I'm beginning to learn cohomology for cyclic groups in preparation for use in the proofs of global class field theory (using ideal-theoretic arguments). I've seen the proof of the long exact sequence and of basic properties of the Herbrand quotient, and I've started to look through how these are used in the proofs of class field theory.
So far, all I can tell is that the cohomology groups are given by some ad hoc modding out process, then we derive some random properties (like the long exact sequence), and then we compute things like $H^2(\mathrm{Gal}(L/K),I_{L})$, where $I_L$ denotes the group of fractional ideals of a number field $L$, and it just happens to be something interesting for the study of class field theory such as $I_K/\mathrm{N}(I_L)$, where $L/K$ is cyclic and $\mathrm{N}$ denotes the ideal norm. We then find that the cohomology groups are useful for streamlining the computations with various orders of indexes of groups.
What I don't get is what the intuition is behind the definitions of these cohomology groups. I do know what cohomology is in a geometric setting (so I know examples where taking the kernel modulo the image is interesting), but I don't know why we take these particular kernels modulo these particular images. What is the intuition for why they are defined the way they are? Why should we expect that these cohomology groups so-defined have nice properties and help us with algebraic number theory? Right now, I just see theorem after theorem, I see the algebraic manipulation and diagram chasing that proves it, but I don't see a bigger picture.
For context, if $A$ is a $G$-module where $G$ is cyclic and $\sigma$ is a generator of $G$, then we define the endomorphisms $D=1+\sigma+\sigma^2+\cdots+\sigma^{|G|-1}$ and $N=1-\sigma$ of $A$, and then $H^0(G,A)=\mathrm{ker}(N)/\mathrm{im}(D)$ and $H^1(G,A)=\mathrm{ker}(D)/\mathrm{im}(N)$. Note that this is a slight modification of group cohomology, i.e. Tate cohomology, which the cohomology theory primarily used for Class Field Theory. Group cohomology is the same but with $H^0(G,A) = \mathrm{ker}(N)$. The advantage of Tate cohomology is that it is $2$-periodic for $G$ cyclic.
Best Answer
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 $x^2+y^2=1$ $$x=\frac{u^2-v^2}{u^2+v^2}, \quad y=\frac{2uv}{u^2+v^2}$$.