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 $[2]\times [n]$, 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 $n\leq 15$.
 A: 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.
A: 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  $2$.
