Isomorphism $U(p,q)/U(1)=SU(p,q)/Z_{n}$ I have a little experience in Lie groups, so I have met the strange isomorphism:
$$U(p,q)/U(1)=SU(p,q)/Z_{n}$$
Here $U(p,q)$ is a set of complex $n\times n$ matrices ($p+q=n$), which satifies the following condition:
$$
U^{\dagger}I_{p,q}U=I_{p,q}
$$,
with $I_{p,q}=$ $ \left(\begin{matrix}
 I_{p}& 0\\ 
 0&-I_{q} 
\end{matrix}\right)$.
I have no idea how to prove this. Can you help me?
 A: The following is a general method of showing $H/K \cong G/L$ when $H \leq G$: you define $f:H \to G/L : h \mapsto hL$ and show that $\ker(f)=K$ and $\operatorname{im}(f) = G/L$. Equivalently, that $H \cap L = K$ and that $HL = G$.

Let $f: \newcommand{\SU}{\operatorname{SU}}\SU(p,q) \to \newcommand{\U}{\operatorname{U}}\U(p,q)/\U(1): g \mapsto g \U(1)$ be the natural map.
If $f(g) = 1$ then $g \in \U(1) \cap \SU(p,q) = Z_n$. That is, if $g$ is a scalar matrix $\lambda I_n$ (to be in $\U(1)$) and $\lambda^n = 1$ (to have determinant 1 to be in $\SU(p,q)$) then $\lambda$ is an $n$th root of unity.
Hence $\ker(f) = Z_n$, so $\SU(p,q)/Z_n \cong \operatorname{im}(f) \leq \U(p,q)/\U(1)$.
Now let $g \in U(p,q)$ have determinant $\lambda^n$ (using that every number in $\mathbb{C}^\times$ has at least one $n$th root). By the defining property of $U(p,q)$ we have $|\lambda|=1$, so that $(\lambda I_n)^\dagger = \tfrac{1}{\lambda} I_n$. Hence $\lambda I_n \in U(1)$. The notice $g \cdot ( \tfrac{1}{\lambda} I_n) \in \SU(p,q)$, and $g \cdot ( \tfrac{1}{\lambda} I_n) U(1) = g U(1)$, and so $f(g \cdot ( \tfrac{1}{\lambda} I_n) ) = gU(1)$.
Hence $\operatorname{im}(f) = \U(p,q)/\U(1)$.
