I have some little questions motivated by the following: I'm given the set

$$ U(1)\times SU(n) = \left\{ (g,h) \hspace{0.2cm} | \hspace{0.2cm} g \in U(1) \wedge H \in SU(n) \right\} $$

and the following map

$$f: (g,h) \to gh $$

It is easy to see that $gh \in U(n)$ since $(gh)^{\dagger}(gh)=h^{\dagger}h=\mathbb{I}=hh^{\dagger}=(gh)(gh)^{\dagger}$ and det$(gh)=e^{in\theta}$ I want to show that $f$ is an homomorphism onto $U(n)$, but I'm having some trouble when interpretiong the homomorphism part

To see that $f$ is onto $U(n)$, let $y \in U(n)$ and $a$=det$(y)$. Let $x=e^{i\theta}z$ where $z \in SU(n).$ As always I can choose that $a=e^{in\theta}$ and


where $R$ is a rotation, does this show that $f$ is onto $U(n)$?

Now the homomorphism part


which is the group operation in $U(1)\times SU(n)$, is this?

$$(g,h)\star(g',h')=(gg',hh') $$

And in general, if $A$ and $B$ are groups, with operations $\cdot$, $\star$ it is true that $A \times B$ is a group with the following operation $\bullet$?

$$(a,b)\bullet(a',b') =(a\cdot a',b\star b')$$

Final question: I found that the kernel of $f$ is

$$\ker f=\{ e^{i2\pi k/n} \mathbb{I} \}_{k=0}^{n-1}$$

by imposing that $e^{i\theta}z=\mathbb{I}$ and taking determinants, I found that there are only $n$ different values of $\theta$ allowed and guessed (I'm not sure how to prove that $z$ have to be like that) $z=\mathbb{I}$

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    $\begingroup$ it is surely true that it is a group, with identity given by $(id_A,id_B)$, inverse of $(a,b)$ given by $(a^{-1}_A,b^{-1}_B)$ and associativity granted by associativity of components. $\endgroup$ – Federica Maggioni Apr 17 '13 at 10:08
  • $\begingroup$ Yes, the last one was an easy question after all. Can you answer the other two? $\endgroup$ – Jorge Lavín Apr 17 '13 at 13:46
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    $\begingroup$ Is $ \left\{ U(1)\times SU(n) \right\}=\left\{g,h \right\} $ a typo? By $ \left\{ U(1)\times SU(n) \right\} $, do you mean $U(1)\times SU(n)$, and why would that set only have two elements $g,h$? $\endgroup$ – Alex Provost Apr 17 '13 at 14:04
  • $\begingroup$ @Silencer It is a very bad and wrong choice of notation sorry $\endgroup$ – Jorge Lavín Apr 17 '13 at 14:26
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    $\begingroup$ You can't simply say "let $z$ be any element in $SU(n)$ and let $\theta$ be any angle". Certainly these elements need to depend on the $y \in U(n)$, yes? $\endgroup$ – Alex Provost Apr 18 '13 at 19:39

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