TheoremStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)audited 2026-08-28

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Pólya's enumeration theorem

Statement

Let a finite group $G$ act on a finite set $X$ , and let $C$ be a finite colour set with $m = ∣ C ∣$ . Then the number of $G$ -orbits of colourings $X \to C$ is

$Z_{G} (m, m, \dots, m) = \frac{1}{∣ G ∣} \sum_{g \in G} m^{\sum_{d \geq 1} j_{d} (g)} .$

Facts & Assumptions

Given: a finite group action $G ↷ X$ and a finite colour set $C$ with $m = ∣ C ∣$ .

[L1]

Cauchy-Frobenius orbit counting averages the fixed-point counts of the acting group (Cauchy-Frobenius orbit counting: $∣ G ∣ ∣ X / G ∣ = \sum_{g \in G} ∣ X^{g} ∣$ for a finite group action).

[L2]

A group element with cycle counts $j_{d} (g)$ fixes exactly $m^{\sum_{d} j_{d} (g)}$ colourings (Fixed colourings factor by cycle type).

Proof

technique · direct

1.1

Apply [L1] to the induced action of $G$ on the colouring set $C^{X}$ . The number of colouring orbits is therefore $\frac{1}{∣ G ∣} \sum_{g \in G} ∣ {f : X \to C : g \cdot f = f} ∣$ .

2.1

Replace each fixed-colouring count in step 1.1 by the formula from [L2]. This gives $\frac{1}{∣ G ∣} \sum_{g \in G} m^{\sum_{d} j_{d} (g)}$ .

step 1.1L2

3.1

By the definition of the cycle index, substituting $s_{d} = m$ for every $d$ turns $\frac{1}{∣ G ∣} \sum_{g \in G} \prod_{d \geq 1} s_{d}^{j_{d} (g)}$ into the sum of step 2.1. Hence the orbit count is $Z_{G} (m, m, \dots, m)$ .

step 2.1∎

Depends on

Used by

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ben Lynn, Polya Theory: The Cycle Index Polynomial (standard reference, not scraped)