Existence of non-Cayley Haar graphs

Symmetries of the honeycomb toroidal graphs

“…, each vertex i j ( , ) is adjacent to i j ( , ± 1) and in addition i j ( , ) is adjacent to i j ( + 1, + 1) for all even j (see Figure 2 where the isomorphic graphs GPr(4), HTG(4, 4, 2), HTG (1,16,7), and HTG(2, 8, 2) are depicted). These graphs might also be called the split wreath graphs as n GPr( ) can be obtained from the well-known wreath graph W n ( ) = Cay( × ; {(0, ±1), (1, ±1)}) n 2 by performing the "splitting" construction with respect to the cycle decomposition consisting of all the "natural" 4-cycles of W n ( ) (see [11,Construction 11] for details).…”

“…Let us also point out that, since the index 2 abelian subgroup of of course acts semiregularly with two orbits on the vertex‐set of , the graphs are also so‐called bi‐Cayley graphs on abelian groups (see, e.g., [4,15,16]). Moreover, as the two orbits of are independent sets, they are also so‐called Haar graphs (see, e.g., [6,7]) of abelian groups. One could thus make use of some known results, in particular those of [4] where symmetries of edge‐transitive bi‐Cayley graphs were studied extensively.…”

Symmetries of the Honeycomb toroidal graphs

“…Moreover, as the two orbits of x, y are independent sets, they are also so-called Haar graphs of abelian groups. One could thus make use of some results from [3,5,6,14,15]. Nevertheless, as we will see, in this special setting a direct approach works just fine.…”

“…It is now simply a matter of determining all HTG graphs with this property. Given the special structure of the HTG graphs this is fairly easy to do (for instance, considering a red-blue-red 3-path yields n = 6 or m ≤ 2, and then considering a red-green-red 3-path yields that m > 2 only for HTG (3,6,3)), but instead of describing the argument we simply rely on a more general result from the literature. Namely, it follows from [3,Lemma 5.3] or [11,Theorem 1] (but see also [9]) that the graph Γ must be the Heawood graph, the Pappus graph or the Möbius-Kantor graph (it can easily be verified that the Desargues graph is not a HTG graph).…”