Every connected $\omega$-stable group has a zero element?

Question

Let $G$ be a connected $\omega$-stable group and $p$ its unique generic. Let $a$ be a realization of $p$, $G\prec G_1$ an elementary extension containing $a$ and $q$ the non forking extension of $p$ to $G_1$. So $p=tp(a/G)$, $q=tp(a/G_1)$.

Since $p$ is generic, so is $q$. Hence, $Stab(q)=G_1$, or in other words, $\forall x\in G_1\, xa=a$. Hence there is also such element in $G$.

Where did I go wrong?

Yatir

Answer 1

I'm not very familiar with the theory of $\omega$-stable groups, but I think you've made two mistakes:

1. You've assumed that $a\models q$, for which there is no reason ($a$ is just any realization of $p$, of which $q$ is an extension, but by no means the only one). In fact, this is impossible, as any element of $G_1$ has isolated type over $G_1$ and generic type can't be isolated (because isolated types have rank $0$ and the group is infinite, and so it has a nonzero rank).
2. Another mistake, though possibly easily fixed, is that you tacitly assumed that $G_1$ is connected. Maybe you can do that, but I don't know that and you've certainly not said so.