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Automorphism Groups with Cyclic Commutator Subgroup and Hamilton Cycles
, 1997
"... . It has been shown that there is a Hamilton cycle in every connected Cayley graph on each group G whose commutator subgroup is cyclic of primepower order. This paper considers connected, vertextransitive graphs X of order at least 3 where the automorphism group of X contains a transitive subgroup ..."
Abstract

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. It has been shown that there is a Hamilton cycle in every connected Cayley graph on each group G whose commutator subgroup is cyclic of primepower order. This paper considers connected, vertextransitive graphs X of order at least 3 where the automorphism group of X contains a transitive subgroup G whose commutator subgroup is cyclic of primepower order. We show that of these graphs, only the Petersen graph is not hamiltonian. Key words: graph, vertextransitive, Hamilton cycle, commutator subgroup 1 Introduction Considerable attention has been devoted to the problem of determining whether or not a connected, vertextransitive graph X has a Hamilton cycle [1], [8], [14]. A graph X is vertextransitive if some group G of automorphisms of X Preprint submitted to Discrete Mathematics 5 December acts transitively on V (X). If G is abelian, then it is easy to see that X has a Hamilton cycle. Thus it is natural to try to prove the same conclusion when G is "almost abelian." Recalling ...
Hamiltonicity of cubic Cayley graphs
 J. Europ. Math. Soc
"... Following a problem posed by Lovász in 1969, it is believed that every connected vertextransitive graph has a Hamilton path. This is shown here to be true for cubic Cayley graphs arising from groups having a (2, s, 3)presentation, that is, for groups G = 〈a, ba 2 = 1, b s = 1, (ab) 3 = 1, etc. 〉 ..."
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Following a problem posed by Lovász in 1969, it is believed that every connected vertextransitive graph has a Hamilton path. This is shown here to be true for cubic Cayley graphs arising from groups having a (2, s, 3)presentation, that is, for groups G = 〈a, ba 2 = 1, b s = 1, (ab) 3 = 1, etc. 〉 generated by an involution a and an element b of order s ≥ 3 such that their product ab has order 3. More precisely, it is shown that the Cayley graph X = Cay(G, {a, b, b −1}) has a Hamilton cycle when G  (and thus s) is congruent to 2 modulo 4, and has a long cycle missing only two vertices (and thus necessarily a Hamilton path) when G  is congruent to 0 modulo 4. 1 Introductory remarks In 1969, Lovász [21] asked whether every connected vertextransitive graph has a Hamilton path, thus tying together, through this special case of the Traveling Salesman Problem, two seemingly unrelated concepts: traversability and symmetry of graphs. Lovász problem is, somewhat misleadingly, usually referred to as the Lovász conjecture, presumably in view of the fact that, after all these years, a connected vertextransitive graph without a Hamilton path is yet to be produced. Moreover, only four connected vertextransitive graphs (having at least three vertices) not possessing a Hamilton cycle are known to exist: the Petersen graph, the Coxeter graph, and the two graphs obtained from them by replacing each vertex with a triangle. All of these are cubic graphs, suggesting perhaps that no attempt to resolve the above problem can bypass a thorough analysis of cubic vertextransitive graphs. Besides, none of these four graphs is a Cayley graph. This has led to a folklore conjecture that every Cayley graph is hamiltonian. This problem has spurred quite a bit of interest in the mathematical community. In spite of a large number of articles directly and indirectly related to this subject (see