Past exam of the groups and representations course of the University of Oxford / 2024 / 4 / b / Solution

First prove surjectivity. For any $A\in U(n)$, its determinant has modulus one. Choose $z\in U(1)$ with $z^n=\det A$ and define $U=z^{-1}A$. Then $U^{†}U=I$ and $\det U=z^{-n}\det A=1$, so $A=f(z,U)$.

Next, $f(z,U)=I$ is equivalent to $U=z^{-1}I$. This lies in $SU(n)$ precisely when $z^n=1$. Therefore the kernel of a group homomorphism is the central cyclic groupIts generator is $(\zeta ,{\zeta }^{-1}{I}_n)$, which has order $n$. The inverse phases in the two factors are essential: this is an explicitly embedded subgroup, not a quotient by scalar matrices in just one factor. The first isomorphism theorem now givesThis also holds as a Lie group isomorphism: the induced map is smooth and a continuous bijection from a compact quotient space to the Hausdorff unitary group, and its differential is the Lie algebra isomorphism $(it,X)\mapsto itI+X$. For $n=1$ the kernel of a group homomorphism is trivial, consistently with $SU(1)=\{1\}$.