branched covering factors through a primitive one I'm struggling with an assertion I found in an article I'm reading. 
A projective complex curve $X$ is rationally uniformized by radicals if there exists a branched covering $X\to \mathbb{P}^1$ such that its Galois group is soluble. Now, the part that gives me problems:
"We recall only results about galois groups in the generic case, that is, groups of coverings $f:X\to \mathbb{P}^1$ where $X$ is general in $M_g$. Any such covering factors as a primitive covering (a covering that does not factor nontrivially) $h:X\to \mathbb{P}^1$ followed by a covering $k:\mathbb{P}^1 \to \mathbb{P}^1$".
I don't see why this factorization is possible, and in general how it works. Also, there is a correspondence between primitive coverings and primitive galois groups, is there an easy way to see that? 
 A: I think I have an answer to my question. It is not always true that we have a factorization as above, however I managed to prove the following:
The monodromy group of a branched covering is a primitive permutation group if and only if the covering is primitive.
Moreover, it is always possible to factor a branched covering $\pi:X\to Y$ in a primitive covering $h:X\to X'$ followed by a covering $k:X' \to Y$, a solvable covering factors through a solvable primitive covering and the composition of solvable coverings is solvable.
Suppose we have a branched covering $\pi: X\to Y$ with imprimitive monodromy subgroup $G$. After discarding branch points of $Y$ and the ramification points in $X$, we get a topological covering whose fiber is divided in blocks $T$ under the action of $G$. Consider now the subgroup of $G$ given by 
\begin{equation}
H=\bigcap\limits_{h\in G}G_{hT}
\end{equation}
where $G_T=\left\lbrace g\in G | gT=T\right\rbrace $ is the stabilizer of the block $T$. Hence $H$ contains the elements of $G$ that fix each block of the block system of $G$. Now the covering $X\to Y$ factors through the space of orbits $X/H$, and the monodromy group of $X\to X/H$ is exactly $G/H$. 
Conversely, if the covering factors nontrivially through $X\to X' \to Y$, we can divide the general fiber of $\pi$ into blocks containing the preimages of the same point via $X\to X'$, and these blocks must be preserved under the action of $G$ on $X$.\
Now let $K$ be the function field of $Y$, $f$ the algebraic element such that $K(X)=K(f)$, and $L:K$ the splitting field of $K(X):K$. The factorization through a primitive covering takes place if it is possible to find a maximal intermediate field $K'$, $K\subset K' \subset K(X)$. This is always possible, since we have a finite number of intermediate fields between $K$ and $K(f)$. By a keystone result in Galois theory, the group $G(L:K')$ is a subgroup of $G$, hence it is solvable if $G$ is.
Recall that a group $G$ is solvable if and only if, for any $H$ normal subgroup, $H$ is solvable together with $G/H$. Together with the construction in the first part of the proof, this implies that the composition of solvable morphisms is solvable.
