If $G$ is a finite cyclic group acting faithfully on $\mathbb{P}^1_\mathbb{C}$, must it have exactly two fixed points? If $G$ is a finite (nontrivial) cyclic group acting faithfully on $\mathbb{P}^1_\mathbb{C}$ by holomorphic automorphisms, must it have exactly two fixed points?
I believe this should follow from the Hurwitz formula, but it's possible I've made a mistake. Is this well known? Are there other proofs of this which don't use the Hurwitz formula?
 A: Thanks to Kenny Wong, the answer is an easy one.
The fixed points of a cyclic group are the same as the fixed points of a generator, which is just a mobius transformation given by $z\mapsto (az+b)/(cz+d)$, or alternatively given by a matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ representing an element of $PGL_2(\mathbb{C})$. By looking at the Jordan canonical form, we see that if the matrix has finite order, then it is diagonalizable, which is to say that it has two linearly independent eigenvectors - ie, two fixed points.
Alternatively, if $\alpha$ is an automorphism, then since $PGL_2(\mathbb{C})$ is sharply 3-transitive, if $\alpha$ fixes at least 3 points, then it must be the identity. If it fixes $\le 1$ points and has finite order then the quotient map $$\mathbb{P}^1_\mathbb{C}\rightarrow\mathbb{P}^1_\mathbb{C}/\langle\alpha\rangle$$
is ramified at $\le 1$ points, but $\mathbb{P}^1_\mathbb{C}/\langle\alpha\rangle$ is a smooth compact Riemann surface, hence also isomorphic to $\mathbb{P}^1_\mathbb{C}$, and thus the map above restricts to an unramified cover of $\mathbb{P}^1_\mathbb{C} - \{\text{$\le 1$ points}\}$, which cannot happen since both $\mathbb{P}^1_\mathbb{C}$ and $\mathbb{A}^1_\mathbb{C}$ have trivial fundamental groups.
Note that the finite order assumption is crucial, since the matrix $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ representing the automorphism $z\mapsto z+1$ has only one fixed point, but this is okay since it has infinite order, and hence is not the generator of a finite cyclic group.
A: Here's an alternative point of view.
It's a fact from algebraic geometry that isomorphisms of $P^n$ are in $PLG_n$ (they are all linear), so what we have is a projective representation of $G$. 
Let $g \in G$ be a generator. It's action can be represented by $cA$ for some matrix $A$ acting on $V = \mathbb{C}^2$ and any $c \in \mathbb{C}^*$. We know that $A^n = d I$, for some $d \in \mathbb{C}^*$. So if $c$ is some $nth$ root of $1 /d$, $cA$ is a lift of the action of $g$ that defines a representation of $G$ on $\mathbb{C}^2$, and the projectiviation of this representation is the original action.
Since $G$ is abelian, we get a decomposition $V = \mathbb{C}^2 = W' \oplus W$, for $W'$ and $W$ 1 dimensional representations of $G$. If they have the same character, then the induced action on $P^1$ is trivial. Hence we assume that they have different characters.
Then it follows from the uniqueness of the isotypic decompositions that $W'$ and $W$ are the only proper fixed subspaces of $V$. So the representation has exactly two fixed lines, hence the projective representation has exactly two fixed points.
