LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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Brauer homomorphism is conjugation equivariant

Statement

For $g \in G$ and $a \in (k G)^{P}$ , write $^{g} a = g a g^{- 1}$ and $^{g} P = g P g^{- 1}$ . Then $^{g} a \in (k G)^{^{g} P}$ and ${Br}_{^{g} P} (^{g} a) =^{g} {Br}_{P} (a)$ .

Facts & Assumptions

Given: $G, P, g, a$ as stated.

[F1]

The map retains centralizer coefficients. (Brauer homomorphism for a p subgroup)

Proof

technique · direct

1.1F1

For $p \in P$ , conjugating $^{g} a$ by $g p g^{- 1}$ gives $g p a p^{- 1} g^{- 1} = g a g^{- 1}$ , proving the required fixed-algebra membership. Also $x \in C_{G} (P)$ holds exactly when $g x g^{- 1} \in C_{G} (g P g^{- 1})$ , by conjugating each commutation equation.

2.1F1step 1.1∎

Writing $a = \sum_{x} a_{x} x$ , either order of conjugation and truncation yields $\sum_{x \in C_{G} (P)} a_{x} g x g^{- 1}$ . This is the required equality; no coefficient is changed.

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

Brauer homomorphism for a p subgroup

Used by

Dependency tree · two levels

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Sources

Webb, A Course in Finite Group Representation Theory, §§11.3, 11.6 and 12.3–12.5, especially pp.240–245 (standard reference, not scraped)