Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedverified 2026-09-26 (gpt-6-sol)
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FALSE: every block has one ordinary and one Brauer irreducible character

Statement

Every block contains exactly one ordinary irreducible character and exactly one irreducible Brauer character.

Facts & Assumptions

Given: The cyclic group Cp at the prime p.

[L1]

Blocks partition the ordinary and Brauer irreducible characters (Blocks partition the ordinary and Brauer irreducible characters).

[L2]

After ordering by blocks, the decomposition matrix is block diagonal (After block ordering, the decomposition matrix is block diagonal).

Refutation

technique · direct
1.1givenalgebra

The group algebra kCp has only one irreducible Brauer character, namely the trivial one: in characteristic p, a generator acts on every simple module as 1, since (g−1)p=0. Over a splitting field of characteristic 0, Cp has p distinct ordinary irreducible characters, all linear. Each has a stable rank-one lattice, whose reduction is a one-dimensional kCp-module and hence the trivial Brauer character. Thus every ordinary character has decomposition number 1 in the unique Brauer column.

2.1L1L2step 1.1

By [L1] and [L2], a nonzero decomposition number joins an ordinary character to a Brauer character in the same block. Step 1.1 therefore places all p ordinary irreducibles in the single block containing the trivial Brauer character. For any prime p≥2, that block contains more than one ordinary irreducible.

3.1step 2.1∎

Therefore the statement is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources