Maschke and Schur

As usual, ordering of presenting this material and level of generality are proving to be difficult decisions. For my purposes, I don’t want to do things as generally as they can be done. But on the other hand, most of the proofs are no harder in the general case, so it seems pointless to avoid generality.

First we prove Maschke’s Theorem. Note that there are lots of related statements and versions of what I’m going to write. This says that if G is a finite group and k is a field whose characteristic does not divide the order of the group, then kG is a left semisimple ring.

Proof: We’ll do this using the “averaging operator.” Let’s use the version of semisimple that every left ideal is a direct summand. Let I be a left ideal of kG. Then since kG can be regarded as a vector space over k, I is a subspace. So there is a subspace, V, such that kG=I\oplus V. We are done if it turns out that V is a left ideal.

Let \pi : kG\to I be the projection. (Since any u=i + v uniquely, define \pi(u)=i.) Now it is equivalent to show that \pi is a kG-map, since then it would be a retract and hence I would be a direct summand. Unfortunately, it is not a kG-map. So we’ll force it to be one by averaging.

Let D:kG\to kG by D(u)=\frac{1}{|G|}\sum_{x\in G}x\pi(x^{-1}u). By our characteristic condition, |G|\neq 0.

Claim: Im(D)\subset I. Let u\in kG and x\in G, then \pi(x^{-1}u)\in I by definition, and I a left ideal, so x\pi(x^{-1}u)\in I. So since I an ideal, the sum is in I which shows the claim.

Claim: D(b)=b for all b\in I. This is just computation, x\pi(x^{-1}b)=xx^{-1}b=b, so D(b)=\frac{1}{|G|}(|G|b)=b.

Claim: D is a kG-map. i.e. we want to prove that D(gu)=gD(u) for g\in G and u\in kG. Here the averaging pays off:

\displaystyle gD(u)=\frac{1}{|G|}\sum_{x\in G}gx\pi(x^{-1}u)
\displaystyle = \frac{1}{|G|}\sum_{x\in G} gx\pi(x^{-1}g^{-1}gu)
\displaystyle = \frac{1}{|G|}\sum_{y=gx\in G} y\pi(y^{-1}gu)
= D(gu).

Thus we have proved Maschke’s Theorem.

The other tool we’ll need next time is that of Schur’s Lemma: Let M and N be simple left R-modules. Then every non-zero R-map f:M\to N is an iso. And End_R(M) is a division ring.

Proof: \ker f\neq M since it is a non-zero map. And so ker f=\{0\} since it is a submodule, so we have an injection. Likewise, im f is a submodule, and hence must be all of N, so we have surjection and hence an iso. The other part of the lemma is just noting that since every map in End_R(M) that is non-zero is an iso, it has an inverse.

Next time I’ll talk about how some of these things relate to representations.


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