Consider the regular icosahedron. We want to compute number of ways to color the faces of the icosahedron with 2 colors.
My attempt :
Let's compute cycle index and use Polya's enumeration theorem.
We need to consider group of isometries of icosahedron. There is a $e$ (identical map) it produce $z_{1}^{20}$. Now there exists 4 types of rotation on $72^\circ$, $144^\circ$, $216^\circ$, $288^\circ$ (rotation around the axis connecting opposite vertices). So it produce $24z_5 ^4$. Now we may consider rotations around axis connecting middle points of opposites edges. They produce $15z_2^{10}$. And there are 2 types of rotation around axis connecting middle of opposites faces. They produce $20z_1^2 z_3^6$. So $P_{G} = \frac{1}{60}(z_1^{20}+24z_5^4 + 15z_2^{10}+20z_1^2 z_3^6)$. Substituting $z_i = 2$ gives us $P_G(2) = 17824$.
I have two questions :
1) Should we consider other types of symmetries ?
2) How can we decide the answer true or not?
UPD :
The OEIS tells me the answer is correct. So the second question is performed.
If you are only considering colorings equivalent up to rotation and not reflection, then these are the only types of symmetries you need to consider.
You can see this by counting: you've already accounted for $1+24+15+20=60$ symmetries, but in fact this must be the total: if we rotate an icosahedron into a copy occupying the same space, there are twelve places we can carry a given vertex, times five ways we can rotate the icosahedron once we fix the location of that vertex. So we know there can't be any other types of rotations.
As an aside, you can see fairly easily that performing "moves" consisting of any one of these kinds of rotation (besides the identity) will let you obtain any rotation of the icosahedron you want. From this, we conclude that every conjugacy class of the group generates the whole group, and hence that it is a simple group.