Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 77 3 Solution Created 2026-10-03 Updated 2026-10-14
For a finite group action , Burnside's lemma statesCount pairs with in two ways. Counting first by gives the displayed sum. Counting first by gives . For an orbit , the orbit-stabilizer theorem gives , so each orbit contributes exactly . Division proves the formula.
Use colors , with variables , on the underlying finite set . The weight of a coloring is . It is constant on each coloring orbit. The pattern inventory is , one weight for each orbit, including its multiplicity when different orbits have the same content.
If is the number of length- cycles of on , define the cycle indexIn the symmetric-function cycle-indicator convention, write for its substitution . A fixed coloring must be constant on every cycle of . Coloring one length- cycle therefore has weight sum , 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 is specialized before comparison. The identity holds for all finite and hence stably for arbitrarily many colors.
For the last assertion, the permutation character on satisfies . A pair is fixed in the diagonal action on 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 , the diagonal is one orbit; rank two means exactly one further orbit on distinct ordered pairs, which is double transitivity.