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

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.)

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