Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 3 / 8D / Solution

The internal direct product theorem states that if $H,K⊴G$, $H\cap K=\{1\}$, and $HK=G$, then $G\cong H\times K$. Indeed, for $h\in H,k\in K$, the commutator $[h,k]$ lies in both $H$ and $K$, so it is trivial. The map $(h,k)\mapsto hk$ is consequently a homomorphism; the intersection condition makes it injective and $HK=G$ makes it surjective. Conversely, the two factors in a direct product satisfy these conditions.

In odd dimension, $-I$ is central and has determinant $-1$. Every $Q\in O(3)$ has a unique expression $Q=(-I{)}^{\epsilon }R$ with $R\in SO(3)$, soBy contrast, $SO(2)\times C_2$ is abelian, while reflections in $O(2)$ conjugate a rotation to its inverse. Thus $O(2)\ncong SO(2)\times C_2$.

The center of the unitary group consists exactly of scalar unitary matrices:The determinant gives a surjective homomorphismwhose kernel is the special unitary group $SU(2)$.

Nevertheless $U(2)$ is not isomorphic to $SU(2)\times \mathbb{T}$. The center of $U(2)$ has exactly one nonidentity element of order two, namely $-I$. The center of the proposed product is $\{\pm I\}\times \mathbb{T}$, which has three nonidentity elements of order two. Since an isomorphism preserves the center and element orders,