Solution
= Solution
Let $\phi\in Z^2(G,\mathbb Q)$. Part iii gives $d\psi=|G|\phi$. Since division by $|G|$ is possible in the <rational numbers>,
$$
\phi=d\left(\frac{\psi}{|G|}\right)
$$
is a <two-coboundary>. Hence
$$
\boxed{H^2(G,\mathbb Q)=0}.
$$
This is a degree-two instance of the vanishing of finite-group cohomology when the group order is invertible on the coefficient module.