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

Question

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.

Answer 1

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.

Answer 2

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.