Is $SU(1) = U(1)$ or the trivial group?

Question

The $S$ denotes usually determinant $1$ and therefore I'm not sure how one defines it in the one-dimensional case. Is $SU(1)$ actually the same as $U(1)$, because it makes no sense to talk about determinant=$1$ in the one-dimensional case? Or is $SU(1)$ the trivial group with just one element, $1$?

Answer 1

Ah okay, I get the question. The determinant of a $1$-dimensional matrix is equal to the only element of the matrix, $\det([a])=a$. So the only $1\times1$ matrix with determinant $1$ is the identity matrix, $[1]$, and that's the only element of $SU(1)$, which is therefore trivial.