ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$C_{4}$ shows that divisibility of irreducible degrees by $∣ G ∣$ is not an equivalence

Example

The divisor $2$ of $∣ C_{4} ∣ = 4$ is not the degree of any irreducible complex character of $C_{4}$ . So the theorem $χ (1) ∣ ∣ G ∣$ is a necessary condition for irreducible degrees, not a characterization.

Facts & Assumptions

Given: The cyclic group $C_{4}$ .

[F1]

A cyclic group of order $4$ exists and is finite (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

[F2]

Every irreducible representation of a finite abelian group over a splitting field is one-dimensional (Every irreducible representation of a finite abelian group over a splitting field is one-dimensional).

[F3]

Irreducible complex character degrees divide the group order (The degree of an irreducible complex character divides $∣ G ∣$ ).

Verification

technique · direct

1.1

By [F1], $C_{4}$ is a finite cyclic group of order $4$ , hence finite abelian.

F1given

2.1

Over $C$ , every irreducible representation of the finite abelian group $C_{4}$ has degree $1$ by [F2]. Therefore no irreducible complex character of $C_{4}$ has degree $2$ , even though $2 ∣ 4$ .

F2step 1.1algebra

3.1

This shows that [F3] is not an equivalence: a divisor of $∣ G ∣$ need not occur as an irreducible degree.

F3step 2.1∎

Depends on

Used by

Every divisor of |G| is an irreducible character degree False statement

Dependency tree · two levels

21 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

Peter Webb, A Course in Finite Group Representation Theory, Section 3.5 (standard reference, not scraped)

C4C4​ shows that divisibility of irreducible degrees by ∣G∣∣G∣ is not an equivalence