Application of Burnside's Theorem, Colorings

[saubbie]

Homework Statement

I have multiple problems, all dealing with Burnside's Theorem, perhaps help on one would help explain the others. How many ways may the faces of a cube be colored using 3 colors, up to symmetry of the cube. And, how many ways may the faces of a dodecahedron be colored using 5 colors up to symmetry.

Homework Equations

Burnside's Theorem: # colorings= 1/(group of symmetry)*Sum of number of fixed colorings

The Attempt at a Solution

The symmetry groups are S4 and A5 respectively. The part that I get hung up on is the number of fixed colorings. I know for the identity permutation on the cube, all colorings are fixed, but I don't know how the other rotations work. Please help!

 

[mighty2000]

Hi there,

I can definitely help you with this problem! Burnside's Theorem is a useful tool for counting colorings up to symmetry, but it can get tricky when dealing with more complex shapes like a cube or dodecahedron. Let's take a look at the cube problem first.

To start, we need to determine the size of the symmetry group for a cube. The cube has 6 faces, so we have 6 rotations (identity, 90 degree rotation about each of the 3 axes, and 180 degree rotation about each of the 3 axes). We also have 8 rotations that involve flipping the cube (180 degree rotation about each of the 4 diagonals). Finally, we have 3 rotations that involve rotating the cube by 120 degrees about a face. This gives us a total of 24 rotations, which means our symmetry group is S4.

Now, for the number of fixed colorings, we need to think about how each rotation affects the coloring. For the identity rotation, all colorings will be fixed. For the 90 degree rotations, each face will be rotated to a different face, so no colorings will be fixed. For the 180 degree rotations, each face will be rotated to itself, so all colorings will be fixed. For the flips, each face will be flipped to a different face, so no colorings will be fixed. Finally, for the 120 degree rotations, each face will be rotated to a different face, so no colorings will be fixed.

Using Burnside's Theorem, we have:

# colorings = 1/24 * (6 + 0 + 6 + 0 + 0) = 1/4 * 12 = 3

So there are 3 ways to color the faces of a cube using 3 colors, up to symmetry.

Now, let's tackle the dodecahedron problem. The dodecahedron has 12 faces, so we have 12 rotations (identity, 72 degree rotation about each of the 5 axes, and 144 degree rotation about each of the 5 axes). We also have 20 rotations that involve flipping the dodecahedron (180 degree rotation about each of the 10 edges). This gives us a total of 32 rotations, which means our symmetry group is A5.

Using the same reasoning as before, we have:

# colorings