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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}

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  1. Loser66
    • one year ago
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    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
    • one year ago
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    @Michele_Laino

  3. anonymous
    • one year ago
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    \[ker \phi = \{t \in \mathbb R ~~~|~~~ \phi (t) =1\}\]because the identity in the multiplicative group is 1

  4. Loser66
    • one year ago
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    Got it, thanks, how about the next "how"? :)

  5. Michele_Laino
    • one year ago
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    no, since the unit element of T is 1

  6. Loser66
    • one year ago
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    @Michele_Laino same idea!!, he puts it under the quotient group concept

  7. Michele_Laino
    • one year ago
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    second part: we have: \[\Large {e^{i2\pi n}} = 1,\quad n \in \mathbb{Z}\]

  8. Loser66
    • one year ago
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    Got it too. Thanks you guys

  9. Michele_Laino
    • one year ago
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    :)

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