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?
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4$U(N)\simeq SU(N)\times U(1)/{\mathbb Z}_N$, because the factorization into $SU(N)$ and phase is not unique. You can pass $N$-th roots of unity between the two. – mike stone Jun 08 '23 at 12:48
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Possible duplicate: https://physics.stackexchange.com/q/169087/2451 – Qmechanic Jun 08 '23 at 16:10