Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29
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The regular character gives a second proof of the sum-of-squares formula

Statement

Let G be a finite group and let χ1,…,χr be its irreducible complex characters, of degrees ni=χi(1). Then

∑i=1rni2=∣G∣.

Facts & Assumptions

Given: A finite group G with irreducible characters χ1,…,χr, the regular representation C[G], and the regular character χreg.

[F1]

The regular character is ∣G∣ at 1 and 0 elsewhere (The regular character is ∣G∣ at 1 and 0 away from 1).

[F2]

The multiplicity of an irreducible summand is an inner product: mi=⟨χreg,χi⟩ (The multiplicity of an irreducible summand is a character inner product).

[F3]

Characters add on direct sums: χV⊕W=χV+χW (Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[A1]

The degree of a character is its value at 1, χi(1)=ni.

Proof

technique · direct
1.1F1F2A1given

By [F2], the regular representation decomposes as C[G]≅⨁imiVi with mi=⟨χreg,χi⟩. By [F1] and the definition of the inner product this is 1∣G∣⋅∣G∣⋅χi(1)=χi(1)=ni, because the only nonzero term is at g=1.

2.1F1F3step 1.1algebra∎

Characters add on direct sums by [F3], so evaluating both sides of the decomposition of step 1.1 at 1 gives χreg(1)=∑imini. By [F1] the left side is ∣G∣, and step 1.1 gives mi=ni, so ∣G∣=∑ini2.

Depends on

Used by

Dependency tree · two levels

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Sources