MHB How Does Burnside's Lemma Calculate the Number of Distinct Colored Cubes?

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Q: How many different colored cubes are there in which each face is colored either red or white or blue?

I know we have to use Burnside's Lemma to solve this. (1/|G|)* Σ f(g) where f(g) are the number of fixed points of g. I also know that one of the values will be 3^6, but that is where I get stuck. Can someone please give an explanation as to how to find the rest of the equation? Thank you!
 
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The first thing you need to do is figure out what "$G$" is (in this case, it is $S_4$), and then figure out what $X$ is (the set that $G$ is acting upon). In this case it is the configuration space of all $3^6$ possible colorings of the cube, where we keep track of which face got which color (some of these we would not be able to tell apart under a physical rotation in the real world).

Indeed the number of fixed points of the identity is $3^6$, since every face remains unchanged.

Suppose we rotate about the line between the mid-point of the top face and the mid-point of the bottom face, say by 90 degrees. We can then ask: how may fixed points does this have?

Well, the only way this can happen is if the rotated faces (there are 4 of these) are all the same color, all red, or all white, or all blue. For each of these 3 possibilities, we are free to chose the top and bottom faces any way we choose. Since we have $3^2$ choices for top and bottom faces, this gives $3^3$ fixed points of $X$ for this rotation.

The same is true for any such rotation about two opposite faces. We have six such rotations, so a second term is:

$6\cdot 3^3$.

Your task, now, is to investigate the other three types of rotations (a 180 degree face rotation, about a vertex (120 degrees), and about two opposite edges (180 degrees)).
 
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