Group cohomology intuition Group cohomology is one thing I don't seem to get my head around.
I understand $H^0$ as being the "fixed points" but when it comes to anything higher I have no idea what the notions are meant to be capturing...if anything.
For example I can see that 1-cocycles are homomorphisms with a specific property but that property doesn't seem to mean anything to me...and I think it should!
Is there an intuition or example that explains the meaning or is it just something that "stuck" and seems to work in lots of places?
 A: Group cohomology can be defined very naturally in a purely topological way. The definition of $1$-cocycles is not random, or due to historical accident.
More specifically, given a group, $G$, the Eilenberg-Maclane space $X=K(G,1)$ is defined which has $\pi_1(X)=G$, and $\pi_{\geq 2}(X)=0$. This is well-defined up to homotopy-type if you assume it is a CW-complex. Since the cohomology functor is invariant under homotopy equivalence, the groups $H^i(X)$ are well-defined abelian groups associated with the original group $G$, and this is what we call group cohomology. Now when you take coefficients in a $\mathbb Z[G]$ module $M$, this is actually equivalent to taking cohomology with ``twisted coefficients" of $K(G,1)$.
A: To add to George's answer above, 1st Galois cohomology group has a natural interpretation as the set of classes of principal homogeneous spaces for a group.
Let $G$ be an algebraic group defined over $k$. Let $K/k$  be a Galois extension. The $K$-points of $G$ form a $\mathrm{Gal}(K/k)$-group (or a $\mathrm{Gal}(K/k)$-module, if $G$ is commutative). A principal homogeneous $K$-space is a $K$-variety $X$ with a free transitive action of $G$. There is a bijective correspondence between $k$-isomorphism classes of principal homogeneous $K$-spaces and cocycles of $H^1(\mathrm{Gal}(K/k), G(K))$.
Let $X$ be a p.h.s. Pick a point $x \in X$ and act on it by $\sigma \in \mathrm{Gal}(K/k)$. Since $G$ acts on $X$ freely and transitively there is a unique $g_\sigma$ such that $g_\sigma \cdot x = \sigma(x)$. One checks that $\sigma \mapsto g_\sigma$ define a 1-cocycle. 
Conversely, given a 1-cocyle $\{g_\sigma\}$, consider a disjoint union $\sqcup G \times \{\sigma\}_{\sigma \in \mathrm{Gal}(K/k)}$ and define an action of $\mathrm{Gal}(K/k)$ on it: $\sigma(x,\tau)=(g_\sigma \cdot x, \sigma\tau)$. There is a natural $G$ action on the disjoint union, and the factor by the action $\mathrm{Gal}(K/k)$ inherits the action of $G$ (a highbrow term for what is happening is "Galois descent"), turning it into a p.h.s.
Moreover, if $G$ is commutative, the group law on $H^1(\mathrm{Gal}(K/k), G(K))$ has a natural geometric interpretation in terms of principal homogeneous spaces. This group is also known as Weil-Chatelet group. I recommend the original Weil's article on the subject: A. Weil, "On algebraic groups and homogeneous spaces". Amer. J. Math. , 77 (1955) pp. 493–512
A: [At Mariano's suggestion, I'll copy my answer from MathOverflow.]
Here is a completely elementary example which shows that group cohomology is not empty verbiage, but can solve a problem ("parametrization of rational circle") whose statement has nothing to do with cohomology.
Suppose you somehow know that for a finite Galois extension $k\subset K$ with group $G$ the first cohomology group  $H^1(G,K^*)$ is zero : this is the homological version of Hilbert's Theorem 90 ( you can look it up in Weibel's book on homological algebra, pages 175-176).
If moreover $G $ is cyclic with generator $s$, this implies  that an element of $K$
has norm one if and only if it can be written $\frac{a}{s(a)}$ for some $a\in K$.
Consider now the quadratic extension $k=\mathbb Q \subset K=\mathbb Q(i)$ with generator $s$ of $Gal(\mathbb Q (i)/\mathbb Q)$  the complex conjugation.The statement above says that $x+iy\in \mathbb Q(i)$ satisfies $x^2+y^2=1$ iff 
$x+iy=\frac{u+iv}{s(u+iv)}=\frac{u+iv}{u-iv}=\frac{u^2-v^2}{u^2+v^2}+i\frac{2uv}{u^2+v^2}$ for some $u+iv\in \mathbb Q (i)$  . 
So we have obtained from group cohomology the well-known parametrization for the rational points of the unit circle  $x^2+y^2=1$
 $$x=\frac{u^2-v^2}{u^2+v^2}, \quad y=\frac{2uv}{u^2+v^2}$$
