Proving that every torsion-free Abelian group $(A, +)$ can be linearly ordered

abelian-groupsgroup-theorylogicorder-theory

Show that every torsion-free Abelian group (A, +) can be linearly ordered so
that
$$(a < b) ∧ (c \le d) → (a+c < b+d).$$
(Hint: First show this for finitely generated groups. Then use compactness.)

I got this question in logic class and I have yet to partake in a group course, my proffesor gave me this hint:
Any finitely generated torsion-free Abelian group is a direct sum of a finite number of groups of integers ($Z$,+) (which I don't understand how can it help me here..).

I am just wondering if my attempt makes sense:
given $A$ which is a finite generated torsion-free Abelian group let $$(g_1,g_2,….,g_n)$$
be $A$'s basis, let $v$ belong to $A$ then exist
$$\beta_1,\beta_2,\ldots,\beta_n \in F$$
such that
$$v = \sum_{i=1}^n g_i*\beta_i$$
lets create a function from $A$ to $R^n$ lets call her $f$ such that for any $v$ in $A$
$$f(v) = (b_1,b_2,\ldots,b_n),$$
I am sending v to its coordinate vector.
Then I will create a linear order $R$ as follows:
$$aRb = f(a) \le f(b).$$
I have two probeloms: first lets say I define $\le$ as Product order relation, then does it work for a vector with n coordinates? and if it does and I managed to find a relation that linearly orders a finite generated group $A$ then how does compactness theorm help me? Can I claim that given B which is torsion-free Abelian group such that
$A \subseteq B$, since $A$ is general finite generated group who has a model for that linear order then by compactness therom $B$ also has a model?

Thx in advance any help would be appreciated!

Best Answer

To show that it is possible to choose basis (so that your $f$ is well-defined and coefficients can be added as integers) is equivalent to showing that group is isomorphic to $\mathbb Z_+^n$.

Product order relation isn't linear, it's only partial. However, if you multiple linearly ordered sets (for example, multiple copies of $\mathbb Z$ as in your case) it's easy to define linear order on them: we say that $(a_1, \ldots, a_n) < (b_1, \ldots, b_n)$ if either for some $j$ we have $a_i = b_i$ for $i < j$ and $a_j < b_j$ - in other words, on the first position where $a$ and $b$ differ, $a$ is less then $b$, for example $(0, 1) < (1, 0)$. This is called lexicographic order.

Using compactness, we can prove that for group $G$, if all finitely generated subgroups of $G$ can be linearly ordered, so can be $G$. Given that finitely generated subgroups of torsion-free abelian group are, of course, torsion-free finitely generated abelian groups, this shows necessary result.

Assume all finitely generated subgroups of $G$ can be linearly ordered. Consider some theory strong enough to speak about groups and their ordering (for example, ZF), add to it as a constant $G'$ and for every element $g \in G$ add constant $g'$.

  1. $G'$ is a group that can be linearly ordered.
  2. For every element $g \in G$, add axiom $g' \in G$.
  3. For every elements $x, y \in G$, add axiom $x'y' = (xy)'$.
  4. For every different elements $x, y \in G$, add axiom $x' \neq y'$.

By compactness, this new theory has a model if when we add any finite number of this axioms the result theory will have a model. However, if we add only finite number of our axioms, they use only finite number of elements, and this theory will have a model - just use subgroup of $G$ generated by these elements for $G'$.

So, our theory has a model. Take such model, and notice that subset of $G'$ consisting of all $g'$ is a group isomorphic to $G$, and it can be linearly ordered by order induced from $G'$.

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