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. The quotient map $P\to \mathrm{hd}(P)$ is a projective cover because its kernel is $\mathrm{rad}(P)$, which is superfluous by [L2]. Write the semisimple head as a direct sum of simple modules $\mathrm{hd}(P)\cong S_1\oplus \cdots \oplus S_r$. Taking the direct sum of the projective covers of the $S_i$ gives another projective cover of $\mathrm{hd}(P)$. By uniqueness in [L1], that direct sum is isomorphic to $P$. Since $P$ is indecomposable, one must have $r=1$. Thus $\mathrm{hd}(P)$ is simple.

Step 1.1 constructs an indecomposable projective from each simple module, and step 2.1 shows 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.