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这个专题主要介绍有限群的线性表示。粗略地说，我们把有限群中的元素看作是某个有限维线性空间上的（可逆）线性变换，从而来研究群的性质。本文主要介绍群表示论的最基础的知识，并介绍 Maschke 定理，这个定理在群表示论的教材中往往一开始就会碰到，虽然证明并不难，但它在有限群的表示论中起着至关重要的作用。
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.
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.
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).
The group algebra is semisimple if and only if .
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.
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 .