Proposition 3

Let $A$ be a commutative ring, $M$ an $A$-module, $G$ a finite group acting on $M$, and $A'$ an $A$-module. Assume that the order $q$ of $G$ is invertible in $A$. Let $M^G$ be the set of elements of $M$ invariant under $G$. Then the canonical homomorphism from $M^G \otimes_A A'$ to $M \otimes_A A'$ defines an isomorphism from $M^G \otimes_A A'$ onto the module $(M \otimes_A A')^G$ of elements of $M \otimes_A A'$ invariant under $G$.

Indeed, let $Q$ be the projection of $M$ onto $M^G$ defined by $Q(x) = q^{-1} \sum_{g \in G} g(x)$ for all $x \in M$. If $i$ denotes the canonical injection of $M^G$ into $M$, $Q \circ i$ is the identity map of $M^G$, so $(Q \otimes 1_{A'}) \circ (i \otimes 1_{A'})$ is the identity map of $M^G \otimes_A A'$. Since $Q \otimes 1_{A'} = q^{-1} \sum_{g \in G} (g \otimes 1_{A'})$, the image of $i \otimes 1_{A'}$ is $(M \otimes A')^G$. On the other hand, $i \otimes 1_{A'}$ is injective by what has been said before.