Where can I found an explanation of group cohomology from the point of view of invariants? I heard once that we can view group cohomology as the right derived functor quantifying precisely (i.e. by the usual long exact sequence) how much the functor of "taking the invariants" is not right exact. That is, given a short exact sequence of $G$-modules:
$$
0\to A \to B \to C \to 0\;,
$$
the corresponding sequence:
$$
0\to A^G \to B^G \to C^G \to 0
$$
fails to be exact at $C^G$, i.e. the map $B^G \to C^G$ is not surjective.
This is very intuitive, because it means that in a quotient ($C$) the group $G$ acts "less freely" (I'm sure there are ways to phrase this more precisely).
In all my courses, though, group cohomology was always introduced differently, either with group-ring modules, Ext and Tor, or with the bar resolution. Of course all ways are mathematically equivalent, but I particularly like the "invariant point of view" that I described above. 
Is there a text or paper where I can find such a treatment (i.e. group cohomology constructed from the long exact sequence given by the functor "taking the invariants")? Also, how does one construct that functor in practice?
Thanks!
 A: From what you say, I guess that the definition you have been given is that $H^n(G,A) = Ext^n_{\mathbb{Z}[G]}(\mathbb{Z},A)$. 
By definition, this is the $n$th derived functor of $A\mapsto Hom_{\mathbb{Z}[G]}(\mathbb{Z},A)$. But since $\mathbb{Z}$ has trivial $G$-action, $Hom_{\mathbb{Z}[G]}(\mathbb{Z},A) = A^G$, because a morphism $\mathbb{Z}\rightarrow A$ is determined by the image of $1$, which must be $G$-invariant. So the functor $A\mapsto A^G$ is naturally isomorphic to $A\mapsto Hom_{\mathbb{Z}[G]}(\mathbb{Z},A)$, and $H^n(G,-)$ is indeed the $n$th derived functor of $A\mapsto A^G$.
As for the bar resolution, it is given by the following "trick" : it is well-known that $Ext^n_R(X,Y)$ can be seen either as deriving $Hom_R(X,-)$ and applying the derived functor to $Y$, or as deriving $Hom_R(-,Y)$ and applying to $X$. Hence $H^n(G,A)$ is also obtained by deriving $Hom_{\mathbb{Z}[G]}(-,A)$ and applying the derived functor to $\mathbb{Z}$. Which means that $H^n(G,A)$ can be calculated from a single projective resolution of $\mathbb{Z}$ as a $\mathbb{Z}[G]$-module (instead of finding an injective resolution of every possible $A$). The canonical choice is the free resolution : $\dots\rightarrow\mathbb{Z}[G^n]\rightarrow\dots\rightarrow \mathbb{Z}[G]\rightarrow \mathbb{Z}\rightarrow 0$, where $\mathbb{Z}[G^{n+1}]\rightarrow \mathbb{Z}[G^n]$ is given by $(g_0,\dots g_n)\mapsto \sum_i (-1)^i (g_0,\dots \hat{g_i},\dots,g_n)$.
You can check that the resulting complex $Hom(\mathbb{Z}[G^n],A)$ with the induced differential is isomorphic to the bar resolution, so the homology of this complex gives $H^n(G,A)$.
You may want to check Gille and Szamuely's "Central simple algebras and Galois cohomology" for instance, their point of view is similar to that.
