Is the First Isomorphism Theorem Applicable to this Complex Number Group?

[Mr Davis 97]

Homework Statement

$(\mathbb{C}^\times,\cdot)/\mu_m\cong (\mathbb{C}^\times,\cdot)$ for any integer $m\geq 1$, where $\mu_m=\{z\in \mathbb{C} \mid z^m=1\}$.

Homework Equations

The Attempt at a Solution

Here is my idea. Consider the map $f: \mathbb{C}^{\times} \to \mathbb{C}^{\times}$ where $f(z) = z^m$. Then certainly $\ker(f) = \mu_m$. But since this kernel is nontrivial, that means that the map can't be surjective (since if f were surjective we would have to have a trivial kernel). This would mean that we couldn't establish the result using the first isomorphism theorem. Where am I going wrong?

 

[fresh_42]

Homework Statement

$(\mathbb{C}^\times,\cdot)/\mu_m\cong (\mathbb{C}^\times,\cdot)$ for any integer $m\geq 1$, where $\mu_m=\{z\in \mathbb{C} \mid z^m=1\}$.

Homework Equations

The Attempt at a Solution

Here is my idea. Consider the map $f: \mathbb{C}^{\times} \to \mathbb{C}^{\times}$ where $f(z) = z^m$. Then certainly $\ker(f) = \mu_m$. But since this kernel is nontrivial, that means that the map can't be surjective ...

Here.

... (since if f were surjective we would have to have a trivial kernel). This would mean that we couldn't establish the result using the first isomorphism theorem. Where am I going wrong?

Let us test surjectivity. Given any number $z=r\cdot e^{i\varphi}$ we construct a number by setting $s=\sqrt[m]{r}$ in $\mathbb{R}$ and $\psi =\varphi / m$. Now what is $f(z')=f(s\cdot e^{i \psi})\,?$

The isomorphism theorem still holds. The difference to what you are used to is the fact that we have a really big group: uncountable infinite. Now you divide by a finite group with $m$ elements. It simply doesn't make a difference to those uncountable infinitely many elements (cp. https://en.wikipedia.org/wiki/Covering_group).

 

[Mr Davis 97]

Here.
Let us test surjectivity. Given any number $z=r\cdot e^{i\varphi}$ we construct a number by setting $s=\sqrt[m]{r}$ in $\mathbb{R}$ and $\psi =\varphi / m$. Now what is $f(z')=f(s\cdot e^{i \psi})\,?$

The isomorphism theorem still holds. The difference to what you are used to is the fact that we have a really big group: uncountable infinite. Now you divide by a finite group with $m$ elements. It simply doesn't make a difference to those uncountable infinitely many elements (cp. https://en.wikipedia.org/wiki/Covering_group).

So you're saying that if we have a map $f: \mathbb{C}^{\times} \to \mathbb{C}^{\times}$ that is surjective, this doesn't necessarily imply that it is injective (and injectivity doesn't imply surjectivity)? So this is different than what the situation would be with finite groups?

 

[fresh_42]

So you're saying that if we have a map $f: \mathbb{C}^{\times} \to \mathbb{C}^{\times}$ that is surjective, this doesn't necessarily imply that it is injective (and injectivity doesn't imply surjectivity)? So this is different than what the situation would be with finite groups?

Yes. We have a surjective map, a non-trivial kernel, and as to the isomorphism theorem a group which is isomorphic to a homomorph image of itself. With finite groups bijective, injective (with an equal number of elements) and surjective (with an equal number of elements) are the same. We just have to few elements for tricks. But here we have $m\cdot |\mathbb{C}^\times|=|\mathbb{C}^\times|$.

When it comes to complex numbers, a professor of mine used to call for the image of a radish cut into a spiral. (I don't know whether there is a radish among the vegetables on these pictures, but the principle is the same: https://spinner66.com/products/94207) She meant it as a description of the complex logarithm, but anyway, it's close.