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 $\operatorname{Res}_H^GM$ is projective. Then
$$
\operatorname{Ind}_H^G\operatorname{Res}_H^GM
$$
is 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
$$
\boxed{M\text{ is }RG\text{-projective}
\quad\Longleftrightarrow\quad
\operatorname{Res}_H^GM\text{ is }RH\text{-projective}.}
$$

Solution

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

If $[G:H]$ is invertible in $R$, every $RG$-module is relatively H-projective. An $RG$-module is then projective exactly when its restriction to $RH$ is projective.


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