Is it possible for two non-isomorphic groups to satisfy the same first-order sentences and be equicardinal? My question is the same as the title. A proof or a counterexample would be nice.
 A: 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. 
A: My favorite 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. 
