Show that the set of one-to-one holomorphic maps $\Bbb{C}\setminus\{a,b,c\} \to \Bbb{C}\setminus\{a,b,c\}$ forms a finite group. 
Let $\Omega = \mathbb{C}\setminus\{a, b, c\}$ be the complement of three distinct points in the complex plane. Show that the set of one-to-one holomorphic maps $f : \Omega \to \Omega$ forms a finite group. Is the order of this group independent of the choice $\{a, b, c\}$?

By Casorati-Weierstrass, none of the points can be essential singularities for $f$. However, it is possible that at least one of the points is a pole, for example $f : \mathbb{C}\setminus\{-1, 0, 1\} \to \mathbb{C}\setminus\{-1, 0, 1\}$, $z \mapsto \frac{1}{z}$.   For this choice of points, we also have $z \mapsto z$, $z \mapsto -z$, and $z \mapsto -\frac{1}{z}$, so $\mathbb{Z}_2\times\mathbb{Z}_2 \leq \operatorname{Aut}(\mathbb{C}\setminus\{-1, 0, 1\})$. Furthermore, for any $a, b, c$ on a straight line, we can conjugate by a rotation, a translation, and a rescaling to get a corresponding subgroup. If I had to guess, the group will depend on whether or not $a, b, c$ are collinear.
Any hints on how to proceed would be very much appreciated.
 A: First normalise the situation. Since the structure of the automorphism group is preserved under biholomorphisms, we can map $\Omega$ to $\Omega' = \mathbb{C}\setminus \{0,1,c'\}$ per $z\mapsto \frac{z-a}{b-a}$.
Now consider the situation from the viewpoint of the Riemann sphere, $\Omega' = \widehat{\mathbb{C}} \setminus \{0,1,c',\infty\}$.
Each of the four points is a removable singularity of $f\in \operatorname{Aut}(\Omega')$ when that is considered as a holomorphic map to the sphere, so $f$ extends to an automorphism of the sphere. Hence
$$\operatorname{Aut}(\Omega') = \left\{f\in \operatorname{Aut}(\widehat{\mathbb{C}}) : f(\{0,1,c',\infty\}) = \{0,1,c',\infty\}\right\},$$
and we see that $\operatorname{Aut}(\Omega')$ can be viewed as a subgroup of $S_4$.
Now investigate what restrictions the location of $c'$ imposes on the permutations of the four points that arise from automorphisms of the sphere.

We always have the four automorphisms
$$T_1 \colon z \mapsto z;\quad T_2\colon z \mapsto \frac{c'}{z};\quad T_3\colon z\mapsto \frac{z-c'}{z-1};\quad T_4\colon z \mapsto c'\frac{z-1}{z-c'}$$
of $\Omega'$. The three non-identity automorphisms among them swap the four points in pairs, and each point in $\{0,1,c',\infty\}$ is mapped to $c'$ by one of these four.
Thus, if $f$ is any automorphism of $\Omega'$, $T_i\circ f$ is an automorphism of $\Omega'$ that leaves $c'$ fixed for exactly one $i\in \{1,2,3,4\}$. So it suffices to see which Möbius transformations permuting $\{0,1,\infty\}$ have $c'$ as a fixed point. Ignoring the identity, we have


*

*$z \mapsto \frac{1}{z}$, which swaps $0$ and $\infty$ and has the fixed points $1$ and $-1$, so is an automorphism of $\Omega'$ if and only if $c' = -1$;

*$z \mapsto 1-z$, which swaps $0$ and $1$ and has the fixed points $\infty$ and $\frac{1}{2}$, so is an automorphism of $\Omega'$ if and only if $c' = \frac{1}{2}$;

*$z\mapsto \frac{z}{z-1}$, which swaps $1$ and $\infty$ and has the fixed points $0$ and $2$, so is an automorphism of $\Omega'$ if and only if $c' = 2$;

*$z\mapsto \frac{1}{1-z}$, which cyclically permutes $0 \mapsto 1 \mapsto \infty \mapsto 0$, and its inverse $z \mapsto 1 - \frac{1}{z}$, which have the fixed points $e^{\pm \pi i/3}$, so are automorphisms of $\Omega'$ if and only if $c' = e^{\pi i/3}$ or $c' = e^{-\pi i/3}$.
We see that generically, $\operatorname{Aut}(\Omega')$ is a Klein four-group, but for some special situations, the automorphism group is larger.
If $c' \in \left\{ -1, \frac{1}{2}, 2\right\}$, there is one nontrivial automorphism of $\Omega'$ having $c'$ as a fixed point [an involution], and $\operatorname{Aut}(\Omega')$ is (isomorphic to) the dihedral group ($D_4$ or $D_8$, depending on the used nomenclature) of the rigid motions of a square.
If $c' = e^{\pm \pi i/3}$ (then the three points $0,1,c'$ form an equilateral triangle), there is a nontrivial automorphism of period $3$ of $\Omega'$ having $c'$ as a fixed point, and then $\operatorname{Aut}(\Omega')$ is (isomorphic to) the alternating group $A_4$ of even permutations of $\{0,1,c',\infty\}$.
