On a Conjecture Concerning the Petersen Graph
Robertson has conjectured that the only 3-connected internally 4-connected graph of girth 5 in which every odd cycle of length greater than 5 has a chord is the Petersen graph. We prove this conjecture in the special case where the graphs involved are also cubic. Moreover, this proof does not require the internal-4-connectivity assumption. An example is then presented to show that the assumption of internal 4-connectivity cannot be dropped as an hypothesis in the original conjecture.
We then summarize our results aimed toward the solution of the conjecture in its original form. In particular, let $G$ be any 3-connected internally-4-connected graph of girth 5 in which every odd cycle of length greater than 5 has a chord. If $C$ is any girth cycle in $G$ then $N(C)\backslash V(C)$ cannot be edgeless, and if $N(C) \backslash V(C)$ contains a path of length at least 2, then the conjecture is true. Consequently, if the conjecture is false and $H$ is a counterexample, then for any girth cycle $C$ in $H$, $N(C) \backslash V(C)$ induces a nontrivial matching $M$ together with an independent set of vertices. Moreover, $M$ can be partitioned into (at most) two disjoint non-empty sets where we can precisely describe how these sets are attached to cycle $C$.