Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 138 / 4 / b / ii / Solution

If $M$ is projective over $RG$, then its restriction is projective over $RH$ because $RG$ is a free right $RH$-module and a free $RG$-module restricts to a free $RH$-module.

Conversely, suppose ${\mathrm{Res}}_H^{G}M$ is projective. Thenis projective over $RG$. Part (i) says that $M$ is a direct summand of this induced module, so $M$ is projective. Hence projectivity detected on a subgroup of invertible index gives