Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedaudited 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
1.1F1algebra

Let C4=⟨g⟩ act on X through the quotient map g↦τ. Then the two even powers of g act as the identity and the two odd powers act as τ, so ZC4(X)=14(2s14+2s22)=12(s14+s22).

1.2F1algebra

Let V4={1,a,b,ab} act on X through a quotient V4↠{1,τ} with kernel of size 2. Then again two group elements act as the identity and two act as τ, so ZV4(X)=14(2s14+2s22)=12(s14+s22).

2.1step 1.1step 1.2∎

The groups C4 and V4 are not isomorphic, but steps 1.1 and 1.2 give the same cycle index. Therefore the displayed claim is false.

Depends on

Used by

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