 Home

Codex Wiki

List of universities

University of Cambridge

Course of the University of Cambridge

Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2021

Paper 151

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

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

 Ancestors (12)

 Discussion (0)

 Articles by others on the same topic (0)

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

 Ancestors (12)

 Discussion (0)  Subscribe (1)

 Articles by others on the same topic (0)

 Discussion (0)