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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

$_{A} A ≅ ⨁_{S \in Irr (A)} P (S)^{\oplus \dim_{k} S},$

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 $k$ .

[L1]

Finite-dimensional modules decompose uniquely into indecomposables (Finite-dimensional kG-modules decompose as finite direct sums of indecomposables uniquely up to order and isomorphism).

[L2]

Indecomposable projectives correspond to simple heads (Indecomposable projective kG-modules correspond to simple modules through taking the head).

[L3]

The quotient $A / J (A)$ is semisimple (For a finite-dimensional algebra, the Jacobson radical is nilpotent and the quotient by it is semisimple).

[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

1.1L1L2givenalgebra

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 $_{A} A ≅ ⨁_{S} P (S)^{\oplus m_{S}}$ for uniquely determined multiplicities $m_{S} \geq 0$ .

2.1L3F1step 1.1algebra

Modding out by the radical preserves direct sums and sends each $P (S)$ to its simple head $S$ . Hence $A / J (A) ≅ ⨁_{S} S^{\oplus m_{S}}$ . 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 $M_{n_{S}} (k)$ , and the left regular module of $M_{n_{S}} (k)$ is the simple column module repeated $n_{S}$ times. That simple module has $k$ -dimension $n_{S}$ , so $m_{S} = \dim_{k} S$ .

3.1step 2.1∎

Substituting the multiplicities from step 2.1 gives the displayed decomposition.

Depends on

Used by

Dependency tree · two levels

27 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter Webb, A Course in Finite Group Representation Theory (23 Feb 2016 draft) (standard reference, not scraped)