Character group of $\mathbb{C}^*$ Let $\mathbb{C}^* = \mathbb{C}-\{0\}$ be the group of multiplication. Then why the character group $G := Hom(\mathbb{C}^*, \mathbb{C}^*)$ is $\mathbb{Z}$?
 A: Well, it isn't. Let me help you classify the elements of $G$, which should make classifying $G$ simple.
There are a few properties that all group homomorphisms $f:\Bbb C^*\to\Bbb C^*$ have. In particular:


*

*$\forall z\in\Bbb C^*,$ $|f(z)|=f(|z|)$

*$\forall z\in\Bbb C^*,$ $f(\overline z)=\overline{f(z)}$

*$\forall z\in\Bbb C^*,$ $z\in\Bbb R\implies f(z)\in\Bbb R$

*$\forall z\in\Bbb C^*,$ $z>0\implies f(z)>0$


These traits will come in handy.
Some of the elements of $G$ are obvious:


*

*$z\mapsto 1$

*$z\mapsto z$

*$z\mapsto z^{-1}$


The first of these will be the identity in the character group (call it $f_0$). The other two will be inverses of each other (call them $f_1,f_{-1},$ respectively). More generally, for any $n\in\Bbb Z,$ let $f_n$ be the map $\Bbb C^*\to\Bbb C^*$ given by $z\mapsto z^n.$ It is readily seen that each $f_n\in G$ and that $f_n=(f_1)^n$ for all $n\in\Bbb Z$. Furthermore, $n\mapsto f_n$ is a ready one-to-one homomorphism $\Bbb Z\to G.$ Let's see if we can prove that every element of $G$ is one of these functions $f_n,$ so that $n\mapsto f_n$ is actually an isomorphism. First, we prove a few preliminary results.

Lemma: Given $f\in G,$ its restriction $g$ to the unit circle is a character of the unit circle. In particular, there is some $n\in\Bbb Z$ such that, for all $u$ in the unit circle, we have $g(u)=u^n.$

Proof Outine: That $g$ is a character of the unit circle is readily shown. For the other, consider the kernel of $g.$ It is necessarily a multiplicative subgroup of the unit circle. If infinite, then it will be dense in the unit circle, so by continuity of characters, we have that $g$ is identically $1$ (so the $n$ in question is $n=0$). If the kernel is finite, then it consists of the $n$th roots of unity for some positive integer $k,$ and it is not difficult to see that the desired $n$ is either $n=k$ or $n=-k$.

Lemma: Given $f\in G,$ its restriction $g$ to the positive reals is a character of the positive reals. In particular, there is some real $\alpha$ such that, for all $t>0,$ $g(t)=t^\alpha$.

Proof Outline: Let $\alpha=\log_2 g(2).$ It is readily shown by induction that for all non-negative integers $m,$ we have $g(2^m)=2^{m\alpha},$ from which it follows for all negative integers $m,$ as well. Next, observe that for any integer $m$ and any positive integer $n$ coprime with $m,$ we have $$2^{m\alpha}=g(2^m)=g\left(\left(2^{\frac mn}\right)^n\right)=g\left(2^{\frac mn}\right)^n,$$ so that $$2^{\frac mn\alpha}=\left(2^m\right)^\frac1n=\left(g\left(2^{\frac mn}\right)^n\right)^\frac1n=g\left(2^{\frac mn}\right),$$ and so $g(2^q)=2^{q\alpha}$ for all rational $q$. By continuity of character, we have $g(2^s)=2^{s\alpha}=(2^s)^\alpha$ for all real $s,$ and so $g(t)=t^\alpha$ for all $t>0.$

Now, let's see what the general form of a character of $\Bbb C^*$ is. Let $f\in G,$ and let $n,\alpha$ as in the Lemmas above. Then for all $z\in\Bbb C^*,$ we have $$f(z) = f(|z|)f(z/|z|) = |z|^\alpha(z/|z|)^n = z^n|z|^{\alpha-n}.$$
Since $z\mapsto z^w$ is ill-defined for non-integer $w,$ we find that the elements of $G$ are of the form $f(z)=|z|^tz^n$ for some $t\in\Bbb R,n\in\Bbb Z.$
