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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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The regular module is a direct sum of the projective covers of the simple modules, with the split-field multiplicities

Statement

Let A=k[G] with k a splitting field for the finite group G. Then the left regular module decomposes as

AASIrr(A)P(S)dimkS,

where P(S) is the projective cover of the simple module S.

Facts & Assumptions

Given: The finite group algebra A=k[G] over a splitting field k.

[L2]
[F1]

A splitting field is one over which the simple endomorphism rings are scalars (A splitting field for a finite group: every irreducible representation has scalar endomorphism ring).

Proof

technique · direct
1.1

By [L1], the regular module decomposes as a finite direct sum of indecomposable projective modules. By [L2], each summand is the projective cover P(S) of a unique simple module S, so AASP(S)mS for uniquely determined multiplicities mS0.

L1L2givenalgebra
2.1

Modding out by the radical preserves direct sums and sends each P(S) to its simple head S. Hence A/J(A)SSmS. By [L3], the quotient is semisimple. Since k is a splitting field by [F1], Wedderburn-Artin writes the semisimple algebra as a product of matrix algebras MnS(k), and the left regular module of MnS(k) is the simple column module repeated nS times. That simple module has k-dimension nS, so mS=dimkS.

L3F1step 1.1algebra
3.1

Substituting the multiplicities from step 2.1 gives the displayed decomposition.

step 2.1

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