We focus in this chapter on facts about group algebras that are not true for finite-dimensional algebras in general. The results are a mix of general statements and specific examples describing the representations of certain types of groups. At the beginning of the chapter, we summarize the properties of projective modules for p-groups and also the behavior of projective modules under induction and restriction. Toward the end, we show that the Cartan matrix is symmetric (then the field is algebraically closed) and also that projective modules are injective. In the middle, we describe quite explicitly the structure of projective modules formany semidirect products, and we do this by elementary arguments. It shows that the important general theorems are not always necessary to understand specific representations, and it also increases our stock of examples of groups and their representations. Because of the diversity of topics, it is possible to skip certain results in this chapter without affecting comprehension of what remains. For example, the reader who ismore interested in the general results could skip the description of representations of specific groups between Example 8.2.1 and Theorem 8.4.1.

The Behavior of Projective Modules under Induction,Restriction, and Tensor Product

We start with a basic fact about group algebras of p-groups in characteristic p.

Theorem 8.1.1.Let k be a field of characteristic p and G a p-group. The regular representation is an indecomposable projective module that is the projective cover of the trivial representation. Every finitely generated projective module is free. The only idempotents in kG are 0 and 1.

Proof. We have seen in 6.12 that kG is indecomposable and it also follows from 7.14. By Nakayama's Lemma, kG is the projective cover of k. By 7.13 and 6.3, every indecomposable projective is isomorphic to kG. Every finitely generated projective is a direct sum of indecomposable projectives, and so is free. Finally, every idempotent e ∈ kG gives a module decomposition kG = kGe ⊕ kG(1 − e). If e ≠ 0 then we must have kG = kGe, so kG(1 − e) = 0 and e = 1.