Identifying a quotient group (NBHM-$2014$)

Question

Let $\mathbb C^*$ denote the multiplicative group of non-zero complex numbers and let $P$ denote the subgroup of positive real numbers. Identify the quotient group.

My thought $$\frac{\mathbb C^*}{P}=\{P,-P,iP,-iP\}.$$ Is it right?

@GerryMyerson : I do not understand your idea.... :O could you please explain a bit more... — , Jan 25 '14 at 14:55
@TattwamasiAmrutam : why do you think the quotient group is same as what you have written... it would be better if you can try writing that partially... — , Jan 25 '14 at 14:57
@PraphullaKoushik Partially?? I am not even sure about the correctness of what I have written.. — tattwamasi amrutam, Jan 25 '14 at 14:58
I know that if you are sure of the correctness you would not ask :D.. I am asking for your intuition.... — , Jan 25 '14 at 15:00
@TattwamasiAmrutam: Is ... positive real numbers... above the same as $\mathbb R^*_{>0}$? — Mikasa, Jan 25 '14 at 15:03
I guess Gerry is thinking that if none of these four cosets contains a complex cube root of $1$ then the conjectured solution cannot be corret. — Derek Holt, Jan 25 '14 at 15:13
The quotient essentially collapses each open ray from 0 to a point. There is one such ray for each direction from 0. Can you visualize the resulting quotient? — MPW, Jan 25 '14 at 15:26

score 13 · Accepted Answer · edited Jan 25 '14 at 16:03

13

I think you can set this map $$f:\mathbb C^*\to U,~~z=a+ib\mapsto \frac{a}{|z|}+i\frac{b}{|z|}$$ wherein $U=\{z\in\mathbb C^*\mid|z|=1\}$. Show this map is surjective with $\ker f=\mathbb R^*_{>0}$.

edited Jan 25 '14 at 16:03

Gerry Myerson

185,413

answered Jan 25 '14 at 15:08

Mikasa

67,942

4

For short: $f(z)=z/|z|$. – user26857 Jan 02 '17 at 20:16

score 1 · Answer 2 · edited Jan 13 '20 at 17:06

1

One another way,

The map $f:\mathbb{C}^* \to \mathbb{C}^*$ defined by $f(z=r.e^{i\theta})=1.e^{i\theta}$ is a homomorphism with kernel being the set of positive reals.

By the Fundamental theorem of homomorphism we can say that $\mathbb{C}^*/P$ is isomorphic to $f(\mathbb{C}^*)=S^1$. where $S^1=\{z\in\mathbb{C} | |z|=1\}$.

edited Jan 13 '20 at 17:06

Community

1

answered Jan 02 '17 at 08:40

Siddhant Trivedi

2,338

score -1 · Answer 3 · answered Feb 14 '14 at 12:14

I am not sure if this is in any way different than B.S's way. i am just omitting maps and trying to see just as a quotient.

We see $x+iy\in \mathbb{C}$ as $(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})$

Now, As $\sqrt{x^2+y^2}\in \mathbb{R}$ when we see $(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})$ modulo real numbers we would get :

$$(\sqrt{x^2+y^2})(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})\equiv (\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}}) \text { mod }\mathbb{R}$$

And then we have $|(\dfrac{x}{\sqrt{x^2+y^2}}+i\dfrac{y}{\sqrt{x^2+y^2}})|=\sqrt{(\dfrac{x}{\sqrt{x^2+y^2}})^2+(\dfrac{y}{\sqrt{x^2+y^2}})^2}=1$

So,for any $z\in \mathbb{C}$ we have $z=|z|(z')$ for $|z|\in \mathbb{R}$ and $|z'|=1$.

Thus $\mathbb{C}^*/P=\{z\in \mathbb{C} : |z|=1\}$

Identifying a quotient group (NBHM-$2014$)

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