# Hyperoctahedral group and odd permutations?

I apologize in advance for possibly using faulty terminology as I am a group theory novice. I am interested in looking at Hyperoctahedral groups when viewed as permutations. According to Wikipedia, when $n=2$, one can obtain any of the permutations of the square (https://en.wikipedia.org/wiki/Hyperoctahedral_group), because one of it's group elements is an odd permutation (i.e the cycle on 4 elements). However, there is no odd permutation in the group when $n=3$. Does this hold for larger values of $n$ as well?

• This is a bit of an advanced topic for a group theory novice, no? – theREALyumdub May 3 '18 at 21:41
• It's for a research problem in a different area (coming from CS), that may have some connections here depending on the answer. – Nizbel99 May 3 '18 at 21:42
• might I ask what the somputer science problem is, if it's roughly simple to explain to an undergrad mathematics major? – theREALyumdub May 3 '18 at 22:45
• The thing about permutations being odd or even is that this requires us to have identified them as permutations in the first place. And any group can be embedded in the group of even permutations of a suitable set, which makes this distinction only meaningful when we are discussing a specific way to realize everything as permutations. In this case, are you considering the group as a subgroup of the symmetric group on $2n$ elements? – Tobias Kildetoft May 4 '18 at 6:28
• As mentioned in a comment on the answer, you may want to be more specific about what you are looking for. I do know a decent amount about these groups, as they are Coxeter groups of type $B$, but I am not at all sure what would constitute an answer to your question. – Tobias Kildetoft May 7 '18 at 6:57

I'm not sure what exactly you are stating when you say that there is no odd permutation in the case $n = 3$, mostly because a group is not just a collection of permutations, it can have a lot of different structure.
Anywho, the wikipedia page also says the hyper octahedral group with $n = 3$ is isomorphic to $S_4 \times S_2$, and thus contains the element $( (1, 2, 3, 4) , e )$, which you might consider in some sense to be an odd permutation.
• Thank you for your answer! What I am trying to say is: when you look at the "Illustrations of rotoreflections" for the cube ($n =3$) - en.wikipedia.org/wiki/Octahedral_symmetry - they all induce permutations with an even number of inversions. However this is not the case for $n=2$. But based on what you have said, it sounds like the answer should be a no. – Nizbel99 May 3 '18 at 22:00
• Here they have an exhaustive list for $n=3$: en.wikiversity.org/wiki/Full_octahedral_group – Nizbel99 May 3 '18 at 22:04
• Sylow's theorem says that there is a subgroup of order $p^n$, not an element of that order. There is a group of order $4$ with no element of order $4$. – Derek Holt May 3 '18 at 22:36
• The question is, i think, the following one. Fix $n\ge 4$. Let $X=HC(n)=\{-1,+1\}^n$ be the cartesian product of $n$ copies of the (multiplicatively written) group with two elements. The group of signed permutations $W$, acts on this hypercube $HC(n)$, $(w,x)\to wx$, and thus each $w$ induces a permutation, we write then $\rho:W\to S(X)$ for this group morphism $w\to \rho(w)$, from $W$ to the symmetric = permutation group of the symbols in $X$. From $S(X)$ we have a sign morphism to $\pm 1$. Question: Is $W\to S(X)\to \pm 1$ surjective - for a specific $n$? For which $n$? – dan_fulea May 3 '18 at 22:58