Let a regular tetrahedron be colored with \(C\) available colors.
- Count colorings of its four vertices (equivalently, its four faces) up to the full symmetry group.
- Count colorings of its six edges up to the full symmetry group.
- How do the answers change if only orientation-preserving rotational symmetries are allowed?
Solution
The full symmetry group of a tetrahedron is \(S_4\), of order \(24\), acting naturally on the four vertices; the rotational subgroup is \(A_4\), of order \(12\).
1. Vertices or faces under all symmetries.The cycle types in the natural action of \(S_4\) are\[ 1^4,\quad 2\,1^2,\quad 2^2,\quad 3\,1,\quad 4, \]with multiplicities \(1,6,3,8,6\). A permutation with \(k\) cycles fixes exactly \(C^k\) colorings. Burnside’s lemma gives\[ \frac{C^4+6C^3+3C^2+8C^2+6C}{24} = \boxed{\frac{C^4+6C^3+11C^2+6C}{24}}. \]Equivalently this is \(\binom{C+3}{4}\), since the full \(S_4\) action only remembers the multiset of four colors.
2. Edges under all symmetries.Identify the six edges with the two-element subsets of \(\{1,2,3,4\}\). The five conjugacy types above induce respectively \(6,4,4,2,2\) cycles on the edge set. Hence Burnside gives\[ \boxed{\frac{C^6+9C^4+14C^2}{24}}. \]
3. Rotations only.The subgroup \(A_4\) consists of the identity, three double transpositions, and eight \(3\)-cycles.For vertices/faces the corresponding cycle counts are \(4,2,2\), so the number of rotational orbits is\[ \boxed{\frac{C^4+11C^2}{12}}. \]For edges the cycle counts are \(6,4,2\), giving\[ \boxed{\frac{C^6+3C^4+8C^2}{12}}. \]