# When does a LCA group not contain a (closed) infinite cyclic subgroup?

If $G$ is an LCA (locally compact abelian) group, is there any 'nice' sufficient (or preferably necessary and sufficient) criteria for when $G$ does not contain a closed (and hence discrete in the subspace topology) infinite cyclic subgroup?

An easy necessary condition comes from the usual decomposition theorem that any LCA group $G$ can be expressed as $G = \mathbb{R}^n \times H$ for some $H$ which contains a compact-open subgroup, so we know that any such $G$ must contain a compact-open subgroup, but this is obviously nowhere near sufficient.

EDIT: Another sufficient condition is that $G$ is topologically torsion, that is, every element is contained in a compact subgroup.

• I guess that this is true iff the dual group doesn't have $\mathbb{S}^1$ as a factor or something like that. – Mark Jul 21 '11 at 18:18
• I don't see this since $\mathbb{R}$ contains $\mathbb{Z}$ as a closed subgroup, but is self-dual. If however, we are considering groups without copies of $\mathbb{R}$ inside, then this, via duality, would say that the only way $\mathbb{Z}$ can occur (closed) in such a group is as a direct factor. Is this the case? – Iian Smythe Jul 21 '11 at 19:06
• You'r completely right, I was overlooking that not everything splits. I deleted my previous comment, but please have a look at my final answer. – Marc Palm Jul 21 '11 at 19:34
• The correct statement that $\mathbb{Z}$ maps into $A$, iff $\widehat{A}$ maps onto $\mathbb{T}$. Pontryagin duality switches the arrow;) Sorry for the confusion -.- – Marc Palm Jul 21 '11 at 20:33
• This holds iff $G$ is the union of its compact open subgroups, iff $G$ is the filtering union of its compact open subgroups. The latter condition (for $G$ not necessarily abelian) has various names in the literature, including "topologically locally finite" (yuk!), "locally elliptic", "elliptic". – YCor Jan 15 '14 at 22:10

In general, you have for a compactly generated group $G = \mathbb{R}^n \times \mathbb{Z}^n\times K$, with $K$ compact. And there is no way to embed $\mathbb{Z}$ discretely in something compact, see Deitmar-Echterhoff Principles of harmonic Analysis on page 96. These are a reasonably nice family of groups, because the Haarmeasure is $\sigma$ finite, if(f) the group is compactly generated, I guess (?).

For Lie type abelian group on page 97, you have $G = \mathbb{R}^n \times \mathbb{T}^n\times D$, with $D$ discrete abelian. $D$ is here finitely generated, iff $G$ is compactly generated.

Seeing Mark Schwarzmann comment, you can move back and forth between the above descriptions via Pontryagin duality $\widehat{\mathbb{R}^n} \cong \mathbb{R}^n$, $\widehat{\mathbb{Z}^n} \cong \mathbb{T}^n$ and $\widehat{K}=D$. So essentially you want something like $K=D$ finite, but note there are examples like $D =\mathbb{Q} /\mathbb{Z}$, whose dual is the profinite completion of $\mathbb{Z}$.

Copied from Locally compact abelian groups: Corollary 7.54 of Hoffman and Morris "The Structure of Compact Groups" does the rest of the job: if $A$ is an LCA group, then each neighborhood of the identity contains a compact subgroup $K$ such that $A/K≅\mathbb{R}^m×\mathbb{T}^n×D$, where D is a discrete abelian group.

Theorem: $\mathbb{Z}$ does not embed discretly in a locally compact abelian group, iff there exists a compact subgroup $K$ with $A/K = \mathbb{T}^n \times D$ with $D$ discrete abelian consisting only elements with finite order.

• Yes, this makes sense for the compactly-generated case. – Iian Smythe Jul 21 '11 at 19:08
• Now, I found a way to give a general answer. – Marc Palm Jul 21 '11 at 19:32
• Thanks, I think this looks right to me, but I will have to think about it just a bit more. – Iian Smythe Jul 21 '11 at 20:09

Actually, a different solution occurred to me:

Claim: $\mathbb{Z}$ does not embed into an lca group $L$ iff $L$ is topologically torsion (i.e. every element is contained in a compact subgroup).

Proof: If $\mathbb{Z}$ does not embed into $L$, then for each $x\in L$, the closure of $\langle x\rangle$ is compact, since it is monothetic. The converse follows from the fact mentioned by pm that the infinite cyclic group cannot embed in any compact group.

• what is monothetic? – Marc Palm Jul 21 '11 at 21:25
• monothetic = has a dense copy of $\mathbb{Z}$. – Mark Jul 22 '11 at 4:13