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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.

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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).

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  • $\begingroup$ Is $R[G]^H\cong R[G/H]$ given by $\sum_{g\in G}r_gg \mapsto \sum_{s\in S}a_s sH$ where $S$ is a set of coset representatives for $H$? Also when $H$ is infinite I can't see why a non-zero element of $R[G]$ has finite support and $P^H$ is zero. Regards $\endgroup$
    – eddie
    Apr 22 '16 at 10:09
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    $\begingroup$ Yes, that is the isomorphism (except that $a_s$ needs to be replaced with $r_s$). By definition, elements of $R[G]$ are finite sums, which means that $r_g = 0$ for all but finitely many $g$. (In other words, elements of $R[G]$ have finite support.) When $H$ is infinite, we have $R[G]^H = \{0\}$ (because elements of $R[G]$ have finite support). Since $P^H$ is a submodule of $(R[G]^H)^n$ for some $n$, this implies that $P^H = \{0\}$. $\endgroup$ Apr 22 '16 at 16:52

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