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Let S be a Subgroup of $\mathbb{C}^*$ with S = {$z \in \mathbb{C} : z \bar z = 1$}. Prove that $\mathbb{C}^*/S$ is isomorphic to $\mathbb{R}^*_{+}$. Where $\mathbb{C}^*/S$ is the left coset. I know that I have to show that this function is a bijective Group homomorphism but I'm currently trying to figure out what our function looks like.

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Hint: Take $\mathbb C^* \to \mathbb{R}^*_{+}$ given by $z \mapsto |z|$. Prove that it is a surjective homomorphism with kernel $S$.

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  • $\begingroup$ And that would be enough? why? $\endgroup$ – JonDoeMaths Dec 15 '16 at 16:55
  • $\begingroup$ @JonDoeMaths, yes, by the isomorphism theorem. $\endgroup$ – lhf Dec 15 '16 at 17:33
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The modulus function is a homomorphism between $\mathbb C^*$ with multiplication as the operation and the positive real numbers, again with multiplication as the operation.

To see this, notice that $|wz| = |w| \cdot |z|$ for all $w,z \in \mathbb C$.

The first isomorphism theorem says that if $f : G \to H$ is a group homomorphism, $f(G)$ denotes the image of $G$ under $f$, and $\ker f$ is the kernel of $f$, then: $$G \cong f(G)/\ker f$$

The identity element of $(\mathbb R^+,\times)$ is $1$, so $\ker f = \{z \in \mathbb C \backslash \{0\} : |z|=1\}$, i.e. $\ker f = S$.

The image of $f$ is all of $\mathbb R^+$ since $\mathbb R^+ \subset \mathbb C^*$ and for any $x \in \mathbb R^+$, $|x|=x$.

It follows that $\mathbb R^+ \cong \mathbb C^*/S$.

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