Green induction has one distinguished summand

The input $U$ is relatively $P$-projective by transitivity from $Q\le P$. F4 gives ${\mathrm{Res}}_H^G{\mathrm{Ind}}_H^{G}U\cong U\oplus Y$, with $Y$ relatively $\mathcal{Y}$-projective. The exclusion assertion of F4 makes $U$ non-$\mathcal{Y}$-projective. Thus its multiplicity in this restriction is exactly one.

Under the inherited AC assumption, decompose ${\mathrm{Ind}}_H^{G}U={⨁}_j{M}_j$. Comparing restrictions with 1.1 by F7, exactly one index, say $j=1$, has $U\mid {\mathrm{Res}}_H^G{M}_1$. For $j\mathrm{\ne }1$, every indecomposable of ${\mathrm{Res}}_H^G{M}_j$ occurs in $Y$, so this restriction is $\mathcal{Y}$-projective. Each $M_j$ is relatively $Q$-projective, and vertex containment permits a vertex $T_j\le Q$ after conjugation in $G$.

For $j\mathrm{\ne }1$, F3 retains a $T_j$-vertex summand in ${\mathrm{Res}}_H^G{M}_j$, since $T_j\le Q\le H$. This summand is $\mathcal{Y}$-projective by 2.1, so F6 gives $T_j{\le }_H\mathcal{Y}$. F2 gives $T_j{\le }_G\mathcal{X}$. Transitivity and conjugation of induction therefore make $M_j$ relatively projective for a member of $\mathcal{X}$. Their direct sum is $\mathcal{X}$-projective.

Put $M=M_1$ and $T=T_1$. If $M$ were $\mathcal{X}$-projective, its restriction would be $\mathcal{Y}$-projective by F4, contrary to its summand $U$ from 2.1. Consequently $T\le Q\le P$ is not $G$-contained in $\mathcal{X}$: such containment would make $M$ $\mathcal{X}$-projective by transitivity. Thus $T\in \mathcal{Z}$. Apply F5 to $M$. Its unique non-$\mathcal{Y}$ restriction summand has vertex $T$ and must be $U$, by 1.1 and 2.1. Since vertices of $U$ are $H$-conjugate, $T$ is $H$-conjugate to $Q$; hence $Q$ itself is a $G$-vertex of $M$.

The other summands are $\mathcal{X}$-projective by 3.1, so their vertices cannot be in $\mathcal{Z}$, by vertex containment and F1. In particular none is isomorphic to $M$, proving multiplicity one and uniqueness. We already have $U\mid {\mathrm{Res}}_H^{G}M$. When $H=G$, induction is identified with $U$ by $g\otimes u\mapsto gu$, so the complement is zero. When $P=1$ this same case is forced. The zero input is excluded; a zero complement is allowed. These statements prove the result. [F1, F6, step 2.1, step 3.1, step 3.2] QED