Q:

$UP = _fP$ is a projective $\mathbb{Z}S$-module.

$S$ is a subgroup of $G$ and f is the inclusion map. As an $\mathbb{Z}S$-module, $\mathbb{Z}G$ is free. It follows that if $P$ is a projective $\mathbb{Z}G$-module, then $UP = _fP$ is a projective $\mathbb{Z}S$-module. Here $U$ is the change of groups functor.
I have no clue how $UP = _fP$ is projective. Any help would be appreciated!

A:

$P$ is a direct summand of some free $\Bbb ZG$-module $F$, so regarded
as $\Bbb ZS$ modules, $P$ is a still a direct summand of $F$, so it suffices
to prove that $F$ is projective as a $\Bbb ZS$-module. As $F$ is a direct sum
of a bunch of copies of $\Bbb ZG$, it suffices to prove that
$\Bbb ZG$ is a projective $\Bbb ZS$-module.
But $\Bbb ZG$ is a free $\Bbb ZS$-module. Write $G$ as a disjoint union of cosets
$Sg_i$. Then the $g_i$ are free generators of $\Bbb ZG$ as a left $\Bbb ZS$-module.

