Loser66
  • Loser66
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}
Mathematics
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SOLVED
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katieb
  • katieb
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Loser66
  • 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 \(S^1 =\{z:|z|=1\} \cong \mathbb R /\mathbb Z\)
Loser66
  • Loser66
@Michele_Laino
anonymous
  • anonymous
\[ker \phi = \{t \in \mathbb R ~~~|~~~ \phi (t) =1\}\]because the identity in the multiplicative group is 1

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Loser66
  • Loser66
Got it, thanks, how about the next "how"? :)
Michele_Laino
  • Michele_Laino
no, since the unit element of T is 1
Loser66
  • Loser66
@Michele_Laino same idea!!, he puts it under the quotient group concept
Michele_Laino
  • Michele_Laino
second part: we have: \[\Large {e^{i2\pi n}} = 1,\quad n \in \mathbb{Z}\]
Loser66
  • Loser66
Got it too. Thanks you guys
Michele_Laino
  • Michele_Laino
:)

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