The trivial action gives
$$
H^1(D_n,\mathbb Z)=\operatorname{Hom}(D_n,\mathbb Z)=0
$$
because $D_n$ is finite. By part b(ii),
$$
H^2(D_n,\mathbb Z)
\cong\operatorname{Hom}(D_n,\mathbb Q/\mathbb Z)
\cong\operatorname{Hom}(D_n^{\mathrm{ab}},\mathbb Q/\mathbb Z).
$$
In the <abelianization> the relation $aba=b^{-1}$ becomes $b=b^{-1}$. Hence
$$
D_n^{\mathrm{ab}}\cong
\begin{cases}
C_2,&n\text{ odd},\\
C_2\times C_2,&n\text{ even}.
\end{cases}
$$
Each finite cyclic group is naturally isomorphic to its character group in $\mathbb Q/\mathbb Z$, so
$$
\boxed{H^2(D_n,\mathbb Z)\cong
\begin{cases}
\mathbb Z/2\mathbb Z,&n\text{ odd},\\
(\mathbb Z/2\mathbb Z)^2,&n\text{ even}.
\end{cases}}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 151 / 4 / c / i / Solution

 Solution

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