Flag-transitive non-symmetric 2-designs with (r, λ) = 1 and alternating socle

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1 Flag-tranitive non-ymmetric -deign with (r, λ = 1 and alternating ocle Shenglin Zhou, Yajie Wang School of Mathematic South China Univerity of Technology Guangzhou, Guangdong , P. R. China lzhou@cut.edu.cn Submitted: Sep 4, 014; Accepted: Mar 5, 015; Publihed: Apr 14, 015 Mathematic Subject Claification: 05B5, 0B5 Abtract Thi paper deal with flag-tranitive non-ymmetric -deign with (r, λ = 1. We prove that if D i a non-trivial non-ymmetric -(v, k, λ deign with (r, λ = 1 and G Aut(D i flag-tranitive with Soc(G = A n for n 5, then D i a -(6, 3, deign, the projective pace P G(3,, or a -(10, 6, 5 deign. Keyword: non-ymmetric deign; automorphim group; flag-tranitive; alternating group 1 Introduction Thi paper i inpired by a paper of P. H. Ziechang [17] on flag-tranitive -deign with (r, λ = 1. He proved in [17, Theorem] that if G i a flag-tranitive automorphim group of a -deign with (r, λ = 1 and T i a minimal normal ubgroup of G, then T i abelian, or imple and C G (T = 1. From [6,.3.7(a](ee alo Lemma 3 below we know that if G Aut(D i flag-tranitive with (r, λ = 1 then G act primitively on P. It follow that G i an affine or almot imple group. So it i poible to claify thi type of deign by uing the claification of finite primitive permutation group, epecially for the cae of G i almot imple, with alternating ocle. A -deign with λ = 1 i alo called a finite linear pace. In 001, A. Delandtheer [5] claified flag-tranitive finite linear pace, with alternating ocle, ee Lemma 4 below. Recently, in [16], Zhu, Guan and Zhou have claified flag-tranitive ymmetric deign with (r, λ = 1 and alternating ocle. Thi paper i a continuation of [5, 16] and a contribution to the cae where D i a non-ymmetric -deign. Supported by the NSFC (No and the NSF of Guangdong Province (No.S the electronic journal of combinatoric ( (015, #P.6 1

2 A -(v, k, λ deign D i a pair (P, B where P i a v-et and B i a collection of b k- ubet (called block of P uch that each point of P i contained in exactly r block and any -ubet of P i contained in exactly λ block. The number v, b, r, k, λ are parameter of the deign. It i well known that bk = vr, b v, and o r k. The complement D of a -(v, k, λ deign D = (P, B i a -(v, v k, b r + λ deign (P, B, where B = {P \ B B B}. A -(v, k, λ deign i ymmetric if b = v (or equivalently, r = k, otherwie i non-ymmetric. Thi paper deal only with non-trivial non-ymmetric deign, thoe with < k < v 1. So that we have b > v and r > k. An automorphim of D i a permutation of P which leave B invariant. The full automorphim group of D, denoted by Aut(D, i the group coniting of all automorphim of D. A flag of D i a point-block pair (x, B uch that x B. For G Aut(D, G i called flag-tranitive if G act tranitively on the et of flag, and point-primitive if G act primitively on P. A deign D i antiflag tranitive if G Aut(D act tranitively on the et {(x, B x / B} P B of antiflag of D. It i eaily known that G Aut(D i flag-tranitive on D if and only if G i antiflag tranitive on D. Flag-tranitivity i one of many condition that can be impoed on the automorphim group G of a deign D. Lot of work have been done on flag-tranitive ymmetric -deign, ee [11, 1, 13, 15], for example. Although there exit large familie of non-ymmetric - deign, le i known when D i non-ymmetric admitting a flag-tranitive automorphim group. The aim of thi paper i to claify the flag-tranitive non-ymmetric -deign with (r, λ = 1, whoe automorphim group i almot imple with an alternating group a ocle. Thi can be viewed a a firt tep toward a claification of non-ymmetric -deign with (r, λ = 1. The main reult of thi paper i the following. Theorem 1. Let D be a non-ymmetric -(v, k, λ deign with (r, λ = 1, where r i the number of block through a point. If G Aut(D i flag-tranitive with alternating ocle, then up to iomorphim (D, G i one of the following: (i D i a unique -(15, 3, 1 deign and G = A 7 or A 8. (ii D i a unique -(6, 3, deign and G = A 5. (iii D i a unique -(10, 6, 5 deign and G = A 6 or S 6. Thi, together with [16, Theorem 1.1], yield the following. the electronic journal of combinatoric ( (015, #P.6

3 Corollary. If D i a -(v, k, λ deign with (r, λ = 1, which admit a flag-tranitive automorphim group G with alternating ocle, then D i a -(6, 3, deign, a -(10, 6, 5 deign, the projective pace P G(3, or P G (3,. The paper i organized a follow. In Section, we introduce ome preliminary reult that are important for the remainder of the paper. In Section 3, we complete the proof of Theorem 1 in three part. Preliminarie Lemma 3. ([6,.3.7(a] Let D be a -(v, k, λ deign with flag-tranitive automorphim group G. If (r, λ = 1 then G i point-primitive. The following reult due to A. Delandtheer [5] give the claification of flag-tranitive finite linear pace with alternating ocle. Lemma 4. Let S be a finite non-trivial linear pace having an automorphim group G which act flag-tranitively on S. If A n G Aut(A n with n 5, then S = P G(3, and G = A 7 or A 8 = P SL4 (. Lemma 5. Let D be a -(v, k, λ deign. Then (i bk(k 1 = λv(v 1. (ii r = λ(v 1. In particular, if (r, λ = 1 then r v 1 and (r, v = 1. k 1 Proof. Counting in two way triple (α, β, B, where α and β are ditinct point and B i a block incident with both of them, give (i. Part (ii follow from the baic equation bk = vr and Part (i. Lemma 6. If D i a non-ymmetric -(v, k, λ deign and G i a flag-tranitive pointprimitive automorphim group of D, then (i r > λv, and G x 3 > λ G, where x P ; (ii r λd i, where d i i any ubdegree of G. Furthermore, if (r, λ = 1 then r d i. Proof. (i The equality r = λ(v 1 implie λv = r(k 1+λ < r(r 1+λ = r r+λ, k 1 and by r > k > λ we have r > λv. Combining thi with v = G : G x and r G x give G x 3 > λ G. Part (ii wa proved in [4]. Lemma 7. ([9, p.366] If G i A n or S n, acting on a et Ω of ize n, and H i any maximal ubgroup of G with H A n, then H atifie one of the following: (i H = (S k S l G, with n = k + l and k l (intranitive cae; (ii H = (S k S l G, with n = kl, k > 1 and l > 1 (imprimitive cae; the electronic journal of combinatoric ( (015, #P.6 3

4 (iii H = AGL k (p G, with n = p k and p prime (affine cae; (iv H = (T k.(outt S k G, with T a nonabelian imple group, k and n = T k 1 (diagonal cae; (v H = (S k S l G, with n = k l, k 5 and l > 1 (wreath cae; (vi T H Aut(T, with T a nonabelian imple group, T A n and H acting primitively on Ω (almot imple cae. Remark 8. Thi lemma doe not deal with the group M 10, P GL (9 and P ΓL (9 that have A 6 a ocle. Thee exceptional cae will be handled in the firt part of Section 3. Lemma 9. [10, Theorem (b(i] Let G be a primitive permutation group of odd degree n on a et Ω with imple ocle X := Soc(G, and let H = G x, x Ω. If X = A c, an alternating group, then one of the following hold: (i H i intranitive, and H = (S a S c a G where 1 a < 1 c; (ii H i tranitive and imprimitive, and H = (S a S c/a G where a > 1 and a c; (iii H i primitive, n = 15 and G = A 7. Lemma 10. [7, Theorem 5.A] Let G := Alt(Ω where n := Ω 5, and let be an integer with 1 n. Suppoe that, K G ha index G : K < ( n. Then one of the following hold: (i For ome Ω with < we have G ( K G { } ; (ii n = m i even, K i imprimitive with two block of ize m, and G : K = 1 ( n m ; or (iii one of ix exceptional cae hold where: (a K i imprimitive on Ω and (n,, G : K = (6, 3, 15; (b K i primitive on Ω and (n,, G : K, K = (5,, 6, 5 :, (6,, 6, P SL (5, (7,, 15, P SL 3 (, (8,, 15, AGL 3 (, or (9, 4, 10, P ΓL (8. Remark 11. (1 From part (i of Lemma 10 we know that K contain the alternating group G ( = Alt(Ω \ of degree n + 1. ( A reult imilar to Lemma 10 hold for the finite ymmetric group Sym(Ω which can be found in [7, Theorem 5.B]. We will alo need ome elementary inequalitie. Lemma 1. Let and t be two poitive integer. (i If t > 7, then ( +t > t 4 > t. the electronic journal of combinatoric ( (015, #P.6 4

5 (ii If 6 and t, then ( 1(t 1 > 4( t implie (t 1 > ( + 1 4( t. (iii If t 6 and, then ( 1(t 1 > 4( t implie ( 1t > 4( t+1. (iv If t 4 and 3, then ( +t > t implie ( +t+1 > (t + 1. (i It i neceary to prove that ( +t ( > t 4 hold. Since t > 7 then t+7 > t 4. Proof. 7 [ +t], it follow that ( +t (ii Suppoe that 6, t and ( 1(t 1 > 4( t. Then 7 ( ( t t ( (t 1 = ( 1(t 1 t 1 > 4 t 1 = ( t Combing thi with the fact ( t 1 ( > 1 give (ii. (iii Suppoe that t 6, and ( 1(t 1 > 4( t. Then ( ( t t + 1 ( ( 1t = ( 1(t 1 1 > 4 1 = t + 1 Combing thi with the fact (1 t+1 1 ( 5 7 > 1 give (iii. (iv Suppoe that ( +t > t. Then ( ( + t t ( + t + 1 = > ( + t + 1 t = ( + t + 1t (t + 1. (t + 1 (t + 1 (t Thi, together with the inequality ( + t + 1t > (t + 1 3, yield (iv. 3 Proof of Theorem 1 Throughout thi paper, we aume that the following hypothei hold. Hypothei: Let D be a non-ymmetric -(v, k, λ deign with (r, λ = 1, G Aut(D be a flag-tranitive automorphim group G with Soc(G = A n. Let x be a point of P and H = G x. By Lemma 3, G act primitively on P. So that H i a maximal ubgroup of G by [14, Theorem 8.] and v = G : H. Furthermore, by the flag-tranitivity of G, we have that b divide G, r divide H, and r > v by Lemma 6(i. Suppoe firt that n = 6 and G = M 10, P GL (9 or P ΓL (9. Each of thee group ha exactly three maximal ubgroup with index greater than, and their indice are preciely 45, 36 and 10. By uing the computer algebra ytem GAP [8], for v = 45, 36 or 10, we will compute the parameter (v, b, r, k, λ that atify the following condition: r v 1; (1 < k < r; ( the electronic journal of combinatoric ( (015, #P.6 5

6 We obtain three poible parameter (v, b, r, k, λ a follow: b = vr k ; (3 bk(k 1 λ = v(v 1 ; (4 (r, λ = 1; (5 r H. (6 (10, 30, 9, 3, ; (10, 18, 9, 5, 4; (10, 15, 9, 6, 5. Now we conider the exitence of flag-tranitive non-ymmetric deign with above poible parameter. For the parameter (10, 15, 9, 6, 5, we will conider the exitence of it complement deign which ha parameter (10, 15, 6, 4,. Suppoe that there exit a -(10, k, λ deign D with flag-tranitive automorphim group G, where the block ize k i 3, 5 or 4, and λ =, 4 or repectively. Here v = 10. Let P = {1,, 3, 4, 5, 6, 7, 8, 9, 10}, the group G = M 10, P GL (9 or P ΓL (9 a the primitive permutation group of degree 10 acting on P, ha the following generator repectively ([, p.88]: M 10 = (1, 6, 10, 9, 3, 8, 4, 5(, 7, (1, 7,, 6, 5, 9, 4, 10(3, 8, P GL (9 = (1,, 3, 4, 5, 6, 7, 8, 9, 10, (1,, 5, 8, 9, 7, 4, 10, 6, 3, P ΓL (9 = (1,, 3, 4, 5, 6, 7, 8, 9, 10, (1, 8, 6,, 9, 5, 3, 10(4, 7. There are totally ( v k k-element ubet of P. For any k-element ubet B P, we calculate the length of the G-orbit B G where G = M 10, P GL (9 or P ΓL (9 repectively. By uing GAP [8], we found that B G > b for any k-element ubet B. So G cannot act block-tranitively on D, a contradiction. Now we conider G = A n or S n with n 5. The point tabilizer H = G x act both on P and the et Ω n := {1,,, n}. Then by Lemma 7 one of the following hold: H i primitive in it action on Ω n ; H i tranitive and imprimitive in it action on Ω n ; H i intranitive in it action on Ω n. The proof of Theorem 1 conit of three ubection. 3.1 H act primitively on Ω n. Propoition 13. Let D and G atify Hypothei. Let the point tabilizer act primitively on Ω n. Then D i a -(6, 3, deign or the projective pace P G(3,. the electronic journal of combinatoric ( (015, #P.6 6

7 Proof. Suppoe firt that r i even, ince r v 1 then v i odd. Thu by Lemma 9, v = 15 and G = A 7, and then r = 14. The poible parameter (v, b, r, k, λ uch that < k < r and (r, λ = 1 are (15, 35, 14, 6, 5, (15, 1, 14, 10, 9. If there i a deign D with parameter (15, 1, 14, 10, 9, then the complement deign D with parameter (15, 1, 7, 5, alo exit. However, by [3, Theorem 5.], we know that -(15, 5, deign doe not exit. So the parameter (15, 1, 14, 10, 9 cannot occur. Now aume that (v, b, r, k, λ = (15, 35, 14, 6, 5. Let P = {1,, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14, 15}. Suppoe that there exit a -(15, 6, 5 deign D with flag-tranitive automorphim group A 7. By [, p.89], we know A 7 = (1, 4, 7, 10, 13(, 5, 8, 11, 14(3, 6, 9, 1, 15, (, 6, 3, 8, 1, 4, 9(5, 7, 13, 11, 10, 14, 15. There are totally 5005 = ( element ubet of P. For any 6-element ubet B P, uing GAP [8], we calculate the length of the A 7 -orbit B A 7. It follow that B A 7 > 35 for any 6-element ubet B. So A 7 cannot act block-tranitively on D, a contradiction. Then r i odd. Let p be an odd prime divior of r, then (p, v = 1 according to Lemma 5(ii. Thu H contain a Sylow p-ubgroup R of G. Let g G be a p-cycle, then there i a conjugate of g belong to H. Thi implie that H acting on Ω n contain an even permutation with exactly one cycle of length p and n p fixed point. By a reult of Jordan [14, Theorem 13.9], we have n p. Therefore n p n, p G, and o p r. It follow that r i either a prime, namely n, n 1, n, or the product of two twin prime, namely r = (n n. Moreover, ince the primitivity of H acting on Ω n and H A n implie that v [ n+1 ]! by [14, Theorem 14.], combining thi with r > v give r > [ n+1]!. Therefore, (n, r = (5, 5, (5, 15, (6, 5, (7, 5, (7, 7, (7, 35, (8, 7 or (13,143. From Lemma 5, 6, the fact v [ n+1 ]! and [b, v] G, where the condition [b, v] G i a conequence of v G and b G, we obtain 3 poible parameter (v, b, r, k, λ which lited in the following: (6, 10, 5, 3,, (15, 1, 7, 5,, (15, 35, 7, 3, 1. Cae (1: (v, b, r, k, λ = (6, 10, 5, 3,. By [, p.7, p.36], we know that there i up to iomorphim a unique -(6, 3, deign D = (P, B where P = {1,, 3, 4, 5, 6}; B = {{1,, 3}, {1,, 5}, {1, 3, 4}, {1, 4, 6}, {1, 5, 6}, {, 3, 6}, {, 4, 5}, {, 4, 6}, {3, 4, 5}, {3, 5, 6}}. Here G = A 5, S 5, A 6 or S 6, the tabilizer G x = D 10, AGL 1 (5, A 5 or S 5 repectively. Aume firt that G = S 5, A 6 or S 6. A the primitive permutation group of degree 6, S 5, A 6 or S 6 i 3-, 4- or 6-tranitive repectively, o G i 3-tranitive. If we chooe B = {1,, 3}, then B G = 0 > b = 10, a contradiction. the electronic journal of combinatoric ( (015, #P.6 7

8 Hence G = A 5. Without lo of generality, aume that A 5 = (4(56, (13(456. Let B = {1,, 3} be a block. Then B A 5 = B. Now G 1 = (4(56, (35(46 = D 10, and B G 1 = {{1,, 3}, {1,, 5}, {1, 3, 4}, {1, 4, 6}, {1, 5, 6}}, which i the G 1 -orbit on B containing B. So that G 1 i tranitive on 5 block through 1, note that G i alo tranitive on P, and hence D i flag-tranitive. Cae (: (v, b, r, k, λ = (15, 1, 7, 5,. Thi can be ruled out by [3, Theorem 5.]. Cae (3: (v, b, r, k, λ = (15, 35, 7, 3, 1. Here λ = 1, by [5] or Lemma 4, we know that D i the projective pace P G(3, and G = A 7 or A 8 = P SL4 (. For completene, the tructure of the deign and the proof of flag-tranitivity are given below. Let P = {1,, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14, 15}, the group G = A 7 or A 8, a the primitive group of degree 15 acting on P, ha the following generator repectively ([, p.89]: A 7 = (1, 4, 7, 10, 13(, 5, 8, 11, 14(3, 6, 9, 1, 15, (, 6, 3, 8, 1, 4, 9(5, 7, 13, 11, 10, 14, 15, A 8 = (1,, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14, 15, (1, 5(6, 13(7, 8(10, 1. There are totally 455 = ( element ubet of P. For any 3-element ubet B P, uing GAP [8], we calculate the length of the G-orbit B G. It follow that up to iomorphim there exit a unique -(15, 3, 1 deign D = (P, B where or B = {{1,, 13}, {1, 4, 5}, {1, 6, 11}, {4, 7, 8}, {1, 7, 9}, {4, 9, 14}, {1, 3, 10}, {7, 10, 11}, {9, 1, 13}, {4, 10, 1}, {, 7, 1}, {, 9, 15}, {4, 6, 13}, {1, 8, 14}, {10, 13, 14}, {1, 1, 15}, {, 4, 11}, {7, 13, 15}, {5, 10, 15}, {3, 5, 1}, {, 5, 6}, {3, 9, 11}, {11, 14, 15}, {3, 4, 15}, {5, 7, 14}, {6, 9, 10}, {5, 11, 13}, {3, 8, 13}, {6, 8, 15}, {5, 8, 9}, {3, 6, 7}, {6, 1, 14}, {, 8, 10}, {, 3, 14}, {8, 11, 1}}. Let B = {1,, 13} be a block. Then it i eaily known that B G = B. Now G 1 = (, 6, 3, 8, 1, 4, 9(5, 7, 13, 11, 10, 14, 15, (3, 8(4, 11(5, 6(7, 9(10, 14(1, 15 = P SL3 ( G 1 = (, 5, 8, 3, 15, 11, 7(4, 14, 10, 1, 6, 9, 13, (3, 14(7, 1(8, 10(9, 15 = AGL 3 ( with G = A 7 or A 8 repectively. Then B G 1 containing 7 block: {1,, 13}, {1, 6, 11}, {1, 3, 10}, {1, 4, 5}, {1, 8, 14}, {1, 7, 9}, {1, 1, 15}. So that G 1 i tranitive on r block through 1, note that G i alo tranitive on P, and hence G i flag-tranitive and D = P G(3,. 3. H act tranitively and imprimitively on Ω n. Propoition 14. Let D and G atify Hypothei. Let the point tabilizer act tranitively and imprimitively on Ω n, then D i a -(10, 6, 5 deign with Soc(G = A 6. the electronic journal of combinatoric ( (015, #P.6 8

9 Proof. Suppoe on the contrary that Σ := { 0, 1,..., t 1 } i a nontrivial partition of Ω n preerved by H, where i =, 0 i t 1,, t and t = n. Then v = = ( t ( (t 1 ( t 1 1 (... 3 t! ( ( (t ( ( 1 1. (7 Moreover, the et O j of j-cyclic partition with repect to X (a partition of Ω n into t clae each of ize i an union of orbit of H on P for j =,..., t (ee [5, 15] for definition and detail. (1 Suppoe firt that =, then t 3, v = (t 1(t 3 5 3, and d j = O j = 1 ( ( t j 1 j ( t = j 1 j We claim that t < 7. If t 7, then it i eay to know that v = (t 1(t > t (t 1, and a r divide d = t(t 1 it follow that t(t 1 r, hence v > t (t 1 r which i a contradiction. Thu t < 7. For t = 3, 4, 5 or 6, we calculate d = gcd(d, d 3 which lited in Table 1 below. Table 1: Poible d when = t n v d d 3 d In each cae r d which contradict to the fact r > v. ( Suppoe econd that 3, then O j i an orbit of H on P, and d j = O j = ( t ( j j 1 ( = j t j. In particular, d = ( t ( ( 1 = t and r d. Moreover, from ( i 1 1 = i 1 i i ( 1 > i 1, for i =, 3,..., t, we have v > ( 1(t 1. Then 1 1 ( 1(t 1 < v < r 4 ( t, and o ( 1(t 1 < 4 ( t. (8 We will calculate all pair (, t atifying the inequality (8. Since ( 1(t 1 = 5 > 4( t = , i.e. the pair (, t = (6, 6 doe not atify the inequality (8 but the electronic journal of combinatoric ( (015, #P.6 9

10 atifie the condition of Lemma 1 (ii and (iii. Thu, we mut have < 6 or t < 6. It i not hard to get 3 pair (, t atifying the inequality (8 a follow: (3,, (3, 3, (3, 4, (3, 5, (3, 6, (3, 7, (3, 8, (3, 9, (4,, (4, 3, (4, 4, (4, 5, (4, 6, (5,, (5, 3, (5, 4, (6,, (6, 3, (6, 4, (7,, (7, 3, (8,, (8, 3, (9,, (10,, (11,, (1,, (13,, (14,, (15,, (16,, (17,. For each (, t, we calculate the parameter (v, b, r, k, λ atifying Eq.(7, Lemma 5, 6 and r d. Then we obtain five poible parameter (v, b, r, k, λ correponding to (, t are the following: (.1 (, t = (3, : (10, 30, 9, 3,, (10, 18, 9, 5, 4, (10, 15, 9, 6, 5; (. (, t = (5, : (16, 55, 5, 6, 1, (16, 150, 5, 1, 4. Cae (.1: (, t = (3,. Then n = 6, v = 10 and G = A 6 or S 6. Let P = {1,, 3, 4, 5, 6, 7, 8, 9, 10}, the group G = A 6 or S 6, a the primitive permutation group of degree 10 acting on P, ha the following generator repectively ([, p.88]: A 6 = (1, 10, 4, 7, 5(, 8, 6, 9, 3, (1, 3, 4, 5, 7(, 10, 9, 8, 6, S 6 = (1, 8, 6(, 3, 7, 9, 10, 5, (1, 8, 9, 3, 5, 6(, 7, 4. Aume firt (v, b, r, k, λ = (10, 30, 9, 3, or (10, 18, 9, 5, 4, and there exit a -(10, k, λ deign D with flag-tranitive automorphim group G, where k = 3 or 5, and λ = or 4, repectively. There are totally ( v k k-element ubet of P. For any k-element ubet B P, we calculate the length of the G-orbit B G where G = A 6 or S 6 repectively. By uing GAP [8], we found that B G > b for any k-element ubet B. So G cannot act block-tranitively on D, a contradiction. Therefore, (v, b, r, k, λ = (10, 15, 9, 6, 5. Here G = A 6 or S 6 act tranitively on P. There are totally 10 = ( element ubet of P. For any 6-element ubet B P, uing GAP [8], we calculate the length of the G-orbit B G. It follow that up to iomorphim there exit a unique -(10, 6, 5 deign D = (P, B where B = {{1,, 3, 4, 5, 8}, {1,, 6, 7, 8, 10}, {3, 4, 5, 6, 7, 10}, {4, 5, 6, 8, 9, 10}, {1,, 3, 6, 9, 10}, {1,, 4, 5, 7, 9}, {1, 3, 4, 6, 7, 9}, {, 5, 6, 7, 8, 9}, {, 3, 4, 8, 9, 10}, {1, 3, 5, 7, 8, 10}, {, 3, 5, 7, 9, 10}, {1, 3, 5, 6, 8, 9}, {, 3, 4, 6, 7, 8}, {1,, 4, 5, 6, 10}, {1, 4, 7, 8, 9, 10}}. Let B = {1,, 3, 4, 5, 8} be a block. Then it i eaily known that B G = B. Now or G 1 = (, 9, 10, 3(4, 7, 8, 5, (, 5, 8, 10(3, 7, 6, 4 = 3 : 4 (, 10, 8, 3, 9, 4(5, 6, 7, (3, 9, 7, 8(4, 10, 5, 6 = 3 : D 8 with G = A 6 or S 6 repectively. Then B G 1 containing 9 block: {1,, 3, 4, 5, 8}, {1,, 4, 5, 7, 9}, {1, 3, 5, 7, 8, 10}, {1, 4, 7, 8, 9, 10}, {1, 3, 5, 6, 8, 9}, {1,, 6, 7, 8, 10}, {1,, 3, 6, 9, 10}, {1,, 4, 5, 6, 10}, {1, 3, 4, 6, 7, 9}. the electronic journal of combinatoric ( (015, #P.6 10

11 So that G 1 i tranitive on the block through 1, note that G i alo tranitive on P, and hence D i flag-tranitive. Cae (.: (, t = (5,. Then n = 10, v = 16 and G = A 10 or S 10. For the parameter (16, 55, 5, 6, 1 we have λ = 1, it can be ruled out by Lemma 4. Suppoe that (v, b, r, k, λ = (16, 150, 5, 1, 4. Since G i flag-tranitive, then it i block-tranitive and point-tranitive. So that G mut ha ubgroup with index 16 and 150. By uing Magma [1] we know that G ha 16 ubgroup with index 16. Let B B, o that G : G B = b = 150 and G B = 1096 or 419 with G = A 10 or S 10 repectively. Clearly, G : G B < ( Then by Lemma 10 and [7, Theorem 5.B], one of 3 cae hold. If Cae (i hold, there exit ome Ω 10 = {1,,..., 10} with < 4. We have A 7 G B by Remark 11. However, A 7 = 7!/ doe not divide G B, a contradiction. Cae (ii can be ruled out by G : G B = ( 10 5 and Cae (iii can be ruled out by n = 10. Hence, G ha no ubgroup with index 150, a contradiction. So that G = A10 or S H act intranitively on Ω n. Propoition 15. Let D and G atify Hypothei. Then the point tabilizer cannot be intranitive on Ω n. Proof. Suppoe on the contrary that H act intranitively on Ω n. We have H = (Sym(S Sym(Ω \ S G, and without lo of generality aume that S = < n/ by Lemma 7(i. By the flag-tranitivity of G, H i tranitive on the block through x, and o H fixe exactly one point in P. Since H tabilize only one -ubet of Ω n, we can identify the point x with S. A the orbit of S under G conit of all the -ubet of Ω n, we can identify P with the et of -ubet of Ω n. So that v = ( n, G ha rank + 1 and the ubdegree are: n 0 = 1, n i+1 = ( i ( n i, i = 0, 1,..., 1. (9 Firt, we claim that 6. Since r n i for any ubdegree n i of G by Lemma 6(ii and n = (n i a ubdegree of G by (9, then r (n. Combining thi with r > v, we have (n > ( n. Since the condition < n equal to < t := n, we have ( + t t >. (10 Combining it with Lemma 1 (i, we get 6. Cae (1: If = 1, then v = n 5 and the ubdegree are 1, n 1. The group G i (v -tranitive on P. Since < k v, G act k-tranitively on P. Then b = B = B G = ( n ( k for every block B B. From the equality bk = vr we can obtain n ( k n k = nr. On the one hand, by Lemma 5(ii we have r n 1, it follow that k k n(n 1. On the other hand, by < k n, we have n i k i+ > k i+1 the electronic journal of combinatoric ( (015, #P.6 11

12 for i =,..., k 1. So that ( n k = n(n 1 n k k 1 n 3 k n k + 1 > n(n 1, a contradiction. Cae (: If =, then the ubdegree are 1, ( n (, (n, and G i a primitive (n rank 3 group acting on P. By Lemma 6(ii, r, (n = n, n, or (n with n (mod 4, n 1 (mod 4, or n 3 (mod 4 repectively. Aume firt that r n n(n 1. Then = v < r ( n, which i impoible. Aume that r (n. From Lemma 6(i, λv = λn(n 1 < r (n which force λ = 1. It follow from Lemma 4 that D = P G(3, and G = A 7 or A 8. Thi contract the aumption that v = ( n = 1 or 8. Hence r (n. From above analyi we know that r i even. By Lemma 6(i, λv = λn(n 1 < r 4(n which implie 1 λ 7. Recall that (r, λ = 1, o that λ i odd. Since Lemma 4 implie that λ 1, we aume that λ = 3, 5 or 7 in the following. Let r = (n u follow that 8 > 8(n n(n 1 5, we get k = λ(n+1 + 1, and b = vr 4 k for ome integer u. From Lemma 6(i, we have 4(n. It u > λu which force u = 1. Therefore, r = (n. By Lemma = 4n(n 1(n λ(n+1+4, where λ = 3, 5 or 7. > λn(n 1 If λ = 3, then (3n + 7 4n(n 1(n ince b N. And ince (3, 3n + 7 = 1, we have (3n n(n 1(n. Recall that n 3 (mod 4, and from 3 b = 36n(n 1(n 3n + 7 = 1n 64n n 3n + 7 N we have (3n + 7 (111 n, o n = 7, 11, 15, 91, 119 or 171. Similarly, if λ = 5, then 5 b = 100n(n 1(n = 0n 96n + 13 n+1917 N. Now 5n+9 5n+9 the fact (5n + 9 (n and n 3 (mod 4 imply n = 15, 99, 135 or 11. If λ = 7, then 7 b = 196n(n 1(n = 8n 18n n N. Now the fact 7n+11 7n+11 (7n + 11 (87 n and n 3 (mod 4 imply n = 7, 11, 7, 55, 17 or 187. For each value of n, we compute the poible parameter (n, v, b, r, k atifying Lemma 5 and the condition b > v which lited in the following. λ = 3 : (7, 1, 30, 10, 7, (11, 55, 99, 18, 10, (15, 105, 10, 6, 13, (91, 4095, 10413, 178, 70, (119, 701, 18054, 34, 91, (171, 14535, 37791, 338, 130. λ = 5 : (15, 105, 130, 6, 1, (99, 4851, 7469, 194, 16, (135, 9045, 14070, 66, 171, (11, 155, 34815, 418, 66. λ = 7 : (55, 1485, 1590, 106, 99, (17, 8001, 8890, 50, 5, (187, 17391, 19499, 370, 330. Aume firt that (n, v, b, r, k = (7, 1, 30, 10, 7 and there exit a -(1, 7, 3 deign D with flag-tranitive automorphim group G = A 7 or S 7. Let P = {1,, 3,..., 1}, the the electronic journal of combinatoric ( (015, #P.6 1

13 group G = A 7 or S 7, a the primitive permutation group of degree 1 acting on P, ha the following generator repectively ([, p.830]: A 7 = (1,, 3, 4, 5, 6, 7(8, 9, 10, 11, 1, 13, 14(15, 16, 17, 18, 19, 0, 1, (1, 4(, 11(3, 9(5, 15(7, 0(8, 13(10, 1(1, 14, S 7 = (1,, 3, 4, 5, 6, 7(8, 9, 10, 11, 1, 13, 14(15, 16, 17, 18, 19, 0, 1, (1, 1, 4, 10(, 9, 11, 3(5, 1, 15, 14(7, 13, 0, 8(17, 18. There are totally ( element ubet of P. For any 7-element ubet B P, we calculate the length of the G-orbit B G where G = A 7 or S 7 repectively. By uing GAP [8], we found that B G > 30 for any 7-element ubet B. So G cannot act block-tranitively on D, a contradiction. For all other poible parameter (n, v, b, r, k lited above, G = A n or S n. On the one hand, by the block-tranitivity, we have b = G : G B where B B. On the other hand, ince G : G B = b < ( n 3 for each cae, by Lemma 10 and [7, Theorem 5.B], it i eaily known that G ha no ubgroup of index b, a contradiction. Cae (3: Suppoe that 3 6. Now for each value of, uing the inequality (10 and Lemma 1(iv, we know that t (and hence n i bounded. For example, let = 3, ince ( > 3 48, we mut have 4 t 47 by Lemma 1(iv, and o 7 n 50. The bound of n lited in Table below. Here the lat column denote the arithmetical condition which we ued to ruled out each line. Table : Bound of n when 3 6 t n Reference 3 4 t 47 7 n 50 (1-(4, Lemma t 14 9 n 18 Lemma 6 5 6, 7, 8, 9 11, 1, 13, 14 Lemma Lemma 6 Note that v = ( n, and n1 = ( n, n = (n are two ubdegree of G acting on P. Therefore, the 6-tuple (v, b, r, k, λ, n atifie the following arithmetical condition: (1-(4, (r, λ = 1, r > v (Lemma 6(i, and r d, where d = gcd(n 1, n. (11 If = 3, by uing GAP [8], it output five 6-tuple atifying above arithmetical condition: (364, 1001, 33, 1, 1, 14, (1540, 3135, 57, 8, 1,, (4960, 7440, 87, 58, 1, 3, (19600, 19740, 141, 140, 1, 50, (1540, 1596, 57, 55,,. The four parameter with λ = 1 can be ruled out by Lemma 4. For the parameter (1540, 1596, 57, 55,,, we have λ =, G = A or S. By the block-tranitivity, the electronic journal of combinatoric ( (015, #P.6 13

14 b = G : G B = 1596 where B B. However, ince G : G B = 1596 < ( 4, by Lemma 10 and [7, Theorem 5.B], it i eaily known that A and S ha no ubgroup of index So the parameter (1540, 1596, 57, 55,, cannot occur. If = 4, 5 or 6, by uing GAP [8], there are no parameter (v, k, n atifying the condition. For example, if = 6, then n = 13, v = 1716, d = 7. It follow that r 7 by (10, then r < v which i impoible. If = 5, then n = 11, 1, 13 or 14, v = 46, 79, 187 or 00 and d = 6, 7, 8 or 9, repectively. It i eay to check that r < v for every cae. Thi i the final contradiction. Propoition finih the proof of Theorem 1. Acknowledgement The author wih to thank the referee for their valuable comment and uggetion which lead to the improvement of the paper. Reference [1] W. Boma, J. Cannon, C. Playout. The MAGMA algebra ytem I: The uer language. J. Symb. Comput., 4:35-65,1997. [] C. J. Colbourn, J. H. Dinitz (Editor. Handbook of Combinatorial Deign. Chapman & Hall/CRC, Boca Raton, FL, 007. [3] W. S. Connor, Jr.. On the tructure of balanced incomplete block deign. Ann. Math. Statitic, 3:57-71, 195. [4] H. Davie. Flag-tranitivity and primitivity. Dicrete Math., 63:91-93, [5] A. Delandtheer. Finite flag-tranitive linear pace with alternating ocle. Algebraic Combinatoric and Application (Gößeintein, 1999, page 79-88, Springer, Berlin, 001. [6] P. Dembowki. Finite Geometrie. Springer-Verlag, New York, [7] J. D. Dixon, B. Mortimer. Permutation Group. Springer-Verlag, [8] The GAP Group, GAP-Group, Algorithm, and Programming. Verion 4.4, [9] M. W. Liebeck, C. E. Praeger and J. Saxl. A claification of the maximal ubgroup of the finite alternating and ymmetric group. J. Algebra, 111: , [10] M. W. Liebeck, J. Saxl. The primitive permutation group of odd degree. J. London Math. Soc., 31(:50-64, [11] C. E. Praeger, S. Zhou. Imprimitive flag-tranitive ymmetric deign. J. Combin. Theory Ser A, 113: , 006. [1] E. O Reilly Regueiro. On primitivity and reduction for flag-tranitive ymmetric deign. J. Combin. Theory Ser A, 109: , 005. the electronic journal of combinatoric ( (015, #P.6 14

15 [13] D. Tian, S. Zhou. Flag-tranitive point-primitive ymmetric (v, k, λ deign with λ at mot 100. J. Combin. De., 1:17-141, 013. [14] H. Wielandt. Finite Permutation Group. Academic Pre, New York, [15] S. Zhou, H. Dong. Alternating group and flag-tranitive triplane. De. Code Cryptogr., 57: , 010. [16] Y. Zhu, H. Guan and S. Zhou. Flag-tranitive ymmetric deign with (k, λ = 1 and alternating ocle. Submitted, 014. [17] P. H. Ziechang. Flag tranitive automorphim group of -deign with (r, λ = 1. J. Algebra, 118: , the electronic journal of combinatoric ( (015, #P.6 15