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.
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.