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Let $G$ be a connected compact separable Hausdorff metric group, which is monothetic, i.e., has a dense subgroup generated by a single element. Such a group is necessarily Abelian.

Question:

Can the cardinality of the set of all closed subgroups of $G$ be uncountable?

Thank you.

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    $\begingroup$ 1. Yes: A compact connected nonabelian Lie $G$ has uncountably many maximal tori (all conjugate). $\endgroup$ – Francois Ziegler Jun 6 '18 at 17:23
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    $\begingroup$ 2. No change, as compact connected semisimple Lie groups have 2-generated dense subgroups (Kuranishi’s theorem). $\endgroup$ – Francois Ziegler Jun 6 '18 at 17:28
  • $\begingroup$ Well, this is embarrassing.. Of course, that's right. Thanks a lot for the comment. In fact, what I meant in the second question was that $G$ is monothetic. I hope you don't mind if I modify the question accordingly. $\endgroup$ – Bedovlat Jun 6 '18 at 17:33
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Yes. Take $G=\prod_{i<\omega}S^1$. It admits a metric which is compatible with the group action. An element $x=(e^{i\pi a_0},e^{i\pi a_1},\ldots,e^{i\pi a_n},\ldots)$ generates a dense subgroup if the set $\{1,a_0,a_1,\ldots,a_n,\ldots\}$ is $\mathbb{Q}$-linearly independent.

For every subset $I\subseteq\omega$ you have a closed subgroup $G=\prod_{i\in I}S^1$.

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    $\begingroup$ Thanks for the answer. Maybe you meant "dense subgroup" in the second line. This is a concise and complete answer to my question. $\endgroup$ – Bedovlat Jun 7 '18 at 7:23
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Not only the answer is yes (as indicated in Adam's answer), but there's a full picture:

$G$ is a compact abelian group with a homomorphism with dense image $\mathbf{Z}\to G$. "With dense image" means an epimorphism in the category of locally compact abelian groups. This means that its Pontryagin dual $\hat{G}$ is a discrete abelian group comes with a monomorphism into the circle group $\mathbf{R}/\mathbf{Z}$, where monomorphism simply means injective homomorphism. In other words, $G$ is obtained by considering the Pontryagin dual of any subgroup of the circle group endowed with the discrete topology. That $G$ is metrizable means that $\hat{G}$ is countable.

Pontryagin duality yields a bijection between the set of closed subgroups of $G$ and the set of subgroups of $\hat{G}$.

Now discrete abelian groups with countably many subgroups were characterized by Boyer (1956) and are pretty rare: these are abelian groups $H$ with a finitely generated subgroup $A$ such that $H/A$ is isomorphic to $\bigoplus_{p\in I}C(p^\infty)$, where $I$ is a finite set of primes (no multiplicity allowed). Here $C(p^\infty)$ is the Prüfer (quasi-cyclic) group $\mathbf{Z}[1/p]/\mathbf{Z}$.

So it's enough to consider any group not of this form, and embeddable in the circle group. For a subgroup $H$ of the circle group, there are three ways to fail to satisfy the above criterion:

  1. to have infinite $\mathbf{Q}$-rank: for instance, one considers free $\mathbf{Z}$-module of infinite countable rank: this yields $G$ to be the infinite-dimensional torus as in Adam's answer;
  2. to have $p$ torsion for infinitely many primes $p$. For instance, for some infinite set $J$ of primes $H=\bigoplus_{p\in J}\mathbf{Z}/p\mathbf{Z}$, and then $G=\prod_{p\in J}\mathbf{Z}/p\mathbf{Z}$.
  3. to have twice the same prime occurring in the Prüfer part. This holds if $H\simeq\mathbf{Z}[1/p]^2$, so $G=\hat{H}\simeq S^2$ where $S$ (the Pontryagin dual of $\mathbf{Z}[1/p]$) is called "solenoid" (unlike the last two previous examples it's not straightforward here that $G$ has uncountably many closed subgroups, since $S$ itself has only countably many ones, but still it's true).
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  • $\begingroup$ Thanks for the answer. Adam's answer was the first to appear, and it was complete, so it is accepted. But this answer of yours is, of course, more substantial and instructive. Given the opportunity, I would upvote this answer several times. $\endgroup$ – Bedovlat Jun 7 '18 at 7:25
  • $\begingroup$ The problem is that $G$ is supposed to be connected. This means $\hat G$ has to be torsion-free. Must contain no infinitely divisible elements. What about a pure subgroup of rank $2$ of $\mathbb{Z}^\wedge_2$. I don't know. If the dual of such a group is not connected then I am ready to bet that my example is the only one. $\endgroup$ – Adam Przeździecki Jun 7 '18 at 10:18
  • $\begingroup$ @AdamPrzeździecki it's not a problem, since I gave a more general answer. As you say, connectedness corresponds to the case when the dual is torsion-free and this is the case in the first and third of my examples. (And I don't understand your second sentence. Anyway my example 3 is connected and distinct from yours.) $\endgroup$ – YCor Jun 7 '18 at 11:03
  • $\begingroup$ To sum up, the only compact abelian groups with countably many subgroups are those $A$ lying in an exact sequence $0\to\mathbf{Z}_n\to A\to T\to 0$, where $T$ is a virtual torus (connected compact Lie group with $T_0$ a torus) and $\mathbf{Z}_n=\underleftarrow\lim_k\mathbf{Z}/n^k\mathbf{Z}$ for some $n\ge 1$ ($\simeq\prod_{p|n}\mathbf{Z}_p$). $\endgroup$ – YCor Jun 7 '18 at 11:50
  • $\begingroup$ @YCor Yes, you are right. The troubles I was worried about yield solenoid-like structures, still connected. To sum up, $G$ is connected iff $\hat G$ is torsion-free. $\endgroup$ – Adam Przeździecki Jun 7 '18 at 13:04

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