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?
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.

It might be easier to visualize as just rotating a hypercube in one dimension, since they are defined as the group of symmetries of those objects. They all are composed of squares, so just rotate one of the squares four times, and there will be an element of order four (that rotation).
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.

It might be easier to visualize as just rotating a hypercube in one dimension, since they are defined as the group of symmetries of those objects. They all are composed of squares, so just rotate one of the squares four times, and there will be an element of order four (that rotation).