Defect zero blocks are simple algebras

If the defect is zero, [F1] makes each simple $B$-module relatively $1$-projective. The finite counit witness in [F3] splits it from $kG{\otimes }_{k}M$, a finite free module, so it is projective. Finite-dimensional $B$ has a simple quotient (choose a proper left ideal of maximal dimension), hence [F4] gives a full matrix algebra.

A full matrix algebra $M_n(k)$ has the single simple column module $k^n$, which is the left ideal generated by a matrix diagonal unit and therefore projective. More generally existence of any projective simple gives the matrix assertion by [F4]; its field is $k$ by [F6]. This agrees with the single-factor semisimple description in [F5].

Conversely suppose $B=M_n(k)$. Its enveloping algebra $B{\otimes }_k{B}^{\mathrm{o}\mathrm{p}}$ is $M_{n^2}(k)$: the action of $E_{ij}\otimes E_{kl}^{\mathrm{o}\mathrm{p}}$ sends $E_{rs}$ to ${\delta }_{jr}{\delta }_{sk}{E}_{il}$, giving all matrix units on the basis $E_{rs}$. Thus its module $B$ is projective. Under the inversion anti-isomorphism $\ast :kG\to kG$, this enveloping algebra is the central factor cut out by $b\otimes b^{\ast }$ in $k[G\times G]$, since the second group factor acts on the right by its inverse. A projective module for this factor is projective for the whole algebra because the factor is a direct summand. Consequently the block bimodule is relatively $1$-projective, and its vertex is $1$; [F2] identifies this as defect zero.