prove that there are 9099 different ways of colouring the faces of a dodecahedron red, white or blue. (this is from Amstrong's "group and symmetry") attempt: Since the question refers to the faces of the dodecahedron, i know we need the cycle index of the group of the icosahedron, acting on its vertices. There is a table for that. but since this is a group theory problem at a introductory level, I need a clear and specific reasoning. any help would be appreciated!

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    $\begingroup$ related: math.stackexchange.com/questions/53859/coloring-dodecahedron $\endgroup$ – vadim123 May 4 '13 at 2:28
  • $\begingroup$ I saw the link you provided before. but I need a specific proof. thanks $\endgroup$ – Alex May 4 '13 at 2:43
  • $\begingroup$ If you saw that link before, I think it would be appropriate a) to provide that link yourself, so people don't waste your and their time explaining things to you that you already know, and b) describe how you tried to apply that answer and where you got stuck. $\endgroup$ – joriki May 4 '13 at 2:53
  • $\begingroup$ If you're coloring the faces of a dodecahedron, why do you need the cycle index of the group of the icosahedron? $\endgroup$ – Gerry Myerson May 4 '13 at 3:59
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    $\begingroup$ @Gerry: Because the faces of the dodecahedron correspond to the vertices of the icosahedron. $\endgroup$ – joriki May 4 '13 at 14:33

We compute the cycle index of the permutation group of the faces. In order to answer this question it is best to work with an image of the dodecahedron like the one at Wikipedia. Look at the image and rotate it in your mind. By considering the properties of the object as they appear by visual inspection, we see that there are three types of symmetries in addition to the identity: rotations about an axis passing through the centers of two opposite faces, rotations about an axis passing through opposite vertices and 180 degree rotations that flip two opposite edges, mapping each onto itself.

The identity constributes the following term to the cycle index: $$ a_1^{12}.$$

There are six pairs of opposite faces and four rotations for each of these which fix the two opposite faces and create two five-cycles, giving $$ 6 \times 4 \times a_1^2 a_5^2 = 24 a_1^2 a_5^2.$$

There are ten pairs of opposite vertices and two rotations for each of these which create two three-cycles at the two vertices. The two rotations create two three-cycles among the faces not adjacent to the two vertices, giving $$ 10 \times 2 \times a_3^4.$$

There are fifteen pairs of opposite edges and the 180 degree rotations about the plane passing through them partition everything into two-cycles, giving $$ 15 \times a_2^6.$$

It follows that the cycle index of the permutation group $G$ of the faces is $$ Z(G) = \frac{1}{60} \left( a_1^{12} + 24 a_1^2 a_5^2 + 20 a_3^4 + 15 a_2^6\right).$$

Now evaluating $Z(G)$ at $X_1 + X_2 + \cdots + X_n$ and setting $X_1 = X_2 = X_3 = \ldots = X_n = 1$, we obtain the following sequence of values: $$1, 96, 9099, 280832, 4073375, 36292320, 230719293, 1145393152, 4707296613, 16666924000.$$ Of course this provides the generating functions as well, e.g. for two colors we get $${X_{{1}}}^{12}+{X_{{1}}}^{11}X_{{2}}+3\,{X_{{1}}}^{10}{X_{{2}}}^{2}+5\,{X_{{1}}}^{9}{X_{{2}}}^{3}+12\,{ X_{{1}}}^{8}{X_{{2}}}^{4}+14\,{X_{{1}}}^{7}{X_{{2}}}^{5}+24\,{X_{{1}}}^{6}{X_{{2}}}^{6}\\+14\,{X_{{1}}}^{ 5}{X_{{2}}}^{7}+12\,{X_{{1}}}^{4}{X_{{2}}}^{8}+5\,{X_{{1}}}^{3}{X_{{2}}}^{9}+3\,{X_{{1}}}^{2}{X_{{2}}}^ {10}+X_{{1}}{X_{{2}}}^{11}+{X_{{2}}}^{12}.$$

Substituting into the cycle index we obtain the explicit formula $$\frac{1}{60} \left(n^{12} + 24 n^4 + 20 n^4 + 15 n^6\right) ={\frac {1}{60}}\,{n}^{12}+\frac{1}{4}\,{n}^{6}+{\frac {11}{15}}\,{n}^{4}.$$

Remark, Nov 12 2018. Obviously when we only seek a count rather than a classification we do not need to substitute $n$ variables into the cycle index. It is sufficient to use Burnside with the substitution $a_q = n.$ We obtain the sequence OEIS A000545.

  • $\begingroup$ Nice. As noted in the comments on the original question, the numerical sequence is tabulated at the OEIS. $\endgroup$ – Gerry Myerson May 4 '13 at 23:48
  • $\begingroup$ Gerry Myerson , take me to that OASIS ( what's the link ) $\endgroup$ – Randin Nov 12 '18 at 12:53

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