Let $G$ be a finite group; denote by $\mathbb{Z}_2$ the cyclic group of order $2$. Let $\pi: G \rightarrow \mathbb{Z}_2$ be a non-trivial group homomorphism. Let M be the $G$ representation $\mathbb{Z}_2 \times \mathbb{Z}_2$ with action given by \begin{align} g[(a,b)] = \begin{cases} (a,b)& \quad \text{if $\pi(g) = 0$}\\ (b,a)& \quad \text{if $\pi(g) = 1$}. \end{cases} \end{align}
By sendingLet \begin{align} f: M &\rightarrow \mathbb{Z}_2\\ f(a,b) & \mapsto a+b \end{align}
$f$ maps a cocycle2-cocycle in $t \in Z^2(G,\mathbb{Z}_2)$$Z^2(G,M)$ to $(t,t) \in Z^2(G,M)$ we see thata 2-coycle in $Z^2(G,\mathbb{Z}_2)$. I want an example where the image of $f$ is not a subgroup ofcoboundary.
Examples I have checked:
$H^2(\mathbb{Z}_2, M) = 0$ for $Z^2(G,M)$$\pi = \text{id}$.
What's an example of a groupI have also checked numerically that $G$$H^2(\mathbb{Z}_2 \times \mathbb{Z}_2, M)$ for $\pi(a,b) = a+b$ and normalized cocycle in $Z^2(G,M)$ which is not in the subgroup$\pi(a,b)=a$ also only have images that are coboundaries under $Z^2(G,\mathbb{Z}_2)$?$f$.