Subgroup of the unit circle under complex multiplication

Let $T=\{z:|z|=1 \mathbin{\text{and}} z \in \mathbb C\}$ be the unit circle in the complex plane, considered as a topological group under complex multiplication and the usual topology.

Show that every subgroup of $T$ is either dense in $T$ or finite.

How to prove the denseness?

Hint: The map $\theta \mapsto e^{i \theta}$ is a continuous epimorphism from $(\mathbb{R},+)$ to $(T ,\times)$ and a subgroup of $(\mathbb{R},+)$ is either dense in $\mathbb{R}$ or of the form $\alpha \mathbb{Z}$ for some $\alpha \in \mathbb{R}$.

The above classification of the subgroups of $(\mathbb{R},+)$ is a classical result and can be shown in the following way:

• Let $H$ be a subgroup of $(\mathbb{R},+)$ and let $\alpha= \inf \{ h \in H \mid h>0\}$.
• If $\alpha=0$, show that $H$ is dense in $\mathbb{R}$.
• If $\alpha >0$, show that $H= \alpha \mathbb{Z}$.

HINT: First note that $T=\{e^{i\theta}:\theta\in[0,2\pi)\}$, and that $e^{i\theta}\cdot e^{i\varphi}=e^{i(\theta+\varphi)}$. Then show that if $\theta=2r\pi$ for some rational $r\in[0,1)$, then $e^{i\theta}$ generates a finite cyclic subgroup of $T$; that’s pretty straightforward once you write $r$ as a fraction. Use this to show that if a subgroup $H$ of $T$ contains only numbers of the form $e^{2i\pi r}$ with $r\in[0,1)$ rational, and if moreover $H$ is infinite, then $\{r\in[0,1):e^{2i\pi r}\in H\}$ contains rational numbers with arbitrarily large denominators when written in lowest terms; then use its proof to show that this implies that $H$ is dense in $T$.

Finally, show that if $\theta=2r\pi$ for some irrational $r\in[0,1)$ is irrational, then $e^{i\theta}$ generates an infinite, dense subgroup of $T$. For this last step you can adapt the argument in this answer, where a very similar result is proved.

IF $U \subset T$ in an infinite subgroup THEN $U$ is dense in $T$

PROOF: For any (large) positive integer N , select any N distinct elements of U,

$u_1, u_2, u_3, ..., u_N$

These N points divide $T$ into N arc segments with their corresponding angles, and the sum of those angles is equal to $2\pi $. Thus, you must be able to find two points $u_j$ and $u_k$ satisfying the following two conditions:

(1) $u_j$, $u_k$ are adjacent to each other

(2) The angle defined by $u_j$ and $u_k$ is less than or equal to $2\pi /N$.

So the element $v = u_j u_k ^{-1}$ is also no further away from 1 than $2\pi /N$.

The points $v, v^2, v^3, ..., v^M$ will 'traverse' the circle, and, for some M, must also complete the 'orbit' (cover the $2\pi$ distance).

Now if $z$ is any point on the Circle Group, it must be no further away than $\pi/N$ from some point $v^k$.

QED