/l$ are not multiples of $2$ because $g$ and $h$ are nontrivial rotations.
The problem has now been transformed into a problem about the group of rigid motions of the Euclidean plane. To get a contradiction, I'll show that the subgroup generated by $g$ and $h$ is infinite, by giving an inductive definition of a sequence 
$$g=g_0, h = g_1, g_2, \ldots
$$
of elements in this subgroup such that each $g_i$ is a nontrivial rotation about some point $\mathcal P_i$ and the distance $d(\mathcal O, \mathcal P_i)$ is strictly increasing. We've already started the induction with $g=g_0$, $\mathcal O = \mathcal P_0$, $h=g_1$, $\mathcal P = \mathcal P_1$.
Suppose by induction that $g = g_0,\ldots,g_n$, $\mathcal O = \mathcal P_0,\ldots,\mathcal P_n$ have already been constructed. Draw the circle $C$ centered on $\mathcal O$ passing through $\mathcal P_n$. Let $L$ be the radial segment of that circle connecting $\mathcal O$ to $\mathcal P_n$. Consider $g_n$, which fixes the point $\mathcal P_n$. Letting the exponent $i \in \mathbb Z$ vary, consider the rotated segments $g_n^i(L)$, each having one endpoint at $\mathcal P_n$. Because $g_n$ is a nontrivial rotation, there exists an exponent $i \in \mathbb Z$ such that $g_n^i$ has rotation angle in the interval between $\pi/2$ and $3\pi/2$. It follows that the segment $g_n^i(L)$ lies entirely outside the circle $C$ (except for its endpoint $\mathcal P_n$). Let $\mathcal P_{n+1}$ be the opposite endpoint of $g_n^i(L)$. This point $\mathcal P_{n+1}$ is the center of rotation of the element
$$g_{n+1} = g_n^i \circ g_0 \circ g_n^{-i}
$$
That element has rotation angle equal to the rotation angle of $g_0$ which is not $\pi$ times a multiple of $2$. And by construction, $\mathcal P_{n+1}$ is further from $\mathcal O$ than $\mathcal P_n$.
