Given a group $G$ and $G$-modules $M,N$ with $M$ $\mathbb{Z}$-free then it's well known that 
$$Ext\_{\mathbb{Z}G}^i(M,N) \cong H^i(G,Hom(M,N))$$
for all $i \ge 0$ (a reference is Brown, Cohomology of Groups, Proposition 2.2). 


But what happens if $M$ is not $\mathbb{Z}$-free ? Is it still possible to express $Ext\_{\mathbb{Z}G}^i(M,N)$ by cohomology groups of $G$ (maybe in form of a spectral sequence) ? 



Can Ext over a group ring always be expressed as group cohomology ? 
There is a long exact sequence 



> 
> $$0 \to H^1(G,Hom(M,N)) \to Ext\_{\mathbb{Z}G}^1(M,N) \to \cdots $$
>  $$\begin{array}{lll}
> \cdots & \to & H^i(G,Hom(M,N)) \to Ext\_{\mathbb{Z}G}^i(M,N) \newline 
> & \to & H^{i-1}(G,Ext\_{\mathbb{Z}}^1(M,N))\to H^{i+1}(G,Hom(M,N)) \to \cdots
> \end{array}$$
> 
> 
> 


For, as pointed out by Will and Mariano, there is a spectral sequence 
$$H^i(G, Ext^j\_\mathbb Z (M,N)) \Rightarrow Ext\_{\mathbb ZG}^{i+j}(M,N)$$
and since the projective dimension of $\mathbb{Z}$ is one, the spectral sequence takes the form 
$$E\_2 = \quad 
\begin{array}{ccccc}
\vdots & \vdots & \vdots & \vdots & \newline 
0 & 0 & 0 & 0 & \cdots \newline 
\bullet & \bullet & \bullet & \bullet & \cdots \newline 
\bullet & \bullet & \bullet & \bullet & \cdots 
\end{array}$$
Now the relation $E\_\infty =E\_3 = \ker(d\_2)/\text{im}(d\_2)$ yields the exact sequence 
$$0 \to \ker(d\_2^{i-2,1}) \to E\_2^{i-2,1} \to E\_2^{i,0} \to H^i \to \ker(d\_2^{i-1,1}) \to 0$$
where $H^i=Ext\_{\mathbb ZG}^i(M,N)$ is the abutment. 


**Remark:** The statement remains true if we replace $\mathbb{Z}$ by an hereditary commutative ring $R$, $\mathbb{Z}G$ by an augmented $R$-projective $R$-algebra $A$ and $H^\ast(G,-)$ by $Ext\_A^\ast(R,-)$. 


