Branching rule from symmetric group $S_{2n}$ to hyperoctahedral group $H_n$ Embed the hyperoctahedral group $H_n$ into the symmetric group $S_{2n}$ as the centralizer of the involution $(1, 2) (3, 4) \cdots (2n-1, 2n)$ (cycle notation). Label representations of $S_{2n}$ by partitions of $2n$ and label representations of $H_n$ by pairs of partitions whose sizes add up to $n$ in the standard way. I am looking for a combinatorial description of the branching rule from $S_{2n}$ to $H_n$. This should be in the literature, but I couldn't find it.
Second small question: Is it possible to have a different embedding of $H_n$ into $S_{2n}$ so that the rule changes? I'd guess probably, but didn't think hard about it.
 A: Okay I found it in the article by Koike and Terada, "Littlewood's formulas and their application to representations of classical Weyl groups". Here is the theorem. Define $d^\nu_{\lambda, \mu}$ as the coefficients in the product of plethysms:
$\displaystyle (s_\lambda \circ h_2)(s_\mu \circ e_2) = \sum_\nu d^\nu_{\lambda, \mu} s_\nu$
where $s_\lambda$ is a Schur function, $h_2$ is the complete homogeneous symmetric function of degree 2, and $e_2$ is the elementary symmetric function of degree 2. Then the restriction of the representation $\chi(\nu)$ of $S_{2n}$ to $H_n$ is $\sum_{\lambda, \mu} d^\nu_{\lambda, \mu} \chi(\lambda, \mu)$.
Richard Stanley has informed me that a more general statement appears as Theorem A2.8 of Enumerative Combinatorics Vol. 2 (set $k=2$). This formula gives the coefficients for representations of $H_n$ indexed by bipartitions where one of the partitions is the empty set. The general case comes from the fact that $\chi(\lambda, \mu)$ is induced from a parabolic subgroup $H_k \times H_{n-k}$ of $H_n$ and that the map ch takes induction to product of characters.
A: This is an answer to the second question: I ran an experiment with $S_6$ (which was the best guess due to the famous "oddness" of 6). There are two subgroups in $S_6$ isomorphic to this involution centraliser (up to conjugation of course) and the irreducible characters of $S_6$ decompose differently upon restriction to these two subgroups - even up to automorphism of $H_n$. E.g. there is a 5-dimensional character whose one restriction has a 2-dimensional summand in its decomposition, but the other restriction doesn't.
In $S_8$, there is only one conjugacy class of subgroups isomorphic to $H_4$ and that's where the computing power of my little laptop ends.
