Complex Analysis and Mobius Transformation.

[BrainHurts]

Homework Statement

If $\phi \in \mathcal{M}$ (group of all linear fractional transformations or Mobius Transformations has three fixed points, then it must be the identity. (The proof should exploit the fact that $\mathcal{M}$ is a group.

The Attempt at a Solution

Hi all,

So the book I'm reading is "Complex Made Simple" by David C. Ullrich. and this is Exercise 8.5 if anyone has done this.

Anyway this is what I was thinking.

Let $\phi_j \in \mathcal{M}$ where $j = 1,2$ and let $\phi_j(z_i) = z_i$ for $i = 1,2,3$

So we see that $z_i = \phi_{j}^{-1}(z_i)$, we know that $\phi_{j}^{-1}$ since their elements of the group $\mathcal{M}$

My question would be this, if I define a map $\phi = \phi_{2}^{-1} \circ \phi_{1}$ and show that that $\phi_1 = \phi_2$ is that enough to show that $\phi$ is the identity?

 

[BrainHurts]

I think it is because the identity element and the inverse element is unique.