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Let us consider the set $S^1=\{z∈C: |z|=1\}$. Show that there is no non-trivial connected subgroup of $S^1$.

I know that $S^1$ denotes the unit circle in $R^2$. But how can I show mathematically that there is no non-trivial connected subgroup of $S^1$. I want a simple proof. I think one way to prove above it is suffices to show that the continuous map $R→S^1$ given by $x→e^{ix}$ is surjective. Then continuous image of a connected space is connected. And $R$ is a connected space.Then I think it follows from it. Please help me to prove the above content.

abcdmath
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    You seem to just be quoting random things that have nothing to do with the question. A circle minus a point is not disconnected. And nowhere does the question ask about the circle being connected (so your last few sentences are pointless). – Steve D Feb 15 '18 at 21:18
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    Your argument shows that $S$ itself is connected, but doesn’t give the desired fact. – Lubin Feb 15 '18 at 21:19
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    If the group misses one point of the circle it also misses its square root (which is a new point because $1$ is never missed). If the group is not finite, then it is dense in the circle, therefore it has points in between the missing point and its square root, on both arcs that are in between. –  Feb 15 '18 at 21:19
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    @abcdmath If you cut a circle in one point, there is still one piece. – Cheerful Parsnip Feb 15 '18 at 21:25
  • Yah thank you. I have edited my mistake. – abcdmath Feb 15 '18 at 21:29

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Assume $G\subset S^1$ is a proper non-trivial subgroup. So there is an $x_1\in S^1-\{1\}$ which is not in $G$. This means that also some square root $x_2$ of $x_1$ cannot be in $G$, because otherwise $x_1=x_2^2\in G$.

Define $\theta_i:=\arg(x_i)\in[0,2\pi)$. W.l.o.g. assume that $\theta_1<\theta_2$. We define the two sets

\begin{align} V_1&:=G\cap \{\exp(i\theta)\mid \theta\in(\theta_1,\theta_2)\},\\ V_2&:=G-V_1=G\cap \{\exp(i\theta)\mid \theta\in(\theta_2,2\pi) \cup [0,\theta_1)\}. \end{align}

If $G$ is finite, then it is clearly disconnected (note: it has at least two elements). But if $G$ is infinite, then it is dense in $S$. Hence, both of the sets $V_i$ are non-empty, open (w.r.t. $G$) and disjoint, and we have $G=V_1\cup V_2$. This means $G$ is not connected.

M. Winter
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  • $V_1$ or $V_2$ could be empty. That is why you need to treat the case of $G$ finite, differently (by just saying that doh it is disconnected). –  Feb 15 '18 at 21:38
  • @jfKeys You are right! I will think about how to formulate this nicely. Thank you. – M. Winter Feb 15 '18 at 21:39
  • You don't need to say almost anything as long as you are fine invoking that the infinite subgroups are dense. Otherwise you need to prove that too. And the finite case is clear. –  Feb 15 '18 at 21:41
  • @jfKeys Yeah, I mainly thought about an easy density-argument. Or maybe I just leave it and assume it's folklore. – M. Winter Feb 15 '18 at 21:43