Group Theory: Coloring

I am really stuck on what to do with this problem. We have started burnside's theorem and this is suppose to be a warm-up exercise.

How many ways can the vertices of a regular tetrahedron be colored using at most 3 colors?

• Are there any restrictions on how the colors can be assigned? If it is completely arbitrary then the answer is just that you can pick an arbitrary one out of three colors for each vertex, giving you $3^4=81$ colorings. – ajd Mar 20 '14 at 18:20
• Detailed examples of the Cauchy-Frobenius-Burnside-Redfield-Pólya theorem have been posted here in the past; you may want to look at one of them: 1 2 3 4 5 – MJD Mar 20 '14 at 19:30

I'm assuming that colorings are considered the same if they can be obtained from each other by rotating the tetrahedron. I'm also assuming that the colorings that are obtained by reflection are not considered the same. If these assumptions are correct, then the relevant group is $A_4,$ the group of rotational symmetries of the tetrahedron. Added: It turns out that for three colors, it doesn't matter whether reflections are included in the symmetry group or not - you get the same answer either way. If reflections are included, the relevant symmetry group is $S_4.$
For each of the $12$ elements of $A_4,$ you need to figure out the number of colorings that are fixed under the action of that group element. For example, consider rotation by $120^\circ$ about the vertical axis. For a coloring to be fixed by that rotation, the three base vertices must be colored the same. Hence there are nine such colorings since there are three choices for the color of the top vertex and three choices for the color of the base vertices.
Added: Once you've got the number of colorings fixed by each of the group elements, Burnside's Theorem states that you need to compute the average number of fixed colorings over all of the group elements. (So, if the group is $A_4,$ add up the number of fixed colorings for each of the $12$ group elements, and divide the total by $12.$)