Automorphism of $\mathbb{C}\setminus \{0,1\}$.

Question

I'm looking for an easy way to define this group.

"Here, $f$ is an automorphism of $\mathbb{C}\setminus \{0,1\}$ means $f$ is a biholomorphism between $\mathbb{C}\setminus \{0,1\}$ and $\mathbb{C}\setminus \{0,1\}$."

I have these theorems that might be useful for this problem.

1. $f \in \text{Aut} \left (\mathbb{C} \right )$ $\Leftrightarrow$ $\exists a,b\in \mathbb{C}$ with $a \neq 0$ s.t $f(z)=az+b$.
2. $f \in \text{Aut} \left (\mathbb{C}\setminus \{0\} \right )$ $\Leftrightarrow$ $\exists a \neq 0$ s.t either $f(z)=az$ or $f(z)=\frac{1}{az}$.
3. $f \in \text{Aut} \left (\mathbb{C}\setminus \{1\} \right )$ $\Leftrightarrow$ $\exists a \neq 0$ s.t either $f(z)=a(z-1)+1$ or $f(z)=\frac{1}{a(z-1)}+1$.

So far, I showed that $0,1$ are not essential singularities of $f$ with Casorati-Weierstrass Thm and the Open-Mapping Thm. But, I cannot proceed anymore from here.

For any once-punctured (at $z_0$) complex plane, letting another function $g$ be a shifted function of $f \in \text{Aut} \left (\mathbb{C}\setminus \{z_0\} \right )$ so that $g \in \text{Aut} \left (\mathbb{C}\setminus \{0\} \right )$ would define the automorphism group of that punctured plane easily (with Thm 2), but for a twice-punctured plane like $\mathbb{C}\setminus \{0,1\}$ I can't think of any easy method like I did for an once-punctured plane.

Would there be a way to find this automorphism group just using those three Thms?

Answer 1

You already know that $0$ and $1$ are not essential singularities. In the same way (or by considering $1/f$) one can see that $\infty$ is not an essential singularity either.

So $f$ is holomorphic in $\mathbb{C}\setminus \{0,1\}$ with removable singularities or poles at $0, 1, \infty$. It follows that $f$ is a rational function. Since the restriction of that rational function to $\mathbb{C}\setminus \{0,1\}$ is injective, the degree must be equal to one.

So $f$ is a rational function of degree one (aka “Möbius transformation”), which maps $\{ 0, 1, \infty \}$ onto itself. There are exactly six such functions, one for each permutation of $(0, 1, \infty)$.

$z, 1-z, 1/z, 1/(1-z), z/(z-1), (z-1)/z$