Question

Need examples of problems and proofs being solved surrounding group cohomology

Answer 1

Let G = Z/nZhσi be a nite cyclic group of order n with generator σ,
and let A be a G-module. Let
NG =
X
g∈G
g =
Xn−1
i=0
σ
i = 1 + σ + · · · + σ
n−1
be the norm element of Z[G], which we also view as a map NG : A → A. Then
H
i
(G, A) =



AG i = 0
ker NG/(σ − 1)A i = 1, 3, . . . ,
AG/NGA i = 2, 4, . . .
Proof. We denote NG by N. We have the following periodic projective (actually free) reso-
lution of the trivial G-module Z, where is the augmentation map, characterized by σ 7→ 1
and Z-linearity.
· · · Z[G] Z[G] Z[G] Z[G] Z 0
N σ−1 N σ−1
It is immediate to check that this is a chain complex, and not terribly hard to check that
it is in fact exact. Then we apply HomZ[G](−, A) and drop the Z term to obtain a complex
whose homology is Hi
(G, A).
0 HomZ[G](Z[G], A) HomZ[G](Z[G], A) HomZ[G](Z[G], A) · · ·
(σ−1)∗ N∗
(σ−1)∗
Since HomZ[G](Z[G], A) ∼= A via φ 7→ φ(1), we have an isomorphism of chain complexes
0 HomZ[G](Z[G], A) HomZ[G](Z[G], A) HomZ[G](Z[G], A) · · ·
0 A A A · · ·
(σ−1)∗
∼=
N∗
∼=
(σ−1)∗
∼=
σ−1 N σ−1
The maps on the bottom row of A's are determined by commutativity of this diagram, and
thinking about the exact description of the isomorphism HomZ[G](Z[G], A) ∼= A says that they must be as written. Thus
H
i
(G, A) =



ker(σ − 1) i = 0
ker N/(σ − 1)A i = 1, 3, . . .
ker(σ − 1)/NA i = 2, 4, . . .
Since the G-action is determined by the action of σ, we see that ker(σ − 1) = AG, hence the
result.

(##)

Let G = Z/nZhσi be a nite cyclic group and A a trivial G-module. Then
H
i
(G, A) =



A i = 0
nA i = 1, 3, . . . ,
A/nA i = 2, 4, . . .
where nA is the n-torsion subgroup of A.
Proof. Since A is a trivial module, AG = A, and the norm element just acts by multiplication
by |G| = n on A, and (σ −1)A = 0, so the result is immediate from the previous calculation.
As a somewhat interesting application of the previous calculations, we have a cohomology
calculation for a matrix group.