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Let $(G,+)$ be an abelian group and "$<$" be a transitive relation on $G$ such that for every $x,y \in G$ exactly one of $x <y , y<x , x=y$ holds and $x < y \implies x+z < y+z , \forall z \in G$ . Let us call such a relation on $G$ a strict linear group order . Can we characterize all torsion-free abelian groups which admits infinitely many strict linear group orders ? Can we give at least separate necessary and sufficient conditions ?

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Note that any ordering on $G$ induces an ordering on $G\otimes\mathbb{Q}$, by saying $x/n<y/m$ for $n,m\in\mathbb{Z}_+$ iff $mx<ny$. Conversely, any ordering on $G\otimes\mathbb{Q}$ induces an ordering on $G$ by restriction, and the original order on $G\otimes\mathbb{Q}$ can be recovered from the order on $G$ as above. So the set of orderings on $G$ is in bijection with the set of orderings on $G\otimes\mathbb{Q}$.

So we may assume $G$ is a $\mathbb{Q}$-vector space. Now the answer is easy. If $G$ has dimension $0$ there is only one possible order, and if $G$ has dimension $1$ there are only two possible orders (the usual order on $\mathbb{Q}$ and its reverse). If $G\cong\mathbb{Q}^2$, it has infinitely many possible orders: for any $q\in\mathbb{Q}$, take the lexicographic order with respect to the basis $(1,0)$ and $(q,1)$. Explicitly, $(a,b)$ is positive in these orders iff either $a>qb$ or $a=qb$ and $b>0$ (this makes it clear the orders are distinct). Finally, if $G$ has dimension greater than $2$, choose a two-dimensional subspace $V$ and a complement $W$ so that $G=V\oplus W$. There are infinitely many orders on $V$, and combining them lexicographically with any order on $W$, we get infinitely many orders on $G$.

Thus a $\mathbb{Q}$-vector space has infinitely many orders iff it has dimension greater than $1$. It follows that a torsion-free group has infinitely many orders iff it has rank greater than $1$.

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  • $\begingroup$ The way you seem to write elements of $G \otimes_{\mathbb Z} \mathbb Q$ ; is it true that $G \otimes_{\mathbb Z} \mathbb Q \cong S^{-1}G $ where $S=\mathbb Z \setminus \{0\}$ ? $\endgroup$ – user456828 Aug 8 '17 at 16:56
  • $\begingroup$ Yes, they are naturally isomorphic. In general, for any commutative ring $R$, a multiplicatively closed subset $S\subseteq R$, and an $R$-module $M$, there is a natural isomorphism $M\otimes_R S^{-1}R\cong S^{-1}M$. $\endgroup$ – Eric Wofsey Aug 8 '17 at 17:01

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