Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Induction of an inertia constituent is irreducible

Statement

Let G be finite, N⊴G, θ∈Irr⁡(N), and I=IG(θ). If W is an irreducible complex I-module lying over θ, then W∣N is θ-isotypical and X=Ind⁡IGW is irreducible. Its θ-isotypical component is the identity-coset copy of W, consisting of the covariant functions supported on I.

Facts & Assumptions

Given: The groups, modules, characters, and hypotheses in the statement. All representations here are finite-dimensional complex left representations.

[F1]

For an irreducible module of a finite group, restriction to a normal subgroup has one orbit of constituents. (Normal restriction has one orbit of constituents).

[F2]

An N-submodule of a complex G-module is the direct sum of its intersections with the normal isotypical components. (Translation permutes normal isotypical components).

[F3]

Evaluation on a finite left transversal identifies an induced function module with the direct sum of its coset-supported copies of the inducing space. (A left transversal identifies Ind⁡HGW with a direct sum of [G:H] copies of W).

Proof

technique · direct
1.1F1given

Apply the orbit result inside I: every element of I fixes θ, so W∣N is a positive number of copies of θ.

2.1F3step 1.1algebra

Use a left transversal T containing 1, and write X=⨁t∈TXt, where Xt consists of functions supported on tI. For n∈N, covariance gives (nf)(t)=f(n−1t)=f(t(t−1n−1t))=(t−1nt)f(t). Thus evaluation makes Xt a module of type tθ. These types are distinct for distinct tI, so the Xt are exactly the normal isotypical components, and X1≅W as an I-module.

3.1F2step 2.1given∎

If Y⊆X is a nonzero G-submodule, the intersection decomposition gives Y∩Xt≠0 for some t. Left translation by t−1 carries Xt onto X1 and preserves Y, so Y∩X1≠0. This intersection is I-stable, hence equals X1 by irreducibility of W. Translating back fills every Xt, giving Y=X. This includes I=G (one block) and I=N (induction of the chosen normal type).

Depends on

Used by

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