Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My question is the same as the title. A proof or a counterexample would be nice.

share|cite|improve this question
Free groups have the same first-order theory. So these work, no? The proof is non-trivial. Look up the web page of Zlil Sela for the gory details. – user1729 Jul 4 '14 at 21:49
@user1729: A small nitpick: All free groups of finite rank $\ge 2$. – studiosus Jul 5 '14 at 4:06
@user1729 Equicardinal means the same cardinality. – user107952 Jul 5 '14 at 18:36
up vote 5 down vote accepted

Let $T$ be the theory whose axioms are all sentences true in the integers. Here the language has symbols for $0$, $1$, and for addition and multiplication.

This theory has a countable non-standard model. It is obtained in the usual way, by adding a constant symbol $a$ and axioms that say $a$ is different from $0$, $\pm 1$, and so on. Now use Compactness and Lowenheim-Skolem.

Let $G$ be the additive group of the integers, and let $H$ be the additive group of the non-standard model.

share|cite|improve this answer
+1, but why include multiplication at all? Just start from the theory of $(\mathbb Z,{+})$ and adjoin a non-standard element using the same compactness argument. – Henning Makholm Jul 5 '14 at 14:52
Because the example is more or less instantly recognizable. – André Nicolas Jul 5 '14 at 14:54

My favorate example is the theory of torsion free divisible abelian groups. All of these are first order properties, but with infinitely many axioms. A little reflection leads one to see that such a group is in fact a vector space over the rationals, and thus its isomorphism type is determined by its dimension. For dimensions $1,2,\ldots \aleph_0$ the groups are all countable but nonisomorphic satisfying the same sentences. In uncountable cardinalities the groups size and dimension are equal showing that two such uncountable groups of the same cardinality are isomorphic, which in turn implies that this theory is complete.

share|cite|improve this answer
Why down vote ? It is a perfectly good answer ? – Rene Schipperus Jul 4 '14 at 23:44
it gives exactly what the poster asked – Rene Schipperus Jul 4 '14 at 23:45
The example is a good one. Conceivably the downvoter did not understand. But it is hard to imagine that someone who knows what the words mean does not understand. – André Nicolas Jul 5 '14 at 1:45
Thank you, i feel better now. – Rene Schipperus Jul 5 '14 at 1:49
It is a perfectly good answer. Sometimes people hit the wrong up/down button (which can easily happen on a smartphone). – studiosus Jul 5 '14 at 4:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.