TITLE: Group cohomology vs. topological cohomology in the case of non-trivial action
QUESTION [7 upvotes]: When A is an abelian group with trivial G-action (G being a discrete group) we get that Hn(G,A)≅Hn(BG,A). Is there a similar connection between group cohomology and topological cohomology if A is a non-trivial ℤG-module? What can we say in that case?

REPLY [4 votes]: For any space $X$ with $\pi_1X=G$  and any $ZG$ module $A$, define $C_*(X;A)$ to be $C_*(\tilde{X})\otimes_{ZG}A$ and define $C^* (X;A)$ to be 
$Hom_{ZG}(C_*(\tilde{X}),A)$, where  $C_*(\tilde{X})$ is the
(cellular, or singular) chain complex of the universal cover of $X$ viewed as a free $ZG$ complex. (supressing notation involving left v.right action). Then take co/homology. It makes no difference whether the action is trivial or not. If $A$ has a trivial action, then $C_*(\tilde{X})\otimes_{ZG}A=C_*(X)\otimes A$ and $Hom_{ZG}(C_*(\tilde{X}),A)=Hom(C_*(X),A)$.  Taking $X=BG$ answers your question, assuming you like topology and  define group homology in terms of $BG$
If you prefer to start with an algebraic definition of group cohomology (eg bar construction) then you need some argument that relates the algebra to a cell structure for $BG$: but you need this argument whether or not your coefficients are twisted.