Proving an isomorphism without a defined function

Question

Let $G$ be the group of nonzero complex numbers under multiplication and let $N$ be the set of complex numbers of absolute value $1$ (that is, $a + bi \in N$ if $a^2 + b^2 = 1$). Show that $G/N$ is isomorphic to the group of all positive real numbers under multiplication.

No answer please

Suppose that the group of real numbers is $\mathbb R$.

I'm a bit confused about proving isomorphism since there isn't any defined function that maps between $G/N$ and $\mathbb R$.

I've tried to approach this by proving that group $G$ is homomorphic to $G/N$, using the the quotient group of $G$ by $N$, $N(x+y)$ for an $X$ and $Y$ in $G/N$ where $X = Nx$ and $Y = Ny$, as the function $g:G \to G/N$, but now I'm not sure what function is mapping between $G/N$ and $\mathbb R$. I know I need to first prove that $f:G/N\to\mathbb R$ is homomorphic, but how is that possible if $f$ isn't even defined?

Answer 1

To prove the isomorphism, you have to provide the function $f$ (the one you say is not defined).

Hint: the quotient yields an equivalency between complex numbers that lie in the same circumference (two complex numbers with the same modulus differ only in argument and their quotient is a complex number with modulus $1$ and with that difference as argument). The classes under this equivalency are all circumferences centered at $0$. Which is the natural way of assigning one positive real number to each of this classes?

Answer 2

The comments are there, who knows if they helped you out. Now, you can try the following: define

$$\phi:\Bbb C^*\to\Bbb R^+\;,\;\;\phi\left(|z|e^{it}\right):=|z|$$

Now prove what's needed: it is surjective, what its kernel is, etc. and use the first isomorphism theorem