Would the rotation group of a cube change if the changes to the edges were focused on instead of the vertices?

Question

Currently learning abstract algebra and was wondering about rotation groups. I understand that the rotation group of a given cube is $24$. I was wondering if that number would change if the edges were kept track of instead of the vertices? My gut instinct says it should be the same. Since the edges and vertices are obviously linked, there's no way for an edge to move without at least one of the vertices it connects to moving as well. That seems to suggest that they would be the same. The groups of both the rotations per the edges and per the vertices should even be isomorphic since the two groups would be homomorphic and bijective.

Applying that same logic, if I define the cube as being hollow, with a face inside and outside the cube, and kept track of the $12$ faces ($6$ inside, $6$ outside), wouldn't that also end up be isomorphic? If I kept track of just the outside faces it couldn't be bijective could it?

I'm still learning all this though so I wanted someone to confirm my logic so that I'm not barking up the wrong tree so to speak.

Answer 1

The edges' graph of a cube is a cuboctahedron as can be easily seen by joining the midpoints of neighbour edges of a cube.

You are reasoning on the group of rotations leaving the cube invariant. I will consider here a larger group (in fact with $48$ elements instead of $24$) by including also symmetries wrt to planes which is the group of all isometries (either direct, with determinant $1$ as your rotations, or non-direct, with determinant $-1$, the symmetries I just introduced).

The isometry group of the cube is the octahedral group ($O_3$) (see here).

The isometry group of the cuboctahedron is isomorphic to it. This is established as Theorem B5 in this document.

This isometry result can as well be obtained by reviewing a sufficient number of different isometries leaving the cube invariant and see that each one leaves as well the cuboctahedron invariant.

Remark: one could be convinced that the octahedron play a rôle here because the neighborhood graph of the faces of a cube is an octahedron.