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 Incoming links: Orbit count for simultaneous colorings

Past exam of the options I course of the University of Oxford / 2007 / A / A1 / Solution 2026-10-31

Burnside's lemma counts the orbits of a finite group action:

∣Ω/ G ∣ = \frac{1}{∣ G ∣} \sum_{g \in G} ∣ Fix (g) ∣.

(1)

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 is

\frac{6 ^{4} + 8 \cdot 6 ^{2} + 3 \cdot 6 ^{2}}{12} = 141 .

(2)

Allowing 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 becomes

\frac{6 ^{4} + 6 \cdot 6 ^{3} + 3 \cdot 6 ^{2} + 8 \cdot 6 ^{2} + 6 \cdot 6}{24} = 126 .

(3)

Equivalently, 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.

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