As the title suggests, my question is does $U(n)\simeq SU(n)\times U(1)$ for general $n$?
If so, how does one prove it?
If not, is at least $U(n)\supset SU(n)\times U(1)$? What is the group $\frac{U(n)}{SU(n)\times U(1)}$ in that case?
Asked
Active
Viewed 73 times
