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Does there exist a lattice-ordered group of rational rank $1$?
This is true for totally ordered group.

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Totally ordered groups are in particular lattice-ordered ones. – boumol Jan 11 '13 at 2:24
@ Boumol, thanks. Sorry, I did not mention in question but I am looking lattice ordered but not the totally ordered. – Rajnish Jan 11 '13 at 2:47
Have you thought about considering the following? Take the group structure of your totally ordered group, and as lattice structure take some non total order (compatible with the group operation). This has to work. – boumol Jan 11 '13 at 3:13
By "rational rank 1", do you simply mean that the underlying group is isomorphic to $\mathbb{Q}$? – Todd Trimble Jan 13 '13 at 2:39
For abelian groups, I'd expect "rational rank 1" to mean that the group is isomorphic to a non-zero subgroup of $\mathbb Q$. For non-abelian groups, I won't make a guess. – Andreas Blass Jan 13 '13 at 21:16
up vote 2 down vote accepted

If "rational rank 1" simply means that the underlying group is isomorphic to $\mathbb{Q}$, then every such lattice-ordered group structure $(\mathbb{Q}, \preceq)$ is totally ordered, in fact isomorphic to $\mathbb{Q}$ with its usual order $\leq$.

By a classical result (see the first sentence here), any abelian lattice-ordered group ($l$-group) $G$ can be embedded (by an $l$-group homomorphism) into a direct product of totally ordered non-trivial abelian groups. By projecting to any one of the factors, there exists a totally ordered $l$-group $G'$ and a non-trivial $l$-group homomorphism $f: G \to G'$. By an epi-mono factorization, we may assume $f$ is surjective.

In the case $G = \mathbb{Q}$, any group quotient $G'$ is divisible, and it is well-known that an $l$-group $G'$ is torsionfree (see e.g. here, corollary 1.2.6 page 16). But divisible torsionfree abelian groups are rational vector spaces, so in this situation the quotient $G'$ is isomorphic to $\mathbb{Q}$ and $f: G \to G'$ is a group isomorphism. But $f$ is also a lattice homomorphism. Since bijective lattice homomorphisms are lattice isomorphisms, $f: G \to G'$ is a lattice isomorphism, and hence $G$ is totally ordered, as desired.

Finally, a totally ordered $l$-group structure $(\mathbb{Q}, \preceq)$ is given by two submonoids $P, N = -P$ whose intersection is $\{0\}$ and whose union is $\mathbb{Q}$. To show $(\mathbb{Q}, \preceq)$ is isomorphic to $(\mathbb{Q}, \leq)$, we need only show that $P$ contains every multiple $\frac{m}{n}p$ where $p \in P$ is any chosen non-zero element and $0 < m, n$. But clearly the submonoid generated by $\frac{m}{n}p$ contains $0 \neq mp \in P$. Thus $\frac{m}{n}p \in N$ is impossible. This completes the proof.

Edit: Following on Andreas's first interpretation of "rational rank 1" (subgroup $B \subseteq \mathbb{Q}$), the same results hold. For again, we have a surjective $l$-group homomorphism $f: B \to B'$ to a totally ordered $l$-group $B'$, where $B'$ is nontrivial and torsionfree. If $A = \ker(f)$, we have $\mathrm{rank}(B) = \mathrm{rank}(A) + \mathrm{rank}(B')$, where $\mathrm{rank}(B') = 1$ since $B'$ is torsionfree. Thus $\mathrm{rank}(A) = 0$, meaning $A$ is a torsion subgroup of a torsionfree group, meaning $A = 0$. Thus $f: B \to B'$ is a group isomorphism, and we deduce as before it is a lattice isomorphism, so $B$ is totally ordered. We deduce that $B$ is isomorphic to $B$ with its standard lattice structure, by essentially the same argument as in the last paragraph before this edit.

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@Trimble, thank you for the nice idea. – Rajnish Jan 13 '13 at 23:04
Is there ordered structure on the group of rational numbers $\mathbb Q$ which is not totally ordered? If the answer is yes, is that torsion free or not? – Rajnish Jan 14 '13 at 2:30
Certainly there are many partial orders on $\mathbb{Q}$ compatible with the group structure; all you need is a submonoid $P \subseteq \mathbb{Q}$ with the property that $x \in P$ and $-x \in P$ implies $x = 0$. For example, $P = \mathbb{N}$ would work: define $x \leq y$ to mean there is a nonnegative integer $n$ such that $y = n + x$. "Torsion free" refers only to group structure, not order structure; $\mathbb{Q}$ is of course torsion free. – Todd Trimble Jan 14 '13 at 2:41

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