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Having $n$ colors, use the lemma to find a formula for the number of ways to color the edges of the cube.

Here is what I have so far: The Burnside lemma says that $\displaystyle |X/G| = \dfrac{1}{|G|} \sum_{g \in G} |X^g|$ where $G$ is a finite group, $X^g$ is the set of elements fixed by $g \in G$, and $|x/G|$ is the number of orbits. Now, using $n$ colors, we have the following:

1 identity element leaving $n^6$ elements of $X$ unchanged.

6 90-degree face rotations each leaving $n^3$ elements unchanged.

3 180-degree face rotations each leaving $n^4$ elements unchanged.

8 120-degree vertex rotations each leaving $n^2$ elements unchanged.

6 180-degree edge rotations each leaving $n^3$ elements unchanged.

Using the above results and the formula from Burnside's lemma, we obtain $\dfrac{n^6+6 \cdot n^3 + 3 \cdot n^4 + 8 \cdot n^2 + 6 \cdot n^3}{24} = \dfrac{n^6 + 3n^4 + 12n^3 + 8n^2}{24}$.

I think I did it right but wanted to check with you guys. Thank you!

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  • $\begingroup$ Did you mean that you're coloring the 6 faces or the 12 edges? $\endgroup$ Mar 19, 2013 at 4:40

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The algorithm is correct. But note that you are coloring edges. What you did is not.

For example, for identity, there are actually $n^{12}$ elements fixed. Because there are $12$ edges and each edge has $n$ colors.

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  • $\begingroup$ Oh, wow. Thank you guys for pointing out my mistake. I'm working on it now and will ask further questions if I have any. Any help for "edges" will be appreciated! $\endgroup$
    – Math Damon
    Mar 19, 2013 at 5:22
  • $\begingroup$ I got $\dfrac{n^{12} + 6n^3 + 3n^6 + 8n^4 + 6n^7}{24}$ but when I try $n=2$ or $n=3$, the results seem too large. Thanks for your help. $\endgroup$
    – Math Damon
    Mar 19, 2013 at 8:54

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