Maschke lemma

Statement

Let be a finite group and be a field. If then is a semisimple ring.

Proof

It suffices to check that any finite dimensional (finitely generated) -module is semisimple.

Let be a finite dimensional -module (that satisfies the conditions of the lemma). Let be a submodule. To prove is semisimple, we need a submodule such that

We can look at the action (that is, forget about , and use ), therfore, is a -vector space, with a subspace. Thus, splits as a vector space along the inclusion , that is

as vector spaces using a convienient basis. Thus we have a -linear map

and thus

The idea for the rest of the proof is to now take this projection as vector spaces and “fix it” using an average over the group with the Reynolds operator. Consider the map

Claim: is -linear. Let and .

Note is a projection since for ,

where the second line comes from the fact that is a submodule, so it is invariant under the action. Note: This also shows why in , since then we would need to divide by .

Therefore, we can take , and is semisimple.

Examples

• For a field with characteristic 0, the is not a restriction. That means is semisimple for and , for any group .

• (Non-example:) , , then the ring is not semisimple, not in .

• Let , be a smooth manifold, and is a finite group that acts on . This makes smooth functions a left -module using the action

then by Maschke’s lemma, where are simple submodules. This is the main idea behind Fourier analysis.

Importance

Given the conditions needed, i.e. the characteristic of the

Worked examples

Example 1 Let , then with .
Then we can see that

since each of these parts is invariant under the action of because

and

therefore, acts as

Example 2 Let , then . Consider

therefore, the action is trivial on the span. This gives a decomposition

where is 2-dimensional. Consider

then, this is a vector space complement, but it cycles through the indices, so it a submodule complement as well. has a basis (we can call them and respectively). Note that

and

therefore, acts on with the following matrix

The statement that that is semisimple says that the matrix has eigenvalues. The characteristic polynomial

has roots . Therefore, if is a splitting field of then is not simple. Otherwise it is.