For each positive integer $n$, write $S\_n$ for the symmetric on $n$-letters. Suppose that $m | n$ is a proper divisor of $n$, and write $n = km$. Consider the element


$$\displaystyle u(m,n) = \underbrace{\begin{pmatrix} 1 & 2 & \cdots & k\end{pmatrix} \cdots \begin{pmatrix} n-k+1 & n-k+2 & \cdots & n \end{pmatrix}}\_m$$


Let $C(u)$ be the centralizer of $u$ in $S\_n$. The orbit of $u$ under conjugation by $S\_n$ is the set of elements in $S\_n$ with the same cycle-type as $u$, and the size of the orbit is readily computed to be 


$$\displaystyle \binom{km}{k} (k-1)! \binom{k(m-1)}{k} (k-1)! \cdots \binom{2k}{k} (k-1)! (k-1)! \frac{1}{m!} = \frac{n!}{k^m \cdot m!},$$


so by the orbit-stabilizer theorem, it follows that $C(u)$ is a group of cardinality $k^m \cdot m!$. 


Is there a name for these groups? Did anyone study them in particular in the literature? I am interested in knowing when such groups (or their subgroups) can be realized as the Galois group of a polynomial of degree $n$. 


For example, Lehmer's polynomial 


$$\displaystyle x^{10} + x^9 - x^7 - x^6 - x^5 - x^4 - x^3 + x + 1$$


has Galois group order $1920$ (see [Is Lehmer's polynomial solvable?](https://mathoverflow.net/questions/225055/is-lehmers-polynomial-solvable)), and the corresponding value for the size of $C(u)$ in this case is $2^5 \cdot 5! = 3840$. It can be shown that the Galois group isomorphic to a subgroup of $C(u)$ (and in fact can be embedded by some natural ordering).



What do we know about these subgroups of $S_n$?
If I understand your notation correctly, these subgroups are wreath products $C\_{k} \wr S\_{m}$, where $C\_{k}$ is a cyclic group of order $k$ (acting as a single $k$-cycle). These groups are a semidirect product of a "base group" which is a direct product of $m$ cyclic subgroups of order $k$, and a group $S\_{m}$ which permutes the cyclic factors of the base group in the natural manner.


For many purposes, in studying the representation theory of $S\_{n}$, it is often necessary for inductive purposes to understand the representation theory of these wreath products, so they are well-studied subgroups.


When considering Galois groups, one way which such subgroups of $S\_{n}$ can occur as (overgroups of) Galois groups of irreducible polynomials is when we have an irreducible polynomial $f(x) \in \mathbb{Q}[x]$ of degree $n$ with no real roots such that all roots of $f(x)$ have the same absolute value. For then complex conjugation is central in the Galois group of $f(x)$ and the Galois group is isomorphic to a subgroup of $C\_{2} \wr S\_{\frac{n}{2}}$.


