Show that a RAAG acts on a tree of valence 4, acting transitively on vertices, if at least one pair of vertices is not joined by an edge.
I'm trying first to prove that every such RAAG $G$ acts on a Tree. Let $a$ and $b$ be two vertices that aren't joined by an edge, then the group $F$ generated by these two vertices is free. I'm thinking of using the quotient $G/F,$ but I don't know how. Is this even a group? Is it free?
Any help would be appreciated
Let $T$ be the Cayley graph of the subgroup $F$ generated by $a$ and $b$ (wrt these two generators). Since $F$ is free on $\{a, b \}$, $T$ is a tree of valence $4$, on which $F$ already acts by an action which is transitive on the vertices. We need to extend this action to an action of $G$. In other words, we need to extend a morphism from $F$ to $Isom(T)$ to a morphism from $G$ to $Isom(T)$. We do this by sending all the generators of $G$ different from $a$ and $b$ to the identity. Using the presentation of $G$, we see that this gives a well-defined morphism, corresponding to a well-defined action of $G$ on $T$, which is transitive on the vertices, since its restriction to $F$ is.