A Short Course on Galois Cohomology - William Stein - University of ...

for all s ∈ G, that is, s(ba) = ba and thus ba ∈ B G .
Then as ∼ (ba) −1 s(ba) and ba ∈ B G . But a ∈ A so ρ(ba) = ρ(b) = c, so
ba ↦→ c and ba ∈ B G hence c ∈ ρ0(B G ).
Now, we prove exactness at H 1 (G, A). Suppose that ι1 sends [as] to
the trivial element of H 1 (G, B). Then there exists a b ∈ B such that as ∼
b −1 s(b) for all s ∈ G. Thus [as] = δ(ρ(b)), which shows that ker(ι1) ⊂ Im(δ).
For the other direction, note that δ(c) is given by as = b −1 s(b) for some b
such that b ↦→ c. But this means that ι1(as) ∼ 1.
Finally, we prove exactness H 1 (G, B). First, note that as ∈ H 1 (G, A)
maps to s ↦→ ρ1(ι1(as)) = 1, since rho◦ι = 1, showing that Im(ι1) ⊂ ker(ρ1).
For the other direction, let bs ∈ H 1 (G, B) such that (s ↦→ ρ1(bs)) ∼ 1. Then
there exists a c ∈ C such that ρ1(bs) = c −1 s(c). We may modify by a lift of
c so that ρ1(bs) = 1 for all s. It then follows that bs ∈ A for all s, so that
[bs] ∈ Im(ι1).
12 Homology
In this section, we let G be a group, A an abelian G-module and set DA =
〈s.a − a : a ∈ A, s ∈ G〉, a subgroup of A.
We first observe that DA is in fact a G-module,
t.(s.a − a) = t.s.a − t.a = (tst −1 ).(t.a) − t.a ∈ DA.
Thus, A/DA is also a G-module, with trivial G-action.
Definition 33. We define H0(G, A) = A/DA, which is the largest quotient
of A with trivial G-action. For q ≥ 1, we also let
Hq(G, A) = Tor Z[G]
q (Z, A)
so that Hq(G, −) are the left-derived functors of H0(G, −).
Given a short exact sequence 0 → A → B → C → 0, as before we obtain
a long exact sequence
· · · → H1(G, A) → H1(G, B) → H1(G, C)
δ
−→ H0(G, A) → H0(G, B) → H0(G, C) → 0.
Example 34. For any group G, we have H1(G, Z) = G/G ′ , the abelianisation
of G, where G ′ = 〈aba −1 b −1 : a, b ∈ G〉 is the commutator subgroup.
See Serre §VII.4 for a proof.
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