Show that the subgroup of Mobius maps for which $f(z_1)=z_1$ and $f(z_2)=z_2$ is isomorphic to $\mathbb{C}^*=(\mathbb{C}^*\setminus{\{0\}} ,\times)$. Show that the subgroup of Mobius maps for which $f(z_1)=z_1$ and $f(z_2)=z_2$ is isomorphic to $\mathbb{C}^*=(\mathbb{C}^*\setminus{\{0\}} ,\times)$.
If we consider $z_1=0$ and $z_2=\infty$, then $f$ must have the form $f(z)=az,a \in \mathbb{C}^*$. Then to define the isomorphism, we map $f$ to $a$.
But in general, how can we define the isomorphism?
 A: Say $G$ is the group of Mobius transformtions that fix $z_1$ and $z_2$ and $H$ is the group that fixes $0$ and $\infty$. Then $G$ is isomorphic to $H$:
Fix $\psi$ with $\psi(0)=z_1$ and $\psi(\infty)=z_2$. Then $\phi\in G$ if and only if $\psi^{-1}\phi\psi\in H$.


Moral: If $G$ is a group of bijections of $X$ then conjugation in $G$ is really just a "change of variables" or "substitution".


A: Let $z_1, z_2 \in \mathbb C^*$ with $z_1 \neq z_2$. Let
$$f(z) = \frac{a z + b}{c z + d}$$
for some $a,b,c,d \in \mathbb C$ with $ad - bc \neq 0$ and suppose $f(z_1) = z_1$ and $f(z_2) = z_2$.
Consider first the case in which $z_1 \in \mathbb C$ and $z_2 = \infty$. Then $f(z_2) = z_2$ implies that
$$f(z) = \frac a d z + \frac b d$$
where $a, d \neq 0$. Then $f(z_1) = z_1$ implies that
$$f(z) = \frac a d z - \left ( \frac a d - 1 \right ) z_1.$$
Therefore we map $f$ to $\frac a d \in \mathbb C \setminus \{ 0 \}$.
Consider now the case in which $z_1, z_2 \in \mathbb C$. Then $f(z_1) = z_1$ and $f(z_2) = z_2$ imply that
$$c z_i^2 - (a - d) z_i - b = 0 \qquad \text{for } i = 1, 2.$$
If $c = 0$ then also $a - d = 0$ and $b = 0$, so we get the identity $f(z) = z$. Otherwise, by Vieta's formulas,
$$\frac {a - d} c = z_1 + z_2 \qquad {-\frac b c} = z_1 z_2$$
from which we get
$$f(z) = \frac{[(c + 1) z_2 - z_1] z - c z_1 z_2}{c z - [(c + 1) z_1 - z_2]}.$$
Now, it is tempting to map $f$ to $c$, but this doesn't work. Since the identity has $c = 0$, in order to map the identity to $1$ we must instead map $f$ to $c + 1 \in \mathbb C \setminus \{ 0 \}$.
You can check that these maps give you the required isomorphism.
