A graph is *vertex transitive* if $x \mapsto y$ by an automorphism.


A graph is *generously vertex transitive* if $x \mapsto y \mapsto x$ by an automorphism.


Simple facts:


* GVT $\rightarrow$ unimodular. Just plug in the definition of unimodular to see why.
* Cayley $\not \rightarrow$ GVT. Free product $\mathbb Z\_2 \* \mathbb Z\_3$ is an example.
* GVT $\not \rightarrow$ Cayley. Petersen graph is a finite example.


Question:
**Is there an infinite generously vertex transitive connected graph with finite degree which is not a Cayley graph?**


(Before edition, it was also asked: "Is there a common name for this property?", answered by Chris Godsil in the comments.)


Thanks!



A generously vertex transitive graph which is not Cayley?
Take the graph product $G = P \times \mathbb{Z}$ of the Petersen graph with the infinite path graph. This is clearly infinite, finite-degree, and generously vertex-transitive.


Then we have two distinguishable types of edge, namely $P$-edges (ones which belong to $5$-cycles) and $\mathbb{Z}$-edges (ones which do not). This means that the automorphisms of $G$ must preserve the obvious product structure, and can uniquely be written as compositions of automorphisms of $P$ with automorphisms of $\mathbb{Z}$.


If $G$ were a Cayley graph of some group $H$, then there are three generators corresponding to $P$-edges and two corresponding to $\mathbb{Z}$-edges. The actions of the $P$-generators and $\mathbb{Z}$-generators must, respectively, preserve the $\mathbb{Z}$-coordinate and $P$-coordinate of each vertex.


Hence the $P$-generators and $\mathbb{Z}$-generators each generate disjoint normal subgroups of $H$ (and $H$ is their internal direct product). The normal subgroup generated by the $P$-generators is sharply vertex-transitive on $P$, implying that the Petersen graph is a Cayley graph.


Contradiction.


So $G$ has all the properties you requested.


