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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}
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}

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0The 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 \(S^1 =\{z:z=1\} \cong \mathbb R /\mathbb Z\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ker \phi = \{t \in \mathbb R ~~~~~~ \phi (t) =1\}\]because the identity in the multiplicative group is 1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Got it, thanks, how about the next "how"? :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, since the unit element of T is 1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino same idea!!, he puts it under the quotient group concept

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1second part: we have: \[\Large {e^{i2\pi n}} = 1,\quad n \in \mathbb{Z}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Got it too. Thanks you guys
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