False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-09-05

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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 $C_{p}$ 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

1.1

The group algebra $k C_{p}$ has only one irreducible Brauer character, namely the trivial one, because $C_{p}$ is a $p$ -group. But over characteristic $0$ , the cyclic group $C_{p}$ has $p$ distinct ordinary irreducible characters.

givenalgebra

2.1

By [L1], all of those characters are distributed among the blocks of $C_{p}$ , and [L2] shows that the decomposition data are organized blockwise. Since there is only one Brauer irreducible, some block contains more than one ordinary irreducible character.

L1L2step 1.1

3.1

Therefore the statement is false.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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. Miquel Martinez, Modular Representation Theory of Finite Groups (standard reference, not scraped)
Tudor Ciurca, Representation Theory (standard reference, not scraped)