Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
Did you mean that you're coloring the 6 faces or the 12 edges? –  Sammy Black Mar 19 '13 at 4:40
add comment

1 Answer 1

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.

share|improve this answer
    
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! –  Math Damon Mar 19 '13 at 5:22
    
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. –  Math Damon Mar 19 '13 at 8:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.