Showing $U(n)/Z(U(n))=SU(n)/Z(SU(n))$ I was working on the following problem from Stillwell's Naive Lie Theory.

Prove that $U(n)/Z(U(n))=SU(n)/Z(SU(n))$.

It was shown earlier in the text that $Z(U(n))=\{ e^{i\theta} \textbf{1}: \theta \in \mathbb{R} \} \cong S^{1}$ and $Z(SU(n))=\{ \omega \textbf{1}: \omega^{n}=1 \text{ and } \omega \in \mathbb{C} \}$.  We have that $\textbf{1}$ denotes the identity matrix.
When thinking about this problem, I first considered the case where $n=2$. We have that unitary matrices in $U(2)$ are of the form
\begin{equation*}
e^{i\theta}
\begin{bmatrix}
\alpha &-\beta \\ \bar{\beta} & \bar{\alpha} 
\end{bmatrix}
\end{equation*}
We have that the relation which sends $e^{i\theta}
\begin{bmatrix}
\alpha &-\beta \\ \bar{\beta} & \bar{\alpha} 
\end{bmatrix}$ to $
\begin{bmatrix}
\alpha &-\beta \\ \bar{\beta} & \bar{\alpha} 
\end{bmatrix}$ seems to be a well defined function from $U(2) \rightarrow  SU(2)/Z(SU(2)$ since the center $Z(SU(2))$ consists of the matrices $\pm \textbf{1}$. This function even is a homorphism that  has $Z(U(2))$ as it's kernel. From here we can conclude that $U(2)/Z(U(2))=SU(2)/Z(SU(2))$
$\textbf{However, I am having trouble generalizing from here}$
 A: By the short exact sequence $$1\to SU(n)\to U(n)\xrightarrow{\det}S^1\to 1$$ $SU(n)$ is a normal subgroup of $U(n)$, and now identify  $U(1)$ to $e^{i\theta}I$, where $I$ is the $n\times n$ identity matrix and $\theta\in[0,2\pi)$, and therefore $U(1)$ is a subgroup (and even normal) of $U(n)$.
Now, we have 
(1) $SU(n)\cap U(1)=\{e^{ik2\pi/n}:0\le k\le n-1\}\cong \mathbb Z_n$;
(2) $SU(n)\cdot U(1)=U(n)$, since for each $A\in U(n)$, let $z=\det (A)$, so $A=(A/\sqrt[n]{z})\cdot (\sqrt[n]{z}\cdot I)\in SU(n)\cdot U(1)$.
By the 2nd Isomorphism Theorem for groups, we have $$(SU(n)\cdot U(1))/U(1)\cong SU(n)/(SU(n)\cap U(1)).$$ Combine the facts in (1) and (2), we have therefore proved $$U(n)/U(1)\cong SU(n)/\mathbb Z_n$$
@user135520 tried the special case $n=2$, it is right. But should be careful that the map should be $$U(2) \rightarrow  SU(2)/\{\pm 1\}$$ $$e^{i\theta}
\begin{bmatrix}
\alpha &-\beta \\ \bar{\beta} & \bar{\alpha} 
\end{bmatrix} \mapsto
\{
\begin{bmatrix}
\alpha &-\beta \\ \bar{\beta} & \bar{\alpha} 
\end{bmatrix}, \begin{bmatrix}
\alpha &\beta \\ \bar{\beta} & -\bar{\alpha} 
\end{bmatrix}\}$$ 
For this problem, there is still a different solution in here, who shows that $$U(n) \cong (SU(n) \times U(1))/
\mathbb{Z}_{n}.$$ But to use this isomorphism to solve our problem, it should be careful to take $U(1)$ from right hand side as a product to the left hand side as a quotient.
A: I tried to construct a map, and I didn't see any problems.
For any element in $U(N)$, it can be written $e^{\mbox{i}\theta}M$, where $M\in SU(N)$. Consiering $e^{\mbox{i}\theta_1}\times e^{\mbox{i}\theta_2}M=e^{\mbox{i}(\theta_1+\theta_2)}M$, the element in $\frac{U(N)}{Z(U(N))}$ is of the form $\{e^{\mbox{i}\theta}M\}$, where $\theta\in[0,2\pi]$. Analyze similarly, we can get the element in $\frac{SU(N)}{Z(SU(N))}$ is of the form $\{
    M,e^{\mbox{i}2\pi/N}M,\dots,e^{\mbox{i}2\pi(N-1)/N}M\}$. And we can construct an isomorphic map $\phi$:
\begin{align*}
    \begin{aligned}
        \phi:\{e^{\mbox{i}\theta}M\} \mapsto \{
            M,e^{\mbox{i}2\pi/N}M,\dots,e^{\mbox{i}2\pi(N-1)/N}M\}
    \end{aligned}
\end{align*}
It can be easily checked that the $\phi$ is a surjective homomorphism. And the kernel of $\phi$ is $\{e^{\mbox{i}\theta}I\}$.
