Group Homomorphism Involving with The circle group in C^x

Question

If U = {${z\in C^x | |z|=1}$}, how do I show that $C^x$/U $\cong$ $\Bbb R^+$.

I know that the Fundamental Theorem of Group Homomorphisms has to come into play, given a function $$f:C^x-> \Bbb R^+$$ where f(z) = |z|.

Answer 1

I take it that our OP means what is more usually written "$\Bbb C^\times$" by "$\Bbb C^x$" and that his

$\Bbb R^+ = \{ r \in \Bbb R \mid r > 0 \}, \tag 1$

so that $\Bbb R^+$ is the multiplicative subgroup of positive reals. Then the map he calls

$f(z) = \vert z \vert, \; f: \Bbb C^\times \to \Bbb R^+ \tag 2$

obeys

$f(z_1 z_2) = \vert z_1 z_2 \vert = \vert z_1 \vert \vert z_2 \vert = f(z_1) f(z_2), \tag 3$

which shows it is a group homomorphism $\Bbb C^\times \to \Bbb R^+$; it is clearly surjective, since for any positive real $\alpha$ there is some $z \in \Bbb C^\times$ with $\vert z \vert = \alpha$. Also, we have

$\ker f = \{z \in \Bbb C^\times \mid \vert z \vert = 1 \} = U; \tag 4$

thus, by the usual theorem(s) of elementary group theory, $f$ induces an isomorphism

$\tilde f: \Bbb C^\times / U \cong \Bbb R^+, \tag 5$

and that's how it it shown.