Is the unit circle the only infinite compact subgroup of the multiplicative group of non-zero complex numbers? Let $G$ be an infinite subgroup of $(\mathbb C \setminus \{0\},.)$ such that $G$ is compact as a subset of $\mathbb C$ , then is it true that $G=S^1$ ? 
I know that $G \subseteq S^1$ ; and since any closed subset of $S^1$ is compact in complex plane , so basically I am asking whether any infinite subgroup of $S^1$ which is closed in the complex plane , must necessarily be $S^1$ .  
 A: Consider the continuous surjective homomorphism $f: \mathbb R \to S^1$ as $f(x)=e^{2\pi i x} , \forall x \in \mathbb R$ . 
Then for any subgroup $H$ of $S^1$ , $f(f^{-1}(H))=H$ , where $f^{-1}(H)$ is a subgroup (as $f$ is a group homomorphism) of $(\mathbb R,+)$ . So , $f^{-1}(H)$ is either cyclic or dense in $\mathbb R$ . If $f^{-1}(H)$ is dense , then $f$ being a continuous surjection , $f(f^{-1}(H))$ is dense , i.e. $H$ is dense in $S^1$ . Otherwise , $f^{-1}(H)$ is cyclic i.e. $f^{-1}(H)=b\mathbb Z$ for some $b \in \mathbb R$ . Two cases appear . 
CASE I : $ b \in \mathbb Q$ : Say $b=p/q$ , where $p \in \mathbb Z , q\in \mathbb N$ . Then for any $n \in \mathbb Z$ , $n=kq+r ; 0\le r<q$ ,
then $e^{2\pi i bn}=e^{2\pi i(p/q)(kq+r)}=e^{2\pi i r/q } ; 0\le r<q$ . So then
$H=f(f^{-1}(H))=f(b\mathbb Z)=\{e^{2\pi i bn}:n\in \mathbb Z\} \subseteq \{e^{2\pi i r/q } : 0\le r <q\} $ , hnece $H$ is finite . 
CASE II : $b$ is irrational
Then $\mathbb Z+b\mathbb Z$ is dense in $\mathbb R$ and noticing $\mathbb Z \subseteq ker f$ (actually it is equal , but we don't need that ) , we get 
$H=f(f^{-1}(H))=f(b\mathbb Z)=f(\mathbb Z+b\mathbb Z)$ , and since $f$ is continuous surjection , so $H=f(\mathbb Z+b \mathbb Z)$ is dense in $S^1$ .
So we see that any subgroup of $S^1$ is either finite or dense in $S^1$ , hence if it is closed and infinite , it should be whole of $S^1$ . 
A: Table of contents:


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*A sketchy outline considering cases of whether or not $G$ contains a nonroot of unity.

*An elaboration on the claim about what happens if $G$ contains a nonroot of unity.

*An alternative approach not depending on whether or not $G$ contains a nonroot of unity.



1.
If $a\in G$ is not a root of unity, then $\{a^n:n\in\mathbb N\}$ is dense in $S^1$.  Since $G$ is closed, this implies $G=S^1$.
On the other hand, if every $a\in G$ is a root of unity, and $G$ is infinite, then the set of $n$ such that $G$ has a primitive $n$th root of unity is infinite, and the union of the sets of all powers of such roots is dense, containing $n$ equally spaced points on the circle for arbitrarily large $n$.  Again, density and closedness implies $G=S^1$.

2.
Elaboration on the first claim:
Suppose $a\in G$ is not a root of unity.  Take $a=e^{i\theta}$ with $0<|\theta|<\pi$.  Taking powers of $a$, let $k\in \mathbb N$ be the smallest exponent greater than $1$ such that $a^k$ passes $1\in S^1$, going counterclockwise if $\theta>0$, otherwise clockwise. Then either $a^k$ or $a^{k-1}$ has less than half the arclength to $1$ as $a$, that is, we can write it as $a_2=e^{i\theta_2}$ with $0<|\theta_2|<\dfrac{|\theta|}{2}$.   Continuing this with $a_2\in G$ (also not a root of unity), we get an $a_3$, a power of $a_2$, such that $a_3=e^{i\theta_3}$ and $0<|\theta_3|<\dfrac{|\theta_2|}{2}$. Continuing indefinitely we get a sequence $a_1=a, a_2,a_3,\ldots$ of powers of $a$ getting arbitrarily close (but not equal) to $1$.  If $|a_n-1|<\varepsilon$, then powers of $a_n$ cover $S^1$ with consecutive points with distance less than $\varepsilon$.  

3.
Because $G$ is infinite and $S^1$ is compact, $G$ has an accumulation point, and thus $G$ contains unequal but arbitrarily close elements.  For all $\varepsilon>0$, there exist $a,b\in G$ such that $0<|a-b|<\varepsilon$.  Then $c=ab^{-1}$ is in $G$ and $0<|c-1|<\varepsilon$.  Consecutive powers of $c$ cover the circle with points less than $\varepsilon$ apart.  Hence $G$ is dense, and assuming it is closed, it must be $S^1$.

3.b. 
Elaborating by request on showing that density follows from the above:
To show that $G$ is dense in $S^1$ is equivalent to showing that for all $x\in S^1$, for all $\varepsilon>0$, there exists $g\in G$ such that $|x-g|<\varepsilon$.  So, given such $x$ and $\varepsilon$, let $c=c(\varepsilon)$ be as above.  Either $x$ is a power of $c$ or there are consecutive powers $c^n$ and $c^{n+1}$ with $x$ lying between them, hence $|x-c^n|<|c^{n+1}-c^n|<\varepsilon$.
A: Since $G$ is infinite and $S^1=S$ is compact, there is a sequence of distinct points $(z_n)$ in $G$ that converges to some $z\in S.$ Because $G$ is compact, $z\in G.$ The sequence $z_nz^{-1}$ is then a sequence of distinct points in $G$ converging to $1.$ Thus given any neighborhood $U$ of $1,$ there is a point of $G$ in $U\setminus \{1\}.$
Let $\epsilon > 0.$ From the above it follows that there exists $w\in G$ such that $w=e^{it}$ for some $t\in (-\epsilon,\epsilon), t\ne 0.$ Let $N>2\pi/\epsilon$ be an integer. Define $A = \{w,w^2,\dots, w^N\}.$ Then $A\subset G,$ and every point in $S$ lies within $\epsilon$ of some point of $A$ (measuring distance in arc length).
Now a subset $D$ of $S$ is dense in $S$ iff given $\epsilon>0$ and a point $z\in S,$ there exists $w\in D$ within $\epsilon$ of $z.$ We have shown $G$ has this property. Thus $G$ is dense in $S,$ i.e., $\overline G = S.$ Because $G$ is compact, $G =\overline G,$ hence $G=S.$
A: Using the topological isomorphism $S^1 \cong \mathbb{R}/\mathbb{Z}$ and lifting the situation to $\mathbb{R}$, you are asking that, if $\tilde{G}$ is a subgroup which intersects $[0,1]$ infinitely, then is $\tilde{G}$ dense in $\mathbb{R}$?
The answer is yes, since for such $\tilde{G}$, one can always find elements $g\in \tilde{G}$  with arbitrary small absolute value and then by looking at the subgroup $g\mathbb{Z} \subset \tilde{G}$, one can conclude that $\tilde{G}$ is indeed dense in $\mathbb{R}$; it follows that $G$ is dense in $S^1$ too.
