Bourbaki

Lemma 1

Tag 00LB, permanent. Cite it as [Bourbaki, Tag 00LB].

Algebra, VIII, p. 355, § 18. Simple Algebras over a Finite Field

Let L be a field, G a finite group, and χ\chi a nontrivial homomorphism from G to the multiplicative group L∗L^*. We have ∑x∈Gχ(x)=0\sum_{x\in G}\chi (x) = 0.

By assumption, there exists an element aa of G such that χ(a)≠1\chi (a)\not= 1; since multiplication by aa is a permutation of G, we have

∑x∈Gχ(x)=∑x∈Gχ(ax)=χ(a)∑x∈Gχ(x) \sum_{x\in G}\chi (x) =\sum_{x\in G}\chi (ax) =\chi (a)\sum_{x\in G}\chi (x)

this gives Lemma 1.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

Cites

Nothing.

Transcribed from the printed edition, 2023, Springer Nature. Not translated.