TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Equivalence classes of degree-one representations are exactly homomorphisms $G \to k^{\times}$ ; equivalently they factor through $G / G^{'}$ , and they form an abelian group

Statement

Let $k$ be a field and let $G$ be a group.

choosing a basis of a degree-one representation of $G$ over $k$ produces a homomorphism $χ : G \to k^{\times}$ , and every such homomorphism produces a normalized degree-one representation on the one-dimensional space $k$ ;
two degree-one representations are equivalent if and only if they produce the same homomorphism, so equivalence classes of degree-one representations are in bijection with the homomorphisms $G \to k^{\times}$ ;
equivalently, those equivalence classes are in bijection with the homomorphisms that factor through the abelianisation quotient $G / G^{'}$ ;
under pointwise multiplication and inversion of the corresponding homomorphisms, these equivalence classes form an abelian group.

Facts & Assumptions

Given: A field $k$ and a group $G$ .

[L1]

A degree-one representation is a finite-dimensional representation whose underlying vector space has dimension $1$ (A finite-dimensional representation $ρ : G \to GL (V)$ over a field, and its degree).

[L2]

The units $k^{\times}$ of the field form a group under multiplication (The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times}$ is a group under multiplication; $0 \in R^{\times}$ only in the zero ring, Left inverse, right inverse, and invertible element of a monoid), and field multiplication is commutative (Field).

[L3]

Every homomorphism from $G$ to an abelian group factors uniquely through the quotient $G / G^{'}$ (The derived subgroup is characteristic and the abelianization is universal, Monoid homomorphism and group homomorphism).

[L4]

Two representations are equivalent when there is an invertible intertwiner between them (Intertwiners, the spaces ${Hom}_{G} (V, W)$ and ${End}_{G} (V)$ , equivalent representations, and faithful representations).

Proof

technique · constructive

1.1

Let $ρ : G \to GL (V)$ be a degree-one representation. By [L1], choose a basis vector $v$ of the one-dimensional space $V$ . For each $g \in G$ there is a unique scalar $χ (g) \in k$ with $ρ (g) v = χ (g) v$ , and because $ρ (g)$ is invertible, that scalar lies in $k^{\times}$ .

L1L2givenconstruct

2.1

The representation law gives $χ (g h) v = ρ (g h) v = ρ (g) ρ (h) v = χ (g) χ (h) v$ , so $χ (g h) = χ (g) χ (h)$ . Thus $χ : G \to k^{\times}$ is a group homomorphism. Under the identification $V ≅ k$ sending $v$ to $1$ , the representation becomes the normalized action $g \cdot a = χ (g) a$ on the one-dimensional space $k$ .

step 1.1L2givenalgebra

3.1

Conversely, if $χ : G \to k^{\times}$ is a group homomorphism, define $ρ_{χ} (g) : k \to k$ by $ρ_{χ} (g) (a) = χ (g) a$ . Each $ρ_{χ} (g)$ is invertible with inverse multiplication by $χ (g)^{- 1}$ , and the homomorphism law for $χ$ gives $ρ_{χ} (g h) = ρ_{χ} (g) ρ_{χ} (h)$ . So $ρ_{χ}$ is a degree-one representation.

step 2.1L2givenconstruct

4.1

If two degree-one representations are equivalent, an invertible intertwiner sends a chosen basis vector of one space to a nonzero scalar multiple of a basis vector of the other, so the character extracted in step 2.1 is unchanged; conversely, if two degree-one representations produce the same homomorphism $χ$ , then step 2.1 identifies each of them with the same normalized representation $ρ_{χ}$ of step 3.1, so they are equivalent by [L4]. Therefore equivalence classes of degree-one representations are in bijection with homomorphisms $G \to k^{\times}$ .

step 2.1step 3.1L4givenalgebra

5.1

Because $k^{\times}$ is abelian by [L2], [L3] identifies the same equivalence classes with the homomorphisms factoring through $G / G^{'}$ . If $χ, ψ : G \to k^{\times}$ are homomorphisms, then $(χ ψ) (g) : = χ (g) ψ (g)$ and $χ^{- 1} (g) : = χ (g)^{- 1}$ are again homomorphisms. Thus the equivalence classes of degree-one representations form an abelian group under pointwise multiplication and inversion of their corresponding homomorphisms.

step 4.1L2L3discharge-construct∎

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

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Sources

Peter Webb, A Course in Finite Group Representation Theory, Example 1.1.2 and Proposition 4.2.1 (standard reference, not scraped)

Equivalence classes of degree-one representations are exactly homomorphisms G→k×G→k×; equivalently they factor through G/G′G/G′, and they form an abelian group