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?



Maximal subgroups of a generalized symmetric group
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 $0$.


