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

Let

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

where

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

$\alpha_{p,i}=D(p,S_i)=D(p,Q_i)=\mu_{p,i}$.

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

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.