2015
DOI: 10.1093/qmath/hav010
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On the Brauer Indecomposability of Scott Modules

Abstract: Abstract. Let k be an algebraically closed field of prime characteristic p, and let P be a p-subgroup of a finite group G. We give sufficient conditions for the kG-Scott module Sc(G, P ) with vertex P to remain indcomposable under the Brauer construction with respect to any subgroup of P . This generalizes similar results for the case where P is abelian. The background motivation for this note is the fact that the Brauer indecomposability of a p-permutation bimodule is a key step towards showing that the modul… Show more

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Cited by 11 publications
(12 citation statements)
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“…Therefore by (14), Res N C (M 1 ) = Sc(C, ∆P 0 ), so that by the definition of M 1 , we finally have that Res N C (M(∆Q)) = Sc(C, ∆P 0 ). Since ∆Q acts trivially on M(∆Q), this is the equivalent to say that Res N ∆Q•C (M(∆Q)) = Sc(C, ∆P 0 ).…”
Section: Case (I)mentioning
confidence: 91%
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“…Therefore by (14), Res N C (M 1 ) = Sc(C, ∆P 0 ), so that by the definition of M 1 , we finally have that Res N C (M(∆Q)) = Sc(C, ∆P 0 ). Since ∆Q acts trivially on M(∆Q), this is the equivalent to say that Res N ∆Q•C (M(∆Q)) = Sc(C, ∆P 0 ).…”
Section: Case (I)mentioning
confidence: 91%
“…Hence, Sc(C, ∆P 0 ) | Res N C (M 1 ) and we can write that (14) Res N C (M 1 ) = Sc(C, ∆P 0 ) X for a kC-module X.…”
Section: Scott Modules With Wreathed 2-group Verticesmentioning
confidence: 99%
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“…These theorems in a sense generalize [11][12][13][14], and there are results on Brauer indecomposability of Scott modules also in [15,21]. Notation 1.3.…”
Section: Introduction and Notationmentioning
confidence: 90%