A classification of pentavalent arc-transitive bicirculants

Abstract

A bicirculant is a graph admitting an automorphism with two cycles of equal length in its cycle decomposition. A graph is said to be arc-transitive if its automorphism group acts transitively on the set of its arcs. All cubic and tetravalent arc-transitive bicirculants are known, and this paper gives a complete classification of connected pentavalent arc-transitive bicirculants. In particular, it is shown that, with the exception of seven particular graphs, a connected pentavalent bicirculant is arc-transitive if and only if it is isomorphic to a Cayley graph Cay(D 2n , {b, ba, ba r +1 , ba r 2 +r +1 , ba r 3 +r 2 +r +1 }) on the dihedral group D 2n = a, b | a n = b 2 = baba = 1 , where r ∈ Z * n such that r 4 + r 3 + r 2 + r + 1 ≡ 0 (mod n).

On Certain Edge-Transitive Bicirculants

“…Similarly, the classification of tetravalent arc-transitive bicirculants is obtained by combining the results of [11,12,13]. Recently, the classification of pentavalent arc-transitive bicirculants was also obtained [1,2].One of the most important steps in these classifications is the classification of the arc-transitive bicirculants (which for these valences actually coincide with the edge-transitive ones), for which at least one of the induced subgraphs on the two orbits of the corresponding semiregular automorphism is a cycle. The corresponding graphs for valences 3, 4 and 5 are called generalized Petersen graphs, Rose window graphs and Tabačjn graphs, respectively.…”

Symmetry properties of generalized graph truncations

“…Remark. It is well known that there exist infinitely many 5-valent arc-transitive graphs (see for instance [2]), and so Theorem 6.3 implies that there exist infinitely many cubic vertex-transitive graphs of girth 5.…”

Arc-transitive cyclic and dihedral covers of pentavalent symmetric graphs of order twice a prime

“…Proof. If |Ω| = 2 or 4, then G ≤ S 2 ∼ = AGL (1,2) or G ≤ S 4 ∼ = AGL(2, 2), respectively. Let |Ω| ≥ 6 and write N := soc(G).…”

“…By Proposition 3.4, Γ ∼ = K 6 for p = 3, K 5,5 for p = 5, or CD p for 5 | (p − 1). If Γ ∼ = K 6 then Γ ∼ = K 6,6 − 6K 2 or I 12 by [29, Theorem 1.1] (also see the proof in [2,Theorem 3.6]). In the following, we assume that p ≥ 5.…”

“…By Lemmas 4.2 -4.5, Γ ∼ = CD np or CGD i mp e ×p (1 ≤ i ≤ 5), as required. Now, assume L/K ∼ = Z 2 5 . Then p = 5 and B/K ∼ = Z 2 5 Z 8 .…”