A simple reason that $\text{Tor}(\mathbb{C}^{*}) \neq S^1.$

Question

Let $\mathbb{C}^{*}$ denote the group of nonzero complex numbers under multiplication, and $S^{1} \subset \mathbb{C}^{*}$ the subgroup of complex numbers of length one. Torsion elements of $\mathbb{C}^{*}$ are called roots of unity.

Show that $\text{Tor}(\mathbb{C}^{*}) \subset S^1.$ Now give a simple reason that $\text{Tor}(\mathbb{C}^{*}) \neq S^1.$

My question is:

1- I know from here Torsion subgroup of $\mathbb{C}^\times$ that the torsion elements are the roots of unity, but I do not know how to prove that $\text{Tor}(\mathbb{C}^{*}) \subset S^1.$ could anyone help me in writing a rigorous proof for that, please?

2- What is a simple reason that $\text{Tor}(\mathbb{C}^{*}) \neq S^1$?

EDIT: My definition of $S^{1}$ is $\{ z \in \mathbb{C^{*}\ :\ |z|=1 }\}$

Answer 1

For a point $z=re^{i\theta}$ with $r>0$ and $\theta\in \Bbb R$ is torsion element if and only if $z^n=1$ for some $n\geq 1$. In particular, $z$ is torsion element implies $|z|^n=|z^n|=1$, so that $r=|z|=1$. Therefore, $z=re^{i\theta}$ is torsion if and only if $z=e^{i\theta}$ for some $\theta\in \Bbb R$ with $z^n=e^{in\theta}=1$. Now, we know that $e^{i\theta}=\cos(n\theta)+i\sin(n\theta)=1\iff\cos(n\theta)=1$ and $\sin(n\theta)=0\iff \theta=\frac{2m\pi}{n}$ for some $m\in \Bbb Z$.

Combining all these a non-zero element $z$ is torsion if and only if $z=e^{i\frac{2m\pi}{n}}$ for some $n\in \Bbb N$ with $m\in \Bbb Z$.

Answer 2

Here is an algebraically concise way to answer your questions:

1. In general, if $f \colon G \to G'$ is a morphism of abelian groups it must map the source torsion subgroup to the target torsion subgroup, in other words $f[\mathrm{T}(G)] \leqslant \mathrm{T}(G')$. In your particular case, the absolute value map $z \mapsto |z|$ implements a morphism between the multiplicative groups $\left(\mathbb{C}^{\times}, \cdot\right)$ and $\left(\mathbb{R}^{\times}_{+}, \cdot\right)$, which means that it must map $\mathrm{T}\left(\mathbb{C}^{\times}\right)$ into the torsion subgroup of $(0, \infty)$. However, this latter group is totally ordered (by the standard order) and thus has trivial torsion. This means that $\mathrm{T}\left(\mathbb{C}^{\times}\right) \subseteq |\bullet|^{-1}[\{1\}]=\mathbb{U}$ (my notation for the circle group) and therefore that $\mathrm{T}\left(\mathbb{C}^{\times}\right)=\mathrm{T}\left(\mathbb{U}\right)$.
2. As an abelian group (even as a topological group), the unit circle $\mathbb{U}$ is isomorphic to the quotient $\mathbb{R}/\mathbb{Z}$ of the additive group $(\mathbb{R}, +)$ (this is intimately related to the complex exponential). It is very easy to show that $\mathrm{T}\left(\mathbb{R}/\mathbb{Z}\right)=\mathbb{Q}/\mathbb{Z}$, in other words the torsion subgroup of this latter quotient group is a proper subgroup. This entails the fact that the inclusion $\mathrm{T}(\mathbb{U}) \subset \mathbb{U}$ is proper (strict), otherwise the circle group $\mathbb{U}$ would itself be torsion (and we have seen it can't be, since it is isomorphic to a group strictly containing its torsion subgroup).