Cyclic or elementary abelian Covers of K 4 Yan-Quan Feng Mathematics, Beijing Jiaotong University Beijing 100044, P.R. China Summer School, Rogla, Slovenian 2011-06
Outline 1 Question 2 Main results 3 Automorphisms of K 4 4 Ideas for the proof of main results 5 References
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.
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).
Y.-Q. Feng, J.H. Kwak / Journal of Combinatorial Theory, Series B 97 (2007) 627 646 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.
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.
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.
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.
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 ).
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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. 3 Let p 17. We aim to deduce a contradiction.
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. 3 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.
Conditions: V (X) = 4p 2 and P A 1 Thus, np + 1 > 1. p 17 np + 1 18, A 192m np + 1 192 (m = p or p 2 ).
Conditions: V (X) = 4p 2 and P A 1 Thus, np + 1 > 1. p 17 np + 1 18, A 192m np + 1 192 (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.
Conditions: V (X) = 4p 2 and P A 1 Thus, np + 1 > 1. p 17 np + 1 18, A 192m np + 1 192 (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.
Conditions: V (X) = 4p 2 and P A 1 Thus, np + 1 > 1. p 17 np + 1 18, A 192m np + 1 192 (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.
Conditions: V (X) = 4p 2 and P A 1 Thus, np + 1 > 1. p 17 np + 1 18, A 192m np + 1 192 (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.
Conditions: V (X) = 4p 2 and P A 1 p = 19 and n = 5 A : N A (P) = np + 1 = 96 4 19 2 3 2 3 A X is 4-arc-transitive A = 2 5 3 19 2 or 2 6 3 19 2.
Conditions: V (X) = 4p 2 and P A 1 p = 19 and n = 5 A : N A (P) = np + 1 = 96 4 19 2 3 2 3 A X is 4-arc-transitive A = 2 5 3 19 2 or 2 6 3 19 2. 2 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.
Conditions: V (X) = 4p 2 and P A 1 p = 19 and n = 5 A : N A (P) = np + 1 = 96 4 19 2 3 2 3 A X is 4-arc-transitive A = 2 5 3 19 2 or 2 6 3 19 2. 2 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. 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].
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