13
$\begingroup$

Definition. For an infinite structure $\mathcal{A}$ and $cl : P(dom(\mathcal{A})) \longrightarrow P(dom(\mathcal{A}))$ , we say that $(\mathcal{A}, cl)$ is a structure carrying an $\omega$-homogeneous pregeometry if the following holds:

(a) $(\mathcal{A}, cl)$ is a pregeometry,

(b) $dim(\mathcal{A})$ is the same as the cardinality of $\mathcal{A}$,

(c) If $A \subseteq \mathcal{A}$ is finite and $a, b \in \mathcal{A} \setminus cl(A)$ , then there is $f\in Aut(\mathcal{A}/A)$ such that $f(a) = b$ and for all $B \subseteq \mathcal{A}$, $f(cl(B)) = cl(f(B))$.

Theorem. Suppose that $(G, cl)$ is a group carrying an $\omega$-homogeneous pregeometry. Then either $G$ is commutative or unstable.

Question. Are there non-commutative groups that carry an $\omega$-homogeneous pregeometry?

Giving references is appreciated.

$\endgroup$
6
$\begingroup$

This question has been around for some time.

Connections of homogeneous pregeomtries, quasiminimal structures and regular types have been studied in a recent article of Pilay and Tanovic. They show that the generic type of a homogeneous pregeometry is strongly regular (and generically stable) and conversely a global strongly regular type induces a homogeneous pregeomtry (Theorem 3). Earlier they analyse regular groups and ask if every regular group is commutative (which a variant of your question), see the question after Theorem 2.

As far as I know, the question is still open.

$\endgroup$
  • $\begingroup$ Dear Levon, Thank you for your answer. $\endgroup$ – Mostafa Mirabi Sep 16 '14 at 4:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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