Orbit-Stabilizer-Theorem and colourings of the cube

Question

This question is about colourings of the cube with 6 colours.

It is well known that one can colour a cube in 30 non-equivalent ways using 6 colours. The way to see this is to observe that a cube can be rotated into 24 different positions which means that for one colouring there are 23 others that are equivalent to it. Then total number of colourings $6!$ divided by $24$ yields $30$.

Talking in terms of group actions the set the group acts on is the set of colourings and the group is the set of isometries of the cube. The 24 coulourings that are rotaitonally equivalent form an orbit of a given colouring and the group of isometries acting on the cube.

One can use Burnside's lemma to find the number of orbits.

But I was wondering if the Orbit-Stabilizer-Theorem also applied. The theorem states that $\mid G \mid$ (= number of isometries of the cube = $48$) is equal to the stabilizer of a given element (=colouring) times the size of its orbit (=$24$).

Now the problem I have is that $\frac{48}{24}$ equals two and not $30$ as it should. Basically, using this I have calcualted that the stabilizers have size $2$ but I think that's wrong too because obviously the only isometry that leaves a given side alone is the identity.

So my questions are:

1. Is it possible to use the orbit-stabilizer-theorem to count ways to colour a cube?

2. What's the size of a stabilzer in this example?

Answer 1

The rigid motions of the cube forms a group of order $24$, as you observed. You can consider all $6!$ colorings (which are distinct but not necessarily equivalent), as the set on which the rigid motions of the cube act on. Then, two colorings are equivalent if and only if they belong to the same orbit. For any coloring, the stabilizer is the trivial subgroup of the rigid motions group, since you have to keep all faces intact (which is the trivial rigid motion) in order to stabilize the coloring. Thus, by the Orbit-Stabilizer theorem, the size of each orbit is: $$\frac{|G|}{|\text{Stab}(x)|} = \frac{24}{1}=24$$ Since all orbits are equal in size, the number of orbits i.e. the number of non-equivalent colorings is: $$\frac{|S|}{|\text{Orb(x)}|} = \frac{720}{24} = 30$$ This is exactly your first argument, where you had implicitly understood that the stabilizer is trivial.