Finite morphism from a smooth projective curve. Let $k$ be an algebraically closed field and $C$ be a smooth projective curve over $k$. Let $p$ be a prime number. Does there exists a finite morphism $f : C \to \mathbb{P}^1$ such that the degree of the Galois closure of the field extension $k(\mathbb{P}^1) \to k(C)$ is coprime to $p$ ? 
 A: Following Jason Starr's suggestion, let $C$ be a very general genus $2$ curve, forming a covering of degree $n$ of $\mathbb P^1$. Without loss of generality there is no intermediate curve between $C$ or $\mathbb P^1$ (it would have to have genus $0$ or $1$. In the first case, we may simplify by replacing $\mathbb P^1$ with that curve, and in the second case we get an elliptic factor in the Jacobian, contradicting "very general".)
Let $G \subseteq S_n$ be the Galois group of the covering. As Jason pointed out, it is solvable. It is also transitive and primitive. Because $G$ is solvable, it has "soluble socle", hence $n=p^d$ is a prime power and $G$ acts as a subgroup of the group of affine transformations of $\mathbb F_{p^d}$ by a fragment of the O'Nan-Scott theorem (The argument goes that because $G$ is solvable, it has some normal subgroup $J$ isomorphic to $\mathbb F_p^d$ for some $d$. $J$ must act transitively, or else the action would not be primitive - the cover would factor through the cover defined y the orbits of $J$. Hence the $n$ points acted on are isomorphic to $\mathbb F_p^d$, and $G$ acts by transformations that normalize the group of translations, i.e. affine transformations).
Now the local monodromy around each ramification point is an affine transformation, hence has at most $p^{d-1}$ fixed points. Because the ramification at the non-fixed points has order at least $3$, the local contribution to Riemann-Hurwitz is at least $(2/3) (p^{d} - p^{d-1})$. Because the curve is very general, the number of ramification points is at least $6$, so that after moving the first three to $0,1,\infty$ we still have three parameters. So the total contribution to Riemann-Hurwitz is at least $4 (p^{d}- p^{d-1})$ and we obtain the inequality
$$-2 = 2-2g  \leq 2 p^d - 4 (p^{-d}  - p^{d-1}) = 4 p^{d-1} -2 p^d $$
$$\left(1-\frac{2}{p} \right) p^d \leq 1 $$
$p$ is at least $3$ so $1-\frac{2}{p}$ is at least $1/3$ so $p^d$ is at most $3$ which is an obvious contradiction as for $3$-coverings every point has full ramification and then we obtain $2-2g \leq 6 - 6 \cdot 2 = - 6$.
Because this is a contradiction, a very general genus $2$ curve cannot be so represented.
