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The twelve rotational symmetries of a tetrahedron act on its six edges. Their edge cycle patterns are six fixed edges for the identity, two three-cycles for each of eight rotational symmetries of order three, and two fixed edges plus two transpositions for each of three half-turns. Burnside's lemma therefore counts edge colourings with $q$ labelled colours by $(q^6+8q^2+3q^4)/12$.

Suppose a finite group acts on two finite position sets $A$ and $B$. Color the first set using $r$ colors and the second using $s$ colors, with both colorings moved by the same group element. If $c_A(g)$ and $c_B(g)$ are the cycle counts of its two induced permutations, the number of simultaneous colorings fixed by $g$ is $r^{c_A(g)}{s}^{c_B(g)}$: each cycle must be monochromatic, and different cycles can be colored independently. Burnside's lemma therefore givesfor the number of orbits. When the two position actions are equivariantly identified, this simplifies to the usual coloring count with $rs$ combined colors. For example, a tetrahedron's face is paired with its opposite vertex, and this pairing commutes with every symmetry.

Pair a tetrahedron vertex with its opposite face. This is an equivariant correspondence under all tetrahedral symmetries, so three face colors and two vertex colors become six combined symbols at each of four positions. Burnside's lemma on the alternating group $A_4$ gives $(6^4+3\cdot 6^2+8\cdot 6^2)/12=141$. Including reflections gives $S_4$ and reduces the count to multisets of four symbols from six, namely $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126$.

For a finite group action $G\curvearrowright X$, Burnside's lemma statesCount pairs $(g,x)$ with $gx=x$ in two ways. Counting first by $g$ gives the displayed sum. Counting first by $x$ gives ${\sum }_x|G_x|$. For an orbit $\mathcal{O}$, the orbit-stabilizer theorem gives $|\mathcal{O}||G_x|=|G|$, so each orbit contributes exactly $|G|$. Division proves the formula.

Use colors $1,\dots ,m$, with variables $x_1,\dots ,x_m$, on the underlying finite set $S$. The weight of a coloring $f:S\to \{1,\dots ,m\}$ is $w(f)={\prod }_{s\in S}{x}_{f(s)}$. It is constant on each coloring orbit. The pattern inventory is $F_G(x)={\sum }_{\mathcal{O}}w(\mathcal{O})$, one weight for each orbit, including its multiplicity when different orbits have the same content.

If $c_r(g)$ is the number of length-$r$ cycles of $g$ on $S$, define the cycle indexIn the symmetric-function cycle-indicator convention, write $Z_G(x)$ for its substitution $t_r=p_r(x)={\sum }_{i=1}^m{x}_i^r$. A fixed coloring must be constant on every cycle of $g$. Coloring one length-$r$ cycle therefore has weight sum $p_r(x)$, independently of other cycles. The weighted Burnside lemma, proved by exactly the same orbit-stabilizer double count with weights, yieldsThis defines both sides in the same alphabet; the raw cycle index in the independent variables $t_r$ is specialized before comparison. The identity holds for all finite $m$ and hence stably for arbitrarily many colors.

For the last assertion, the permutation character on $T$ satisfies $\chi (g)=|{\mathrm{Fix}}_T(g)|$. A pair $(s,t)$ is fixed in the diagonal action on $T\times T$ exactly when both components are fixed. ThereforeThe character is real, so its character inner product with itself and another application of Burnside's lemma giveThis is the rank of a permutation action formula, and in fact does not require transitivity. If the action is transitive and $|T|\ge 2$, the diagonal is one orbit; rank two means exactly one further orbit on distinct ordered pairs, which is double transitivity.

Burnside's lemma counts the orbits of a finite group action:Identify each face with its opposite vertex. Every symmetry induces the same permutation on these two four-element sets, so a permutation with $c$ cycles fixes $3^c{2}^c=6^c$ colorings. The rotational symmetry group is $A_4$, with one identity, eight three-cycles, and three products of two disjoint transpositions. Their cycle counts are four, two and two. Thus the rotation count isAllowing reflections gives the full tetrahedral group $S_4$. In addition there are six transpositions with three cycles and six four-cycles with one cycle. The count becomesEquivalently, 30 of the 141 rotational classes form 15 mirror pairs, while 111 rotational classes are their own mirror class.

The product of fixed-coloring counts is an instance of the orbit count for simultaneous colorings.

Burnside's lemma gives the number of orbits asFor $m=2n-1\ge 3$, the polygon symmetry group contains the $m$ rotations through angles $2\pi j/m$, $0\le j<m$, and $m$ reflections. Each reflection axis passes through a vertex and the midpoint of the opposite edge. These are all the symmetries, so $|G|=2m=\boxed{4n-2}$.

For the pentagon, apply the same group element simultaneously to vertex colours and edge colours. The identity fixes $v^5{e}^5$ colourings. Each of the four nonidentity rotations is a 5-cycle on vertices and also on edges, fixing $ve$ colourings. Each of the five reflections has one fixed vertex and two swapped vertex pairs, and one fixed edge and two swapped edge pairs. It therefore fixes $v^3{e}^3$ colourings. Averaging yieldsThe two colour palettes are chosen independently, but rotations and reflections act on both parts of the same pentagon.

Count the pairs $(g,\omega )$ for which $g$ fixes $\omega$. Counting first by group elements gives ${\sum }_g\mathrm{Fix}(g)$, while counting first by points gives ${\sum }_{\omega }|\mathrm{Stab}(\omega )|$. On a group orbit of size $r$, orbit-stabilizer theorem makes each stabilizer have size $|G|/r$, so that group orbit contributes $|G|$ to the second sum. Dividing by $|G|$ proves Burnside's lemma:A tetrahedron has four vertices. A rotational symmetry is determined by the image of one vertex, with four possibilities, and the cyclic orientation of the three vertices around it, with three possibilities. All these choices are realizable, so its rotational symmetry group of a tetrahedron has order $12$. More explicitly, its elements are the identity, the two nontrivial rotational symmetries around each of four vertex-to-opposite-face axes, and the half-turn about each of three axes through opposite edge midpoints. Thus there is one element of order 1, eight of order 3, and three of order 2. The induced permutations of vertices are the even permutations, identifying the group with $A_4$.

Apply the group action to the set of edge colourings. A colouring is fixed by a rotational symmetry precisely when its colour is constant on each cycle of the induced edge permutation. The identity has six edge cycles and fixes $4^6$ colourings. A vertex-axis rotational symmetry has two cycles of length three, fixing $4^2$. An opposite-edge half-turn fixes the two edges on its axis as sets and exchanges the other four in two pairs, giving four cycles and $4^4$ fixed colourings. ThereforeReflections are excluded because the equivalence in the source uses rotational symmetries.

For a finite group action, Burnside's lemma givesPair each vertex with its opposite face. This correspondence is equivariant under every tetrahedral symmetry. Each pair has six possible combined color states: three face colors and two vertex colors. Thus we are coloring four positions with six symbols; the two kinds of coloring must be counted together.

The group of rotations is the alternating group $A_4$. The identity has four cycles and fixes $6^4$ colorings. Its three half-turns have cycle type $(2,2)$ and each fix $6^2$. Its eight third-turns have cycle type $(3,1)$ and each also fix $6^2$. Consequently the rotation count isAllowing mirror equivalence gives the full tetrahedral symmetry group $S_4$. Now all permutations of the four paired positions are allowed, so a coloring is determined by a multiset of four symbols chosen from six. stars and bars givesFor an independent Burnside check, the five cycle types of $S_4$ give numerator $6^4+6\cdot 6^3+3\cdot 6^2+8\cdot 6^2+6\cdot 6=3024$, whose quotient by 24 is 126. This is paired face and vertex colorings of a tetrahedron.

The rank is the number of orbits on ordered pairs under the diagonal group action. Its fixed-pair count at $g$ is $\chi (g{)}^2$, since the permutation character counts fixed points. Burnside's lemma identifies its average with the orbit count, while real-valuedness makes that average the character inner product of $\chi$ with itself. For a transitive action on at least two points, rank two is equivalent to transitivity on distinct ordered pairs.

For an orbit-invariant weight $w$ on a finite group action $G\curvearrowright X$, the sum of one weight per orbit is $|G{|}^{-1}{\sum }_g{\sum }_{x:gx=x}w(x)$. Double counting gives ${\sum }_x|G_x|w(x)$ before dividing by $|G|$; each orbit contributes $|G|$ copies of its common weight. Thus the usual Burnside's lemma extends to weights in any commutative coefficient ring.