Indecomposable projective kG-modules correspond to simple modules through taking the head

Let $\pi :P\to S$ be the projective cover of a simple module $S$. If $P=P_1\oplus P_2$, then $S=\pi (P_1)+\pi (P_2)$, so simplicity forces one summand, say $\pi (P_1)$, to equal $S$. Then $P=P_1+\ker \pi$, and because $\ker \pi$ is superfluous in a projective cover, $P=P_1$. Hence $P_2=0$, so the projective cover of a simple module is indecomposable.

Now let $P$ be an indecomposable finite-dimensional projective module, and put $J=J(A)$. The quotient map $P\to \mathrm{hd}(P)=P\mathrm{/}JP$ is a projective cover: its kernel is $\mathrm{rad}(P)=JP$ by [F1], and that kernel is superfluous by [L2]. The head is nonzero, since otherwise the superfluity condition applied with the zero submodule would force $P=0$. It is a finite-dimensional module over the semisimple algebra $B=A\mathrm{/}J$ by [L3]. Choose finitely many vector-space generators to make it a quotient of a finite sum of copies of ${}_{B}B$. By [L3] and [F2], that sum is a finite direct sum of simple modules. Its quotient is likewise a finite direct sum of simples: process the summands in order; the image of each is zero or simple, and either lies in the sum of earlier images or meets that sum trivially, so retain exactly the latter images. Thus $\mathrm{hd}(P)\cong S_1\oplus \cdots \oplus S_r$ for some $r\ge 1$.

For each $S_i$, let ${\pi }_i:Q_i\to S_i$ be its finite-dimensional projective cover from [L1], with kernel $K_i$. The finite direct sum $Q={⨁}_i{Q}_i$ is projective by lifting one component at a time, and $\pi ={⨁}_i{\pi }_i:Q\to {⨁}_i{S}_i$ is surjective with kernel $K={⨁}_i{K}_i$. Each $K_i$ is superfluous. If a maximal submodule of $Q_i$ omitted $K_i$, their sum would equal $Q_i$, a contradiction; hence $K_i\subseteq \mathrm{rad}(Q_i)=J{Q}_i$ by [F1]. Therefore $K\subseteq {⨁}_{i}J{Q}_i=JQ=\mathrm{rad}(Q)$, and [L2] makes $K$ superfluous because $Q$ is finite-dimensional. Thus $\pi$ is a projective cover of the head. Uniqueness in [L1] gives $P\cong Q$. Each $Q_i\mathrm{\ne }0$ because it surjects onto $S_i\mathrm{\ne }0$; indecomposability of $P$ therefore forces $r=1$. Thus the head is simple.

Step 1.1 constructs an indecomposable projective from each simple module, and steps 1.2–2.1 show that taking the head of an indecomposable projective returns a simple module. The two constructions are inverse up to isomorphism by uniqueness of projective covers.