Let $G$ be a finite group. We know that the $K$-group $K\_0(QG)$ of the rational group ring $QG$ is a free abelian group generated by the irreducible representations of $G$ over $Q$. Now let $R$ be a subring of the rationals where the order $|G|$ is invertible. What is the relation between $K\_0(RG)$ and $K\_0(QG)$?



Projective modules over non-rational group rings
I would guess that the map on $K\_0$ is an isomorphism, butI could only show the surjectivity right now:


The inclusion of rings $RG\rightarrow QG$ induces a map on $K\_0$. Given a projective $RG$-module - say it is a submodule of $RG^n$ - to its $Q$-span. It is a projective submodule of $QG^n$.


So let us first show that this map is surjective, ie. every projective $QG$ module arises this way. Given any such $P'$ the obvious candidate for a preimage would be $RG^n\cap P'$.


First note that it is as a $R$-module a direct summand of $RG^n$. $R$ is a PID and hence one just has to verify that the quotient is $R$-torsionfree. But $RG^n/(RG^n\cap P')$ embeds into the $Q$-vectorspace $QG^n/P'$ and hence it is $R$-torsionfree.


So we have a section of $R$-modules $s:RG^n/(RG^n\cap P')\rightarrow RG^n$. It need not be a $RG$-map. So let us make it equivariant by setting $s'(x):=\frac{1}{|G|}\sum\_{g\in G}gs(g^{-1}x)$. Note that it is still a section (project down again; it is a $RG$ map).


So we have found a $RG$-complement of $RG^n\cap P'$; hence $RG^n\cap P'$ is a projective $RG$-module. So the map $K\_0(RG)\rightarrow K\_0(QG)$ is surjective.


