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 $K_0(QG)$?
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. 

