False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Every divisor of $∣ G ∣$ is an irreducible character degree

Statement

False claim: if $d$ divides $∣ G ∣$ , then $d$ is the degree of some irreducible complex character of $G$ .

Facts & Assumptions

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

[F1]

In $C_{4}$ , the divisor $2$ of $∣ C_{4} ∣ = 4$ is not an irreducible character degree ( $C_{4}$ shows that divisibility of irreducible degrees by $∣ G ∣$ is not an equivalence).

Refutation

technique · direct

1.1givenalgebra

The integer $2$ divides $∣ C_{4} ∣ = 4$ .

2.1F1step 1.1

But [F1] shows that no irreducible complex character of $C_{4}$ has degree $2$ .

3.1step 2.1∎

Hence not every divisor of the group order occurs as an irreducible character degree.

Depends on

$C_4$ shows that divisibility of irreducible degrees by $|G|$ is not an equivalence

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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)

Every divisor of ∣G∣∣G∣ is an irreducible character degree