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The subgroup $S_1$ (the unit circle) in the multiplicative group of complex numbers without 0 is a normal subgroup. Identify the quotient group $\mathbb{C}^\times / S_1$.

I thought the quotient group was isomorphic to $S_1$ but it's not. My thought was that

$e^{i\theta_1}*e^{i\theta_2} = e^{i\theta_1+i\theta_2}$ and this is bijetive homomorphism. I set $S_1 = ${$e^{i\theta} \mid \theta \in [0,2 \pi)$}.

Is this wrong because the quotient group $ \mathbb{C}^\times / S_1 = ${$r* e^{i\theta} \mid r \geq 0, \theta \in [0,2 \pi)$} ?

Edit:

I think $ \mathbb{C}^\times / S_1 $ = {$r*S_1 \mid r \geq 0 $}

So then the quotient group is isomorphic to $\mathbb{R}_{>0}$ ? because $ \mathbb{C}^\times / S_1$ is all circles about the origin with fixed positive radius?

jun
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2 Answers2

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Every $z\in\mathbb{C}^{\times}$ can be written uniquely as $z=r\cdot \zeta$ where $\zeta\in S^1, \ r\in\mathbb{R}_+$.

Define $f: \mathbb{C}^{\times}\to \mathbb{R}^+$ with $f(r\cdot \zeta)=r$ and show that it is a group epimorphism with $ker(f)=S^1$.

Then use 1st isomorphism theorem

1123581321
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Yep. The elements of your quotient group are all the circles centered at the origin, and the quotient group operation is multiplication of the radii. That makes it isomorphic to the multiplicative group on the (strictly) positive reals.

Arthur
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