Find all kinds of homomorphisms from $(\Bbb Z_n, +)$ to $(\Bbb C^*, \times)$.

Question

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??

Answer 1

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.

Answer 2

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.