Does the Galois group acts transitively on preimages of a branch point of a morphism of curves? This is 21.7.K in Vakil's Foundations of Algebraic Geometry.
Let $C$ be an irreducible smooth projective curve over a field $k$ ($k$ is algebraically closed).  Let $G \subseteq Aut(C)$ be a finite subgroup.  Let $F$ be the subfield of $k(C)$ fixed of $G$ (I think this means that $G$ acts on $k(C)$ so there is a group homomorphism $G \rightarrow Aut(k(C))$, and $F$ is the fixed field of the image). Then, the inclusion $F \rightarrow k(C)$ induces a morphism of curves $f : C \rightarrow C'$, where $C'$ is a smooth projective curve with $k(C) = F$.
Let $p$ be a branch point of $f$, and $q_1, ... q_n =f^{-1}(p)$. Then, does $G$ act transitively on the $q_1, ... q_n$?
I think it has something to do with the fact that since $F$ is the fixed field of the subgroup of $Aut(k(C))$, $k(C) / F$ is Galois. Based on my intuition from Galois coverings of topological spaces, I expect the Galois group to act transitively. But this is a different setting and I'm not so sure about that.
 A: Over an algebraically closed field $k$, there is a category equivalence between smooth projective curves over $k$ (with nonconstant morphisms) and one-dimensional finitely generated function fields over $k$.
In this equivalence, the (closed) points of the curve correspond to (equivalence classes of) discrete nontrivial $k$-valuations of the function field.
The question can thus be rephrased as follows: let $F/K$ be a finite Galois extension, and $v$ a (nontrivial) discrete valuation on $K$. Clearly, the Galois group $G$ acts on the (equivalence classes) of discrete valuations of $F$ that extend $v$ (and this set is not empty; I can elaborate further if needed).
Consider the completion $K_v$: $F \otimes_K K_v$ is a finite reduced $K_v$-algebra, and is thus the product of finite extensions of $K_v$. Every factor corresponds to a valuation on $F$ extending $v$ (again, I can elaborate if needed), compatibly with the Galois action.
So the goal is to show that for any $K$-field $L$, $G$ acts transitively on the connected components of $F \otimes_K L$. Note that topologically, the spectrum of this ring is a finite disjoint union of points.
Let $e \in F \otimes_K L$ be an idempotent which is one on every factor (or connected component of the spectrum) but one, corresponding to the connected component $C$. Then let $e_1 = \prod_{g \in G}{g \cdot e}$ is a $G$-invariant idempotent in $F \otimes L$. But $(F \otimes L)^G=F^G \otimes L=L$, so $e_1$ is zero or one. But $e_1$ can’t be invertible (since $e$ is not), thus it must be zero. But, by construction, if a connected component $C’$ is not in the $G$-orbit of $C$, then $e_1$ is one on the corresponding factor. Thus no such $C’$ exists. QED.
