False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)audited 2026-08-28

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FALSE: nonisomorphic groups acting on finite sets always have different cycle indices

Statement

False claim: if two finite groups are not isomorphic, then every action of the first has a different cycle index from every action of the second.

Facts & Assumptions

Given: the set $X = {1, 2, 3, 4}$ and the permutation $τ = (1 2) (3 4)$ .

[F1]

The cycle index averages the cycle monomials of the acting permutations (The cycle index of a finite permutation group).

Refutation

technique · direct

1.1

Let $C_{4} = ⟨ g ⟩$ act on $X$ through the quotient map $g \mapsto τ$ . Then the two even powers of $g$ act as the identity and the two odd powers act as $τ$ , so $Z_{C_{4}} (X) = \frac{1}{4} (2 s_{1}^{4} + 2 s_{2}^{2}) = \frac{1}{2} (s_{1}^{4} + s_{2}^{2})$ .

F1algebra

1.2

Let $V_{4} = {1, a, b, a b}$ act on $X$ through a quotient $V_{4} ↠ {1, τ}$ with kernel of size $2$ . Then again two group elements act as the identity and two act as $τ$ , so $Z_{V_{4}} (X) = \frac{1}{4} (2 s_{1}^{4} + 2 s_{2}^{2}) = \frac{1}{2} (s_{1}^{4} + s_{2}^{2})$ .

F1algebra

2.1

The groups $C_{4}$ and $V_{4}$ are not isomorphic, but steps 1.1 and 1.2 give the same cycle index. Therefore the displayed claim is false.

step 1.1step 1.2∎

Depends on

The cycle index of a finite permutation group

Used by

C₄ and V₄ can share the same cycle index through nonfaithful actions Counterexample
FALSE: the cycle index of a permutation action determines the abstract group False statement

Dependency tree · two levels

5 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)