For finite groups $G$ of order invertible in $R$, is taking $G$-invariants an exact functor on the category of $R[G]$-modules?

Question

Let $G$ be a finite group of order $n$, and $R$ a ring on which $n$ is invertible. Is taking $G$-invariants an exact functor on the category of $R[G]$-modules?

Answer 1

Here is a slightly different way to think about Nesos's answer. The $G$-invariants functor is naturally isomorphic to the functor $\operatorname{Hom}_{R[G]}(R,-)$ where $G$ acts trivially on $R$, since an element $m\in M$ is invariant iff the $R$-module map $R\to M$ sending $1$ to $m$ preserves the action of $G$. So we want to know whether the functor $\operatorname{Hom}_{R[G]}(R,-)$ is exact.

This Hom-functor is exact iff $R$ is a projective $R[G]$-module. Now note that there is surjection of $R[G]$-modules $R[G]\to R$ which sends each $g\in G$ to $1$, and this is split by the map $R\to R[G]$ which sends $1$ to $\frac{1}{|G|}\sum_{g\in G}g$. So $R$ is a direct summand of the free module $R[G]$, and hence projective. Thus the functor $\operatorname{Hom}_{R[G]}(R,-)$ is exact.

Answer 2

In general taking $G$-invariants is left exact. To show exactness, it suffices to show that it takes surjections to surjections.

Ie, given a surjection of $R[G]$-modules $\varphi : M\rightarrow N$, we want to show that $\varphi^G : M^G\rightarrow N^G$ is also surjective.

As pointed out by darij grinberg, $M^G$ (resp $N^G$) is just the image of the self map $f_{G,M} : M\rightarrow M$ given by $m\mapsto\frac{1}{|G|}\sum_{g\in G}gm$. From the definition of this map it's clear that $\varphi^G \circ f_{G,M} = f_{G,N}\circ\varphi$, but since $f_{G,M},f_{G,N},\varphi$ are surjective (viewing $f_{G,M},f_{G,N}$ as functions $M\rightarrow M^G$ and $N\rightarrow N^G$), it follows that $\varphi^G$ is also surjective.)

Alternatively, if you want to hit this with a hammer, the answer follows from Theorem 7.3.1 of Ribes-Zalesskii's book Profinite Groups.