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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12
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Relative Brauer homomorphisms are transitive

Statement

Let P≤G be a p-subgroup. If Q⊴R⊴P and also Q⊴P, then on (kCG(Q))P one has Br⁡P/R∘Br⁡R/Q=Br⁡P/Q. Also Br⁡P/Q(Br⁡Q(a))=Br⁡P(a) for a∈(kG)P when Q⊴P.

Facts & Assumptions

Given: The p-subgroup P and the normalities in the statement; normality is not assumed transitive.

[F1]

Relative maps truncate from one centralizer to the smaller centralizer on the appropriate fixed domain. (Relative Brauer homomorphism)

[F2]

Brauer maps commute with conjugation. (Brauer homomorphism is conjugation equivariant)

Proof

technique · direct
1.1F1F2

For a∈(kCG(Q))P, truncating to CG(R) leaves a P-fixed element: P normalizes both centralizers by the stated normalities and permutes their basis coefficients, so [F2] applies. Thus the composite has the displayed domain. Since CG(P)⊆CG(R)⊆CG(Q), truncating first to the middle set and then to the smallest retains exactly the same coefficients as truncating directly to the smallest.

2.1F1F2∎

If instead a∈(kG)P and Q⊴P, equivariance puts Br⁡Q(a) in (kCG(Q))P. Truncation to CG(Q) and then CG(P) retains exactly the CG(P) coefficients of a. This proves the second identity, also when any two subgroups agree.

Sources

Webb, A Course in Finite Group Representation Theory, §§11.3, 11.6 and 12.3–12.5, especially pp.240–245. Local argument and conventions as displayed above.

Depends on

Used by

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