# Hyperoctahedral group, preliminaries

finite-groupsgr.group-theorysymmetric-groups

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 $$S(2,3)=<(16),(35),(42),(13)(65),(14)(62),(45)(23)>$$?

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 $$\sigma(-i)=-\sigma(i)$$.