LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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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.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Central p-subgroups lie in every block defect group

Statement

Assume the Axiom of Choice. Let $H$ be a finite group, let $Z \leq Z (H)$ be a $p$ -subgroup, and let $c$ be a block idempotent of $k H$ . Then $Z$ is contained in every defect group of $c$ .

Facts & Assumptions

Given: AC, the finite group, central $p$ -subgroup, and block in the Statement.

[F1]

For a $p$ -subgroup $P \leq H$ , the Brauer map deletes the coefficients outside $C_{H} (P)$ (Brauer homomorphism for a p subgroup).

[F2]

If the Brauer image of a block at $P$ is nonzero, then $P$ is contained in an $H$ -conjugate of every defect group of that block (Defect groups are maximal Brauer support).

[F3]

AC is available (The Axiom of Choice). It is used only to discharge the current published dependency contract behind F2; the displayed finite group argument makes no additional choice.

Proof

1.1

Since $Z$ is central, $C_{H} (Z) = H$ . Consequently F1 gives ${Br}_{Z} (c) = c .$ The primitive idempotent $c$ is nonzero, so F2 says that for every defect group $D$ of $c$ there is an $h \in H$ such that $Z \leq h D h^{- 1}$ .

F1F2F3

2.1

Conjugating this containment by $h^{- 1}$ gives $h^{- 1} Z h \leq D$ . Centrality gives $h^{- 1} Z h = Z$ , and hence $Z \leq D$ . This includes $Z = 1$ and applies to each defect group separately.

step 1.1∎

Depends on

Used by

Brauer subsections and B-subsections Definition
p-sections and Brauer subsections in S3 Example

Dependency tree · two levels

9 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

Craven, The Brauer Correspondence, Lemma 1.2, p. 2 (standard reference, not scraped)
Meierfrankenfeld, MTH 912 Class Notes, sections 6.6–6.7, pp. 156–165 (standard reference, not scraped)