Maximal subgroups of a generalized symmetric group

Question

Let $G$ be the wreath product $C_m\wr S_n$, where $C_m$ is the cyclic group of order $m$ and $S_n$ is the symmetric group on $n$ letters. I would like to understand the maximal subgroups of $G$ up to conjugacy. In particular, how many conjugacy classes are there, and how explicitly can they be described?

Answer 1

The maximal subgroups that contain the base group $C_m^n$ correspond exactly to the maximal subgroups of $S_n$. There is a lot known about those (they subdivide into intransitive maximals, imprimitive amximals, and primitive maximals), and they are known explicitly up to $n$ a few thousand, but there is no hope of complete description for all $n$.

The other maximal subgroups can be described explictly. Any such subgroup will contain $C_{m/p}^n$ for some prime $p$ dividing $m$, so we can assume that $m=p$ is prime.

They all have the structure $N.S_n$ where $N$ is a maximal submodule of $M$ under the action of $S_n$.

When $p$ does not divide $n$, there are two such maximal submodules, of orders $p$ and $p^{n-1}$, which give rise to maximal subgroups with structures $p.S_n$ and $p^{n-1}.S_n$. These extensions are all split. When $p$ is odd, there is just one conjugacy class of each, but when $p=2$, there are two conjugacy classes with structure $p^{n-1}.S_n$. (You get the second class by multiplying an odd permutation in $S_n$ by an element of the module outside of the maximal submodule.)

When $p$ divides $n$, there is a single maximal submodule of order $p^{n-1}$. When $p$ is odd, this gives rise to a single maximal subgroup $p^{n-1}.S_n$, a split extension. When $p=2$, there are two conjugacy classes of maximal subgroups $2^{n-1}.S_n$, one of which is nonsplit.

While the structure of the maximal subgroups follows easily from knowledge of the maximal submodules of the permutation module, which are well-known, the more precise statements about numbers of conjugacy classes depend on information about the cohomology groups $H^1(S_n,M)$ for the various modules involved.

Some more details (added later): Firstly, note that if $M$ is any proper normal subgroup of a group $G$ and $H$ is a maximal subgroup of $G$, then either $M \le H$ or $HM=G$. In this example, let $M$ be the normal subgroup $C_m^n$, and consider those maximal subgroup $H$ that do not contain $M$. These satisfy $HM=G$.

Suppose first that $m$ is not a prime power. So $M = M_1 \times M_2$, where $M_1$ and $M_2$ are both normal subgroups of $G$, and have coprime orders. Then, if $M_1$ is not contained in $H$, we have $HM_1 = G$, and so $(H \cap M)M_1 = M$ which implies that $M_2 \le H \cap M \le H$. So $H$ contains one of $M_1$ and $M_2$, which enables us to reduce to the case when $m$ is a prime power $p^r$.

If $r>1$, then the subgroup $K = C_p^{m/p}$ is normal in $G$, and is equal to the Frattini subgroup of $M$. If $K$ is not contained in $H$, then $KH=G$, so $(M \cap H)K=M$, and hence $M \cap H = M$, contrary to assumption. So $K \le M$.

This enables us to reduce to the case when $m=p$ is prime, and then we can regard $M = C_p^n$ as the permutation module for $S_n$ over the finite field of order $p$.

Again, the assumption that $M$ is not in $H$ implies that $MH=G$, so $H$ has the structure $N.S_n$, where $N = M \cap H$ is a submodule of $M$ (which is equivalent to it being a normal subgroup of $G$). If $N$ was not a maximal submodule, then we would have $N < K < M$ with $K$ a normal subgroup of $G$, and then we would have $H < KH < G$, contradicting maximality of $H$.

So $N$ is indeed a maximal submodule of $M$. The maximal submodules of the permutation modules for $S_n$ are well-known, and the results are not too hard top prove directly, but I am afraid that I do not know a good reference. When $p$ does not divide $n$, it is the same as in characteristic $0$.