Second cohomology of cyclic groups

This article discusses ideas pertaining to the following general problem: given a cyclic group $$G$$ of order $$n$$, and any Abelian group $$A$$, with an action of $$G$$ on $$A$$, find the cohomology group $$H^2(G;A)$$. Equivalently, find all congruence classes of extensions with normal subgroup $$A$$ and quotient group $$G$$.

Let's first recall the relation between cohomology classes and congruence classes of extensions:

A substantially smaller group of cocycles
The group of all 2-cocycles $$G \times G \to A$$ is fairly huge, so the first question is: can be obtain a considerable smaller group of cocycles, that still covers all the cohomology classes? It turns out that we can use the cyclic structure of $$G$$ to do so.

First, let's think in terms of sections of an extension instead of cocycles. For convenience, pick a generator $$g$$ of $$G$$. Then, if $$E$$ is an extension, call a section $$s$$ to be $$g$$-centered if it has the property that for $$0 \le i < n$$, $$s(g^i) = s(g)^i$$. In other words, our section has the property that once we know where it sends $$g$$, we know where it sends everything else.

Further, it's clear that for the corresponding cocycle $$\varphi_s$$, we have, with $$0 \le i,j < n$$:

$$\varphi_s(g^i,g^j) = 0$$ if $$i + j < n$$ and $$= s(g)^n$$ if $$i + j \ge n$$

In other words, $$\varphi_s$$ takes only two values, 0 and $$s(g^n)$$. $$g$$-centered cocycles form a subgroup under addition of the group of 2-cocycles. Moreover, this subgroup maps surjectively to the cohomology group, i.e., every cohomology class has a $$g$$-centered representative.

Moreover, this subgroup is completely described by one element of $$A$$, and hence can also be identified with a subgroup of $$A$$. This leads to two questions:


 * 1) What subgroup of $$A$$ is this? In other words, what elements of $$A$$ can arises as $$s(g)^n$$ for some $$g$$-centered section of some extension? Let's call this subgroup $$B$$.
 * 2) Which of these are equivalent? So, what quotient of this subgroup is actually the cohomology group? Let's call the kernel of the map $$C$$, i.e. $$C$$ is the subgroup comprising those elements that correspond to cohomologically trivial cocycles.

The answers to both questions (1) and (2) depend on the nature of the action. There are situations where B itself is trivial, or very small, and gives a good hold on the cohomology. There are other situations where $$B$$ is very large (for instance, we may even have $$B = A$$) and $$C$$ is also very large.

The subgroup of feasible cocycles
The subgroup comprising cocycles is described very simply, as:

$$B := \{ a \in A \mid g.a = a\}$$

In other words, it is precisely the subgroup of $$A$$ comprising points fixed under the action of $$g$$.

Finding the cohomologically trivial cocycles
This is a somewhat trickier task, and solving it typically requires the use of clever identities. Finding cohomologically trivial cocycles is equivalent to finding all the elements that can be written in the form:

$$a + g.(a) + g^2.(a) + \dots + g^{p-1}(a)$$