This is a basic question (not research level) which has already been asked on SE by someone else but doesn't yet have an answer so I'd like to repost it on MO. 


Let $R$ be a commutative ring with unity and $H$ be a normal subgroup of a finite group $G$. If $P$ is a finitely generated projective left $R[G]$-module then is the submodule of $H$-invariants $$P^H:= \{p\in P:g(p)=p\ \forall g\in H\}$$ a projective left **$R[G/H]$**-module?


There is a $R[G]$-module $Q$ and a positive integer $n$ s.t. $P\oplus Q\cong R[G]^n$. So $P^H\oplus Q^H=(P\oplus Q)^H\cong (R[G]^n)^H\cong (R[G]^H)^n.$ 


Is it true that $R[G]^H\cong R[G/H]?$ 
Or is there is another way to proceed? 


Also is this result true if $P$ is not finitely generated? 


Many thanks.



An invariant submodule of a projective module
If $H$ is finite, then $R[G]^H$ is isomorphic to $R[G/H]$ as an $R[G/H]$-module (because both consist of the elements that are constant on cosets of $H$). However, if $H$ is infinite, then $R[G]^H = \{0\}$ (because elements of $R[G]$ have finite support, and therefore cannot be a nonzero constant on any coset of $H$), whereas $R[G/H] \neq \{0\}$.


Therefore, the proposed argument works whenever $H$ is finite. When $H$ is infinite, $P^H$ is projective because it is $\{0\}$. In both cases, it does not matter whether $P$ is finitely generated (because $R[G]^n$ can be replaced with an infinite direct sum in the argument).


