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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 4 / a / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-151/4/a/iv/solution