Nilpotent 1-factorizations of the complete graph

Abstract

For which groups G of even order 2n does a 1-factorization of the complete graph on 2n veritces exist with the property of admitting G as a sharply vertex-transitive automorphism group? The complete answer is still unknown. Using the definition of a starter in G introduced in [M. Buratti "Abelian 1-factorizations of the complete graph" Europ. J Comb. 2001, pp.291-295], we give a positive answer for new classes of groups; for example, the nilpotent groups with either an abelian Sylow 2-subgroup or a non-abelian Sylow 2-subgroup which possesses a cyclic subgroup of index 2. Further considerations are given in case the automorphism group G fixes a 1-factor

On sharply vertex transitive 2-factorizations of the complete graph

“…naturally arises. Despite the fact that the analogous question for groups of even order and regular 1-factorizations does not seem easy to solve, [11,3,6,15,13], the answer to our question is quite simple if no additional restriction is made. In particular a G-regular 2-factorization in which each factor is fixed by G exists as shown below.…”

“…In some recent papers (see [6,3,13,4]) regular 1-factorizations of K G were studied for several groups G of even order, despite the fact that the existence is not guaranteed for an arbitrary group G. A regular 1-factorization of K G has been proved to be equivalent to the concept of a starter in a group of even order which was introduced in [6].…”

“…Consider the following three cycles of respective lengths 14, 7, 7, and with vertices in G = Z 21 :A = (0,7,3,10,6,13,9,16,12, 19,15,1, 18,4); B =(2,5,8,11,14, 17, 20); C = (0, 1, 3, 11, 16, 6, 12).The stabilizer ofA and B is 3 , i.e., the subgroup of G of order 7. Instead, C has trivial stabilizer.…”

Cyclic Uniform 2-Factorizations of the Complete Multipartite Graph

“…Case 3: Let m = 10 and n = 6, hence t = 1. We obtain [12,18], [11,19], [10,20], [9,21], [8,22], [7,23], [6,24], [5,25], [4,26], [3,27], [2,28]…”

“…Following the proof of Proposition 6.1 we construct the sets necessary to obtain the 2-starter for two choices of m and n. Case 1: Let m = 3 and n = 32, hence t = 1 and v = 5. We obtain B = { [8,12], [6,14], [2,18], [0, 20]} ∪ { [11,13], [7,17], [5,19] [16,30], [14,32], [12,34], [10,36], [6,40], [4,42], [2,44] [17,33], [13,37], [11,39], [9,41], [7,43], [3,47] [2,6], [3,15], [4,9]}, S 2 = (0, 10,9,29 [2,10], [3,9], [4,…”