Subgroup of the unit circle under complex multiplication

Question

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?

Answer 1

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 $H= \alpha \mathbb{Z}$.

Answer 2

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.

Answer 3

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

Answer 4

It has to be shown that each infinite subgroup $G$ of $T = S^1$ is dense in $S^1$.

As in Seirios's answer I shall use the continuous surjection $$p : (\mathbb R, +) \to (S^1,\cdot), p(t) = e^{2\pi i t}$$ which is a group homomorphism with kernel $\mathbb Z$.

It does not suffice to show that a subgroup $H$ of $\mathbb R$ is either dense or has the form $H = \alpha \mathbb Z$ with $\alpha > 0$. If $\alpha$ is irrational, then $p(H)$ is dense in $S^1$, but no proof is given for this fact. We do not need to know this to prove that $G$ is dense in $S^1$, but it will be a by-product.

Here is my proof.

Define $H = p^{-1}(G)$. This is a subgroup of $\mathbb R$. Since $G$ is infinite and $p([0,1]) = S^1$, the set $H' = H \cap [0,1] = p \mid_{[0,1]}^{-1}(G)$ is infinite. Since $[0,1]$ is compact, $H'$ has a cluster point $\tau$. Hence there exists a sequence $(t_n)$ in $H'$ such that all $t_n \ne \tau$ and $t_n \to \tau$. W.l.o.g. we can assume that $t_{n+1} \ne t_n$ for all $n$. Then the $t'_n = \lvert t_{n+1} - t_n \rvert$ form a sequence of positive numbers in $H \setminus \{0\}$ such that $t'_n \to 0$ (note that both $t_{n+1} - t_n$ and $-(t_{n+1} - t_n)$ are contained in $H$).

Now let $(a,b)$ be any open interval. Choose $N$ such that $t'_N < b - a$. Let $m \in \mathbb Z$ be the unique integer such that $mt'_N \le a < (m+1)t'_N$. Then $(m+1)t'_N = mt'_N + t'_N \le a +t'_N < a + b -a = b$, thus $(m+1)t'_N \in (a,b)$. This shows $(a,b) \cap H \ne \emptyset$.

Therefore $H$ is dense in $\mathbb R$ so that

$$S^1 = p(\mathbb R) = p(\overline H) \subset \overline{p(H)} = \overline G . $$

Thus $G$ is dense in $S^1$.

If $H = \alpha \mathbb Z$, then $p(H) = \{e^{2\pi n\alpha i} \mid n \in \mathbb Z \}$. But $e^{2\pi n\alpha i} = e^{2\pi m\alpha i}$ iff $2\pi n\alpha - 2\pi m\alpha = 2\pi k$ for some $k \in \mathbb Z$ iff $(n -m)\alpha = k$. This equation has non-trivial solutions only for rational $\alpha$. For irrational $\alpha$ we must have $n = m$ and $k = 0$. This shows that $p(H)$ is infinite and therefore dense in $S^1$.