I'm writing the answer to my question but I'm not completely sure though
Intro
What do we want to show ?
$$U(n) \simeq \frac{SU(n) \times U(1)}{\mathbb{Z}_{n}}$$
Groups definition
- $U(n)$ = the group of $n\times n$ unitary matrices $\Rightarrow$ $U \in U(n): UU^{\dagger} = U^{\dagger}U = I \Rightarrow \mid det (U) \mid ^{2} = 1$
- $U(1) =$ the group of $1\times 1$ unitary matrices $\Rightarrow U(1) = \lbrace e^{i\varphi} \mid \varphi \in \left[ 0, 2\pi \right] \rbrace$
- $SU(n) =$ the group of $n\times n$ unitary matrices with determinant 1
- $\mathbb{Z}_{n} =$ the cyclic group of n integers modulo n $\Rightarrow$ $\mathbb{Z}_{n} =$ the set $\lbrace 0,1,2,...,n-1 \rbrace$ with the operation of addition modulo n.
\end{itemize}
Isomorphism theorem
We use the first isomorphism theorem:
Let G and H be groups and let $f: G \longrightarrow G'$ be a group homomorphism. Then:
$$Im f \simeq \frac{G}{Ker f} $$
What do we have to do ?
We have to do the following things:
- Find a map $f: SU(n) \times U(1) \rightarrow U(n)$ and show that it's a homomorphism
- Show that $Ker f = \mathbb{Z}_{n}$
- Show that $Im f = U(n)$ which is equivalent to show that f is surjective
We find and homomorphism between $SU(n) \times U(1)$ and $U(n)$
\begin{eqnarray*}
f: && SU(n) \times U(1) \rightarrow U(n) \\
&& (S,e^{i\varphi}) \mapsto Se^{i\varphi}
\end{eqnarray*}
$f$ is a homomorphism if :
\begin{eqnarray*}
f((S_{1},e^{i\varphi _{1}})(S_{2},e^{i\varphi _{2}} )) = f(S_{1},e^{i\varphi _{1}})f(S_{2},e^{i\varphi _{2}} )&& \forall \varphi _{1}, \varphi _{2} \in U(1) \\
&& \forall S_{1},S_{2} \in SU(n)
\end{eqnarray*}
It's easy to show:
\begin{eqnarray*}
f((S_{1},e^{i\varphi _{1}})(S_{2},e^{i\varphi _{2}})) &=& f(S_{1}S_{2},e^{i\varphi _{1}}e^{i\varphi _{2}} ) \\
&=& S_{1}S_{2} e^{i\varphi _{1}} e^{i\varphi _{2}} \\
&=& e^{i\varphi _{1}} S_{1} e^{i\varphi _{2}} S_{2} \\
&=& f(S_{1},e^{i\varphi _{1}})f(S_{2},e^{i\varphi _{2}})
\end{eqnarray*}
We show that $Ker f = \mathbb{Z}_{n}$
$$Ker f = \lbrace (S,e^{i\varphi}) \in (SU(n) \times U(1)) \: \vert \: f(S,e^{i\varphi}) = Se^{i\varphi} = e_{U(n)} = I \rbrace$$
Let's find S
\begin{eqnarray*}
e^{i\varphi}S&=& I \\
S &=& e^{-i\varphi} I\\
det(S) &=& det(e^{-i\varphi} I) \\
1 &=& e^{-in \varphi} \\
1 &=& \cos (n\varphi ) - i \sin (n \varphi)
\end{eqnarray*}
$$
\left\{
\begin{array}{r c l}
\cos (n \varphi ) &=& 1\\
\sin (n \varphi ) &=& 0\\
\end{array}
\right.
\Rightarrow n \varphi = 2k \pi \Rightarrow \varphi = \frac{2k \pi}{n} \quad \text{with} \quad k \in \mathbb{Z}$$
So the matrices S are:
$$S = e^{-i \frac{2 k \pi}{n}} I \quad \text{with} \quad k \in \lbrace 0,1,...,n-1 \rbrace$$
Let's find $e^{i\varphi}$
\begin{eqnarray*}
e^{i\varphi}S&=& I \\
e^{i\varphi}e^{-i \frac{2 k \pi}{n}} I&=& I \\
e^{i\varphi}&=& e^{i \frac{2 k \pi}{n}} \\
\end{eqnarray*}
Ker f
$$Ker f = (e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) \quad \text{with} \quad k \in \lbrace 0,1,...,n-1 \rbrace $$
We show that Ker $f$ is isomorph to $\mathbb{Z}_{n}$
We denote by $\mathbb{Z}_{n}$ the cyclic group of integers modulo n.
Let's not forget that:
The group law $\cdot$ of Ker f is given by:
$(S_{1},e^{i\varphi _{1}})\cdot(S_{2},e^{i\varphi _{2}}) = (S_{1}S_{2},e^{i(\varphi _{1} + \varphi _{2})})$
The group law $\cdot$ of $\mathbb{Z}_{n}$ is given by:
$k_{1} \cdot k_{2} = k_{1} + k_{2}$
\end{itemize}
Let's show that Ker f is isomorph to $\mathbb{Z}_{n}$
$$\phi: Ker f = (S,e^{i\varphi}) \rightarrow \mathbb{Z}_{n} : (e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) \mapsto k$$
$\phi$ is an homomorphism
$$\phi((S_{1},e^{i\varphi _{1}})(S_{2},e^{i\varphi _{2}})) = \phi(S_{1},e^{i\varphi _{1}}) + \phi(S_{2},e^{i\varphi _{2}})$$
\begin{eqnarray*}
\phi((S_{1},e^{i\varphi _{1}})(S_{2},e^{i\varphi _{2}}))&=& \phi(S_{1}S_{2},e^{i(\varphi _{1} + \varphi _{2})}) \\
&=& \phi (e^{-i \frac{2 (k_{1} + k_{2}) \pi}{n}} I,e^{i \frac{2 (k_{1} + k_{2}) \pi}{n}}) \\
&=& k_{1} + k_{2} \\
\phi(S_{1},e^{i\varphi _{1}}) + \phi(S_{2},e^{i\varphi _{2}})&=& (e^{-i \frac{2 k_{1} \pi}{n}} I,e^{i \frac{2 k_{1} \pi}{n}}) + (e^{-i \frac{2 k_{2} \pi}{n}} I,e^{i \frac{2 k_{2} \pi}{n}}) \\
&=& k_{1} + k_{2}
\end{eqnarray*}
$\phi$ is injective
- $\phi$ is injective $\Leftrightarrow \forall \varphi_{1},\varphi_{2} \in \left[ 0, 2\pi \right] \: \text{and} \: \forall S_{1},S_{2} \in SU(n): \phi (S_{1},e^{i\varphi _{1}}) = \phi (S_{2},e^{i\varphi _{2}}) \Rightarrow \varphi_{1} = \varphi_{2} \: \text{and} \: S_{1} = S_{2}$
\begin{eqnarray*}
\phi (S_{1},e^{i\varphi _{1}}) & = & \phi (S_{2},e^{i\varphi _{2}})\\
S_{1}e^{i\varphi _{1}} & = & S_{2}e^{i\varphi _{2}} \\
e^{i(\varphi _{1} - \varphi _{2})} S_{1} & = & S_{2} \\
det( e^{i(\varphi _{1} - \varphi _{2})} S_{1}) & = & det( S_{2} ) \\
e^{in(\varphi _{1} - \varphi _{2})} & = & 1
\end{eqnarray*}
$$\Rightarrow \varphi _{1} = \varphi _{2} \Rightarrow S_{1} = S_{2}$$
- We can also use the fact that $\phi$ is injective if and only if $Ker \phi = \lbrace e_{Ker f} \rbrace = \lbrace (I,1)\rbrace$
$$Ker \phi = \lbrace (e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) \in Ker f \: \vert \: \phi(e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) = e_{\mathbb{Z}_{n}} = 0 \rbrace$$
$$\phi(e^{-i \frac{2 k \pi}{n}} I,e^{i \frac{2 k \pi}{n}}) = k = 0 \Rightarrow Ker \phi = (e^{0} I,e^{0}) = (I,1)$$
$\phi$ is surjective
$\phi$ is injective $\Leftrightarrow \forall k \in \mathbb{Z}_{n}, \exists (e^{i \frac{2 k \pi}{n}} I,e^{-i \frac{2 k \pi}{n}}) \in Ker f : \phi(e^{i \frac{2 k \pi}{n}} I,e^{-i \frac{2 k \pi}{n}}) = k$
Another way to show $\phi$ is surjective is by noting that fact that $\phi$ is an injective map between 2 finites sets with the same number of elements so $\phi$ is surjective
We show that f is surjective
How do we show that $Im f = U(n)$ ?
We want to show that $Im f = U(n)$ where: $ Im f = \lbrace Se^{i\varphi} \vert S \in SU(n), \varphi \in \left[ 0, 2\pi \right] \rbrace$
To do that, we can show that:
$$Im f \subseteq U(n) \quad \text{and} \quad U(n) \subseteq Im f$$
$Im f \subseteq U(n)$
$$Im f \subseteq U(n) \Rightarrow \mid det (Se^{i\varphi}) \mid ^{2} = 1 \quad \forall S \in SU(n), \forall \varphi \in \left[ 0, 2\pi \right] $$
\begin{eqnarray*}
\mid det (Se^{i\varphi}) \mid ^{2} & = & \mid e^{i n \varphi} \mid ^{2} \\
& = & 1 ^{2} \\
& = & 1
\end{eqnarray*}
$U(n) \subseteq Im f$
It's equivalent to show that: $\forall X \in U(n), \exists (Y,z) \in SU(n) \times U(1) \quad \text{such as} \quad f(Y,z) = Yz = X$
\begin{eqnarray*}
X \in U(n)&\Rightarrow& \vert det(X) \vert ^{2} = 1 \Rightarrow det(X) = e^{i\theta} \equiv m \\
&\Rightarrow& X^{\dagger}X = XX^{\dagger} = I \\
\end{eqnarray*}
Let's write X as: $X = zY = m^{1/n} Y$ where $Y = m^{-1/n}X$ and let's show that $Y \in SU(n)$ and $m^{1/n} \in U(1)$:
$Y \in SU(n) \Leftrightarrow YY^{\dagger} = Y^{\dagger} Y = I$ and $det(Y) = 1$
$$YY^{\dagger} = (m^{-1/n} X)(m^{-1/n} X)^{\dagger} = (e^{\frac{i\theta}{n}} X)(e^{\frac{i\theta}{n}} X)^{\dagger} = e^{\frac{i\theta}{n}} e^{\frac{-i\theta}{n}} XX^{\dagger} = I $$
$$det(Y) = det(m^{-1/n} X) = m^{-1} det(X) = m^{-1} m = 1 \Rightarrow Y \in SU(n)$$
- $m^{1/n} = e^{\frac{i\theta}{n}} \in U(1) $
Alright, The first isomorphism theorem (FIT) states that if we have an action $\phi: G\to H$ that is homomorphic, then the $ker(\phi)$ is a normal subgroup of $H$.
First, we need to show that $\phi:\mathbb R\to \mathbb T$ such that $\phi (\theta)=e^{i\pi\theta}$ where $\mathbb T:= \{z\in \mathbb C:\;||z||=1\}$. This can be easily done by letting $x,y\in \mathbb R$, So $\phi(x+y)=e^{i\pi(x+y)}=(e^{i\pi x})e^{i\pi y}=\phi(x)\phi(y)$.
Next, we need to find the kernal of $\phi$. This can be done by noting that $1$ is the identity of $\mathbb T$ so let $$\phi(\theta)=e^{i\pi\theta}=\cos(\pi\theta)+i\sin(\pi\theta)=1+0i.$$
This implies that $\cos(\pi\theta)=1$. As such $\theta = 2k$, $\forall k \in \mathbb Z$. So,
$$ker(\phi)=\{n=2k: k \in \mathbb Z\}$$
Lastly, we need to show that $ker(\phi)$ is a normal subgroup of $\mathbb T$ this is done by showing that $gH=Hg$ if $g\in ker(\phi)$. Thus,
$$\phi(n)H=e^{i\pi n}e^{i\pi\theta}=e^{i(2\pi) k}e^{i\pi\theta}=e^{i\pi\theta}=e^{i\pi\theta}e^{i(2\pi) k}=H\phi(n) $$
as desired.
Best Answer
Let $N = ker(\phi)$ and $K = im(\phi)$, then you're asking when, given an exact sequence $1 \to N \to G \to K \to 1$ is trivial.
I don't include proofs here, as they're found in any basic group theory notes.
In fact you can get away with the first condition. Indeed, if there exists a map $p : G \to N$ which is the identity on $N$, then a section of $\phi$ automatically exists and the isomorphism $G \cong \operatorname{im}(\phi) \times \operatorname{ker}(\phi) = K \times N$ holds. The required isomorphism is $(\phi, p) : G \to K \times N$ (it's not hard to check that this is in fact an isomorphism).