Find $a$ satisfying $x \cup_1 y = \delta a$ when $x,y \in Z^2(G,\mathbb{Z}_2)$

Question

Let $G$ be a finite group. Let $x,y \in Z^2(G,\mathbb{Z}_2)$ be 2-cocycles. Find $a \in C^2(G,\mathbb{Z}_2)$ such that

\begin{align} x \cup_1 y = \delta a. \end{align}

Is there a general solution? Is it possible to know when a solution exists?

Where: \begin{align} [x \cup_1 y](g,h,k) &= x(gh,k)y(g,h) + x(g,hk)y(h,k)\\ \delta a &= a(g,h)+a(gh,k)-a(g,hk)-a(h,k) \end{align}

Answer 1

The $\cup_1$-product is not always a coboundary. For example, for a cocycle $x\in Z^2$, $x\cup_1 x\in Z^3$ is a representative of the class $\operatorname{Sq}^1 x$. This can be non-trivial already for $G=C_2 \times C_2$.

In general, $x\cup_1 y$ (for different cocycles $x, y$) is not even a cocycle, so in that generality there is no such $a$ for trivial reasons.