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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?

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3 Answers 3

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

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  • $\begingroup$ But $|z|$ isn't injective, no? $\endgroup$
    – fxgx
    Oct 21, 2020 at 23:24
  • $\begingroup$ 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 $N$. $\endgroup$
    – Jim
    Oct 21, 2020 at 23:25
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Hint Define $f : \mathbb G \to \mathbb R^+$ by $f(z)=|z|$. SHow that $f$ is an onto homomorphishm, and find its Kernel.

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$G/N$ 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.
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