Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29
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A complex character is irreducible if and only if its self-inner-product is 1

Statement

Let G be a finite group and let χ be a complex character of G. Then χ is irreducible if and only if χ,χ=1.

Facts & Assumptions

Given: A finite group G and a complex character χ=χV of a finite-dimensional representation V, completely reduced as VimiVi.

[F1]

The multiplicity of Vi in V is mi=χ,χi (The multiplicity of an irreducible summand is a character inner product).

[F2]

Characters add on direct sums, so the decomposition VimiVi gives χ=imiχi (Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[A1]

If χ=imiχi and mi=χ,χi, then conjugate-symmetry and linearity in the first argument give χ,χ=imiχi,χ=imiχ,χi=imimi=imi2, because the multiplicities mi are nonnegative integers.

[A2]

A representation is irreducible exactly when it has one irreducible summand with multiplicity 1 and no others.

Proof

technique · direct
1.1

Completely reduce VimiVi with mi0 integers. By [F2] this decomposition gives χ=imiχi, and by [F1] each mi=χ,χi; therefore [A1] gives χ,χ=imi2.

F1F2A1given
2.1

Assume χ is irreducible. Then by [A2] exactly one multiplicity equals 1 and the rest are 0, so the sum in step 1.1 is 1; hence χ,χ=1.

A2step 1.1
2.2

Conversely, assume χ,χ=1. The sum imi2 of nonnegative integers in step 1.1 can equal 1 only when exactly one mi equals 1 and all the others are 0. By [A2], VVi is irreducible, so χ is irreducible.

A2step 1.1algebra
3.1

Steps 2.1 and 2.2 prove the two implications, hence the biconditional.

step 2.1step 2.2

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