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

Theorem: A divisible group is isomorphic to a direct sum \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z}[p^{\infty}]^{(\alpha_p)} \oplus \mathbb{Q}^{(\beta)}. Moreover, the cardinals \alpha_p and \beta 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 G be an abelian group and p be a prime number. Then G[p] := \{g \in G \mid p \cdot g=0 \} is a vector space over \mathbb{Z}_p and we note \dim(G[p]) its dimension. Noticing that the sequence p^nG[p] is decreasing, we define:

D(p,G)= \left\{ \begin{array}{cl} \lim\limits_{n \to + \infty} \dim(p^nG[p]) & \text{if some} \ \dim(p^nG[p]) \ \text{is finite} \\ + \infty & \text{otherwise} \end{array} \right.

Also, let

T(G)= \left\{ \begin{array}{cl} 0 & \text{if} \ G \ \text{is a torsion group} \\ + \infty & \text{otherwise} \end{array} \right.

Our goal is to prove:

Theorem: Let Q_1 and Q_2 be two divisible groups. Then Q_1 and Q_2 are elementary equivalent if, and only if, T(Q_1)=T(Q_2) and D(p,Q_1)=D(p,Q_2) for every prime p.

Proof. Let S_1 (resp. S_2) be a \omega_1-saturated elementary extension of Q_1 (resp. Q_2).  For i \in \{1,2\}, S_i and Q_i are elementary equivalent, so S_i is a divisible group and we can write

S_i= \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z}[p^{\infty}]^{(\alpha_{p,i})} \oplus \mathbb{Q}^{(\beta_i)} and Q_i= \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z}[p^{\infty}]^{(\mu_{p,i})} \oplus \mathbb{Q}^{(\nu_i)}.


S_i'= \bigoplus\limits_{p \in \mathbb{P}} \mathbb{Z}[p^{\infty}]^{(\tilde{\alpha}_{p,i})} \oplus \mathbb{Q}^{(\tilde{\beta}_i)}


\tilde{\alpha}_{p,i} = \left\{ \begin{array}{cl} \alpha_{p,i} & \text{if} \ D(p,Q_i)<+ \infty \\ \aleph_1 & \text{otherwise} \end{array} \right. and \tilde{\beta}_i= \left\{ \begin{array}{cl} \beta_i & \text{if} \ T(Q_i)=0 \\ \aleph_1 & \text{otherwise} \end{array} \right..

Let a_1,\dots,a_n \in S_i' and b \in S_i \backslash S_i'. Let b_j, 1 \leq j \leq r, denote the components of b contained in some direct summand C_j of S_i not in S_i'.

By construction, S_i have strictly more that \aleph_1 direct summands isomorphic to C_j and S_i' exactly \aleph_1. Therefore, permuting C_j with an isomorphic direct summand of S_i' whose intersection with \{a_1,\dots,a_n\} is trivial, we can construct an automorphism \varphi \in \mathrm{Aut}(S_i) such that \varphi(a_j)=a_j for 1 \leq j \leq n and \varphi(b) \in S_i'.

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

Lemma: If D(p,Q_i)<+ \infty then \alpha_{p,i}=\mu_{p,i}=D(p,Q_i); otherwise, \alpha_{p,i} \geq \aleph_1.

Proof. First, if n, k \geq 1 and S= \{(m_1,\dots,m_k) \in \mathbb{N}^k \backslash\{0\} \mid 0 \leq m_j <p \}, let

\Phi_{n,k} : \exists x_1,\dots,x_k \left[ \bigwedge\limits_{j=1}^k (p^n \mid x_j \wedge px_j=0) \wedge \bigwedge\limits_{(m_1,\dots,m_k) \in S} \sum\limits_{j=1}^k m_jx_j \neq 0 \right]

Clearly, for any abelian group G, \dim(p^nG[p]) \geq k is equivalent to G \models \Phi_{n,k}.

Secondly, notice that \dim(p^nQ_i[p])=\dim(p^n \mathbb{Z}[p^{\infty}]^{(\mu_{p,i})} [p])=\mu_{p,i} and in the same way \dim(p^nS_i[p])=\alpha_{p,i}. Therefore,

D(p,Q_i)= \left\{ \begin{array}{cl} \mu_{p,i} & \text{si} \ \mu_{p,i} < \aleph_0 \\ + \infty & \text{otherwise} \end{array} \right. and D(p,S_i)= \left\{ \begin{array}{cl} \alpha_{p,i} & \text{si} \ \alpha_{p,i}<\aleph_0 \\ + \infty & \text{otherwise} \end{array} \right..

Suppose that D(p,Q_i)<+ \infty. Because Q_i and S_i are elementary equivalent, we deduce that


Now suppose that D(p,Q_i)=+ \infty; in particular, \mu_{p,i} \geq \aleph_0. By contradiction, suppose that \mu_{p,i} < \aleph_1, ie. \mu_{p,i} = \aleph_0, and let B be a basis of the \mathbb{Z}_p-vector space p^nS_i[p]; for n large enough, B has cardinality \mu_{p,i} = \aleph_0.

Therefore, since S_i is \omega_1-saturated, the following type over B=\{b_1,b_2,\dots\} is realized:

\displaystyle p(x)= \bigcup\limits_{k \geq 1} \ \bigcup\limits_{0<m_1, \dots, m_k<p} \left\{ p^n \mid x \wedge px=0 \wedge \sum\limits_{i=1}^{k-1} m_ib_i+m_kx \neq 0 \right\}

contradicting the fact that B is a basis. Consequently, \mu_{p,i} \geq \aleph_1. \square

Lemma: If T(Q_i)=0 then \beta_i=0; otherwise, \beta_i \geq \aleph_1.

Proof. Because Q_i and S_i are elementary equivalent, T(Q_i)=T(S_i). So T(Q_i)=0 implies that S_i is torsion-free, hence \beta_i=0.

Suppose that T(Q_i)=+ \infty. By contradiction, suppose that \beta_i \leq \aleph_0 and let B \subset S_i be a set of \beta_i torsion-free \mathbb{Z}-independent elements. Let

\displaystyle p(x)= \bigcup\limits_{n \geq 1} \ \bigcup\limits_{I \subset B \ \text{finite}} \ \bigcup\limits_{(m_b)_{b \in I} \in \mathbb{Z}^I\backslash \{0\}} \left\{ nx \neq 0 \wedge \sum\limits_{b \in I} m_bb + nx \neq 0 \right\}

If we show that p(x) is finitely satisfiable, p(x) will be satisfiable as a type over B in a \omega_1-saturated group, contradicting the maximality of B. However, noticing that B \subset \mathbb{Q}^{(\beta_i)}, it is clear that any finite subset p_0(x) \subset p(x) is satisfiable in \mathbb{Q}^{(\beta_i)} (here \beta_i \geq 1). \square

Suppose that T(Q_1)=T(Q_2) and D(p,Q_1)=D(p,Q_2) for every prime p. According to the two previous lemmas, we deduce that \tilde{\beta}_1=\tilde{\beta}_2 and \tilde{\alpha}_{p,1}= \tilde{\alpha}_{p,2} for every prime p, hence S_1' \simeq S_2'.

Therefore, Q_1 \equiv S_1 \equiv S_1' \equiv S_2' \equiv S_2 \equiv Q_2.

Conversely, suppose that Q_1 and Q_2 are elementary equivalent. It is obvious that T(Q_1)=T(Q_2). Using the formulae \Phi_{n,k} of our first lemma, we deduce also that D(p,Q_1)=D(p,Q_2) for every prime p. \square

Corollary: \mathbb{Q}^n and \mathbb{Q}^m are elementary equivalent for every n,m \geq 1.

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

Claim: Let \mathfrak{U} be a nonprincipal ultrafilter over \mathbb{N}. Then \mathbb{Z} and \mathbb{Z}^{\mathfrak{U}} are non-isomorphic elementary equivalent groups.

Proof. \mathbb{Z} and \mathbb{Z}^{\mathfrak{U}} are elementary equivalent according to Los theorem.

By contradiction, suppose that \mathbb{Z}^{\mathfrak{U}} has a generator \mu= (m_0,m_1,\dots).

If \mu \neq \pm (1,1,\dots), there exists I \in \mathfrak{U} such that m_i \neq \pm 1 for each i\in I. Therefore, n \mu \neq (1,1,\dots) for every n \in \mathbb{Z}.

On the other hand, if \mu= \pm (1,1,\dots) then \langle \mu \rangle= \mathbb{Z} \subsetneq \mathbb{Z}^{\mathfrak{U}} (for example, (1,2,3,\dots) \in \mathbb{Z}^{\mathfrak{U}} \backslash \mathbb{Z}).

Therefore, \mathbb{Z}^{\mathfrak{U}} is not cyclic. \square

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 Q= \mathbb{Q}^n and let \mathfrak{U} be an ultrafilter over \mathbb{N}; according to Los theorem, Q^{\mathfrak{U}} is elementary equivalent to Q and so is a torsion-free divisible group, hence Q^{\mathfrak{U}} \simeq \mathbb{Q}^{(\alpha)} for some cardinal \alpha.

Because \mathrm{card}(Q^{\mathfrak{U}})= \alpha \cdot \aleph_0, it is sufficient to choose an ultrafilter \mathfrak{U} so that \mathrm{card}(Q^{\mathfrak{U}})= 2^{\aleph_0} to deduce that \alpha=2^{\aleph_0}. Then the isomorphic class of Q^{\mathfrak{U}} does not depend on n.

Therefore, \mathbb{Q}^n and \mathbb{Q}^m have isomorphic ultrapowers and hence are elementary equivalent.

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

Definition: Let \mathfrak{U} be a nonprincipal ultrafilter over a set S. We say that \mathfrak{U} is regular if there exists a family \{M_x \mid x \in S\} \subset \mathfrak{P}(S) of finite subsets such that \{x \mid a M_x \} \in \mathfrak{U} for every a \in S.

Lemma: For any infinite cardinal \kappa, there exists a set S of cardinality \kappa and a regular ultrafilter \mathfrak{U} over S.

Lemma: Let \mathfrak{U} be a regular ultrafilter over some set S and let A be an infinite set. Then:

\mathrm{card}( A^{\mathfrak{U}})= \mathrm{card}(A)^{\mathrm{card}(S)}.

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