# how to show that a group is elementarily equivalent to the additive group of integers

Is there any fairly easy way of showing a group is elementarily equivalent to the additive group of the integers?

I've found a simple characterization here: A ‘natural’ theory without a prime model, but the proof in Szmielew's paper is quite long and much more general, while I'm looking for something more elementary.

Specifically, I'd like to show that the subgroup of rationals generated by fractions of the form 1/p for p prime is equivalent to integers, but a more general, relatively simple solution would be appreciated.

edit: As pointed out in the comments, i mean the additive group of rationals (clearly, since for the multiplicative group the fractions would generate the entire group, and it's certainly not equivalent to integers, whether it's multiplicative or additive), and the subgroup can also be characterized as the group of fractions with squarefree denominators, while elementary equivalence is a concept from model theory (as indicated in tags).

szmielew's paper considering equivalence classes of abelian groups can be found here: matwbn.icm.edu.pl/ksiazki/fm/fm41/fm41122.pdf , but it's from the 50's, making it quite hard to read due to outdated language and very apparent lack of modern latex.

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Do you mean subgroup of the rational numbers under multiply? If so, you can apply unique factorization theorem of integers to your question. – Yuchen Liu Apr 22 '12 at 14:31
@jerrysciencemath: We can generate additive groups which are much larger than the rationals which are elementarily equivalent to the integers (e.g. ultrapowers of the integers). – Asaf Karagila Apr 22 '12 at 14:33
@jerrysciencemath: No I think he means the additive group of rational numbers of the form $n/m$ where $m$ is square-free. – Henning Makholm Apr 22 '12 at 15:15
What do you mean by 'elementarily equivalent'? Does it mean isomorphic? – Yuchen Liu Apr 22 '12 at 15:17
i meant the additive group and elementarily equivalent, just the way henning makholm and jim belk have indicated. the article i linked does not contain proof, altough the article (which is quite dated) containing the proof in question is available without subscription, free of charge, right here: matwbn.icm.edu.pl/ksiazki/fm/fm41/fm41122.pdf . – groovin' Apr 22 '12 at 21:56

I believe it can be shown that the axioms of torsion free abelian group + axioms to the effect that $G/nG\cong Z/nZ$ (e.g. for each element $x$ coprime with $n$, multiples of $x$ yield all possible "residues modulo $n$", of which there are exactly $n$) + definitions of new symbols $n\vert\cdot$ for each $n$ are sufficient for q.e., from which completeness of the theory easily follows, but it's quite a tedious task:
For simplicity, assume we also have a symbol for additive inverse function (it does not make much of a difference). The general formula which we're supposed to q.e. is of the form: $\exists x \bigwedge mx=A\wedge \bigwedge mx\neq A\wedge\bigwedge n\vert mx+A\wedge\bigwedge n\nmid mx+A$ (where $A$ is an integer combination of variables other than $x$, and $m,n,A$ are all different but unlabeled to avoid indexitis).
1. If the first bigwedge is nonempty, we can easily q.e. without invoking the $G/nG$ axioms by adding a subformula of the form $n\vert A$, multiplying all other expressions (including the divisors) by $n$ and replacing each occurence of $nx$ by $A$
2. If it is not so, but the third and fourth bigwedge is empty, the formula is simply true (since the group is infinite), so it's equivalent to $x=x$, otherwise for similar reasons we can assume it's empty
3. Using the axioms we can prove a form of chinese remainder theorem sufficient to divide it into a conjunction of formulas of the form $\exists x \bigwedge p^k\vert mx+A\wedge\bigwedge p^k\nmid mx+A$ where $p$ are all the same
4. A little more careful analysis allows us to assume that $k$ are all the same, while $m$ are all of the form $p^c$ with $c<k$.
5. The remaining statement can be interpreted as a statement that allows only some residues modulo $p^k$ and prohibits others; using the $G/nG$ axiom we can see that whether or not is it satisfiable depends solely upon which residues of $A$'s modulo some $p^{k-c}$coincide, and which $A$'s are divisible by some $p^c$