Full Octahedral Group

A regular octahedron has 24 rotational (or orientation-preserving) symmetries, and 48 symmetries altogether. These include transformations that combine a reflection and a rotation. A cube has the same set of symmetries, since it is the polyhedron that is dual to an octahedron.

The group of orientation-preserving symmetries is ''S''₄, the symmetric group or the group of permutations of four objects, since there is exactly one such symmetry for each permutation of the four diagonals of the cube.

Details

Chiral and full (or achiral) octahedral symmetry are the discrete point symmetries (or equivalently, symmetries on the sphere) with the largest symmetry groups compatible with translational symmetry. They are among the crystallographic point groups of the cubic crystal system.

As the hyperoctahedral group of dimension 3 the full octahedral group is the wreath product $S_2 \wr S_3 \simeq S_2^3 \rtimes S_3$,
and a natural way to identify its elements is as pairs $(m, n)$ with m \in direct_product_$S_4_\times_S_2$,_one_can_simply_identify_the_elements_of_tetrahedral_subgroup_''T_d''_as_$a_\in_[0,_4!)$_and_their_inversions_as_$a'$.

So_e.g._the_identity_$(0,_0)$_is_represented_as_$0$_and_the_inversion_$(7,_0)$_as_$0'$.

$(3,_1)$_is_represented_as_$6$_and_$(4,_1)$_as_$6'$.

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But as it is also the Direct product of groups">direct product $S_4 \times S_2$, one can simply identify the elements of tetrahedral subgroup ''T_d'' as $a \in [0, 4!)$ and their inversions as $a'$.

So e.g. the identity $(0, 0)$ is represented as $0$ and the inversion $(7, 0)$ as $0'$.

$(3, 1)$ is represented as $6$ and $(4, 1)$ as $6'$.

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