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

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An induced irreducible complex character is always irreducible

Statement

False claim: if $φ$ is an irreducible complex character of a subgroup $H \leq G$ , then ${Ind}_{H}^{G} φ$ is irreducible.

Facts & Assumptions

Given: The subgroup $H = ⟨ (12) ⟩ \leq S_{3}$ and its trivial character $1_{H}$ .

[F1]

The induced character ${Ind}_{H}^{S_{3}} 1_{H}$ equals $1 + χ_{2}$ , where $χ_{2}$ is irreducible of degree $2$ (Inducing the trivial character of a subgroup of order two in $S_{3}$ gives $1$ plus an irreducible degree-two character).

Refutation

technique · direct

1.1givenalgebra

The trivial character $1_{H}$ of the order-two subgroup $H$ is irreducible because every one-dimensional character is irreducible.

2.1F1step 1.1

But [F1] shows that its induction to $S_{3}$ is $1 + χ_{2}$ , a nontrivial sum of two characters. So the induced character is reducible.

3.1step 2.1∎

This single witness refutes the claim that induced irreducible characters are always irreducible.

Depends on

Inducing the trivial character of a subgroup of order two in $S_3$ gives $1$ plus an irreducible degree-two character

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

Pavel Etingof et al., Introduction to Representation Theory, Section 4.11 (standard reference, not scraped)