an abelian group $\langle G,+,0\rangle $ is orderable if and only if it is torsion-free First of all I confess this is my home work!
But really I don't know how to solve this problem....
I don't even have a hint!
the question is:

An abelian group $\langle G,+,0\rangle$ is orderable if and only if it is torsion-free.

For one side I thought I could write this: if $G$ is an orderable group then there exists $x>0$ and if $G$ is not torsion-free there exists $n$ s.t. $n>0$ and $nx=0$. $x>0$, $x<x$, $x<nx$, $x<0$, this means $x$ can just be $0$. And for $x<0$, $x<x$, $nx<x$, $0<x$,
and also this means the only torsion element in $G$ is $0$.
but $n$ is not in the universe of $G$.
for the other side is it possible to define a topology on torsion-free abelian groups with no loss of generality and then use topology to prove $G$ is orderable?
 A: The forward direction is easy . . .

Suppose $(G,+,0)$ is an ordered abelian group, and suppose $G$ is not torsion-free.

Let $x\in G$ with $x \ne 0$ be such that $nx = 0$, for some integer $n > 1$.

But then $x > 0$ would imply $nx > 0$, contradiction.

Similarly, $x < 0$ would imply $nx < 0$, contradiction.

It follows that an orderable additive abelian group must be torsion-free.

For the reverse direction, here's an outline . . .

Suppose $(G,+,0)$ is a torsion-free abelian group.

Let $H =\mathbb{Q} \otimes_{\Bbb{Z}} G$.

Since $G$ is torsion-free, $G$ can embedded in $H$. 

$\qquad$Embedding torsion-free abelian groups into $\mathbb Q^n$?

Thus, to show $G$ is orderable, it suffices to show that $H$ is orderable.

To order $H$, regard $H$ as a vector space over $\mathbb{Q}$, and let $B$ be a basis for $H$ over $\mathbb{Q}$. 

Order $B$ arbitrarily, and define an order on $H$ by


*
For nonzero $x \in H$, assert $x > 0\;$if and only if $q>0$ where $q \in Q$ is the coefficient of the largest basis element in the unique representation of $x$ with respect to the basis $B$.


In other words, the order on $H$ is lexicographic order with respect to the ordered basis $B$.

Thus, with the order as defined above, $H$ is an ordered abelian group, and hence, $G$ can be ordered via its embedding in $H$.
