Green restriction summand with the same vertex

The six permutations are $1,(12),(13),(23),(123),(132)$. Conjugation sends $(12)$ to $(g(1)g(2))$. To normalize $P$, a permutation must preserve $\{1,2\}$ and fix $3$, hence is $1$ or $(12)$. Thus $N_G(P)=P=H$. A one-dimensional nonzero module is indecomposable, since dimensions of nonzero summands would add to at least two.

Every endomorphism of a one-dimensional trivial module is multiplication by a scalar $a$. For a subgroup $D\le G$, its relative trace is multiplication by $[G:D]a$, since each conjugation in the trace formula acts trivially. For $D=P$, the trace of the identity is $3\cdot 1=1$ in characteristic $2$, proving relative $P$-projectivity. For $D=1$, every trace is $6a=0$, never the identity. The only proper subgroup of $P$ is $1$, so $P$ is a vertex of $M$.

Restriction is the same trivial one-dimensional space. Its trace from $H$ is $1$ on the identity, whereas its trace from $1$ is $2a=0$ for every scalar. Thus it is relatively $H$-projective and not relatively $1$-projective, so its vertex is $P=H$. Its dimension gives precisely one nonzero indecomposable summand and no nonzero complement.

The normalizer condition in 1.1 and the two vertex calculations in 1.2–2.1 verify every hypothesis of F1. Under A1 that corollary identifies the sole restriction summand $k$ as the Green correspondent. No group or vector-space choice was needed for the matrices or traces; AC is only propagated from F1. The nonzero condition is witnessed by ${\dim }_{k}k=1$.