$U(n) \simeq \frac{SU(n) \times U(1)}{\mathbb{Z}_{n}}$ isomorphism I'm trying to proof the following isomorphism
$$U(n) \simeq \frac{SU(n) \times U(1)}{\mathbb{Z}_{n}}$$
So I'm using the first Isomorphism theorem: http://en.wikipedia.org/wiki/Isomorphism_theorem
It's easy to show that the following map is an homomorphism:
$$f: SU(n) \times U(1) \rightarrow U(n): (S,e^{i\varphi}) \mapsto e^{i \varphi} S$$
But I'm having troube to show that:
$$Ker f = \mathbb{Z}_{n}$$
How am I supposed to do that ?
 A: If I assume that x is just a scalar when I'm trying to find Ker f like I did with  Tobias Kildetoft using the nth roots of unity group. Isn't easier to assume directly that x can be written as $e^{i \varphi}$ ?
We show that $Ker f = \mathbb{Z}_{n}$
$$Ker f = \lbrace (S,x) \in (SU(n) \times U(1)) \: \vert \: f(S,x) = Sx = e_{U(n)} = I \rbrace$$
Let's find S
\begin{eqnarray*}
     xS&=& I \\
     S &=&  x^{-1}I
\end{eqnarray*}
Let $x^{-1} = y$. S is a diagonal matrix with $y$ as diagonal entries \
As $S \in SU(2)$ then
\begin{eqnarray*}
     det(S)&=& det(yI) \\
      &=&  y^{n} \\
       &=& 1
\end{eqnarray*}
$n^{th}$ roots of 1
How can we solve this equation ? We want to solve $y^{n} = 1$ with $\lambda \in \mathbb{C}$. Let's write both side of equation in polar form:
$$y^{n} = (r e^{i \theta} )^{n} = r e^{i \theta n}$$
$$1 = (r e^{i \theta} ) = 1 e^{i 0}$$
$y^{n}$ and $1$ are equal if and only if:


*

*$r^{n} = 1$. As r is positif, we have r = 1

*$n\theta = 0 + 2k\pi$ where $k \in \mathbb{Z}$ so $\theta = \frac{2 k \pi}{n} $


We can visualize this geometrically. The points will lie on the unit circle and they will be equally spaced on the unit circle every $\frac{2 \pi}{n}$ radians.
So our equation have $n-1$ different solutions as we go back to where we started once we reach $k = n$
So the $n^{th}$ roots of unity are given by:
$y = e^{i \frac{2 k \pi}{n}}$ with $k = \lbrace 0,1,...,n-1 \rbrace$
So the matrices S are:
$$S = e^{-i \frac{2 k \pi}{n}} I \quad \text{with} \quad k \in \lbrace 0,1,...,n-1 \rbrace$$
It's easy to show that this form a group
Let's find x
\begin{eqnarray*}
    xS&=& I \\
    xe^{i \frac{2 k \pi}{n}} I&=& I \\
    x&=& e^{-i \frac{2 k \pi}{n}} \\
\end{eqnarray*}
Ker f
$$Ker f = (e^{i \frac{2 k \pi}{n}} I,e^{-i \frac{2 k \pi}{n}}) \quad \text{with} \quad k \in \lbrace 0,1,...,n-1 \rbrace $$
