Let $L$ be a countable language and let $T$ be a complete $\omega$-stable $L$ theory (with infinite models). Assume that $T$ contains a definable group (for ease we may as well assume that $x=x$ is a definable group). It is known that there is no infinite descending chain of definable subgroups.
Is it possible to have a infinite chain of descending subgroups that are necessarily not definable? If so are there sufficient/ necessary conditions for such an occurrence? i.e. the chain is definable in stronger logics $L_{\infty\omega}$ etc?
Edit 1: Taking in to account to the comments I have changed it to $\omega$-stable.
Using Ehrenfeucht-Mostowski models, you can easily prove that any theory with infinite models has a model with an infinite descending chain of elementary submodels. So there's no hope of proving the descending chain condition for arbitrary subgroups in any model-theoretic context.
You might ask, given an $\omega$-stable theory $T$ extending the theory of groups, which models of $T$ have infinite descending chains of subgroups. The answer should be "most of them"... Since $\omega$-stable theories have prime models over sets, it's even really easy to get infinite descending chains of elementary submodels. To avoid this, you at least need to assume that every strongly minimal set in the model has finite dimension and there are only finitely many strongly minimal sets up to non-orthogonality.