Show that the group of automorphisms of $\Bbb C_\infty$ generated by the two automorphisms sending $z$ to $\exp(2\pi i/r)z$ and sending $z$ to $1/z$ is a dihedral group of order $2r$, which acts holomorphically and effectively on $\Bbb C_\infty$. Show that there are three branch points to the quotient map.
This is an exercise in Miranda's book Algebraic Curves and Riemann Surfaces, Chapter III.3. Here is my attempt: Let $\zeta=\exp(2\pi i/r)$, $a(z)=\zeta z$, and $b(z)=1/z$. Then $a^r=1, b^2=1$, and $ba=a^{-1}b$, so the subgroup of $\text{Aut}(\Bbb C_\infty)$ generated by $a$ and $b$ is the dihedral group of order $2r$. Next, to show that there are three branch points, we have to show that there are three orbits of order $<2r$.
The orbit of $0$ consists of $\{0,\infty\}$, the orbit of $1$ consists of $\{1,\zeta,\zeta^2,\dots,\zeta^{r-1}\}$, and the orbit of $-1$ consists of $\{-1,-\zeta,-\zeta^2,\dots,-\zeta^{r-1}\}$. These three orbits are distinct if $r$ is odd, so in this case we are done. But if $r$ is even, then the oribts of $1$ and $-1$ are the same. Then what are the three orbits of order $<2r$ in the even $r$ case?
Edit: I've found that this question is related to this one: Exercise of Rick Miranda is wrong? Actions over Riemann sphere, but this one doesn't have an answer.
Best Answer
In general, the three orbits of size $<2r$ are
Any point $w$ not on the unit circle or the antipodal points $\{0,\infty\}$ has an orbit of size $2r$, consisting of the values $\zeta^k w^{\pm1}$. Geometrically, thinking about the dihedral action on the unit circle, the orbit of any point which is a shortest arc of length $\theta$ from an $r$th root of unity is the set of all points which are $\theta$ away, both clockwise and counterclockwise, from a root of unity. This will have size $2r$, unless these two sets of points coincide, so the clockwise and counterclockwise translates of the roots of unity "meet in the middle," which happens precisely at midpoints between roots of unity.