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

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Brauer homomorphism is multiplicative

Statement

The Brauer map ${Br}_{P} : (k G)^{P} \to k C_{G} (P)$ is a unital surjective $k$ -algebra homomorphism.

Facts & Assumptions

Given: A finite group, characteristic- $p$ field, and a $p$ -subgroup $P$ .

[F1]

The Brauer map retains exactly the coefficients centralizing $P$ . (Brauer homomorphism for a p subgroup)

[F2]

Non-singleton orbits of a finite $p$ -group have size divisible by $p$ . (If a finite $p$ -group $P$ acts on a finite set $X$ , then $∣ X ∣ \equiv ∣ X^{P} ∣ (m o d p)$ )

Proof

technique · direct

1.1

Let $a = \sum a_{x} x$ and $b = \sum b_{y} y$ be $P$ -fixed and fix $z \in C_{G} (P)$ . The coefficient of $z$ in $a b$ is $\sum_{x y = z} a_{x} b_{y}$ . Conjugation by $P$ preserves the indexing set, since $z$ centralizes $P$ , and preserves each coefficient product. Each non-singleton orbit contributes its cardinality times one coefficient, which is zero in characteristic $p$ . The singleton pairs are precisely those with both $x, y \in C_{G} (P)$ . Hence this coefficient equals that of $z$ in ${Br}_{P} (a) {Br}_{P} (b)$ .

F1F2

2.1

Equality at each basis element $z$ proves multiplicativity. Projection is $k$ -linear, retains the identity group element, and fixes every element of $k C_{G} (P)$ , which is contained in $(k G)^{P}$ . It is therefore unital and surjective, including when $P = 1$ .

F1step 1.1∎

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

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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)