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Semisymmetric cubic graphs of order \(4p^n\). (English) Zbl 1212.05116

Summary: A regular graph is said to be semisymmetric if its full automorphism group acts transitively on its edge set but not on its vertex set. In this paper we prove that for every prime \(p\) (\(\neq 5\)), there is no semisymmetric cubic graph of order \(4p^n\), where \(n \geq 1\).

MSC:

05C25 Graphs and abstract algebra (groups, rings, fields, etc.)
20B25 Finite automorphism groups of algebraic, geometric, or combinatorial structures
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