Context
This is a sequel to Understanding the relation between central extensions and 2-cocycles and coboundaries
Exercise:
Compute $H^2(C_2;\Bbb Z/2)$ directly from the definition to determine the ways to centrally extend $C_2$ by $C_2$.
My Solution:
Since $c: C_2\times C_2 \to \Bbb Z/2$ and the number of semantically different boolean functions on $2$ variables is $2^{2^2} = 16$, there are 16 possible 2-cochains. That is, $C^2(C_2;\Bbb Z/2)$ has 16 elements. Out of these 16 only 4 satisfy the identity $c(b_2,b_3) - c(b_1 b_2,b_3) + c(b_1,b_2 b_3) - c(b_1,b_2) = 0$ for all possible boolean values $b_1, b_2, b_3$. Let's call these 4 boolean functions $c_0, c_7, c_8, c_{15}$ as per the following table. These are our four 2-cocycles and hence the cardinality of $Z^2(C_2; \Bbb Z/2)$ is 4.
$$\begin{array}{cc|cccccccc} A &B & c_0 & c_1 & c_2 & c_3 & c_4& c_5& c_6 & c_7\\ \hline 0 &0 &0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1& 0 &0 & 0 & 0 & 1 &1 & 1 &1\\ 1 & 0 & 0 & 0 & 1 & 1 & 0 &0 & 1 & 1\\ 1 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1. \end{array}$$
$$\begin{array}{cc|cccccccc} A &B& c_8& c_9& c_{10} & c_{11} &c_{12}& c_{13}& c_{14} & c_{15}\\ \hline 0 &0& 1 & 1 & 1 & 1 & 1 &1 & 1 & 1\\ 0 & 1& 0 &0 & 0 & 0& 1 & 1 & 1 & 1\\ 1 &0 & 0 & 0 & 1 & 1& 0& 0 & 1& 1\\ 1 & 1& 0 &1& 0& 1 & 0 & 1 & 0 & 1. \end{array}$$
Now $d$ is a function from $C_2 \to \Bbb Z/2$ and we know that the set of images of $d(b_2) - d(b_1 b_2) + d(b_1)$ for the four semantically different boolean functions $d_i$'s on a single variable forms the subgroup $B^2(C_2; \Bbb Z/2)$.
The four different possible $d_i$'s are:
$$d_0(0) = 0, d_0(1) = 0$$ $$d_1(0) = 0, d_1(1) = 1$$ $$d_2(0) = 1, d_2(1) = 0$$ $$d_3(0) = 1, d_3(1) = 1$$
And now the corresponding images are:
$$d_0(b_2) - d_0(b_1b_2) + d_0(b_1) = [0 \ 0 \ 0 \ 0]$$
$$d_1(b_2) - d_1(b_1b_2) + d_1(b_1) = [0 \ 0 \ 0 \ 0]$$
$$d_2(b_2) - d_2(b_1b_2) + d_2(b_1) = [1 \ 1 \ 1 \ 1]$$
$$d_3(b_2) - d_3(b_1b_2) + d_3(b_1) = [1 \ 1 \ 1 \ 1]$$
So there are 2 distinct elements in $B^2(C_2; \Bbb Z/2)$ and the cardinality $|B^2(C_2; \Bbb Z/2)| = 2$.
It's clear that the 2 elements in $B^2(C_2; \Bbb Z_2)$ partitions $Z^2(C_2; \Bbb Z/2)$ in two cosets and the index $[Z^2(C_2; \Bbb Z/2):B^2(C_2; \Bbb Z/2)]$ is $2$. Thus, the quotient group $H^2(C_2, \Bbb Z/2)$ has two elements corresponding to two separate equivalence classes of central extensions. This matches with the fact that a finite group containing four elements $\{c_0, c_7, c_8, c_{15}\}$ can only either be the cyclic group or the Klein 4-group.
Question
I discussed this solution with my professor and he said that while this is correct, the same calculation would be much clearer (as well as easier to do in larger cases) if I express it with linear algebra over $\Bbb Z/2$, or over $\Bbb Z/p$ when $A = \Bbb Z/p$. Now I'm not really clear what he means by this.
Could someone please give me some hints about how to frame the solution using linear algebra over $\Bbb Z/2$?
In fact, he also gave me another very related exercise: calculate that $H^2(C_p; \Bbb Z/p)$ has $p$ elements (because it's a 1-dimensional vector space over $\Bbb Z/p$), and then analyze the discrepancy between that and the fact that there are two groups of order $p^2$ rather than $p$ of them. Some hints for this would be helpful too.
