I've been studying the symmetry groups of the platonic solids, and the cube has be confused. I've been considering the four diagonals on the cube. By writing out all permutation in $S_4$, it's clear that almost all permutations of the four diagonals can be reached by one rotational symmetry on the cube, except 3 permutations. These are the three of order $2$, namely $$(1,2)(3,4), (1,3)(2,4), (1,4)(2,3)$$ which is only reached by applying two consecutive rotations on the cube. They are reached by rotations around the axis between two opposite edges. Since the direct symmetry group of the cube is isomorphic to $S_4$, what additional information is gained by considering the reflections?
Consider this sketch: Cube with diagonals. The reflection through the plane in which diagonal $2,4$ live will give me the permutation $(1,3)$, but the same permutation can be reached by rotation around the axis between the edges connecting endpoints of diagonals $1,3$. I'm struggling to see what the reflections do to this group, because it seems that the full symmetry group should be larger than the direct symmetry group, which already is isomorphic to $S_4$.
I also, on the advice of my advisor, considered the symmetries of the cube on its faces. I labeled them as you would see on a six-sided die. By composing the reflection $(1,6)$ and the rotation $(1,4,6,3)$, I got a permutation of order $6$, namely $(3,1,2,4,6,5)$. No permutations of order $6$ appear in the direct symmetry group of the cube when considering the faces, so this seems to me to suggest that the full symmetry group is in fact larger than $S_4$, but I can't figure out exactly how to formualate it.
Any help or tips is greatly appreciated.