CounterexampleConstruction: Literature-sourcedVerification: Literature-sourcedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06

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Cyclic subgroups do not suffice for integral induction

Statement refuted

Every virtual character is an integral linear combination of characters induced from cyclic subgroups.

Facts & Assumptions

[F1]

The cited prerequisite is Frobenius' formula for the character of an induced representation.

Counterexample

Take $G = A_{5}$ . The trivial character $1_{G}$ is not in the integral cyclic induction subgroup.

Given: cyclic subgroups of $A_{5}$ have orders $1, 2, 3,$ or $5$ .

1.1

If $K \leq A_{5}$ is cyclic and $λ$ is linear, then ${Ind}_{K}^{A_{5}} λ (1) = [A_{5} : K]$ , which is respectively $60, 30, 20,$ or $12$ .

F1given

2.1

Every integral combination of such induced characters has even degree at $1$ , whereas $1_{G} (1) = 1$ . Thus $1_{G}$ cannot be such a combination. ∎

step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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

Tammo tom Dieck, Representation Theory, Section 4.6 (standard reference, not scraped)