I have a locally compact abelian group $G$ with the following properties:
- It is connected (therefore divisible) and non-compact;
- It admits a $\mathbb{Q}$-vector space structure and for any $g \neq 0$, the group $\mathbb{Q}\cdot g$ is dense in $G$;
- All its non-trivial closed subgroups are of the form $\mathbb{Z} \cdot x = \langle x \rangle \cong \mathbb{Z}$ for any $x \in G\setminus\{0\}$ and for any such x, the quotient $G/\langle x \rangle$ is isomorphic to the unit circle;
- It admits a natural order by fixing some $x_0 \neq 0$ and saying that $\frac{a_1}{b_1}x_0 \le \frac{a_2}{b_2}x_0$ if and only if $\frac{a_1}{b_1} \le \frac{a_2}{b_2}$. This can be extended through density of $\mathbb{Q}\cdot x_0$ by saying that the limit of a Cauchy net $(x_{\alpha})_{\alpha} \subset \mathbb{Q}\cdot x_0$ is positive iff only a finite number of its terms are non-negative;
- The rays of the form $]a, b[ = \{x \in G: \ a < x < b\}$ are open (it's easy to see that their complements have to be closed by the definition of the order);
- $G$ admits fractional parts in the sense that if $g \in G$, there exists unique $n \in \mathbb{Z}$ and $y \in [0, x_0[$ such that $g = y + nx_0$;
- Any continuous non-trivial homomorphism $\psi: G \to \mathbb{R}$ is necessarily a bijection and any continuous non-trivial homomorphism $\mathbb{R} \to G$ is injective and has dense image.
Note that the first and third properties imply the second one, which in turn implies all the other properties besides the last one (that one can be assumed regardless though).
I'm naturally quite convinced that a group with all these properties will have to be isomorphic to $\mathbb{R}$, since they share way too many things in common (and property 3 itself is probably restrictive enough anyway), but I'm struggling really hard to deal with topological issues. In particular, the topology on G clearly contains the order topology induced by the rays $]a, b[$ but it doesn't seem at all obvious that they would have to be equal.
I thought of trying to turn $G$ into a vector space over the real numbers through density of the rationals, but the problem is that if a sequence of rational numbers $y_n$ converges to a real number $y$, the sequence $y_n \cdot x$ doesn't seem to have any reason to converge in $G$, atleast in general.
I'd appreciate any help on handling this.
Best Answer
The first and third properties suffice. The structure theorem implies that a connected locally compact abelian group $A$ is isomorphic to $\mathbb{R}^n \times K$ where $K$ is compact and connected. Since $K$ is a closed subgroup the third property implies that $K$ is trivial, and since $\mathbb{R}^n$ has $\mathbb{R}^m$ as a nontrivial closed subgroup if $m < n$ then the third property also implies $n = 1$. In fact we only need the first half of the third property.
(Previously I used the second property and asserted that a nontrivial compact group can't have a $\mathbb{Q}$-vector space structure. This is actually false and a counterexample is given by the Pontryagin dual $\text{Hom}(\mathbb{Q}, S^1)$ of $\mathbb{Q}$ regarded as a discrete group; this group is even connected, and is a solenoid.)