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In Wikipedia Unitary Group, it says that $$1\to \operatorname {SU} (n)\to \operatorname {U} (n)\to \operatorname {U} (1)\to 1.$$

However, we see that [this I understand] $\operatorname {U} (n)$ is related to $\operatorname {SU} (n)$ and $\operatorname {U} (1)$ by $$ \operatorname {U} (n)=\frac{\operatorname {U} (1)\times \operatorname {SU} (n)}{\mathbb{Z}_n} $$ where ${\mathbb{Z}_n}= {\mathbb{Z}}/({n \mathbb{Z}})$, a finite Abelian cyclic group of order $n$. This is the case because both $U(1)$ and ${SU} (n)$ shares the same subgroup $${\mathbb{Z}_n}=\{ \exp(\frac{2 \pi i }{n}j) \cdot \mathbb{I}_{n\times n}\}$$ where $j \in \mathbb{Z} \mod n$.

If so, why is this true $$1\to \operatorname {SU} (n)\to \frac{\operatorname {U} (1)\times \operatorname {SU} (n)}{\mathbb{Z}_n} \to \operatorname {U} (1)\to 1?$$ Is ${SU} (n)$ a normal subgroup of ${U} (n)=\frac{\operatorname {U} (1)\times \operatorname {SU} (n)}{\mathbb{Z}_n}$, precisely?

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    $\begingroup$ Some details are given in the answers here. It is simply the semidirect product, hence we obtain a (split) exact sequence. $\endgroup$ Sep 12, 2017 at 19:00
  • $\begingroup$ $SU(n)$ is a kernel of the determinant homomorphism. Is this not enough justification for you? $\endgroup$
    – Wojowu
    Sep 12, 2017 at 19:00
  • $\begingroup$ As I said I already I know $U(n)=[SU( n ) \times U(1)]/Z_n$. I am asking which $SU(n)'$ and which $U(1)'$ they are in the exact sequence $1\to {SU} (n)'\to \frac{{U} (1)\times {SU} (n)}{\mathbb{Z}_n} \to {U} (1)'\to 1?$ Is that ${U} (1)'={U} (1)/Z_N$? or is that $ {SU} (n)'=PSU(n)$? $\endgroup$
    – wonderich
    Sep 12, 2017 at 19:11

2 Answers 2

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The determinant mapping $A \mapsto |A|$ is a group homomorphism $U(n) \to U(1)$. Its kernel is $SU(n)$. So $SU(n)$ is a normal subgroup of $U(n)$.

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  • $\begingroup$ do you happen can comment this as well, math.stackexchange.com/questions/3607058, I believe some version is true? $\endgroup$ Apr 3, 2020 at 22:19
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Note $U(1)\times SU(n)$ maps to $U(n)$ via $(z,M)\mapsto zM$. The kernel $K$ is the set of $(\zeta^{-1},\zeta I)$ where $\zeta^n=1$, and is cyclic of order $n$. Now $U(1)\times SU(n)$ maps onto $U(1)$ by $(z,M)\mapsto z^n$. This kills $K$, so is really a map $\phi$ from $(U(1)\times SU(n))/K$ to $U(1)$. Also, the kernel of $\phi$ is $H/K$ where $H=\{(z,M):z^n=1\}$. Modulo $K$ each element of $H$ has a unique representation $(1,M)$ modulo $K$, and so $H/K\cong\{(1,M):M\in SU(n)\}\cong SU(n)$.

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  • $\begingroup$ Thanks very much. I already I know $U(n)=[SU(n)\times U(1)]/Z_n=[SU(n)×U(1)]/Z_n$. I am asking which $SU(n)′$ and which $U(1)′$ they are in the exact sequence $1\to SU(n)′→U(1)\times SU(n)/Z_n→U(1)′→1$? Is that $U(1)′=U(1)/Z_n$? or is that $SU(n)′=PSU(n)$? $\endgroup$
    – wonderich
    Sep 12, 2017 at 19:27
  • $\begingroup$ @wonderich: I'm not sure what you mean by $SU(n)'$, but $SU(n)$ and $PSU(n)$ are non-isomorphic Lie groups. For example, the first is simply connected and the second is not. In case it helps, the composition $SU(n)\rightarrow SU(n)\rightarrow U(1)\rightarrow [SU(n)\times U(1)]/Z_n$ maps $SU(n)$ Lie isomorphically onto its image in $U(n)$. $\endgroup$ Sep 12, 2017 at 20:02
  • $\begingroup$ do you happen can comment this as well, math.stackexchange.com/questions/3607058, I believe some version is true? $\endgroup$ Apr 3, 2020 at 22:19

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