Assume (iii). Reduction of the projective lattice $W$ is projective, because a direct-summand decomposition of a free $\mathcal OG$-module remains one after tensoring with $k$. Completeness of $\mathcal O$ and <idempotent lifting> show that $\overline W$ is indecomposable: otherwise a nontrivial idempotent of $\operatorname{End}_{kG}(\overline W)$ would lift and split $W$, hence split the simple $KG$-module $M$.

Let $\chi$ be the ordinary irreducible character of $M$. Since $W$ is projective, $\chi$ vanishes on p-singular elements and restricts to the <projective character> $\Phi$ of $\overline W$. Thus
$$
\dim_k\operatorname{End}_{kG}(\overline W)
=\langle\Phi,\Phi\rangle_{p'}
=\langle\chi,\chi\rangle=1.
$$
An indecomposable projective module that is not simple has, besides its identity, a nonzero noninvertible endomorphism obtained by projecting onto its simple head and embedding the isomorphic simple socle. Hence its endomorphism algebra has dimension at least two. Therefore $\overline W$ is simple as well as projective, proving
$$
\boxed{\text{(iii)}\Longrightarrow\text{(v)}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 6 / a / iii / Solution

If a ring is complete with respect to an ideal contained in its <Jacobson radical>, every idempotent in the quotient lifts to an idempotent of the ring. Consequently direct-sum decompositions of a modular reduction lift to a complete integral lattice.

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 Incoming links: Idempotent lifting