A projective simple in a symmetric block forces a matrix block

The central product decomposition $kG=B\times kG(1-b)$ has zero cross-products, so the symmetric form in [F2] restricts nondegenerately to $B$: a vector orthogonal to $B$ is already orthogonal to the other factor, hence is zero. A projective $B$-module is projective over $kG$ because $B$ is a direct summand of $kG$. Thus $S$ is injective over $kG$ by [F3], and also over $B$ by restriction to the block module category.

Decompose the regular $B$-module into indecomposable projectives using [F4] and [F5]. The projective cover of $S$ is $S$ itself; hence the sum $U$ of all summands isomorphic to $S$ is nonzero. Let $V$ be the sum of the other summands. A nonzero map $S\to P$ for an indecomposable summand $P$ is injective and splits because $S$ is injective; it forces $P\cong S$. A nonzero map $P\to S$ is surjective and splits because $S$ is projective, again forcing $P\cong S$. Consequently ${\mathrm{Hom}}_B(U,V)={\mathrm{Hom}}_B(V,U)=0$.

The projection $q:B\to U$ commutes with every $B$-linear endomorphism, because the off-diagonal homomorphism spaces vanish. Such endomorphisms include every right multiplication. A left-module endomorphism is right multiplication by $q(b)$; commuting also with right multiplication makes $q(b)$ central. It is a nonzero central idempotent, so [F1] forces $q(b)=b$ and $V=0$. Thus the regular module is a sum of copies of $S$ and is semisimple. By [F6] and the splitting condition [F7], $B$ is a single matrix algebra over $k$, of size ${\dim }_{k}S$, with exactly one simple module.