TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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The kernel of a complex character agrees with the kernel of any representation affording it

Statement

Let $ρ : G \to GL (V)$ be a finite-dimensional complex representation of a finite group $G$ , with character $χ = χ_{V}$ . Then

$\ker χ = \ker ρ,$

where $\ker ρ$ is the kernel of the group homomorphism $ρ$ and $\ker χ = {g \in G : χ (g) = χ (1)}$ is the kernel of the character.

Facts & Assumptions

Given: A finite group $G$ and a finite-dimensional complex representation $ρ : G \to GL (V)$ with character $χ$ .

[F1]

The kernel of a group homomorphism is the set of elements sent to the identity (The kernel and image of a group homomorphism).

[F2]

The kernel of the character is $\ker χ = {g \in G : χ (g) = χ (1)}$ (The kernel of a complex character).

[F3]

For the character, $χ (1) = \dim V$ , and $∣ χ (g) ∣ = χ (1)$ holds exactly when $ρ (g)$ is a scalar operator (For a complex character, $χ (1) = \dim V$ , $χ$ is a class function, and $∣ χ (g) ∣ \leq χ (1)$ with equality exactly at scalars).

Proof

technique · direct

1.1

If $g \in \ker ρ$ , then $ρ (g) = {id}_{V}$ by [F1], so $χ (g) = tr ({id}_{V}) = χ (1)$ by [F3]. Hence $g \in \ker χ$ by [F2], which proves $\ker ρ \subseteq \ker χ$ .

F1F2F3given

1.2

Conversely, let $g \in \ker χ$ , so $χ (g) = χ (1)$ by [F2]. Then in particular $∣ χ (g) ∣ = χ (1)$ , and the equality clause of [F3] gives $ρ (g) = λ {id}_{V}$ for a scalar $λ$ . Evaluating the character at $g$ and at $1$ with [F3] gives $χ (g) = λ χ (1) = χ (1)$ .

F2F3given

1.3

If $\dim V = 0$ , then $V = {0}$ is the zero space, $ρ (g) = {id}_{V}$ holds for every $g$ , and $χ \equiv 0$ , so $\ker ρ = G = \ker χ$ by [F1] and [F2]; the statement is immediate. Hence assume $\dim V \geq 1$ .

F1F2given

2.1

Since $χ (1) = \dim V \geq 1$ by [F3], the equality $λ χ (1) = χ (1)$ of step 1.2 forces $λ = 1$ ; therefore $ρ (g) = {id}_{V}$ , so $g \in \ker ρ$ by [F1]. Together with step 1.1 and step 1.3 this proves $\ker χ = \ker ρ$ .

F1F3step 1.2step 1.3algebra∎

Depends on

Used by

The normal subgroups of a finite group are exactly the intersections of kernels of irreducible complex characters Theorem

Dependency tree · two levels

12 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

Peter Webb, A Course in Finite Group Representation Theory, Proposition 3.1.1 (standard reference, not scraped)
Shani Meynet and Robert Moscrop, McKay quivers and decomposition, Appendix A.3 (standard reference, not scraped)