For every $R\in SO(4)$,
$$
\iota(R)=\begin{pmatrix}R&0\\0&1\end{pmatrix}
$$
is orthogonal with determinant one. Moreover $\iota(R_1R_2)=\iota(R_1)\iota(R_2)$ and $\iota$ is injective, so these block-diagonal matrices form an $SO(4)$ subgroup of $SO(5)$.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-302/4/b/solution