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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?

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  • $\begingroup$ What, doc, have you tried? $\endgroup$ – dfeuer Sep 25 '13 at 8:38
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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.

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  • $\begingroup$ That's not yet enough, is it? There is still the case of an infinite subgroup containing only elements of the form $e^{2i\pi r}$ with $r$ rational. $\endgroup$ – Magdiragdag Sep 25 '13 at 8:46
  • $\begingroup$ @Peter: The hint doesn’t actually cover every case, but it contains all of the necessary tools. However, the way it’s worded does suggest that it covers every case, so I’ll change it a bit. $\endgroup$ – Brian M. Scott Sep 25 '13 at 8:48
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    $\begingroup$ I'm just trying to point out to the OP that if he proves the two claims in your hint, he isn't yet automatically done. $\endgroup$ – Magdiragdag Sep 25 '13 at 8:51
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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 $H= \alpha \mathbb{Z}$.
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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

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