Rankin, R. A., A Campanological Problem in Group Theory, Math. Proc. Camb. Phil. Soc., 44 (1948) 17--25.  Home/Search   Document Not in Database   Summary   Related Articles   Check

....The classification of groups generated by T of size 2 is nontrivial. Clearly if G is isomorphic to a direct product of two cyclic groups then G is unicursally generated by a subset of size 2. The following theorem is a remarkable result of R. A. Rankin on this case. The result in Rankin s paper [8] is more general than the version we state here, and is somewhat based on ideas of Thompson [16] The end result is very simply stated in group theoretic language, even though the problem and the proof are somewhat combinatorial. We give a proof in section 8. Let #g# denote the subgroup ....

....this in detail in section 7. This was asked in 1741 by a bell ringer John Holt, who was able to construct a method of 4998 permutations, but could not obtain a method of 7 = 5040 permutations. He then (naturally ) queried the existence of such an extent. As in Remark 3, it can be shown (see [8], or section 7) that this question is equivalent to: 3, 4, 5, 6, 7 ) generated unicursally by the two permutations (3 4 6 7 5) and (2 4 7) 3 6 5) The first proof that the answer is no is due to Thompson (1886) 16] with some case by case analysis. An insightful proof was given by Rankin ....

[Article contains additional citation context not shown here]

Rankin, R. A., A Campanological Problem in Group Theory, Math. Proc. Camb. Phil. Soc., 44 (1948) 17--25.

....could not obtain a list of 5041 rows. He then (naturally ) queried the existence of such a list. Phrased in our terminology, the question asks whether S 7 is generated unicursally by the three permutations X = 1 2) 3 4) 5 6) Y = 2 3) 4 5) 6 7) and Z = 1 2) 4 5) 6 7) It can be shown (see [7] [8], 15] that this question is equivalent to asking whether A 6 (acting on 2; 3; 4; 5; 6; 7) is generated unicursally by (3 4 6 7 5) and (2 4 7) 3 6 5) The question was rst shown to have a negative answer by Thompson [12] in 1886, with some case by case analysis. Thompson s result combined with ....

....shown to have a negative answer by Thompson [12] in 1886, with some case by case analysis. Thompson s result combined with Holt s 4999 rows show that unicursal generation questions can be rather delicate, and that it is possible for a subset to nearly generate a group unicursally. Rankin s paper [8] is based somewhat on the ideas of Thompson. For more details see [1] 2] 8] 15] The reader can check that Rankin s theorem applies, and shows that A 6 is not generated unicursally by (3 4 6 7 5) 9 and (2 4 7) 3 6 5) This provides a beautiful proof of Thompson s result and an elegant solution ....

[Article contains additional citation context not shown here]

Rankin, R. A., A Campanological Problem in Group Theory, Math. Proc. Camb. Phil. Soc., 44 (1948) 17-25.

....this in detail in section 6. This was asked in 1741 by a bell ringer John Holt, who was able to construct a method of 4998 permutations, but could not obtain a method of 7 = 5040 permutations. He then (naturally ) queried the existence of such an extent. As in Remark 3, it can be shown (see [3], 7] or section 6) that this question is equivalent to: Question A. Is A 6 (acting on 2; 3; 4; 5; 6; 7) generated unicursally by the two permutations (3 4 6 7 5) and (2 4 7) 3 6 5) The first proof that the answer is no is due to Thompson (1886) 5] with some caseby case analysis. An ....

....is equivalent to: Question A. Is A 6 (acting on 2; 3; 4; 5; 6; 7) generated unicursally by the two permutations (3 4 6 7 5) and (2 4 7) 3 6 5) The first proof that the answer is no is due to Thompson (1886) 5] with some caseby case analysis. An insightful proof was given by Rankin (1948) [3], based somewhat on Thompson s ideas. We present Rankin s proof in section 7, in our special case only. Rankin s result is more general and we state a form of it now, which is more general than our case, but not as general as the result in [3] Most of the ideas of the proof can be found in our ....

[Article contains additional citation context not shown here]

Rankin, R. A., A Campanological Problem in Group Theory, Math. Proc. Camb. Phil. Soc., 44 (1948) 17--25.

....in G if and only if u Gamma1 v or v Gamma1 u is in X . C[G; X ] is always vertex transitive and is connected if and only if X [X Gamma1 generates G. It is an open question whether every Cayley graph is hamiltonian. There are generating sets for which the Cayley digraph is not hamiltonian [Ran48]. This is a special case of the more general conjecture of Lov asz that every connected, undirected, vertex transitive graph has a Hamilton path [Lov70] Results on Hamilton cycles are surveyed in [Als81] for vertex transitive graphs and in [Gou91] for general graphs. A survey of Hamilton cycles ....

....disjoint transpositions is an involution. In perhaps the simplest nontrivial case, when S n is generated by three involutions, it is easy to show that if any two of the generators commute, then the Cayley graph is hamiltonian. Cayley graphs arising in change ringing frequently have this property [Ran48, Whi83]. However if no two of the three involutions commute, it is open whether the Cayley graph is hamiltonian. As a specific example, we have not been able to determine whether there is a Gray code for permutations in which successive permutations may differ only by one of the three operations: i) ....

R. A. Rankin. A campanological problem in group theory. Proceedings of the Cambridge Philosophical Society, 44:17--25, 1948.

....following it by one of the transpositions (1 , 2) 2 , 3) n 1 , n) 2) Such arrangements of permutations were given by Johnson [32] and Trotter [53] in the early 1960 s. Other Hamiltonian circuits through all n permutations (satisfying different constraints) arise in bell ringing ([49], 57] 60] Although several other generalizations of Gray codes have appeared ( 3] 22] 33] 37] 42] 48] 52] we believe our version is new. The theorem is proved in 2, and 3 gives some examples. In particular we give specific Gray codes for all the examples ( 1 , 2 , 3 , 2 , ....

....new. The theorem is proved in 2, and 3 gives some examples. In particular we give specific Gray codes for all the examples ( 1 , 2 , 3 , 2 , 3 , 2 , 3 , 2 (m) in dimensions n 3. It is worth remarking that not all Cayley diagrams for groups contain Hamiltonian circuits. Rankin [49] (see also [60] gives a necessary condition that a certain class of groups must satisfy, and uses it to deduce for example that the Cayley diagram for the alternating group A 6 with generators (2 , 4 , 6 , 5 , 3) and (1 , 6 , 3) 2 , 4 , 5) does not contain a Hamiltonian circuit. The problem of ....

R. A. Rankin, A campanological problem in group theory, Proc. Comb. Phil Soc., 44 (1948), 17-25.

No context found.

R. A. Rankin, A campanological problem in group theory, Proc. Camb. Phil. Soc. 44 (1948) 17--25.

Online articles have much greater impact   More about CiteSeer.IST   Add search form to your site   Submit documents   Feedback  

CiteSeer.IST - Copyright Penn State and NEC