Let $X$ be a curve of genus three which is not hyperelliptic. Then $X$ is trigonal, i.e., there exists a finite morphism $X \to \mathbf P^1$ of degree $3$.


Let $Y\to X \to \mathbf P^1$ be a Galois closure of a trigonal curve. Then, unless $X\to \mathbf P^1$ is the Klein curve, the map $Y\to \mathbf P^1$ is of degree $6$. 


Now, surely the latter map could have non-abelian Galois group, but I don't know any explicit examples. Can somebody give me an explicit example for which it is clear that the Galois group over $\mathbf P^1$ of the Galois closure is non-abelian?


Can we describe all such trigonal curves in the moduli space? How big is the dimension of the locus of curves which are hyperelliptic OR have a trigonal map $X \to \mathbf P^1$ whose Galois closure is cyclic (of degree 6) over $\mathbf P^1$?



Trigonal curves of genus three: can their Galois closure be non-abelian
Turning my comment into an answer:


It is a general fact from Galois thory that a field extension of degree 3 is either Galois with cyclic Galois group, or else its Galois closure has non-abelian Galois group $S\_3$.
So unless your original morphism $X \to {\mathbb P}^1$ is a Galois covering, the Galois closure $Y \to {\mathbb P}^1$ will have non-abelian Galois group.


Assuming you work over $\mathbb C$ (or any other field containing cube roots of unity), Kummer theory tells us that the Galois coverings of ${\mathbb P}^1$ of degree 3 are exactly the superelliptic curves of the form $y^3 = f(x)$ (with map to ${\mathbb P}^1$ given by the $x$-coordinate), where $f$ is a cube-free polynomial and $X$ is taken to be the smooth projective model of the affine curve given by the equation.


