Green indecomposability for index-p integral induction

Put $X={\mathrm{Ind}}_N^{H}L$ and let $I_H(L)=\{h\in H:{}^{h}L\cong L\}$. This is a subgroup containing $N$, so the prime-index hypothesis gives $I_H(L)=N$ or $H$. By F1, $E_0={\mathrm{End}}_{\mathcal{O}N}(L)$ is local. Its residue division ring $E_0\mathrm{/}J(E_0)$ is finite-dimensional over $k$; F3 makes it equal to $k$, because every element satisfies a split polynomial and a division ring has no nonzero zero divisors.

Suppose $I_H(L)=N$. On restriction to $N$, $X$ is the direct sum of the $p$ pairwise nonisomorphic conjugates of $L$. Krull--Schmidt and step 1.1 identify the semisimple quotient of its $N$-endomorphism ring with $k^p$: diagonal entries reduce modulo the local radicals, while every map between distinct indecomposable summands belongs to the categorical radical. Conjugation by a generator of $H\mathrm{/}N$ cyclically permutes these $p$ factors. An $H$-endomorphism idempotent therefore has image $(0,\dots ,0)$ or $(1,\dots ,1)$ in $k^p$. In the first case the idempotent lies in the Jacobson radical and is zero; in the second its complement does, so it is one. Thus $X$ has no nontrivial $H$-endomorphism idempotent and is indecomposable.

Suppose $I_H(L)=H$. Choose isomorphisms among the $p$ conjugate summands. They identify the semisimple quotient of ${\mathrm{End}}_{\mathcal{O}N}(X)$ with $M_p(k)$. The action of a generator of $H\mathrm{/}N$ on this quotient is conjugation by a matrix $T$ that cyclically permutes the $p$ diagonal primitive idempotents. Its $p$th power is scalar, say $T^p=\lambda I$. By F3 choose $\mu \in k$ with ${\mu }^p=\lambda$; in characteristic $p$, $z^p-\lambda =(z-\mu {)}^p$. The cyclic permutation of the diagonal idempotents makes $T$ cyclic of degree $p$, so its Jordan form is one block and $C_{M_p(k)}(T)=k[T]\cong k[z]\mathrm{/}((z-\mu {)}^p),$ a local algebra.

Every $H$-endomorphism of $X$ is an $N$-endomorphism fixed by this conjugation. Hence an $H$-endomorphism idempotent maps to an idempotent in the local centralizer computed in step 2.2, and that image is $0$ or $1$. The kernel of the reduction to $M_p(k)$ lies in the Jacobson radical of the $N$-endomorphism ring; an idempotent in it is zero, and the same argument applied to the complement handles image $1$. Thus again only $0$ and $1$ occur, so $X$ is indecomposable. The two inertia cases are exhaustive. The nonzero hypothesis excludes the zero lattice, and the proof uses only finite decompositions; algebraic closedness is used exactly in steps 1.1 and 2.2, not as a hidden choice principle.