TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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Every irreducible representation of a finite abelian group over a splitting field is one-dimensional

Statement

Let $G$ be a finite abelian group and let $k$ be a splitting field for $G$ . Then every irreducible representation of $G$ over $k$ has degree $1$ .

Facts & Assumptions

Given: A finite abelian group $G$ , a splitting field $k$ for $G$ , and an irreducible representation $ρ : G \to GL (V)$ over $k$ .

[L1]

Over a splitting field, every endomorphism of an irreducible representation is scalar (Over a splitting field, every $G$ -endomorphism of an irreducible representation is scalar).

[L2]

A representation is irreducible exactly when every nonzero invariant subspace is the whole space (Subrepresentations, direct sums of representations, and irreducibility).

Proof

technique · direct

1.1

Because $G$ is abelian, for every $g, h \in G$ one has $ρ (g) ρ (h) = ρ (g h) = ρ (h g) = ρ (h) ρ (g)$ . So each operator $ρ (g)$ commutes with every $ρ (h)$ and therefore lies in ${End}_{G} (V)$ .

L1L2givenalgebra

2.1

By [L1], each $ρ (g)$ is a scalar operator. Hence for every nonzero $v \in V$ , the line $k v$ is stable under every $ρ (g)$ and is therefore a nonzero subrepresentation.

step 1.1L1L2given

3.1

Irreducibility from [L2] forces that nonzero line to be all of $V$ . Therefore $V$ is one-dimensional.

step 2.1L2∎

Depends on

Used by

Over ℂ, a cyclic group of order n has exactly n irreducible representations up to equivalence, represented by the characters g↦ λ with λⁿ=1 Example

Dependency tree · two levels

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Sources

Peter Webb, A Course in Finite Group Representation Theory, Corollary 2.1.7 (standard reference, not scraped)
Pavel Etingof et al., Introduction to Representation Theory, Corollary 1.18 (standard reference, not scraped)