Suppose we have a tiling of the upper half plane by ideal quadrilaterals. Suppose that the principal quadrilateral consists of perpendicular lines at $Re(z)=0$ and $Re(z)=1$, and two circles joining these vertices:[a picture of the set up]
Then, this tiling is dual to the cayley graph of a free group, where the first generator is in the right semicircle adjacent to the principal quadrilateral, and the second generator is located at $(1,1)$. The free group acts on its cayley graph in the usual way, and so it also acts on the hyperbolic plane by isometries.
Homework Problem: determine what mobius transformation corresponds to the action of $x$, the first generator, and likewise for the second.
My Problem: I know that $f(i)=x$ just by definition, but I can't determine a single other point that would be easy to calculate without knowing more about $x$. If $f(\infty) \neq \infty$, since it is inversion, is there a reasonable way to tell what $f(\infty)$ is? I can also see that the circle corresponding to $x^{-1}$ gets mapped to the circle corresponding to $x$ by inversion, and also where the other $3$ sides of the principal quadrilateral are mapped, but I'm unsure how to use this information.
I'm pretty sure that the mobius transformation corresponding to the second generator is $z \mapsto z+1$, since $g(\infty)=\infty$, $g(i)=1+i$ and $g(i+1)=i+2$.
![[a picture of the set up]](https://i.stack.imgur.com/fNDpz.jpg){kind=link}
the action of $x$ is indeed an inversion, and since the tiling is regular, we can see that following the action of $x$ the fundamental quadrilateral has vertices $-1,0,1,\infty$.
In particular, we can check that $-1 \mapsto 1$, $0 \mapsto 0$ and $\infty \mapsto \frac{1}{2}$ so the corresponding mobius transformation is $$\frac{z}{2z+1}$$