Cyclic or elementary abelian Covers of K 4
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1 Cyclic or elementary abelian Covers of K 4 Yan-Quan Feng Mathematics, Beijing Jiaotong University Beijing , P.R. China Summer School, Rogla, Slovenian
2 Outline 1 Question 2 Main results 3 Automorphisms of K 4 4 Ideas for the proof of main results 5 References
3 Question Given a base graph X and a transformation group K, find all non-isomorphic covers such that a group G of automorphisms of X lifts. In this talk, we will give an example by letting X = K 4, K cyclic or elementary abelian, and G an arc-transitive group of K 4. Furthermore, an application of the above result is given to classify cubic symmetric graphs of certain orders.
4 Theorem 1: Let K be a cyclic or an elementary abelian group, and let X be a connected K -covering of the complete graph K 4 whose fibre-preserving subgroup is arc-transitive. Then, (1) if K is cyclic then X is 2-regular and isomorphic to the complete graph K 4, the 3-dimensional Hypercube Q 3, or the generalized Petersen graph P(8, 3). (2) If K is an elementary abelian group Z m p (p prime, m 2), then X is 2-regular and isomorphic to one of EK p 3. Theorem 2: Let X be a connected cubic symmetric graph of order 4p or 4p 2 (p a prime). Then X is isomorphic to the 3-dimensional Hypercube Q 3, the generalized Petersen graph P(8, 3) or P(10, 3), the Dodecahedron (3-regular), or the Coxeter graph C 28 (3-regular).
5 Y.-Q. Feng, J.H. Kwak / Journal of Combinatorial Theory, Series B 97 (2007) Automorphisms of K 4,4 α Fig. = 3. (ab)(cd) The complete graph β = K(bcd) 4 with voltage assignment γ = (bc) φ. A 4 = α, β S 4 = α, β, γ Aut(K 4 ) = S 4 able 2 undamental cycles and their images with voltages on K 4 1 Fundamental cycles: abc, acd, adb φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) 2 Assume that N is the fibre-preserving group and the covering graph is N-arc-transitive. bc z 1 bad z 3 acd z 2 acb z 1 cd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 db z 3 bca z 1 abc z 1 adc z 2 3 Let N project to L. Then K 4 is L-arc-transitive. Thus, L = A 4 or S 4. Hence, α and β lifts.
6 Proof for Theorem 1 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 Table: Fundamental cycles and their images with voltages on K 4 1 K = z 1, z 2, z 3 = Z n or Z m p. L = A 4 = α, β lifts.
7 Proof for Theorem 1 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 Table: Fundamental cycles and their images with voltages on K 4 1 K = z 1, z 2, z 3 = Z n or Z m p. L = A 4 = α, β lifts. 2 α: φ(c) α = φ(c α ), where C ranges over fundamental cycles. Similarly, we can define β and γ.
8 Proof for Theorem 1 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 Table: Fundamental cycles and their images with voltages on K 4 1 K = z 1, z 2, z 3 = Z n or Z m p. L = A 4 = α, β lifts. 2 α: φ(c) α = φ(c α ), where C ranges over fundamental cycles. Similarly, we can define β and γ. 3 Since L lifts, α can be extended to an automorphism of K, say α. Similarly, we can define β and γ if γ lifts.
9 Proof for Theorem 1 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 Table: Fundamental cycles and their images with voltages on K 4 1 K = z 1, z 2, z 3 = Z n or Z m p. L = A 4 = α, β lifts. 2 α: φ(c) α = φ(c α ), where C ranges over fundamental cycles. Similarly, we can define β and γ. 3 Since L lifts, α can be extended to an automorphism of K, say α. Similarly, we can define β and γ if γ lifts. 4 z β 1 = z 2 and z β 2 = z 3 o(z 1 ) = o(z 2 ) = o(z 3 ).
10 K = Z n C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 K = Z n = z 1, z 2, z 3 = z 1 = z 2 = z 3. Let z 1 = 1.
11 K = Z n C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 K = Z n = z 1, z 2, z 3 = z 1 = z 2 = z 3. Let z 1 = 1. 2 Let 1 β = k. Then (k, n) = 1. z β 1 = z 2, z β 2 = z 3, z β 3 = z 1 z 2 = k, z 3 = k 2, k 3 = 1 in Z n.
12 K = Z n C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 K = Z n = z 1, z 2, z 3 = z 1 = z 2 = z 3. Let z 1 = 1. 2 Let 1 β = k. Then (k, n) = 1. z β 1 = z 2, z β 2 = z 3, z β 3 = z 1 z 2 = k, z 3 = k 2, k 3 = 1 in Z n. 3 Let 1 α = l. z1 α = z 3, z3 α = z 1 l = k 2, lk 2 = 1 l = k 2 = l 1 l 2 = 1, k = k 4 = l 2 = 1, l = k 2 = 1.
13 K = Z n C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 K = Z n = z 1, z 2, z 3 = z 1 = z 2 = z 3. Let z 1 = 1. 2 Let 1 β = k. Then (k, n) = 1. z β 1 = z 2, z β 2 = z 3, z β 3 = z 1 z 2 = k, z 3 = k 2, k 3 = 1 in Z n. 3 Let 1 α = l. z1 α = z 3, z3 α = z 1 l = k 2, lk 2 = 1 l = k 2 = l 1 l 2 = 1, k = k 4 = l 2 = 1, l = k 2 = 1. 4 z 1 = z 2 = z 3 = 1. z α 2 = z 1 z 2 z 3 4 = 0 n = 2, 4.
14 K = Z n C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 K = Z n = z 1, z 2, z 3 = z 1 = z 2 = z 3. Let z 1 = 1. 2 Let 1 β = k. Then (k, n) = 1. z β 1 = z 2, z β 2 = z 3, z β 3 = z 1 z 2 = k, z 3 = k 2, k 3 = 1 in Z n. 3 Let 1 α = l. z1 α = z 3, z3 α = z 1 l = k 2, lk 2 = 1 l = k 2 = l 1 l 2 = 1, k = k 4 = l 2 = 1, l = k 2 = 1. 4 z 1 = z 2 = z 3 = 1. z α 2 = z 1 z 2 z 3 4 = 0 n = 2, 4. 5 For n = 2, 4, α, β and γ extends to automorphisms of Z n. It follows that X = Q 3 or P(8, 3) because there is unique connected cubic symmetric graph of order 8 or 16.
15 K = Z m p, m 2 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 K = z 1, z 2, z 3 m = 2 or 3.
16 K = Z m p, m 2 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 K = z 1, z 2, z 3 m = 2 or 3. 2 m = 2 K = Z 2 p = a b. z β 1 = z 2, z β 2 = z 3, z β 3 = z 1 z 1 and z 2 linearly independent. z 1 = a, z 2 = b.
17 K = Z m p, m 2 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 K = z 1, z 2, z 3 m = 2 or 3. 2 m = 2 K = Z 2 p = a b. z β 1 = z 2, z β 2 = z 3, z β 3 = z 1 z 1 and z 2 linearly independent. z 1 = a, z 2 = b. 3 z 3 = ka + lb a = z 1 = z β 3 = kb + l(ka + lb) lk = 1, k + l 2 = 0 in Z p l 3 = 1, k = l 1.
18 K = Z m p, m 2 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 K = z 1, z 2, z 3 m = 2 or 3. 2 m = 2 K = Z 2 p = a b. z β 1 = z 2, z β 2 = z 3, z β 3 = z 1 z 1 and z 2 linearly independent. z 1 = a, z 2 = b. 3 z 3 = ka + lb a = z 1 = z β 3 = kb + l(ka + lb) lk = 1, k + l 2 = 0 in Z p l 3 = 1, k = l 1. 4 z 3 = ka + lb a = z 1 = z3 α = k(ka + lb) l[(k + 1)a + (l + 1)b] k 2 kl l = 1, kl l 2 l = 0 2l = 2, l 2 + l 1 = 0, contradiction.
19 K = Z m p, m 2 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 m = 3 z 1, z 2, z 3 are linearly independent α, β, γ can be extended to automorphisms of K Aut( X) contains a 2-regular subgroup B.
20 K = Z m p, m 2 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 m = 3 z 1, z 2, z 3 are linearly independent α, β, γ can be extended to automorphisms of K Aut( X) contains a 2-regular subgroup B. 2 Suppose that X is 3-arc-transitive. By [2], X is 3-regular. Let A = Aut( X) A : B = 2 and B A.
21 K = Z m p, m 2 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 m = 3 z 1, z 2, z 3 are linearly independent α, β, γ can be extended to automorphisms of K Aut( X) contains a 2-regular subgroup B. 2 Suppose that X is 3-arc-transitive. By [2], X is 3-regular. Let A = Aut( X) A : B = 2 and B A. 3 If p = 2 or 3, by Conder and Dobcsányi [1], X is 2-regular, a contradiction.
22 K = Z m p, m 2 C φ(c) C α φ(c α ) C β φ(c β ) C γ φ(c γ ) abc z 1 bad z 3 acd z 2 acb z 1 acd z 2 bdc z 1 z 2 z 3 adb z 3 abd z 3 adb z 3 bca z 1 abc z 1 adc z 2 1 m = 3 z 1, z 2, z 3 are linearly independent α, β, γ can be extended to automorphisms of K Aut( X) contains a 2-regular subgroup B. 2 Suppose that X is 3-arc-transitive. By [2], X is 3-regular. Let A = Aut( X) A : B = 2 and B A. 3 If p = 2 or 3, by Conder and Dobcsányi [1], X is 2-regular, a contradiction. 4 If p 5 then K Syl p (B) K A K 4 is 3-arc-transitive, a contradiction. Thus, X is 2-regular.
23 Proof for Theorem 2 Theorem 2: Let X be a connected cubic symmetric graph of order 4p or 4p 2 (p a prime). Then X is isomorphic to the 3-dimensional Hypercube Q 3, the generalized Petersen graph P(8, 3) or P(10, 3), the Dodecahedron (3-regular), or the Coxeter graph C 28 (3-regular). 1 Let X be a connected cubic symmetric graph of order 4m, m = p or p 2. Let A = Aut(X). By Tutte [6], A 192m.
24 Proof for Theorem 2 Theorem 2: Let X be a connected cubic symmetric graph of order 4p or 4p 2 (p a prime). Then X is isomorphic to the 3-dimensional Hypercube Q 3, the generalized Petersen graph P(8, 3) or P(10, 3), the Dodecahedron (3-regular), or the Coxeter graph C 28 (3-regular). 1 Let X be a connected cubic symmetric graph of order 4m, m = p or p 2. Let A = Aut(X). By Tutte [6], A 192m. 2 If p < 17, by Conder and Dobcsányi [1], V (X) = 8, 16, 20, 28, which are Q 3, P(8, 3), the Dodecahedron, P(10, 3) or the Coxeter graph C 28.
25 Proof for Theorem 2 Theorem 2: Let X be a connected cubic symmetric graph of order 4p or 4p 2 (p a prime). Then X is isomorphic to the 3-dimensional Hypercube Q 3, the generalized Petersen graph P(8, 3) or P(10, 3), the Dodecahedron (3-regular), or the Coxeter graph C 28 (3-regular). 1 Let X be a connected cubic symmetric graph of order 4m, m = p or p 2. Let A = Aut(X). By Tutte [6], A 192m. 2 If p < 17, by Conder and Dobcsányi [1], V (X) = 8, 16, 20, 28, which are Q 3, P(8, 3), the Dodecahedron, P(10, 3) or the Coxeter graph C Let p 17. We aim to deduce a contradiction.
26 Proof for Theorem 2 Theorem 2: Let X be a connected cubic symmetric graph of order 4p or 4p 2 (p a prime). Then X is isomorphic to the 3-dimensional Hypercube Q 3, the generalized Petersen graph P(8, 3) or P(10, 3), the Dodecahedron (3-regular), or the Coxeter graph C 28 (3-regular). 1 Let X be a connected cubic symmetric graph of order 4m, m = p or p 2. Let A = Aut(X). By Tutte [6], A 192m. 2 If p < 17, by Conder and Dobcsányi [1], V (X) = 8, 16, 20, 28, which are Q 3, P(8, 3), the Dodecahedron, P(10, 3) or the Coxeter graph C Let p 17. We aim to deduce a contradiction. 4 Let P Syl p (A). Then Syl p (A) = np + 1 = A : N A (P). If np + 1 = 1 P A X is a regular covering of K 4, but no such graph X by Theorem 1, a contradiction.
27 Conditions: V (X) = 4p 2 and P A 1 Thus, np + 1 > 1. p 17 np , A 192m np (m = p or p 2 ).
28 Conditions: V (X) = 4p 2 and P A 1 Thus, np + 1 > 1. p 17 np , A 192m np (m = p or p 2 ). 2 np = 191, 95, 47, 23, 63, 31 either p = 191, 47, 31 or 23 and n = 1, or p = 19 and n = 5.
29 Conditions: V (X) = 4p 2 and P A 1 Thus, np + 1 > 1. p 17 np , A 192m np (m = p or p 2 ). 2 np = 191, 95, 47, 23, 63, 31 either p = 191, 47, 31 or 23 and n = 1, or p = 19 and n = 5. 3 Again by Conder and Dobcsányi [1], there is no such X of order 4p.
30 Conditions: V (X) = 4p 2 and P A 1 Thus, np + 1 > 1. p 17 np , A 192m np (m = p or p 2 ). 2 np = 191, 95, 47, 23, 63, 31 either p = 191, 47, 31 or 23 and n = 1, or p = 19 and n = 5. 3 Again by Conder and Dobcsányi [1], there is no such X of order 4p. 4 Thus, V (X) = 4p 2.
31 Conditions: V (X) = 4p 2 and P A 1 Thus, np + 1 > 1. p 17 np , A 192m np (m = p or p 2 ). 2 np = 191, 95, 47, 23, 63, 31 either p = 191, 47, 31 or 23 and n = 1, or p = 19 and n = 5. 3 Again by Conder and Dobcsányi [1], there is no such X of order 4p. 4 Thus, V (X) = 4p 2. 5 n = 1 and p = 191, 47, 31 or 23. Let H = N A (P). Considering the right multiplication action of A on right cosets of H in A A/H A (p + 1)! p H A P H A = p P H A A X P HA is a connected cubic symmetric graph of order 4p, a contradiction.
32 Conditions: V (X) = 4p 2 and P A 1 p = 19 and n = 5 A : N A (P) = np + 1 = A X is 4-arc-transitive A = or
33 Conditions: V (X) = 4p 2 and P A 1 p = 19 and n = 5 A : N A (P) = np + 1 = A X is 4-arc-transitive A = or By Gorenstein [3, pp.12-14], a simple {2, 3, q}-group exists if and only if q = 5, 7, 13 or 17 A is solvable for a minimal normal subgroup N of A, N = Z s r, r = 2, 3, 19.
34 Conditions: V (X) = 4p 2 and P A 1 p = 19 and n = 5 A : N A (P) = np + 1 = A X is 4-arc-transitive A = or By Gorenstein [3, pp.12-14], a simple {2, 3, q}-group exists if and only if q = 5, 7, 13 or 17 A is solvable for a minimal normal subgroup N of A, N = Z s r, r = 2, 3, r = 3 N A v, impossible. r = 2 or 19 X N is a connected cubic 4-arc-transitive graph of order 722 or 38, which is impossible by Conder and Dobcsányi [1].
35 M.D.E. Conder, P. Dobcsǎnyi, Trivalent symmetric graphs on up to 768 vertices, J. Combin. Math. Combin. Comput. 40(2002) M.D.E. Conder, R. Nedela, A refined classification of symmetric cubic graphs, J. Algebra 322(2009) D. Gorenstein, Finite Simple Groups, New York and London: Plenum Press, P. Lorimer, Vertex-transitive graphs: symmetric graphs of prime valency, J. Graph Theory 8(1984), A. Malnič, Group actions, coverings and lifts of automorphisms, Discrete Math. 182(1998), W.T. Tutte, On the symmetry of cubic graphs, Canad. J. Math. 11(1959),
36 Thank you!
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