 Solution

The Ohnishi orderability criterion states that, for each finite list $g_1,\dots ,g_n\ne 1$, there are signs ${\epsilon }_i\in \{-1,1\}$ such that the semigroup of nonempty products of conjugates of $g_i^{{\epsilon }_i}$ does not contain $1$. The conjugates are essential for a two-sided invariant total order. Necessity follows by choosing all signed elements positive: their conjugates and nonempty products are still positive.

For sufficiency, consider the compact product $\{+,-{\}}^{G\setminus \{1\}}$. Impose the closed conditions that an inverse has the opposite sign, that conjugation preserves sign, and that a product of two positive elements is positive whenever it is nonidentity. A positive pair cannot multiply to the identity because of the inverse condition. Each condition uses finitely many coordinates. Given finitely many conditions, collect every nonidentity element appearing in them into a finite list and choose signs using the criterion. These signs satisfy the conditions: an incorrectly negative product would contribute its positive inverse to the generated semigroup, producing identity with the positive factors; an incorrect conjugate or inverse gives the same contradiction. Extend the signs arbitrarily to the remaining coordinates. Thus the finite intersection property and compactness give a global sign assignment. Its positive elements form a conjugation-invariant semigroup $S$ with $S\cap S^{-1}=\varnothing$ and $S\cup S^{-1}=G\setminus \{1\}$. Defining $a<b$ by $a^{-1}b\in S$ gives the required linearly ordered group.

Restriction shows that every finitely generated subgroup of an orderable group is orderable. Conversely, suppose every such subgroup is orderable. If the criterion failed on $g_1,\dots ,g_n$, each of the finitely many sign choices would have an identity witness made of finitely many conjugates. Collect all $g_i$ and all conjugating elements in these witnesses into one finitely generated subgroup $H$. An order on $H$ supplies one positive sign choice, but its corresponding witness is a nonempty product of positive elements in $H$, a contradiction. Hence the criterion holds in $G$.

Finally, a finitely generated subgroup of a torsion-free abelian group is ${\mathbb{Z}}^r$ by the Fundamental theorem of finitely generated abelian groups. Its lexicographic order is two-sided invariant. Applying the local-to-global conclusion proves that every torsion-free abelian group admits the required order.