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 $G/(N\_1\cap N\_2)$.



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> Are there any more maximal normal subgroups in $G$?
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We checked this by computer for $n\leq 15$.



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


