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Let $\mathbb{C}^\times$ be the multiplicative group of complex numbers and $\mathbb{R}_+$ the group of positive ($0\not\in \mathbb{R}_+$) reals under multiplication. The question is, what is $\mathbb{C}^\times/\mathbb{R}_+$?

Going from the definition of a coset, I get that

$$ \mathbb{C}^\times/\mathbb{R}_+=\{g\mathbb{R}_+ : g\in \mathbb{C}^\times\}, $$

which are all the rays from the origin out to infinity. What I mean is that two numbers $z,w\in \mathbb{C}^\times$ are in the same coset iff $\arg z=\arg w$, which can be shown using the polar form for $z$ and $w$ (for example here or here).

However, like in the answers linked above, using one of the group isomorphism theorems, we get that

$$\mathbb{S}^1\cong \mathbb{C}^\times/\mathbb{R}_+,$$

where $\mathbb{S}^1=\{z\in \mathbb{C}:\left|z\right|=1\}$.

I can't reconcile these two answers. The rays clearly form a partition of the set $\mathbb{C}^\times$, while $\mathbb{S}^1$ seems to not. Additionally, I can't think of a isomorphism between the rays and $\mathbb{S}^1$. One can certainly project every point of $\mathbb{C}^\times$ onto $\mathbb{S}^1$, but this is not injective, as far as I know. I think I'm missing something obvious here, but can't figure out what.

imas145
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    Each ray is ${tw:t\in\Bbb R,t>0}$ for a unique $w\in S^1$. – Angina Seng Apr 03 '20 at 14:51
  • @AnginaSeng That's it! I forgot that the rays themselves are the elements of the quotient group. If that's the case, then it's easy to form a isomorphism between the rays and $\mathbb{S}^1$, like in Matt's answer below. – imas145 Apr 03 '20 at 14:55

1 Answers1

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Using your definition, how can we construct an isomorphism with the circle? We can send $$re^{i\theta}\mapsto e^{i\theta}$$ Note actually that since $r\in \mathbb R^+$, we have $$re^{i\theta}\mathbb R^+=e^{i\theta}\mathbb R^+$$ so it's not a stretch to construct this homomorphism.

If you have $g,h\in \mathbb C^\times$, what is $$(g\mathbb R^+)(h\mathbb R^+)?$$ Write $g=r_1e^{i\theta_1}$, $h=r_2e^{i\theta_2}$. Then we have $$r_1e^{i\theta_1}r_2e^{i\theta_2}\mathbb R^+=r_1r_2e^{i(\theta_1+\theta_2)}\mathbb R^+=e^{i(\theta_1+\theta_2)}\mathbb R^+$$ You can see where this is going.

Matt Samuel
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