Over a finite group algebra in defining characteristic, finite-dimensional projective and injective modules coincide

Let $A^{\vee }={\mathrm{Hom}}_k(A,k)$. The symmetric Frobenius form from [L1] identifies $A$ with $A^{\vee }$ as left $A$-modules. An $A$-map $U\to A^{\vee }$ is determined by the scalar function $u\mapsto f(u)(1)$, and any $k$-linear extension of that scalar function from a submodule $U\subseteq V$ to $V$ induces an $A$-linear extension $V\to A^{\vee }$. Thus $A^{\vee }$, and hence $A$, is injective.

If $P$ is projective, [F1] makes it a direct summand of a finite free module $A^n$. Step 1.1 makes $A^n$ injective, and a direct summand of an injective module is injective by [F2]. So every finite-dimensional projective module is injective.

Conversely, let $I$ be injective. Dualizing a split monomorphism into $I$ turns the extension property of [F2] into the lifting property for the right $A$-module $I^{\vee }$, so $I^{\vee }$ is projective over $A^{\mathrm{o}\mathrm{p}}$. By the projective characterization in [F1], $I^{\vee }$ is a direct summand of a finite free right $A$-module. Dualizing back and using the symmetric Frobenius identification $(A^n{)}^{\vee }\cong A^n$, the bidual $I^{\vee \vee }\cong I$ becomes a direct summand of a finite free left $A$-module. Hence [F1] makes $I$ projective.

Steps 2.1 and 2.2 prove the equivalence.