Let $G$ be a finite group, $P$ a Sylow $p$-subgroup and $H$ a subgroup of $G$ containing $P$ such that $P ∩ {^xP} =$ {$1$} for all $x ∈ G$\ $H$. $O$ is a complete DVR.
The action by $P × P$ on $G$ \ $H$ given by left and right multiplication with elements in $P$ is free. Indeed if $u, v ∈ P$ and $x ∈ G \ H$, then the equality $uxv^{−1} = x$ implies $x^{−1}ux = v$, forcing $u = v = 1$ because of the assumption $P ∩ {^xP} = {1}$. Thus $O[G$ \ $ H]$ is projective as an $O(P × P)$-module.
My question is why this module is projective. Thank you!