Geometrical intuition for the action of symmetry group of a cube. Let $G$ be the symmetry group of a cube. It has the group of rotational symmetries $H$($\cong S_4$) as a normal subgroup of index two. 
Now this is the kernel of some action of $G$ on a set of size two. 
My question is: how can I visualise this set of size two as a substructure of the cube? Is it the set of two inscribed tetrahedra? I can't justify to myself rigorously if it is.
 A: Answering my own question - I think this is what's going on.
It is easy to visualise sets acted on by the rotation group $H$ as substructures of the cube because the rotations can be performed on a single physical cube. For example, $H$ acts on the set of three lines connecting centres of opposite faces. This set has symmetry group $S_3$ so the action is not faithful. Since this set of lines is fixed by the identity and the 180 degree rotation about any of the three lines, the kernel of the action is a group with identity and three involutions, i.e. the Klein 4-group.
The problem with the full symmetry group is that essentially it contains the rotations of two different physical cubes (one of them "reflected through the centre"). So if you quotient out by the rotation group, the image is precisely the two different versions of the cube. So there is no "substructure" of the physical cube. What is preserved by the rotation group (i.e. what makes it the kernel of the action) is whether the cube has been reflected or not.
A: The full symmetry group of the cube is $S_4\times\mathbb{Z}_2$, where $S_4$ is the rotational group and $\mathbb{Z}_2$'s nontrivial element is the matrix $-I_3$. It should be obvious $-I_3$ is orientation-reversing (determinant $-1$) but also is a symmetry, and that it commutes with the rotations hence there is a direct product, and the rotational subgroup must have index two so this is the full symmetry group. Thus, the rotational subgroup is the kernel of the action on the two possible orientations of the cube. Keep in mind the $4$ features being permuted by the rotational subgroup are the four so-called space-diagonals between antipodal vertices.
Observe $-I_3$ switches the two inscribed tetrahedra, and we can check manually the rotations preserve them if and only if they are even permutations of the space diagonals. Thus, the kernel of the action on the two tetrahedra is $(A_4\times\{0\})\sqcup(S_4\setminus A_4\times\{1\})$. That is, $(\sigma,\varepsilon)\in S_4\times\mathbb{Z}_2$ is in this kernel if and only if the permutation $\sigma$ and residue $\varepsilon$ have the same parity. This subgroup is isomorphic to $S_4$ via the isomorphism $(\sigma,\varepsilon)\mapsto\sigma$.
