## Loser66 one year ago Explain me, please Show that $$\phi (t) = cis t$$is a group homomorphism of the additive group $$\mathbb R$$ onto the multiplicative group T:={z:|z|=1}

1. Loser66

The usual way is easy to prove and my prof gave us other way but I don't understand it. Please, explain me $$ker \phi = \{t \in \mathbb R ~~~|~~~ \phi (t) =1\}$$ . How? it should be $$\phi(t) =0$$, right? $$\phi (t) =1$$ iff $$t = 2\pi n, ~~n\in \mathbb Z$$ $$ker \phi = \{2\pi n: n\in \mathbb Z\}\cong \mathbb Z$$ hence $$\phi$$ is onto. Again, how? Hence $$S^1 =\{z:|z|=1\} \cong \mathbb R /\mathbb Z$$

2. Loser66

@Michele_Laino

3. anonymous

$ker \phi = \{t \in \mathbb R ~~~|~~~ \phi (t) =1\}$because the identity in the multiplicative group is 1

4. Loser66

Got it, thanks, how about the next "how"? :)

5. Michele_Laino

no, since the unit element of T is 1

6. Loser66

@Michele_Laino same idea!!, he puts it under the quotient group concept

7. Michele_Laino

second part: we have: $\Large {e^{i2\pi n}} = 1,\quad n \in \mathbb{Z}$

8. Loser66

Got it too. Thanks you guys

9. Michele_Laino

:)