Coloring the faces of the regular icosahedron, again... The calculation of the number of ways to color the faces of the regular icosehedron by 2 different colors is given in this link: Coloring the faces of a regular icosahedron with $2$ colors
My question is whether there is another way to do this computation less geometrically knowing that the rotation group of the regular icosehedron is the alternating group $A_5$.
Recall that the number of elements in the conjugacy classes of this group are 1, 20, 15 and 24 and these numbers are the coeffecients of the cycle index polynomial obtained in the above link. But, how can we explain the powers?
And in general, it would be a great assistance if you could give any information about platonic solids's cycle index polynomial computation and their applications.
 A: While the cycle index of the rotation group of a Platonic solid can be connected to the cycle index of solid itself, it cannot fully determine it: The cube and octahedron are duals and therefore have the same rotation group ($S_4$), but the two solids have different cycle index polynomials.  Likewise the dodecahedron and icosahedron which share the rotation group $A_5$.
\begin{align*}
A_5: \quad \frac{1}{60}\left(z_1^5 + 15z_1z_2^2 + 20z_1^2z_3 + 24z_5\right) \\
\text{Dodecahedron:} \quad \frac{1}{60}\left(z_1^{12} + 15z_2^6 + 20z_3^4 + 24z_1^2z_5^2\right) \\
\text{Icosahedron:} \quad \frac{1}{60}\left(z_1^{20} + 15z_2^{10} + 20z_1^2z_3^6 + 24z_5^4\right)
\end{align*}
To derive the icosahedron cycle index from that of $A_5$, you need to think about what five things are being permuted (five tetrahedra inside the icosahedron) and how permuting them affects the faces of the solid.
For good measure, here are the analogous cycle index polynomials for the other dual pair of Platonic solids.
\begin{align*}
S_4: \quad \frac{1}{24}\left(z_1^4 + 3z_2^2 + 6z_1^2z_2 + 6z_4 + 8z_1z_3\right) \\
\text{Cube:} \quad \frac{1}{24}\left(z_1^6 + 3z_1^2z_2^2 + 6z_2^3 + 6z_1^2z_4 + 8z_3^2\right) \\
\text{Octahedron:} \quad \frac{1}{24}\left(z_1^8 + 9z_2^4 + 6z_4^2 + 8z_1^2z_3^2\right)
\end{align*}
In the cube, the four things being permuted are diagonals connecting antipodal vertices.  In the octahedron, it's the four segments connecting the midpoints of opposite faces.  Permuting each of those four things will change the faces of the respective solids differently, thus the different cycle indexes.
At least the cycle indexes for $A_4$ and the tetrahedron are the same.
For a reference, see Biggs Discrete Mathematics (Oxford, 2002) ch. 27, "Symmetry and Counting."  The cycle indexes for the Platonic solids are summarized in Table 27.3.2 on p. 397.
