I am looking for information on the hyperoctahedral group

From what I understand, the hyperoctahedral group is 'the generalized symmetric group' in the case where $m=2$. That is, the hyperoctahedral group = $S(2,n)$.

From what I see about the generalized symmetric group,

$S(1,n)$ permutes the set ${1, 2, 3, \cdots, n}$

ie $$S(1,n)= \hspace{.15cm}\langle(12),(23),\cdots,(n-1 \hspace{.2cm}n)\rangle,$$

It was my intuition that

$S(2,n)$ permutes the set $\{1_a, 2_a, 3_a, \cdots, n_a, 1_b, 2_b, 3_b, \cdots, n_b\}$

ie $$S(2,n)= \hspace{.15cm}\langle(12)_a,(23)_a,\cdots,(n-1\hspace{.2cm}n)_a, (12)_b,(23)_b,\cdots,(n-1\hspace{.2cm}n)_b\rangle$$

but was told this is not the case. Instead, $S(2,n)=\mathbb{Z}_2^n \rtimes S_n$. How then is $S(2,n)$ isomorphic to the signed permutation group?

**–Can someone give me some intuition on how this relates to the symmetries of octahedrons and their $n$-dimensional analogues? (For instance, we know that $S(2,n)=\mathbb{Z}_2^n \rtimes S_n$. Since there are 48 symmetries of the octahedron/cube, I assume this is $\mathbb{Z}_2^3\times S_3$ (ie n=3). How does this relate to the symmetries of the cube?)**

— We used the semi-direct product in our definition (which I have no experience with). Can I list the elements as I would have expected to with the direct product (ie $(\bar{0},\bar{1},\bar{0},(123)) \in S(2,3)$)?

— **Is it true that $S(2,3)=<(16),(35),(42),(13)(65),(14)(62),(45)(23)>$?**

## Best Answer

The hyperoctahedron has $n$ diagonals between opposite vertices, let's number them from $1$ to $n$, and name the vertices of the $i$-th diagonal $+i$ and $-i$ (arbitrarily).

What are the permutations $\sigma$ of the vertices that corresponds to isometries of the hyperoctahedron? The key observation is that opposite vertices are send to opposite vertices, so $\sigma(-i)=-\sigma(i)$. Other than that, there are no restriction: one can permute the diagonals as we want (this gives the $S_n$ factor), and then flip them independently (this gives the $\mathbb Z_2^n$ factor). In short, the hyperoctahedral group consists of all permutations $$\sigma\colon\{\pm 1,\pm 2,\ldots,\pm n\}\to\{\pm 1,\pm 2,\ldots,\pm n\}$$ such that $\sigma(-i)=-\sigma(i)$.

This group is isomorphic to the symmetries of the hypercube (of the same dimension). It just follows from the fact that hypercube and hyperoctahedron are dual. (If you pick a vertex at the center of each facet of an hypercube, you get an hyperoctahedron, and reciprocally.) So isometries of the hypercube coincide with isometries of the hyperoctahedron.

Indeed, one can list the elements as for the direct product (

as a set, the semi-direct product is the cartesian product), however the composition is different. (See definition of semi-direct products.)