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

The image is in the unitary group, since $(zU{)}^{†}(zU)=\bar{z}z{U}^{†}U=I$ for $z\in U(1)$ and $U\in SU(n)$. The direct product of groups has multiplication $(z,U)(w,V)=(zw,UV)$. Scalar matrices commute with every matrix, soThus $f$ is a group homomorphism; in particular $f(1,I)=I$.