Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-28
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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

Let G=⟨g⟩ be a cyclic group of order n. Over C, the irreducible representations of G are all one-dimensional, and up to equivalence they are represented by the n degree-one characters χλ(gm)=λm(λn=1).

Facts & Assumptions

Given: A cyclic group G=⟨g⟩ of order n.

[L1]

Over an algebraically closed field, every endomorphism of an irreducible representation is scalar (Over an algebraically closed field, every endomorphism of an irreducible representation is scalar).

[L2]

A splitting field for G is a field over which every irreducible representation has scalar endomorphism ring, and then every irreducible representation of a finite abelian group is one-dimensional (A splitting field for a finite group: every irreducible representation has scalar endomorphism ring, Every irreducible representation of a finite abelian group over a splitting field is one-dimensional).

[L3]

Equivalence classes of degree-one representations are exactly homomorphisms to the unit group C×, and each such homomorphism has the normalized representative on C (Equivalence classes of degree-one representations are exactly homomorphisms G→k×; equivalently they factor through G/G′, and they form an abelian group).

Verification

technique · direct
1.1L1L2given

By [L1], the field C satisfies the scalar-endomorphism condition of [L2], so it is a splitting field for the finite group G. Since G is abelian, [L2] makes every irreducible complex representation of G one-dimensional.

2.1step 1.1L3givenalgebra

By [L3], every irreducible representation of G is equivalent to a normalized degree-one representation on C, hence to a homomorphism χ:G→C×. Because G=⟨g⟩, such a homomorphism is determined by the value λ=χ(g), and the relation gn=e forces λn=1. Conversely, if λn=1, then χλ(gm):=λm is well defined and multiplicative.

3.1step 2.1L4∎

By [L4], the polynomial tn−1 splits over C and has exactly n distinct roots there, namely the elements of μn(C). So step 2.1 produces exactly n equivalence classes of irreducible complex representations, represented by the normalized characters χλ with λ∈μn(C).

Depends on

Used by

Dependency tree · two levels

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Sources