# Does the hyperoctahedral group have only 3 maximal normal subgroups?

An hyperoctahedral group $G$ is the wreath product of $S_2$ and $S_n$, where $S_{n}$ is the symmetric group on $n$ letters, or in other words the semi-direct product $G=S_2^n\rtimes S_n$, w.r.t. the action on the indices. $G$ comes with a natural action on $\times [n]$, so we consider it as a subgroup of $S_{2n}$.

$G$ has two normal subgroups of index two, the first $N_1$ is the preimage of $A_n$ under the quotient map $G\to S_n$ and the second $N_2$ is the intersection in $S_{2n}$ of $G$ and $A_{2n}$. This immediately gives a third normal subgroup of index two, coming from the diagonal in the Klein group $G/(N_1\cap N_2)$.

Are there any more maximal normal subgroups in $G$?

We checked this by computer for $n\leq 15$.

• It is well-known that the natural permutation module for $S_n$ (with $n>2$) over any field has just two proper nonzero submodules $V_1$ and $V_{n-1}$ of dimensions $1$ and $n-1$. Also $V_1 < V_{n-1}$ when the characteristic of the field divides $n$, and they are disjoint otherwise. The fact that $G$ has exactly three maximal normal subgroups can be deduced from this and the fact that $A_n$ is the only maximal normal subgroup of $S_n$. – Derek Holt Jan 1 '15 at 9:07

Perhaps I should expand my comment into an answer! The group $G$ in question is a split extension $V:S_n$ of the natural permutation module for $S_n$ over the field of order $2$ by $S_n$.

As I said, it is well-known that, at least when $n>2$, the permutation module $V$ (over any field) for $S_n$ over any field has exactly two nonzero proper submodules $V_1$ and $V_{n-1}$ of dimensions $1$ and $n-1$. Perhaps someone else knows a reference for that.

Let $N$ be a maximal normal subgroup of $G$. Since $A_n$ is the only maximal normal subgroup of $S_n$, if $V \le N$ then $N=V:A+n$, so suppose that $V \not\le N$.

Then, by maximality, we must have $NV = G$, and so $N \cap V$ is a submodule of $V$. It is not hard to show by direct computation that $[S_n,V] = V_{n-1}$, and hence $[G,V] = [N,V] = V_{n-1}$, so $V_{n-1} \le N$, and hence $V_{n-1} = N$. So $|G:N| = 2$ and $N$ is one of the two subgroups that you describe.

• To see that those are the only non-zero proper submodules is standard. Take $U$ be an irreducible non-zero submodule not fixed pointwise. Let $\sigma = ( 12 \ldots n )$ and $\tau=(12).$ Sum the images of $u \in U$ under the powers of $\sigma$. Since $U$ contains no non-zero vector with all coordinates equal, every element of $U$ has coordinate sum $0$. Now take $u \in U$ with coordinates 1 and 2 unequal. Then $u−u \tau$ means we can assume that $u_{1}=1,u_{2}=1$,$u_{j}= 0$ for $j>2.$ Take images of $u$ to get a basis of $V_{n−1}$ into $U.$ – Geoff Robinson Jan 2 '15 at 1:31
• Should be $u_{1} =1, u_{2} = -1$ in the above. – Geoff Robinson Jan 2 '15 at 11:33

Addition to the above posts. If $n=2$, then our wreath product $G\cong\text{D}_8$ so it has exactly $3$ normal subgroups of index $2$. If $n>2$, then $(S_2)^n$ is the umique minimal normal subgroup of $G$ so it is contained in all maximal normal subgroups of $G$. It sollows that $G$ has the same number of maximal normal subgroups as $G/(S_2)^n\cong S_n$, so that number is $2$.

• Try to read the question. $S^n$ is not a minimal normal subgroup. – YCor Jul 4 '16 at 0:39