Is ${SU} (n)$ a normal subgroup of ${U} (n)$, precisely?

Question

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?

Answer 1

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 $U(n)$.

Answer 2

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 $H/K\cong\{(1,M):M\in SU(n)\}\cong SU(n)$.