Let be a finite dimensional vector space over a field . Let be a finite group. A linear representation of in is a group homomorphism .

Define the group algebra with obvious addition and multiplication. Then linear representations of are the same things as -modules, since any group homomorphism can be linearly extended to a -algebra homomorphism .

Let , be two representations of . We say a linear map is a morphism of representations of if the diagram

commutes for all . This exactly means that is a -module homomorphism.

## Remark

In fact a representation of can be viewed as a functor from the category consisting of a single object whose morphisms are elements of the group , to the category of -vector spaces, sending a group element to . A natural transformation from to is exactly a linear map such that the diagram above commutes, i.e. a morphism of representations is exactly a natural transformation. Then we say two representations and of are isomorphic if they are naturally isomorphic as functors.

## Definition

We say that a representation is simple (or irreducible) if has no -submodules except .

Note that there are only finitely many irreducible representations up to isomorphism. This is because any irreducible -module is isomorphic to a quotient module of , thus is a composition factor of , and has finitely many composition factors.

If , are two representations of , then we can define their

(1) direct sum

, .

(2) tensor product

,

Suppose that , are two representations of . For any linear map , define its trace to be

Then is a morphism of -modules.

Suppose , are irreducible representations. By Schur's lemma, if and are not isomorphic, then . If , and , then with (by taking trace).

## Theorem[Maschke]

The group algebra is semisimple if and only if .

## Proof

Suppose . Then for any representation , and any -submodule of , let be any subspace of such that . Let be the projection map with respect to this composition. Then is a morphism of -modules and we have . Thus the exact sequence of -modules

splits. Hence as -modules. Therefore, is semisimple.

On the other hand, if , then . Hence is contained in the nilradical of . Thus the Jacobson radical is not 0. So is not semisimple.

## Remark

Note that if is semisimple, then is irreducible if and only if is indecomposable. In this case, every representation of is a direct sum of irreducible representations, by induction on .

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