Topological 2-generation of automorphism groups of countable ultrahomogeneous graphs

TL;DR

This paper investigates the automorphism groups of countable ultrahomogeneous graphs, demonstrating they contain 2-generated dense subgroups and exploring conditions for their construction with high freedom.

Contribution

It proves the existence of 2-generated dense subgroups in automorphism groups of ultrahomogeneous graphs and details methods for constructing such subgroups with flexible elements.

Findings

Automorphism groups of ultrahomogeneous graphs have 2-generated dense subgroups.

Conditions under which a given automorphism can generate a dense subgroup with another automorphism.

High freedom in choosing the second generator, including specific orbit structures.

Abstract

A countable graph is ultrahomogeneous if every isomorphism between finite induced subgraphs can be extended to an automorphism. Woodrow and Lachlan showed that there are essentially four types of such countably infinite graphs: the random graph; infinite disjoint unions of complete graphs $K_n$ with $n\in \mathbb{N}$ vertices; the $K_n$-free graphs; finite unions of the infinite complete graph $K_{\omega}$; and duals of such graphs. The groups $\operatorname{Aut}(\Gamma)$ of automorphisms of such graphs $\Gamma$ have a natural topology, which is compatible with multiplication and inversion, i.e.\ the groups $\operatorname{Aut}(\Gamma)$ are topological groups. We consider the problem of finding minimally generated dense subgroups of the groups $\operatorname{Aut}(\Gamma)$ where $\Gamma$ is ultrahomogeneous. We show that if $\Gamma$ is ultrahomogeneous, then $\operatorname{Aut}(\Gamma)$ has 2-generated dense subgroups, and that under certain conditions given $f \in \operatorname{Aut}(\Gamma)$ there exists $g\in \operatorname{Aut}(\Gamma)$ such that the subgroup generated by $f$ and $g$ is dense. We also show that, roughly speaking, $g$ can be chosen with a high degree of freedom. For example, if $\Gamma$ is either an infinite disjoint unions of $K_n$ or a finite union of $K_{\omega}$, then $g$ can be chosen to have any given finite set of orbit representatives.