Connected subgroups of unit circle

Question

It's well known that any subgroup H of $\mathbb{S}^1$ is either dense or finite. Therefore, if H is compact, it implies that H either is finite or equal to the entire unit circle. My question is whether we can say anything similar in the case where H is connected. Obviously in these circumstances, if it is finite, it has to be trivial, but I'm curious if anything happens when H is dense. Is there any counter-example of such a group that isn't the entire unit circle?

Answer 1

I think what should happen is that if $H$ is a dense connected subgroup of $S^1$, then $H$ is $S^1$. To see this assume for sake of contradiction that we have a dense connected subgroup $H$ of $S^1$ which is not $S^1$. We shall form a disconnection as follows. We can think of $S^1$ as the interval $[0,1]$ (where under the quotient the endpoints are identified). Now consider $H\subset [0,1]$ under this identification. Now we can form a disconnection of $H$ as follows. Since $H$ is dense in $S^1$, but it is not equal to $S^1$ let us take a point $\alpha\notin H$, then we will also have that $\alpha^{-1}\notin H$ these are distinct points (assuming $\alpha\neq 1/2$, in the case $\alpha =1/2$, you can convince yourself that $1/4\notin H$ since if it were then $1/2\in H$ which it isn't). Thus, the important thing is that we have two distinct points $\alpha,\beta\in S^1$ which are not in $H$. Now we have that the subspace topology of $H$ will consist of open intervals in $\mathbb{R}$ intersect $H$. Thus, if we consider the open arcs of $H$ between $(\alpha,\beta)\cap H$ and $(\beta,\alpha)\cap H$ (where by this I mean we split the circle up into the things going counter-clockwise from $\alpha$ to $\beta$ and those going counter-clockwise from $\beta$ to $\alpha$). These will be two non-empty open subsets of $H$ (with the subspace topology) whose union is the entirety of $H$. Thus, we have formed a disconnection contradicting the existence of a dense connected subgroup of $S^1$. Thus, we conclude that if $H$ is a dense connected subgroup of $S^1$, then $H=S^1$.

Answer 2

Since $\mathbb{S}^{1}$ is regarded as a topological group and since $S^1$ is a connected manifold, it is locally path-connected and semi-locally simply connected. If a subgroup $H$ of $S^1$ is also connected then as $G, H$ here are both Lie groups, there is a long exact sequence

$$\dots \to \pi_1(G/H) \to \pi_0(H) \to \pi_0(G) \to \pi_0(G/H)$$

Answer 3

There just aren't very many connected subsets of a circle. The complement of a point in a circle is homeomorphic to $\mathbb{R}$, and any connected subset of $\mathbb{R}$ is a (possibly degenerate) interval. This means that any connected subset $H$ of $S^1$ is an arc: in other words, there are points $a,b\in S^1$ such that $H$ is the arc of $S^1$ obtained by starting from $a$ and going around counterclockwise until you reach $b$ (and $a$ and $b$ themselves may or may not be in $H$).

Now if this $H$ is a subgroup, it must contain $1$. If $H\neq\{1\}$, this means that $H$ contains an entire interval on one side of $1$, since $H$ is an arc. Such an interval then generates the entire group $S^1$ (since every time you take products of elements of the interval with itself its length doubles, until it covers the whole circle), so $H$ must be all of $S^1$.