Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29
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The second orthogonality relation for irreducible complex characters

Statement

Let G be a finite group with irreducible complex characters χ1,…,χr. For g,h∈G,

∑i=1rχi(g) χi(h)‾={∣CG(g)∣,h∈Cl⁡G(g),0,h∉Cl⁡G(g).

Facts & Assumptions

Given: A finite group G, its irreducible characters χ1,…,χr, and elements g,h∈G.

[F1]

The irreducible characters form an orthonormal basis of the class functions cf(G) (The irreducible complex characters form an orthonormal basis of cf(G)).

[F2]

The inner product of class functions is ⟨φ,ψ⟩=1∣G∣∑xφ(x)ψ(x)‾ (The standard inner product on cf(G)).

[A1]

The indicator function δg of the conjugacy class of g, defined by δg(x)=1 for x∈Cl⁡G(g) and δg(x)=0 otherwise, is a class function, and its class has ∣Cl⁡G(g)∣=∣G∣/∣CG(g)∣ elements.

[A2]

If f=∑iciχi in the orthonormal basis of [F1], then ci=⟨f,χi⟩.

Proof

technique · direct
1.1F1A1A2given

The function δg of [A1] is a class function, so by [F1] it expands as δg=∑iciχi with ci=⟨δg,χi⟩ by [A2].

1.2F2A1givenalgebra

By [F2], ⟨δg,χi⟩=1∣G∣∑xδg(x)χi(x)‾. By [A1] the only nonzero terms are at x∈Cl⁡G(g), where δg(x)=1 and χi(x)‾=χi(g)‾ because characters are class functions. Hence ci=∣G∣/∣CG(g)∣∣G∣χi(g)‾=χi(g)‾/∣CG(g)∣.

2.1step 1.1step 1.2algebra

Evaluating the expansion of step 1.1 at h and substituting the coefficients of step 1.2 gives δg(h)=1∣CG(g)∣∑iχi(g)‾χi(h).

3.1A1step 2.1algebra∎

By [A1], δg(h)=1 exactly when h∈Cl⁡G(g), and 0 otherwise. Multiplying the identity of step 2.1 by ∣CG(g)∣ and taking complex conjugates gives ∑iχi(g)χi(h)‾=∣CG(g)∣δg(h), which is the stated formula.

Depends on

Used by

Dependency tree · two levels

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Sources