The object: an abelian group G acts on
A G-module is an abelian group A together with an action of a group G by automorphisms: each g in G gives an automorphism a ↦ g·a, compatibly, so that (gh)·a = g·(h·a) and 1·a = a, and g·(a+b) = g·a + g·b. Equivalently — and this is the slogan worth remembering — A is a module over the group algebra Z[G]. Group cohomology will be nothing but the derived functor of taking invariants.
Three running examples carry the whole track. (1) Trivial action: g·a = a for all g; then A is just an abelian group remembering G is around. (2) For G = Gal(L/K), the additive group L and the multiplicative group L* are G-modules — the action is the natural one of automorphisms on elements of the field. (3) A normal subgroup A inside a group E, acted on by conjugation from a quotient G = E/A, is a G-module when A is abelian. Keep all three in view; each will pay off.
H^0 is invariants, and that is where the trouble starts
Define the invariants A^G = { a in A : g·a = a for all g in G }. This is already the zeroth cohomology: H^0(G, A) = A^G. For the trivial action A^G = A; for L under Gal(L/K), L^G = K by the very definition of Galois extension; for L* under the same group, (L*)^G = K*. Nothing mysterious yet — H^0 just records what the symmetry fixes.
The trouble: the functor A ↦ A^G is only [[left-exact-functor|left exact]]. Apply it to a short exact sequence 0 → A → B → C → 0 of G-modules and you get an exact 0 → A^G → B^G → C^G — but the last map need not be onto. A G-invariant element of C may have no G-invariant preimage in B. That failure of surjectivity is real algebraic content, and the higher cohomology groups H^1, H^2, … are exactly the machine that measures it through a long exact sequence.
G = Z/2Z = {1, s}, A = Z with the sign action s·n = -n.
Invariants: A^G = { n in Z : -n = n } = {0}.
Now the multiplication-by-2 sequence of G-modules
0 -> Z --(x2)--> Z --(mod 2)--> Z/2Z -> 0 (sign action on the two Z's; trivial on Z/2Z)
apply ( - )^G :
0 -> 0 -> 0 -> Z/2Z
The class 1 in (Z/2Z)^G = Z/2Z is invariant but has NO invariant preimage,
because B^G = 0. Surjectivity fails. That missing piece is detected by H^1(G, Z),
the first place the long exact sequence continues.