Relating Ext groups of abelian groups and group cohomology One can define $\mathrm{Ext}$-groups in the category of abelian groups (not $\mathbb{Z}[G]$-modules) and group cohomology in very similar ways.
The second, group cohomology, can be computed in the homotopy category of topological spaces: we can take a group $G$, turn it into a topological space $\mathrm{K}(G,1)$ by geometric realisation, and we have 
$$ \mathrm{H}^n(G,\mathbb{Z}) = \left [ \mathrm{K}(G,1), \mathrm{K}(\mathbb{Z},n) \right ].$$
What about the first? In this case, working in the category of chain complexes of abelian groups, one gets $\mathrm{Ext}$-groups instead (which fits in the above framework by considering the homology of a chain complex to be its homotopy groups).   
I'd like to reconcile this with the principle that considering abelian groups is akin to stabilising: chain complexes of abelian groups correspond to certain kinds of infinite loop spaces, inside the whole category of topological spaces. Is there a stabilisation process that relates $\mathrm{Ext}$-groups of abelian groups and group cohomology of abelian groups that underlies the above picture? 
Consider for instance the cohomology of $\mathbb{RP}^\infty$, and contrast with $\mathrm{Ext}$-groups.
For $i>0$ even:
$$ \mathrm{H}^i(\mathbb{RP}^\infty,M) = M/2M, \qquad \mathrm{Ext}^1(\mathbb{Z}/2\mathbb{Z},M) = M/2M, $$
and for $i>0$ odd:
$$ \mathrm{H}^i(\mathbb{RP}^\infty,M) = M[2], \qquad \mathrm{Hom}(\mathbb{Z}/2\mathbb{Z},M) = M[2]. $$
How does this work in general, for $\mathrm{K}(A,1)$ with $A$ an abelian group (or perhaps even for more general $\mathrm{K}(A,n)$)?
Edit: Removed incorrect objects obtained by the bar construction. The off-by-one in the above two equations is a manifestation of my error, maybe someone can explain how to use bar constructions attached to an adjunction to construct classifying spaces.
 A: As you have noticed, there is an irritating degree shift that pervades this discussion.  Consider the simplicial set $B G$ whose $n$-simplices are $n$-tuples of elements of $G$, with face and degeneracy operators are in the bar construction. (In fact, it is the bar construction induced by the free–forgetful adjunction for $G$-sets.) In other words, $B G$ is the nerve of $G$ considered as a one-object groupoid, and it is well-known that the nerve of a groupoid is always a Kan complex. However, it is not a contractible Kan complex in general: indeed, the geometric realisation of $B G$ is a $K (G, 1)$ space. 
We can then take the décalage of $B G$ to get a simplicial set $E G$ whose $n$-simplices are the $(n + 1)$-tuples of elements of $G$. If we choose the appropriate convention, $E G$ will be the nerve of the groupoid whose objects are the elements of $G$ and whose morphisms $x \to y$ are the elements $g$ such that $g x = y$. In particular, $E G$ is a contractible Kan complex, and we have a Kan fibration $p : E G \to B G$ whose fibre is canonically identified with the underlying set of $G$. In particular this exhibits the geometric realisation of $B G$ as a classifying space for $G$.
Now, consider the associated unnormalised chain complexes. On the one hand, for any abelian group $A$, 
$$H^* (K (G, 1), A) \cong H^* (\mathrm{Hom}_\mathbb{Z} (\mathbb{Z} B G, A))$$
by the isomorphism of singular and cellular cohomology. On the other hand, $\mathbb{Z} E G$ is (by Dold–Kan) a resolution of $\mathbb{Z}$ that is degreewise a free right $\mathbb{Z} G$-module. Thus,
$$H^* (G, A) = \mathrm{Ext}_{\mathbb{Z} G}^* (\mathbb{Z}, A) \cong H^* (\mathrm{Hom}_{\mathbb{Z} G} (\mathbb{Z} E G, A))$$
On the other hand, $\mathbb{Z} E G$ is isomorphic to $\mathbb{Z} B G \otimes_\mathbb{Z} \mathbb{Z} G$, so $\mathrm{Hom}_{\mathbb{Z} G} (\mathbb{Z} E G, A) \cong \mathrm{Hom}_{\mathbb{Z}} (\mathbb{Z} B G, A)$, which proves
$$H^* (G, A) \cong H^* (K (G, 1), A)$$
as is well known.
The business with $\mathrm{Ext}_\mathbb{Z}^*$ is quite different, but there is a curious coincidence worth pointing out. In low degrees, the chain complex $\mathbb{Z} B G$ has differentials given by $(g_0, g_1) \mapsto g_0 - g_0 g_1 + g_1$ and $g \mapsto 0$; so the cochain complex $\mathrm{Hom}_\mathbb{Z} (\mathbb{Z} B G, A)$ has differentials given by $f \mapsto 0$ and $f \mapsto ((g_0, g_1) \mapsto f (g_0) - f (g_0 g_1) + f (g_1))$. Thus:
$$H^1 (K (G, 1), A) \cong \mathrm{Hom} (G, A)$$
This is homotopy-theoretically unsurprising: group homomorphisms $G \to H$ can be canonically identified with pointed homotopy classes of pointed maps $K (G, 1) \to K (H, 1)$, which in turn can be identified with elements of $H^1 (K (G, 1), H)$. 
Is tempting to conjecture that $\mathrm{Ext}^* (G, A) \cong H^{* + 1} (K (G, 1), A)$ in general; sadly, this claim is false. Let $\mathbb{Z} B G [1]$ be the chain complex defined by $\mathbb{Z} B G [1]_n = \mathbb{Z} B G_{n + 1}$. Since the differential $\mathbb{Z} B G_1 \to \mathbb{Z} B G_0$ is zero, we immediately deduce
$$H_* (\mathbb{Z} B G [1]) \cong H_{* + 1} (\mathbb{Z} B G) \cong H_{* + 1} (K (G, 1), \mathbb{Z})$$
and therefore $\mathbb{Z} B G [1]$ is not a free resolution of $H_1 (K (G, 1), \mathbb{Z}) \cong G / [G, G]$ in general! Unsurprisingly, we can find examples where $\mathrm{Ext}_\mathbb{Z}^* (G, A)$ and $H^{* + 1} (K (G, 1), A)$ are genuinely different. For instance, given $n > 0$,
$$\mathrm{Ext}_\mathbb{Z}^n (\mathbb{Z}^{n+1}, \mathbb{Q}) = 0$$
because $\mathbb{Q}$ is an injective $\mathbb{Z}$-module, whereas
$$H^{n+1} (K (\mathbb{Z}^{n+1}, 1), \mathbb{Q}) \cong H^{n+1} (T^{n+1}, \mathbb{Q}) \cong \mathbb{Q}$$
because $K (\mathbb{Z}^{n+1}, 1) \simeq K (\mathbb{Z}, 1)^{n+1}$, $K (\mathbb{Z}, 1)$ is homotopy equivalent to the circle $S^1$, and the torus $T^{n+1}$ is a connected compact orientable manifold of dimension $n + 1$.
