**Notation**. For a ring let be the Jacobson radical of

**Definition**. Recall that if is a field and is a group, then the group algebra has two structures. Firstly, as a vector space over it has as a basis, i.e. every element of is uniquely written as where In particular, as cardinal numbers. Secondly, multiplication is also defined in If and are two elements of then we just multiply in the ordinary fashion using distribution law. To be more precise, we define where

We are going to prove that for every group

**Lemma.** is nil, i.e. every element of is nilpotent.

*Proof.* If is countable, we are done by this theorem. For the general case, let So for some Let Clearly and is countable. So to complete the proof, we only need to show that Write where are the distinct cosets of in Then which means for some right module Now let Since there exists such that We also have for some That now gives us

**Theorem.** for any group

*Proof.* For any define

It’s easy to see that if and only if and for all Now suppose that and let Put By what I just mentioned and for all positive integers By the lemma, there exists such that and Thus which implies that Contradiction!

**Corollary**. If is finite, then is semisimple.

*Proof*. We just proved that So we just need to show that is Artinian. Let

be a descending chain of left ideals of Obviously each is a -subspace of Thus each is finite dimensional because Hence will stablize at some point because and Thus is (left) Artinian and the proof is complete because we know a ring is semisimple if and only if it is (left) Artinian and its Jacobson radical is zero.