Modules and Group Cohomology

Let $A$ be a $G$ module. This means that $A$ is an abelian group equipped with a left action of $G$, i.e., a group homomorphism $G\to \Aut (A)$, where $\Aut (A)$ denotes the group of bijections $A\to A$ that preserve the group structure on $A$. Alternatively, $A$ is a module over the ring $\mathbf{Z}[G]$ in the usual sense of module. For example, $\mathbf {Z}$ with the trivial action is a module over any group $G$, as is $\mathbf{Z}/m\mathbf{Z}$ for any positive integer $m$. Another example is $G=(\mathbf{Z}/n\mathbf{Z})^*$, which acts via multiplication on $\mathbf{Z}/n\mathbf{Z}$.

For each integer $n\geq 0$ there is an abelian group $\H^n(G,A)$ called the $n$th cohomology group of $G$ acting on $A$. The general definition is somewhat complicated, but the definition for $n\leq 1$ is fairly concrete. For example, the 0th cohomology group

$$\H^0(G,A) = \{x \in A : \sigma x = x$$    for all $$\sigma \in G\} = G^A$$

is the subgroup of elements of $A$ that are fixed by every element of $G$.

The first cohomology group

$$\H^1(G,A) = C^1(G,A)/B^1(G,A)$$

is the group of $1$-cocycles modulo $1$-coboundaries, where

$$C^1(G, A) = \{f : G \to A$$    such that $$f(\sigma\tau) = f(\sigma) + \sigma f(\tau)\}$$

and if we let $f_a: G \to A$ denote the set-theoretic map $f_a(\sigma) = \sigma(a)-a$, then

$$B^1(G, A) = \{f_a : a\in A\}.$$

There are also explicit, and increasingly complicated, definitions of $\H^n(G,A)$ for each $n\geq 2$ in terms of certain maps $G \times \cdots \times G \to A$ modulo a subgroup, but we will not need this.

For example, if $A$ has the trivial action, then $B^1(G,A)=0$, since $\sigma a - a = a -a =0$ for any $a\in A$. Also, $C^1(G,A) = \Hom (G,A)$. If $A=\mathbf{Z}$, then since $G$ is finite there are no nonzero homomorphisms $G\to \mathbf{Z}$, so $\H^1(G,\mathbf{Z})=0$.

If $X$ is any abelian group, then

$$A = \Hom (\mathbf{Z}[G], X)$$

is a $G$-module. We call a module constructed in this way co-induced.

The following theorem gives three properties of group cohomology, which uniquely determine group cohomology.

Theorem 11.2.1   Suppose $G$ is a finite group. Then

1. We have $\H^0(G,A) = A^G$.
2. If $A$ is a co-induced $G$-module, then $\H^n(G,A) = 0$ for all $n\geq 1$.
3. If $0\to A \to B \to C \to 0$ is any exact sequence of $G$-modules, then there is a long exact sequence
   $$0 \to \H^0(G,A) \to \H^0(G,B) \to \H^0(G,C) \to \H^1(G,A) \to \cdots$$
   $$\cdots \to \H^{n}(G,A) \to \H^{n}(G,B) \to \H^{n}(G,C) \to \H^{n+1}(G,A) \to \cdots$$

Moreover, the functor $\H^n(G,-)$ is uniquely determined by these three properties.

We will not prove this theorem. For proofs see [Cp86, Atiyah-Wall] and [Ser79, Ch. 7]. The properties of the theorem uniquely determine group cohomology, so one should in theory be able to use them to deduce anything that can be deduced about cohomology groups. Indeed, in practice one frequently proves results about higher cohomology groups $\H^n(G,A)$ by writing down appropriate exact sequences, using explicit knowledge of $\H^0$, and chasing diagrams.

Remark 11.2.2   Alternatively, we could view the defining properties of the theorem as the definition of group cohomology, and could state a theorem that asserts that group cohomology exists.

Remark 11.2.3   For those familiar with commutative and homological algebra, we have

$$\H^n(G,A) = \Ext ^n_{\mathbf{Z}[G]}(\mathbf{Z}, A),$$

where $\mathbf {Z}$ is the trivial $G$-module.

Remark 11.2.4   One can interpret $\H^2(G,A)$ as the group of equivalence classes of extensions of $G$ by $A$, where an extension is an exact sequence

$$0\to A \to M \to G \to 1$$

such that the induced conjugation action of $G$ on $A$ is the given action of $G$ on $A$. (Note that $G$ acts by conjugation, as $A$ is a normal subgroup since it is the kernel of a homomorphism.)