Let $X:=\mathbb{C}^n$, and let the symmetric group $S\_n$ act by permutation of coordinates in the obvious way; let $X\_n:=X/S\_n$ be the quotient by the group action. Now, $X\_n\simeq \mathbb{C}^n$, so we get a map $\pi: X\to X\_n$ with finite fibers. Away from the discriminant $\Delta:=\prod\_{i\lt j}(x\_i-x\_j)$, $\pi$ is a covering map $X\setminus\Delta\to X\_n\setminus\pi(\Delta)$, and this map has monodromy group $S\_n$. In similar fashion, we can let the subgroup $S\_k$ act on the first $k$ coordinates, and let $X\_k$ be the resulting quotient; again away from the discriminant the map $X\to X\_k$ is a covering map and has monodromy group $S\_k$. (We could of course generalise to $S\_\lambda$ for a partition $\lambda$ of $n$, but let's keep it simple.)


Now, $\pi:X\to X\_n$ factors through any of the $X\_k$, so we have maps $X\_k\to X\_n$, and in particular $X\_n\to X\_{n+1}$, and away from the branching locus they are also covering maps.



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> Question: What are the monodromy groups of these covering maps? 
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This seems like the sort of thing someone might have worked out long ago, is anything known about this?



Monodromy of covering map related to symmetric group
For every $k=1,\dots,n-1$, the monodromy group of $q\_k:X\_k\to X\_n$ is isomorphic to the symmetric group $S\_n$; one argument uses the fundamental theorem of Galois theory (or its analogue for topological covering spaces, if you prefer to work topologically).
First of all $q\_1:X\_1\to X\_n$ equals $\pi$, which is a Galois cover with group of deck transformations $S\_n$ by construction. Next, for $k=1,\dots,n$, the morphism $q\_k$ factors $\pi$, with the corresponding fixed subgroup being isomorphic to the copy of $S\_k$ in $S\_n$ that you indicated. Hence also the Galois closure of $q\_k$ factors $\pi$, with the corresponding fixed subgroup $N$ being the intersection of all conjugates of $S\_k$, i.e., the largest normal subgroup of $S\_n$ contained in $S\_k$. For $k=n$, $N$ equals $S\_n$, so assume $k < n$. For $n\geq 5$, of course, $S\_n$ has only the normal subgroups $\{1\}$, $A\_n$ and $S\_n$, so $N$ equals $\{1\}$. Also for $n=2,3,4$, it is straightforward to compute that $N$ equals $\{1\}$. Thus $\pi$ is the Galois closure of $q\_k$, and the Galois group $S\_n/N$ equals $S\_n$. 


