Let $f,g$ be two Möbius transformations with a common fixed point $z_0$. Show that the Möbius transformation $f \circ g \circ f^{-1} \circ g^{-1} $ is either parabolic or the identity.
Möbius transforms with the same fixed points commute, so the identity case is easy enough. But if the second fixed point is not common, I'm confused as to how the given transform can be shown to be parabolic.
WLOG assume the common fixed point is $0$. If the other fixed point of $f$ is $\infty$, then $f(z) = \alpha z$ for some complex $\alpha \ne 0$. If the other fixed point of $g$ is $z_1 \ne \infty$, then $g(z) = (\beta z_1 + \gamma) \dfrac{z}{\beta z + \gamma} $ for some $\beta$ and $\gamma$. Then you can calculate $f \circ g \circ f^{-1} \circ g^{-1}(z)$ explicitly.