Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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If k is algebraically closed and char⁡k∤∣G∣, then ∑i(dim⁡kVi)2=∣G∣

Statement

Let G be a finite group and let k be an algebraically closed field with char⁡k∤∣G∣. If V1,…,Vr is a complete list of the irreducible representations of G over k, up to equivalence, then

∑i=1r(dim⁡kVi)2=∣G∣.

Facts & Assumptions

Given: A finite group G and an algebraically closed field k with char⁡k∤∣G∣.

[L1]

The regular representation decomposes as k[G]≅V1⊕dim⁡kV1⊕⋯⊕Vr⊕dim⁡kVr, where V1,…,Vr are the irreducible representations of G (If k is algebraically closed and char⁡k∤∣G∣, there are finitely many irreducible representations, and each occurs in the regular representation with multiplicity equal to its degree).

[L2]

For a finite group, dim⁡kk[G]=∣G∣ (If G is finite then dim⁡kk[G]=∣G∣).

Proof

technique · direct
1.1L1L2given

By [L1], the regular representation is a direct sum of dim⁡kVi copies of each Vi, and [L2] says its total dimension is ∣G∣.

2.1step 1.1givenalgebra∎

Taking dimensions in step 1.1 gives ∣G∣=dim⁡kk[G]=∑i=1rdim⁡k(Vi⊕dim⁡kVi)=∑i=1r(dim⁡kVi)2. This is the required identity.

Depends on

Used by

Dependency tree · two levels

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Sources