The classification of divisible groups up to isomorphism is well-known:

**Theorem:** A divisible group is isomorphic to a direct sum . Moreover, the cardinals and are uniquely determined.

In this note, we give and prove a similar classification but up to elementary equivalence. As references, we may mention:

- For model theory: David Marker’s book,
*Model Theory: An introduction*. - For abelian groups: Kaplansky’s book,
*Infinite Abelian Groups*. - For a complete classification of abelian groups up to elementary equivalence:
*The Elementary Theory of Abelian Groups*, Paul Eklof and Edward Fischer, Annals of Mathematical Logic, Volume 4, No. 2 (1972) 115-171.

Let be an abelian group and be a prime number. Then is a vector space over and we note its dimension. Noticing that the sequence is decreasing, we define:

Also, let

Our goal is to prove:

**Theorem:** Let and be two divisible groups. Then and are elementary equivalent if, and only if, and for every prime .

**Proof.** Let (resp. ) be a -saturated elementary extension of (resp. ). For , and are elementary equivalent, so is a divisible group and we can write

and .

Let

where

and .

Let and . Let , , denote the components of contained in some direct summand of not in .

By construction, have strictly more that direct summands isomorphic to and exactly . Therefore, permuting with an isomorphic direct summand of whose intersection with is trivial, we can construct an automorphism such that for and .

According to Tarski-Vaught test, we deduce that is an elementary substructure of .

**Lemma:** If then ; otherwise, .

**Proof.** First, if and , let

Clearly, for any abelian group , is equivalent to .

Secondly, notice that and in the same way . Therefore,

and .

Suppose that . Because and are elementary equivalent, we deduce that

.

Now suppose that ; in particular, . By contradiction, suppose that , ie. , and let be a basis of the -vector space ; for large enough, has cardinality .

Therefore, since is -saturated, the following type over is realized:

contradicting the fact that is a basis. Consequently, .

**Lemma:** If then ; otherwise, .

**Proof.** Because and are elementary equivalent, . So implies that is torsion-free, hence .

Suppose that . By contradiction, suppose that and let be a set of torsion-free -independent elements. Let

If we show that is finitely satisfiable, will be satisfiable as a type over in a -saturated group, contradicting the maximality of . However, noticing that , it is clear that any finite subset is satisfiable in (here ).

Suppose that and for every prime . According to the two previous lemmas, we deduce that and for every prime , hence .

Therefore, .

Conversely, suppose that and are elementary equivalent. It is obvious that . Using the formulae of our first lemma, we deduce also that for every prime .

**Corollary:** and are elementary equivalent for every .

In fact, it is easy to find two non-isomorphic elementary equivalent groups, for example:

**Claim:** Let be a nonprincipal ultrafilter over . Then and are non-isomorphic elementary equivalent groups.

**Proof.** and are elementary equivalent according to Los theorem.

By contradiction, suppose that has a generator .

If , there exists such that for each . Therefore, for every .

On the other hand, if then (for example, ).

Therefore, is not cyclic.

However, it is more difficult to find “little” classical non-isomorphic elementary equivalent groups, and the previous corollary gives such an example. We are able to give a proof independent on our classification:

**Proof of corollary.** Let and let be an ultrafilter over ; according to Los theorem, is elementary equivalent to and so is a torsion-free divisible group, hence for some cardinal .

Because , it is sufficient to choose an ultrafilter so that to deduce that . Then the isomorphic class of does not depend on .

Therefore, and have isomorphic ultrapowers and hence are elementary equivalent.

For the ultrafilter, we can take a regular ultrafilter, as described below.

**Definition:** Let be a nonprincipal ultrafilter over a set . We say that is regular if there exists a family of finite subsets such that for every .

**Lemma:** For any infinite cardinal , there exists a set of cardinality and a regular ultrafilter over .

**Lemma:** Let be a regular ultrafilter over some set and let be an infinite set. Then:

.

For details, see Thomas Jech’s book entitled *Set theory*.