Suppose *A* and *B* are abelian groups. I want to find subsets *D* of $A \times A$ such that any 2-cocycle $c:A \times A \to B$ for the trivial action is uniquely determined by what it does on *D*. (Equivalently, if a 2-cocycle is trivial on all of *D* it is the zero cocycle).

I'm not looking for a characterization of all such subsets, just some convenient way of describing some such subsets that makes it easier to list all possible 2-cocycles.

By the way, the 2-cocycle condition states that, for all $g,h,k \in A$:

$$c(g,h + k) + c(h,k) = c(g,h) + c(g + h,k)$$

My first thought was to take a generating set *S* for *A* and set $D = S \times S$, but this doesn't seem to work. The problem appears to be that, say, something like $c(g,h + k)$ cannot be described purely in terms of $c(g,h)$, $c(g,k)$, and $c(h,k)$.

I'm also interested in the corresponding question when we are restricted to the subgroup of the group of 2-cocycles by one or both of the following conditions: (i) it is skew symmetric, i.e., $c(x,y) = -c(y,x)$, and (ii) $c(a,b) = 0$ whenever $a,b$ generate a cyclic subgroup.

For simplicity, please feel free to assume, for instance, that both *A* and B are *p*-groups for the same prime *p*, or even that they are both vector spaces over the field of *p* elements. My actual reason for asking this question involves trying to compute 2-cocycles in the case $p = 2$, but I thought this question might be of general interest.