# bijective homomorphism between complex numbers and reals

Let $$G = (\mathbb{C}\setminus\{0\}, \cdot)$$, y $$N = \{a + bi \in \mathbb{C} \mid a^2 + b^2 = 1\}$$. Show that $$N\triangleleft G$$ and there exists a bijective homomorphism between $$G/N$$ and $$(\mathbb{R}^+, \cdot).$$

$$N \triangleleft G$$ because $$G$$ is abelian and $$N\subset G$$. But I have problems with the homomorphism; I tried to define one using the module of the complex numbers, but clearly is not injective. Can please someone give me a hint?

Try $$z \mapsto |z|$$ as a map $$G/N \to \mathbb R^+$$.

• But $|z|$ isn't injective, no?
– fxgx
Oct 21, 2020 at 23:24
• From $\mathbb C \setminus \{0\}$ to $\mathbb R$ it's not, but remember that you're factoring by $N$. So first show that it's a surjective homomorphism $G \to \mathbb R^+$. Then show that the kernel is exactly $N$.
– Jim
Oct 21, 2020 at 23:25

Hint Define $$f : \mathbb G \to \mathbb R^+$$ by $$f(z)=|z|$$. SHow that $$f$$ is an onto homomorphishm, and find its Kernel.

$$G/N$$ contains equivalence classes modulo the complex argument. You can think of $$h:G\rightarrow G/N$$ as $$h(z)=zN=Nz$$. This is why normality matters!

Therefore:

• Show that multiplication in $$G/N$$ is equivalent to multiplying any two elements from the respective equivalence classes, i.e., any two elements with the respective magnitudes.
• Show that each equivalence class in $$G/N$$ has exactly one positive element.
• Define $$G/N\rightarrow\mathbb{R}^+$$ as the distinct positive element of each equivalence class.