Understanding the relation between central extensions and 2-cocycles and coboundaries

Question

Context

This is basically the lecture note I received on central extensions (we were not introduced to group cohomology before this):

If

$$1 \to A \to G \to B \to 1$$

is a non-trivial central extension (meaning, not a direct product), then it is the opposite type of extension from a semidirect product.

If A is central, then it is of course abelian, and it is easier to discuss it in additive notation. To keep additive and multiplicative groups separate, let's say that A is additive and that exp(A) is a purely formal exponentiation of A in which the group law is changed to multiplication, For instance, you can say that $C_n = \exp(\Bbb Z/n)$. Our exact sequence becomes

$$1 \to \exp(A) \to G \to B \to 1 (*)$$

As in the case of semidirect products, central extensions have a synthetic version and a descriptive version. I'll start with a descriptive version.

As in the case of a semidirect product, we can analyze this by choosing a section of exp(A) in $G$, i.e., a set which has one representative of each coset. Since now the whole concern is that possibly no section is a group complement, we can let $S$ be any section of $\exp(A)$ in $G$ whatsoever. $S$ is still canonically bijective with $B$, and I want a name for the bijection from $B$ to $S$, so I guess I will write $s = \gamma(b)$.

Every element of $G$ is uniquely described as $\gamma(b) \exp(a)$ for some $b$ in $B$ and some $a$ in $A$. Since $A$ is central, you can check that the enitre group law of $G$ is determined by what happens when you multiply two elements of $B$ as elements of $G$. Namely you get

$$\gamma(b_1) \gamma(b_2) = \gamma(b_1 b_2) \exp(c(b_1,b_2))$$

where $c$ is some function from $B \times B$ to $A$. You can think of $c$ as the "correction" to the group law in $G$ as compared to the group law in $B$. (By the way it is a lot like a phase correction in quantum mechanics, and this is not even a coincidence, central extensions of groups are fundamental in QM.)

Anyway, you can check that the associative law in G implies the identity

$$c(b_2,b_3) - c(b_1 b_2,b_3) + c(b_1,b_2 b_3) - c(b_1,b_2) = 0$$

A function $c:B \times B \to A$ in general is called a 2-cochain, while one that satisfies this identity is called a 2-cocycle. Note that set of 2-cochains is a group $C^2(B;A)$ under addition, and the subset of 2-cocycles is a subgroup $Z^2(B;A)$ of the 2-cochains. (In other words, the cocycle constraint happens to be additive.)

The 2-cocycle c is not uniquely determined by the central extension (*), because you can expect a different cocycle $c'$ if you change the section $S$ to some other section $S'$. In this case you can check that $c$ and $c'$ differ as follows for some function $d$ from $B$ to $A$:

$$c'(b_1,b_2) - c(b_1,b_2) = d(b_2) - d(b_1 b_2) + d(b_1)$$

The right side (where here d can be arbitrary) is called a 2-coboundary, and they make a subgroup $B^2(B;A)$ of $Z^2(B;A)$. So, the central extension (*) determines an element of the quotient group

$$H^2(B;A) = Z^2(B;A)/B^2(B;A)$$

This specially constructed abelian group is called the 2nd cohomology of $B$ with coefficients in $A$. It is part of a larger list of abelian groups called the cohomology of $B$, and (although I will skip this digression entirely) it is all a special case of cohomology as it arises in topology.

As with semidirect products, you can also reverse the construction to synthesize an associative group law on $G$, if you are given an abelian group $A$, another group $B$, and a 2-cocycle $c:B \times B \to A$. Moreover, if two cocycles differ by a coboundary, then they will make the same $G$.

In this sense, the set of central extensions of $B$ by $A$ miraculously organizes into another abelian group $H^2(B;A)$. The definition of $H^2(B;A)$ looks very cumbersome at first when $B$ has any appreciable size. However, there are various equivalent definitions for it and various methods to calculate it, and explicit answers are known in a wide range of cases.

So, exercise: Compute $H^2(C_2;\Bbb Z/2)$ directly from the definition to determine the ways to centrally extend $C_2$ by $C_2$. I'll give away the answer that there are two of them.

Questions

1. Although I understand the mathematical definition of $Z^2(B;A)/B^2(B;A)$ from here, it's not clear from here how $Z^2(B;A)/B^2(B;A)$ is equivalent to the central extensions of $B$ by $A$. Could someone please clarify a bit? What does the sentence: "So, the central extension (*) determines an element of the quotient group $H^2(B;A) = Z^2(B;A)/B^2(B;A)$" even mean? How does the central extension $1 \to A \to G \to B \to 1$ determine an element of the given quotient group?

2. I more or less understand what $Z^{2}(B;A)/B^{2}(B;A)$ mathematically means. $Z^2(B;A)$ is the group of functions $c: B \times B \to A$ that satisfies the identity $c(b_2,b_3) - c(b_1 b_2,b_3) + c(b_1,b_2 b_3) - c(b_1,b_2) = 0$ while I think $B^2(B;A)$ is a subgroup that sets up an equivalence relation, namely that $c'$ and $c$ are equivalent if $c'-c$ lies in $B^2(B;A)$. Now the point is I do not know any cohomology as such and I'm not sure how exactly to calculate $Z^2(C_2, \Bbb Z_2)/B^2(C_2, \Bbb Z_2)$ only from first principles (and using the quoted material). Could someone tell me how to proceed with this exercise? Even some hints might be helpful.