Number of isomorphism classes of countable models of a theory

Whether there are countably or uncountably many isomorphism classes of countable models of a given theory depends on the theory: if the theory is strong enough, there will be only countably many isomorphism classes, if the theory is too weak, there will we be uncountably many.

Consider a single binary relation $R$.

Example 1: When the theory requires that all (or no) two objects are related by $R$ (which is a really strong theory), there will be only countably many isomorphism classes of countable models (corresponding to mere cardinalities).

Example 2: When the theory requires nothing (any two objects may be related by $R$ or not, which is a really weak theory), there will be uncountably many isomorphism classes of countable models.

(How) can one "easily" deduce from the axioms of a (first-order) theory, whether there are countably or uncountably many isomorphism classes of countable models? Is there a royal road?

To be specific: I wonder whether there are countably or uncountably many isomorphism classes of countable groups, i.e. models of group theory, which is a theory with one binary function $\circ$ fulfilling some (weak or strong enough?) axioms.

Added: Examples of theories with countably many countable models are welcome!

• Are you just wanting to know if there are uncountably many countable groups? This is true. (There are only finitely many finitely presented groups though.) I think there is a nice example of B.H. Neumann where he produces a continuum of non-isomorphic countable groups. Cannot seem to find it though. Sep 10, 2013 at 15:54
• Found it! Neumann constructed a continuum of non-isomorphic two-generator groups (and two-generator groups are countable). The construction can be found in de la Harpe's book, Topics in Geometric Group Theory. Note that Neumann's result implies that there does not exists a countable group which contains every other countable group, because a countable group can only have countable many finitely generated subgroups (up to isomorphism). Sep 10, 2013 at 16:10

There is no easy way. You could say this is one of the reasons why Vaught's conjecture is still open. One of the goals of Shelah's classification theory is to provide a context where your question can be solved. Notoriously, the countable case is harder to analyze in general that the uncountable case. If a theory (in a countable language) is unstable, then it has $2^\kappa$ non-isomorphic models in size $\kappa$, for any uncountable $\kappa$. The (rough) idea is that in theories that are not stable we can "code" an infinite order (it is not quite that we can define the order in the language of the structure). Naturally, two models whose coded orders are not isomorphic are themselves not isomorphic. But we have great control over what orders we can realize this way, and this gives us as many non-isomorphic models as possible.
The question of how many models a theory has is the spectrum problem. Building on Shelah's results, you can see the full range of possibilities here. For the countable case, the options are finitely many (not two!), $\omega$, $2^\omega$, or (perhaps?) $\omega_1$ (the point here is that this does not depend on $\mathsf{CH}$).