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?



Group cohomology vs. topological cohomology in the case of non-trivial action
This is an example of twisted cohomology. In general, for a generalized cohomology theory (spectrum) E and a space X you can talk about E-twists over X. This is a certain structure on X. Given a particular twist $\tau$, you can then form the $\tau$-twisted cohomology $E^\tau(X)$. This was the subject of a [recent ArXiv paper](http://arxiv.org/abs/1002.3004) by Ando-Blumberg-Gepner. One way to think of this is that E-cohomology of X is naturally graded by the E-twists. (This viewpoint is especially when E is an $E\_\infty$-ring spectrum, in which case twists can be added and there is a cup product-type formula). 


In the case of ordinary cohomology, a twist reduces to a local system on X. A *local system* is the same thing as a functor from the fundamental groupoid of X to the category of abelian groups. This can equivalently be thought of as a locally constant sheaf of abelian groups on X, connecting up with Sammy Black's answer. For more general types of cohomology, these don't correspond to sheaves in the usual sense. If you allow "sheaves of spectra" you can get this work, but that is a difficult and long story. 


In the case that $X = BG$ (with G discrete), then the fundamental groupoid of X is equivalent to the category G, i.e. the category with a single object with automorphisms the group G. Then a local system, i.e. a functor $G \to Ab$ is exactly the same as a G-module. Then the twisted cohomology of $BG$ with this twist is exactly the group cohomology of G with values in the module. There is also a similar homology story. 


