Symmetric and cyclicity-preserving 2-cocycle implies 2-coboundary

In cocycle and coboundary language at the origin
Suppose $$G$$ is a finitely generated abelian group and $$A$$ is an abelian group. Suppose $$c$$ is a 2-cocycle for trivial group action of $$G$$ on $$A$$ satisfying the following two conditions:


 * $$c$$ is a symmetric 2-cocycle for trivial group action, i.e., $$c(g,h) = c(h,g)$$ for all $$g,h \in G$$.
 * $$c$$ is a cyclicity-preserving 2-cocycle for trivial group action, i.e., $$c(g,h) = 0$$ if $$\langle g,h \rangle$$ is a cyclic group.

Then, $$c$$ is a 2-coboundary for trivial group action.

In group extensions language at the origin
Suppose $$E$$ is an abelian group with a subgroup $$A$$ and quotient group $$E/A \cong G$$. Suppose $$G$$ is finitely generated. Suppose there exists a 1-closed transversal $$T$$ of $$A$$ in $$E$$ (i.e., a collection of coset representatives, in other words, $$T$$ intersects each coset at exactly one point, with the property that any power of an element in $$T$$ is also in $$T$$). Then, in fact, $$A$$ is a direct factor of $$E$$ and in particular we can write $$E$$ as an internal direct product $$A \times G$$.

In cocycle and coboundary language away from the origin
Suppose $$G$$ is a finitely generated abelian group and $$A$$ is an abelian group. Suppose $$c_1$$ and $$c_2$$ are 2-cocycles for the action of $$G$$ on $$A$$ such that $$\operatorname{Skew}(c_1) = \operatorname{Skew}(c_2)$$. Then, if $$c_1$$ and $$c_2$$ are both cyclicity-preserving, we must have $$c_1 = c_2$$.