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 $U(2)/Z(U(2))=SU(2)/Z(SU(2))$

$\textbf{However, I am having trouble generalizing from here}$

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.

• @user26857 Thanks for your edition, so for example, if I want to tap "kernel" of a map $\phi$, it should be $\ker \phi$ instead of $ker\phi$? Aug 15, 2015 at 21:13
• Why was the kernel bigger than $U(1)$? How do you obtain that $U(n)/U(1) \cong SU(n)$? shoudn't it be $U(n)/SU(n) \cong U(1)$ from the SES $1 \rightarrow SO(n) \rightarrow U(n) \rightarrow U(1) \rightarrow 1$? Aug 16, 2015 at 14:34
• @user135520 I see it, thanks for your remind. I'll soon get back to you and fix my solution. Aug 16, 2015 at 15:27
• No problem, thanks. Aug 16, 2015 at 16:17
• For a reader unfamiliar with short exact sequences, what is meant by $U(n)\xrightarrow{\det}S^1$, in particular, the "det"? Feb 5, 2018 at 19:00

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 $$\{e^{\mbox{i}\theta}I\}$$.