False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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FALSE: a Brauer character is defined on all elements by the usual trace

Statement

A Brauer character is defined on every element of $G$ and equals the ordinary matrix trace there.

Facts & Assumptions

Given: A field $k$ of characteristic $p$ and the $2$ -dimensional $k C_{p}$ module on which a generator $u$ acts by the unipotent matrix $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ .

[F1]

A Brauer character is defined only on the $p$ -regular elements of $G$ (Brauer character of a finite-dimensional kG-module).

Refutation

technique · direct

1.1

The element $u$ has order $p$ , so it is $p$ -singular. The displayed matrix still has an ordinary trace, namely $2$ in $k$ .

givenalgebra

2.1

But [F1] defines the Brauer character only on $G^{0}$ , and $u \notin G^{0}$ . So the trace value at $u$ is not a Brauer-character value.

F1step 1.1

3.1

Therefore the statement is false.

step 2.1∎

Depends on

Brauer character of a finite-dimensional kG-module

Used by

Ordinary trace on a p-singular unipotent element is not a Brauer-character value Counterexample

Dependency tree · two levels

3 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)