orbits in locally compact group

Question

As everyone knows if $x\in S^1$, then the set $\{ x^n \}$ is either finite or dense. Under which condition is true for any other locally compact group, i.e if $G$ is a locally compact group, and $x\in G$ is a non-trivial element, then the set $x^n$ is either finite or dense.

Answer 1

It is easy to produce groups $G$ in which the set $\{x^n\}$ is finite for every $x\in G$. E.g, let $F$ be any finite group; then $G=F^{\mathbb{N}}$ is a compact, totally disconnected group with this property. So in the OP we may assume that $G$ contains some element generating an infinite cyclic dense subgroup.

I claim that, in this case, $G$ is indeed isomorphic to $S^1$. First, $G$ is locally compact abelian. Here is a result I found in the book by W. Rudin, "Fourier analysis on groups" (Wiley, 1962). Say that a locally compact abelian group is monothetic if it contains a dense cyclic subgroup. Theorem 2.3.2: every monothetic group is either compact or isomorphic to $\mathbb{Z}$ (as a topological group). Coming back to the OP, our group $G$ is monothetic. By the theorem just quoted, $G$ must be compact (as $\mathbb{Z}$ clearly does not satisfy the assumptions of the OP).

Let $\hat{G}$ be the dual of $G$, a discrete, infinite abelian group. By Pontrygin duality, it is enough to prove that $\hat{G}$ is infinite cyclic. Dualizing the assumptions in the OP, we see that every homomorphism $\hat{G}\rightarrow S^1$ either has finite image, or is injective. In particular, $\hat{G}$ is just infinite (infinite, but every proper quotient is finite). It is not very difficult to see that a just infinite, abelian group is infinite cyclic (see McCarthy, Donald, Infinite groups whose proper quotient groups are finite. I. Comm. Pure Appl. Math. 21 1968 545–562).

Answer 2

I don't think ever for a Lie group $G$. If $G$ is not compact then it will have a closed subgroup isomorphic to $\mathbb{R}$, which rules it out. If it is compact, then it will have a proper subgroup isomorphic to $S^1$ which again rules it out.

Answer 3

This is not true even for compact abelian groups. If you take $G = S^1 \times S^1$ and $(x,y) \in G$ is an element such that the orbit of $x$ in $S^1$ is dense and the orbit of $y$ in $S^1$ is finite, then the orbit of $(x,y)$ in $S^1 \times S^1$ will be neither.

Answer 4

For Lie groups, the only one with this property is $S^1$.

To see this, consider a one parameter subgroup given by exponential of a vector in the lie algebra, and the closure of this subgroup should be abelian.

Since the only abelian lie group with the property you are asking is $S^1$, this concludes.

For other groups, I believe it should also be true, but I don't know. The argument above, shows that if the group is not abelian, then it does not hold (since the closure of an orbit is an abelian closed subgroup), but I don't know if abelian topological groups are all known (the ones I know, the only one verifying your property is $S^1$).

Answer 5

These are exactly (a) all locally compact groups in which every element is torsion, and (b) the circle group.

(a) All groups in which every element is torsion work. This involves many discrete groups (which are technically Lie groups!). More generally, a locally compact group with this property has a compact open subgroup with this property. If abelian, this compact group has uniform torsion.

(b) I guess that in spirit, the question is more about examples $G$ in which both cases occur, i.e., there exists an infinite order element, and hence a dense cyclic subgroup. Let me prove that this forces $G$ to be isomorphic to the circle group.

This forces being a compact abelian group $G$. By Pontryagin duality, these are the duals of discrete groups $H$ in which every homomorphism to the circle group is either injective or has a finite image, and both cases occur.

If $H$ has $\mathbf{Q}$-rank $\ge 2$ this is excluded (kill half of a copy of $\mathbf{Z}^2$ and extends the homomorphism. If $H$ has $\mathbf{Q}$-rank $1$, there is a homomorphism with infinite image killing the torsion, so we have a subgroup of $\mathbf{Q}$. In turn, the quotient by an infinite cyclic subgroup is isomorphic to a subgroup of the circle. Hence, $H$ does not satisfy the property unless it is itself infinite cyclic.

If $H$ has $\mathbf{Q}$-rank $0$, i.e., is torsion, then first the condition implies that it has $p$-torsion for only finitely many primes $p$. Write $H=\prod_p H_p$, where $H_p$ is a $p$-group. For some $p$, $H_p$ is infinite. Since $H$ admits an injective homomorphism with dense image into the circle, we see that $H_p$ quasi-cyclic (i.e. isomorphic to the Prüfer group $\mathbf{Z}[1/p]/\mathbf{Z}=\mathbf{Q}_p/\mathbf{Z}_p$). But then it can also be mapped non-injectively with infinite image. So this case is excluded.

In conclusion, the only possibility for $H$ is $\mathbf{Z}$, i.e., the only possibility for $G$ is to be the circle group.