Find all kinds of homomorphisms from $(\Bbb Z_n, +)$ to $(\Bbb C^*, \times)$ and from $(\Bbb Z, +)$ to $(\Bbb C^*, \times)$. Explain why they are the complete collection.

My intuition is:

1) we can construct $\phi:\Bbb Z_n \to \Bbb C^*$ then $Im(\phi)\le n $

Again $\Bbb Z_n/{\ker \phi} \cong Im(\phi)$.

Some finite subgroups of $\Bbb C^*$ are in $S^1$, i.e.,

$\phi:\Bbb Z_n \to S^1$ s.t $x \mapsto e^{2x\frac{\pi}{n}} $, if $n$ is not prime then we can construct more like this also in each case we have the trivial homomorphism.

So are these the only possibilities or there are more?? Again prove whatever your conclusion is.

2) Same like 1) here we can construct $\phi:\Bbb Z \to S^1$ s.t $x \mapsto e^{2x\frac{\pi}{r}} $ , where $r \in \Bbb R \setminus \Bbb Q$. What next now??

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    $\begingroup$ For someone who's been on the site for a few months, posted nearly 50 questions and another 60 answers or so... you show a terrible lack of efforts to give some context to your question. $\endgroup$ – Asaf Karagila Oct 8 '14 at 21:42
  • $\begingroup$ What context do I give more..? Will you explain? $\endgroup$ – Ri-Li Oct 8 '14 at 21:43
  • $\begingroup$ (1) What did you try? (2) What sort of things which are relevant to this problems you might know. Stuff like that. Something much closer to the other question that you posted earlier today, not to something that you might have copied out of a book, or a homework sheet. $\endgroup$ – Asaf Karagila Oct 8 '14 at 21:44
  • $\begingroup$ Yes and I thaught that I can express it in more simpler way. Thats why I posted only the question. $\endgroup$ – Ri-Li Oct 8 '14 at 21:46
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    $\begingroup$ Note that both of these groups are finitely generated. If you specify the value at $1$, you will have the general form of the homomorphism. $\endgroup$ – Cameron Williams Oct 8 '14 at 21:46

Let $\varphi$ such morphism so we have $\varphi(\overline 0)=1$ and let $a=\varphi(\overline 1)$ then we have

$$\varphi(\overline n)=\varphi(n\times \overline 1)=a^n=1\implies a=\exp\left(\frac{2ik\pi}{n}\right),\quad k\in\{0,\ldots,n-1\}$$ Finally we verify easily that the $n$ maps defined by $$\varphi_k(\overline1)=\exp\left(\frac{2ik\pi}{n}\right),\quad k\in\{0,\ldots,n-1\}$$ are morphism of groups.

  • $\begingroup$ for $Z$?? What will be the process?? $\endgroup$ – Ri-Li Oct 8 '14 at 21:56

From the idea of Sami Ben Romdhane For $\Bbb Z$, I have got $\phi (0)=1$ & $\phi (1)=a\in \Bbb C^*$ then $\phi (n)=a^n$. These are all homomorphisms.

  • $\begingroup$ Would you kindly explain how you jumped to the homomorphisms? $\endgroup$ – reflexive May 24 '18 at 13:10

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