Finding a unique Mobius Transformation

Question

Let $z_1, z_2, z_3$ be three distinct points in $\widetilde{\mathbb{C}}$.

(1) show that there is a unique mobius transformation $g$ such that $g(z_1)=0, g(z_2)=1, g(z_3)=\infty$

(2) show that the sub group of the mobius transformations which maps the set $\{z_1, z_2, z_3\}$ to itself is isomorphic to $S_3$, the permutation group of three elements.

Show that the subgroup of the Mobius transformations ofr which $f(z_1)=z_1$ and $f(z_2)=z_2$ is isomorphic to $\mathbb{C}^*=({\mathbb{C}\setminus\{0\},\times })$.

Hi, I kind of solve the first part, I got fuction is

$g(z)=\frac{(z_2-z_3)(z_1-z)}{(z_2-z_1)(z_3-z)}$

But How can I prove it is the unique one to meet the conditions? Also, anyone can give some hints to proceed the rest part. thanks a lot!

Answer 1

You can generalize the uniqueness property to show that there is a unique Möbius Transformation such that $g(z_1)=w_1, g(z_2)=w_2$, and $g(z_3)=w_3$ for distinct $w_1,w_2, w_3$. So the only way $g$ can map $\{z_1,z_2,z_3\}$ to itself is by permutation.

Answer 2

Suppose $f(z)=\frac{az+b}{cz+d}$ satisfies the given conditions. So, $f(z_1)=0$ implies $z_1=-\frac{b}{a}$, $f(z_3)=\infty$ implies $z_3=-\frac{d}{c}$. Now $f(z)=\frac{a(z-z_1)}{c(z-z_3)}$. Finally, $f(z_2)=1$ implies $\frac{a}{c}=\frac{z_2-z_1}{z_2-z_3}$.