CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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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.
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A projective indecomposable module has trivial vertex

Statement

Let $G$ be finite and let $k$ have characteristic $p$ . If $P$ is an indecomposable finite-dimensional projective $k G$ -module, then its vertex is the trivial subgroup $1$ .

Facts & Assumptions

Given: A finite group $G$ , a field $k$ of characteristic $p$ , and an indecomposable finite-dimensional projective $k G$ -module $P$ .

[F1]

Vertices are minimal $p$ -subgroups for relative projectivity (A vertex is a minimal p-subgroup for relative projectivity, and a source is an indecomposable inducing summand there).

[L1]

Vertices exist for indecomposable modules (Vertices exist for indecomposable modules, are conjugate in G, and sources are conjugate by the appropriate normalizer).

[L2]

Induction from the trivial subgroup preserves projectives, and projectives are direct summands of free modules (Restriction and induction along a subgroup preserve projective modules, Equivalent characterizations of projective modules).

Proof

technique · direct

1.1

Choose a finite $k$ -basis of $P$ . It gives a surjection from a finite free $k G$ -module $(k G)^{n}$ onto $P$ , and projectivity splits that surjection. Thus $P$ is a direct summand of $(k G)^{n}$ . But $(k G)^{n}$ is induced from the trivial subgroup, namely from the $n$ -dimensional $k$ -module. Hence $P$ is relatively $1$ -projective.

L2givenalgebra

2.1

By [L1], $P$ has a vertex $Q$ . Since vertices are minimal $p$ -subgroups for relative projectivity by [F1], and step 1.1 shows that $1$ already works, one must have $Q = 1$ .

F1L1step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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

J. MacQuarrie, Modular Representations of Profinite Groups (standard reference, not scraped)
Peter Webb, A Course in Finite Group Representation Theory (23 Feb 2016 draft) (standard reference, not scraped)