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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?), 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).

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    $\begingroup$ You wrote that $C(u)$ has cardinality $k^m\cdot k!$, but I think you meant to write $k^m\cdot m!$ instead. I believe that $C(u)$ is just a wreath product of $m$ copies of a cyclic group of order $k$ by the obvious action of $S_m$. $\endgroup$ Dec 9, 2015 at 18:35
  • $\begingroup$ @JasonStarr you are absolutely right, that was a typo. Thanks for pointing it out $\endgroup$ Dec 9, 2015 at 18:38
  • $\begingroup$ Lehmer's polynomial's Galois group is a subgroup of the wreath product $C_2 \wr S_5$ simply because it's reciprocal, i.e. $\alpha$ is a root of the polynomial if and only if $\alpha^{-1}$ is. (This is easily seen to be equivalent to the polynomials coefficients' being palindromic.) You can easily see that the Galois group of a reciprocal polynomial must respect this partition of the $2n$ roots into $n$ pairs (of a number and its reciprocal). In general, if a permutation group respects this kind of block structure, it's called imprimitive and contained in a wreath product as Geoff describes. $\endgroup$ Dec 10, 2015 at 1:28

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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 $C_{2} \wr S_{\frac{n}{2}}$.

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    $\begingroup$ It might be worth mentioning that for Galois groups imprimitivity (i.e. subgroups of a wreath product) corresponds to decompositions $f(x)$ divides $g(h(x))$ (here $g(x)=x^5-x^4-5*x^3+5*x^2+4*x-3$ and $h(x)=x^9+x^8-x^6-x^5-x^4-x^3-x^2-x+1$), respectively proper subfields of the field defined by the polynomial. $\endgroup$
    – ahulpke
    Dec 10, 2015 at 2:08

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