I am meant to use the fact that Möbius transformations of finite order have exactly two fixed points. Let $G$ be the subgroup. I have observed that if $f,g\in G$ where $f$ fixes $z_0,z_1$, then $f=gfg^{-1}$ fixes $g(z_0),g(z_1)$, so $g$ either fixes $z_0,z_1$ or interchanges them, in which case they are fixed by $g^2$.
Other than noting that $\mathrm{ord}(fg)=\mathrm{lcm}(\mathrm{ord}(f),\mathrm{ord}(g))$, and that maps which fix two points are conjugate to $z\mapsto az$, I'm not sure where to go from here.