k(k 1)(k 2)(p 2) 6(p d.

Transcription

1 BLOCK-TRANSITIVE 3-DESIGNS WITH AFFINE AUTOMORPHISM GROUP Greg Gamble Let X = (Z p d where p s an odd prme and d N, and let B X, B = k. Then t was shown by Praeger that the set B = {B g g AGL d (p} s the block-set of a 3-desgn f and only f the number q 1 (k of collnear trples of ponts of B s q 1 (k = k(k 1(k 2(p 2 6(p d. 2 We gve an explct method for determnng all k such that 3 k < p d /2 and q 1 (k s an nteger. For p = 3 the smallest value of d for whch there s such an ntegral value of q 1 (k s d = 7 and the smallest k 3 gvng ntegral values of q 1 (k are k = 115 and k = 116. We construct many examples of 3-desgns (X, B wth k = 115 or 116 admttng AGL 7 (3 as automorphsm group. 1. INTRODUCTION A t-(v, k, λ desgn s a par D = (X, B where X s a set of v ponts and B s a set of k-subsets of X called blocks such that any t ponts are contaned n exactly λ > 0 blocks. If B contans all k-subsets of X then D s sad to be trval. Let G = Aut D = {g Sym(X B g = B}. That s, G s the set of all permutatons of X that fx B setwse. The group G s sad to be block-transtve f B = B G for B B, where B G = {B g g G}. In ths case we say that D s block-transtve. Smlarly, G s pont-transtve f X = x G for some x X, where x G = {x g g G}. By [2], G block-transtve mples that G s pont-transtve and thus that G s a transtve subgroup of Sym(X. Also by [3, Proposton 1.1] (or see [5], f (X, B G s a blocktranstve t-(v, k, λ desgn and G H Sym(X then (X, B H s a block-transtve t-(v, k, λ desgn for some λ λ. Hence n searchng for non-trval block-transtve t-desgns we may start by consderng maxmal subgroups of Sym(X. As noted by [5],

2 2 Greg Gamble the O Nan-Scott theorem mples that f G s a transtve maxmal subgroup of S v then t s of one of the followng types: 1. mprmtve: G = S c wr S d for some c, d N such that v = cd, c > 1, d > 1; 2. affne: G = AGL d (p for some p, d N such that v = p d, p prme, d 1; 3. product: G = S c wr S d for some c, d N such that v = c d, c 5, d > 1; 4. smple dagonal: G = T d.(out T S d for some d N and some group T such that v = T d 1, d > 1 and T s nonabelan and smple; 5. almost smple: T G Aut T for some nonabelan smple group T. We are concerned wth the second of these cases, G = AGL d (p. We say a group G s t-homogeneous on a set X f t s transtve on the set of t-subsets of X. If G s t- homogeneous then for any k and any k-subset B of X, (X, B G s a t-desgn; and, further, f G s k-homogeneous the desgn s trval. The group AGL d (p s 2-transtve and hence 2-homogeneous for all d, p and AGL d (2 s 3-transtve and hence 3-homogeneous for all d 2. Thus we consder the more nterestng problem of the exstence of block-transtve t-desgns admttng G = AGL d (p wth t 3 for p odd, and n partcular we consder the case where t = 3. The followng lemma s essentally Lemma 1.10(c of [4] (or see [3, Proposton 1.3] or see [5]. It s a corollary of Theorem 2.1, a proof of whch was suppled to the author by Praeger [6] and s reproduced wth mnor modfcatons n the next secton LEMMA. (X, B G s a block-transtve t-desgn f and only f (X, (X \ B G s a block-transtve t-desgn. Ths result allows us to restrct our attenton to desgns for whch 3 = t k < p d /2 where k s the block-sze, as f there s no desgn for a k n ths range then there s certanly no desgn for k satsfyng p d /2 k < p d 2 and, of course, for k {p d 2, p d 1, p d }, all blocks B such that B = k gve rse to trval desgns. For p an odd prme, d 2 and G = AGL d (p Praeger [5, Lemma 2.1] has shown that (X, B G s a block-transtve 3-desgn f and only f the number of collnear trples n the block B s k(k 1(k 2(p 2 q 1 (k = 6(p d, 2 where k = B. We prove ths result n Secton 2. In Secton 2 we prove a number of lemmas whch are needed to prove Theorem 3.1, from whch the followng theorem s a corollary. The theorem shows how to construct the set Ω of all k such that q 1 (k N and 3 k < p d / THEOREM. Let 2 p prme, d 2, q 1 (k = (p d 2, p 2. Let N = pd 2 D k(k 1(k 2(p 2 6(p d 2 and D = = n pε, where p are dstnct prmes, n, ε N and { D f 3D / p d 2, D/3 f 3D p d 2. D =

3 Greg Gamble 3 Suppose u = N p ε Let Ω = { k q 1 (k N and 3 k p d /2 }. Then and t Z satsfy n u t = 1. ( Each element k of Ω has the form k = ( n α u t mod N + ln, where α {0, 1, 2} for 1 n and 0 l < D /2. ( Ω = D.3 n 3. 2 The t of the theorem may be calculated by Algorthm 3.2. An mmedate consequence of ths theorem s an observaton of Praeger [5, p. 196], namely that there are no nontrval block-transtve 3-desgns admttng G = AGL d (3 for d {2,..., 6}, as for each of these values of d the number n of dstnct prme dvsors of N s 1, so that the set Ω s empty. However for d = 7 we fnd Ω s not empty. Praeger [5, p. 196] posed the queston: Is there a block-transtve 3-(3 7, k, λ desgn of ths type? In Secton 4 we show, for at least two values of k that ndeed there are such desgns. 2. PRELIMINARY RESULTS The followng theorem s ncluded manly because the author feels t may eventually lead to new results of non-exstence of t-desgns. It s a self-contaned verson of Corollary 1.4(a and Lemma 1.10(c of Hughes and Pper [4] THEOREM. [6] If D = (X, B s a t-(v, k, λ desgn then ( D = (X, B s a t -(v, k, λ t desgn for all t such that 0 t t where λ t = ( k t ( v t B ; (that s, f 0 t t, C X, C = t then λ t := {B B C B} = ( k t ( v t B s ndependent of the choce of C {C X C = t }; ( f 0 t, D C X, D =, C = t then µ := {B B D = C B} s ndependent of C and the subset D of C, and µ t = λ t = λ t µ = λ j=1+ ( t µ j for 0 < t; j ( D = (X, B s a t-(v, k, µ 0 desgn where B = {B = (X \ B B B}.

4 4 Greg Gamble PROOF. ( λ t = λ s ndependent of C as D s a t-desgn. Suppose 0 < t and λ +1 s ndependent of C {C X C = + 1}. Let C {C X C = }, λ (C := {B B C B} then countng two ways {(x, B {x} C B, x C, B B} = λ (C.(k = (v.λ +1. That s, λ = λ (C = v k λ +1 s ndependent of C {C X C = }. Thus, by nducton, λ s ndependent of C {C X C = } for all such that 0 t. Also λ 0 = {B B B} = B, λ+1 = k v λ. Thus, by nducton, 1 ( k λ = k j λ 0 = ( v j v B, for 0 t. j=0 ( We wrte for dsjont unon. Let 0 t, D C X, D =, C = t, then If = t then D = t and λ = {B B D B} = {B B D C and D B} = {B B D C B}. D C B D = C B. That s, µ t = λ t s ndependent of C and for each choce of C, D = C s unque. Suppose now that 0 < t and that µ j s ndependent of C and the subset D {D C D = j}, for < j t. Let D C, D =, µ (C, D := {B B D = C B}. Then µ (C, D = {B B D C B and C B = } = {B B D C B} \ {B B D C B and C B > } = λ {B B D C B and C B > } t = λ {B B D C B and C B = j} = λ = λ = λ j=+1 t j=+1 t j=+1 t j=+1 {B B D C B and C B = j} E C\D E =j ( t j {B B D E = C B} µ j

5 Greg Gamble 5 s ndependent of C and the choce of D {D C D = }. Thus, by nducton, µ : = {B B D = C B} t ( t = λ µ j for 0 t j j=+1 s ndependent of C {C X C = t} and the choce of D {D C C = }. ( µ 0 = {B B B C = } = {B B C B = (X \ B} = {B B C B } s ndependent of C {C X C = t}. That s, D = (X, B s a t-(v, k, µ 0 desgn. The followng theorem (noted n [5] or see [4, pp ] or [1] allows us to reduce the problem of exstence of desgns to a farly elementary number theory problem THEOREM. If B s a k-subset of X, G Sym(X, Q 1,..., Q m are the G-orbts on t-sets of X, q = Q B for 1 m then (X, B G s a t-desgn q 1 Q 1 = = q m Q m m q j Q j = q m Q for 1 j m. The group G = AGL d (p has 2 orbts on the 3-sets of X = (Z p d namely Q 1 := collnear trples, Q 2 := non-collnear trples, where we say {u, v, w} s a collnear trple f and only f u, v, w are dstnct and w = u + α(v u for some α Z p. Equvalently {u, v, w} s a collnear trple f and only f v u and w = u + α(v u for some α Z p \ {0, 1}. Thus we see that Q 1 = pd.(p d 1.(p 2 3! as for ordered trples (u, v, w there are p d choces for u, leavng p d 1 choces for v and then p 2 choces for α (and hence w. So by Theorem 2.2 q 1 Q 1 = q ( k 1 + q 2 #trples n B = 3 Q 1 + Q 2 #trples n X = ( p d. 3

6 6 Greg Gamble Hence we arrve at the concluson of [5, Lemma 2.1], q 1 = ( k 3 ( p d 3 pd.(p d 1.(p 2 3! = k(k 1(k 2(p 2 6.(p d. 2 Of course, q 1 s an nteger. The followng lemma gves a smpler equvalent condton for the ntegralty of q 1, n terms of an nteger N whch s easy to calculate LEMMA. Let 2 p prme, q 1 = D = (p 2, p d 2. Then k(k 1(k 2(p 2 6.(p d, where d, k N, and let 2 q 1 Z k(k 1(k 2 N Z { where N = pd 2 D f 3D / p D and D d 2, = D/3 f 3D p d 2. PROOF. Frstly, q 1 = k(k 1(k 2(p 2 6(p d 2 = k(k 1(k 2a, 6b where a = (p 2/D, b = (p d 2/D. So q 1 Z f and only f 6b k(k 1(k 2a. Of course, for any k N, 6 k(k 1(k 2. Observe that 2 p prme mples that 2 / (p d 2 and hence 2 / b. Also (a, b = 1. Now we consder two cases. Case 1. Assume 3 / b (or equvalently 3D / p d 2. In ths case we defne D = D, so that N = (p d 2/D = b. Thus q 1 Z f and only f N = b k(k 1(k 2. Case 2. Assume 3 b (or equvalently 3D p d 2. In ths case we defne D = D/3, so that N = (p d 2/D = 3b. As (a, b = 1, and by assumpton 3 b, we have 3 / a. Thus q 1 Z f and only f N = 3b k(k 1(k 2. So we have that, q 1 Z f and only f k(k 1(k 2/N Z. Note that f 3 p d 2 then 3 p 2 (as suppose 3 / p 2, then p ε 2 (mod 3 for some ε {0, 1}, and hence p d ε 2 (mod 3, so that 3 / p d 2. Thus we see that n Lemma 2.3, D s always an nteger (as f 3 b then 3 p d 2 and hence 3 p 2, so that 3 D and D = D/3 Z; and consequently N s at most p d 2. From the followng lemma we have nformaton as to how to construct ntegers k satsfyng the ntegralty condton gven n Lemma 2.3. We show n Algorthm 3.2 how to construct the t of 2.4 ( LEMMA. Let k Z, N = n Then the followng are equvalent. pε, where 2 p are dstnct prmes and n, ε N.

7 Greg Gamble 7 ( k(k 1(k 2 N Z, ( k α (mod p ε for some α {0, 1, 2} : 1 n, ( k n α u t (mod N for some α {0, 1, 2} where u = N p ε n u t = 1. and t Z satsfy PROOF. Any prme dvsor p of N dvdes at most one of k, k 1, k 2, as p 3. k(k 1(k 2 So, for each such that 1 n, f Z then p ε dvdes exactly one of N k, k 1, k 2, and thus k 0, 1 or 2 (mod p ε. Hence ( mples (. ( mples ( s trval. For n = 1, the equvalence of ( and ( s trval, as then u 1 = t 1 = 1. For n 2, the equvalence of ( and (, follows by applcaton of the Chnese Remander Theorem. In the followng lemma we show the unqueness of the expresson gven n Lemma 2.4 ( LEMMA. Let N = n pε, where 2 p are dstnct prmes and n, ε N, and let u = N p ε, t Z satsfy n u t = 1, and α, β {0, 1, 2}. Then n α u t n β u t (mod N α = β : 1 n. PROOF. As n u t = 1 and p ε j j u f j, we have u j t j n u t 1 (mod p ε j j for 1 j n. Now suppose n α u t n β u t (mod N. Then α j α j u j t j n α u t n β u t β j u j t j β j (mod p ε j j j : 1 j n. But α j, β j {0, 1, 2} and p ε j j converse s trval. 3 for 1 j n. Hence α j = β j for 1 j n. The We are now n a poston to prove Theorem 3.1 whch along wth Algorthm 3.2 shows how to explctly construct (and count all k such that q 1 (k s a postve nteger and 3 k < p d /2.

8 8 Greg Gamble 3. INTEGRALITY CONDITIONS The followng theorem shows how to construct the set Ω of all k such that q 1 (k Z and 0 k p d. We show that each k has a unque expresson of a gven form and that they occur n pars (k, k where k = p d k. Ths s not surprsng as, of course, f a block-transtve 3-(p d, k, λ desgn for some λ and admttng AGL d (p exsts then, by Theorem 1.1, ts complementary desgn s a 3-(p d, k, λ desgn for some λ. Below f r r (mod N where 0 r < N then we take (r mod N + s to mean r + s. k(k 1(k 2(p THEOREM. Let 2 p prme, d 2, q 1 (k = 6(p d and D = 2 (p d 2, p 2. Let Ω = { k q 1 (k Z and 0 k p d} and Ω = { k Ω 3 k p d /2 }. Let N = pd 2 D = n pε, where p are dstnct prmes, n, ε N and Suppose u = N p ε D = { D f 3D / p d 2, D/3 f 3D p d 2. and t Z satsfy n u t = 1. Then ( Each element k of Ω s unquely dentfed by an (n + 1-tuple (α 1,..., α n, l, where k = ( n α u t mod N + ln, α {0, 1, 2} for 1 n, 0 l D and, f l = D then the α are all equal. Also, f k Ω corresponds to (α 1,..., α n, l then k = p d k Ω and corresponds to ( Ω = D.3 n 3. 2 PROOF. ( q 1 Z { (2 α1,..., 2 α n, D (l + 1 f α not all equal, (2 α 1,..., 2 α n, D l f α all equal. k(k 1(k 2 N Z (by Lemma 2.3 k n α u t (mod N for some α {0, 1, 2}, 1 n (by Lemma 2.4 ( k = ( n α u t mod N + ln for some α {0, 1, 2}, 1 n, and l Z. Now suppose α {0, 1, 2} for 1 n. Note that f the α are all equal then ( n n α u t = α 1 u t = α 1.

9 Greg Gamble 9 Hence by Lemma 2.5, 0 ( n α u t mod N 2 f and only f the α are all equal. Thus ( n p d 2 = ND α u t mod N + ln ND + 2 = p d α are all equal and l = D. Consequently ( n 0 α u t mod N + ln p d { 0 l < D f α not all equal, 0 l D f α all equal. Thus, each element k of Ω s dentfed by an (n + 1-tuple, (α 1,..., α n, l, where k = ( n α u t mod N + ln, α {0, 1, 2} for 1 n, 0 l D and, f l = D then the α are all equal. Now we show that each k Ω s unquely dentfed by an (n + 1-tuple of ths form. Suppose that k Ω corresponds to both (α 1,..., α n, l 1 and (β 1,..., β n, l 2 then so that ( n ( n α u t mod N + l 1 N = β u t mod N + l 2 N, ( n ( n α u t mod N β u t mod N = (l 2 l 1 N. ( Consderng the left-hand sde of (, we have N < (l 2 l 1 N < N. Hence l 2 l 1 = 0. That s l 1 = l 2. Havng shown that the rght-hand sde of ( s zero. We have ( n α u t mod N = ( n β u t mod N, so that by Lemma 2.5, we have α = β for 1 n. Hence each element k of Ω s unquely dentfed by an (n + 1-tuple (α 1,..., α n, l. Suppose now k = ( n α u t mod N + ln Ω, so that α {0, 1, 2} for 1 n. Observe that the α not all equal mples that 2 < ( n α u t mod N < N, or equvalently that 2 < 2 + N ( n α u t mod N < N. Thus f the α are not all equal then ( n ( ( n 2 + N α u t mod N = 2 + N α u t mod N mod N = ( 2 + N ( n α u t mod N = 2 ( n = (2 α u t mod N. n u t + N n α u t mod N

10 10 Greg Gamble Hence ( n k = p d k = 2 + (p d 2 k = 2 + D N α u t mod N ln ( n 2 + N α = u t mod N + N ( D (l + 1 f α not all equal, (2 α 1 + N(D l f α all equal; ( n (2 α u t mod N + N ( D (l + 1 f α not all equal, = ( n (2 α u t mod N + N(D l f α all equal. So k corresponds to the (n + 1-tuple { (2 α1,..., 2 α n, D (l + 1 f α not all equal, (2 α 1,..., 2 α n, D l f α all equal. Moreover, as each α {0, 1, 2}, we have (2 α {0, 1, 2} for 1 n. When the α are not all equal, we have 0 l < D and hence 1 < D l 1 D 1, so that 0 D (l + 1 < D. On the other hand, when the α are all equal, we have 0 l D and hence 0 D l D. Thus k Ω. ( Ω = D.3 n + 3. By ( elements of Ω come n pars (k, k one of whch s less than p d /2 and the other greater than p d /2. Thus {k Ω k < p d /2} D.3 n + 3 = and hence 2 {k Ω 3 k < p d /2} = D.3 n = D.3 n The followng algorthm based on the Eucldean Algorthm shows how the t of Theorem 3.1 may be constructed for n 2. For n = 1, as noted n the proof of Lemma 2.4 (, we have u 1 = t 1 = 1 and n fact Theorem 3.1 ( shows that Ω s empty. The case n = 0 does not occur as the condton d 2 of Theorem 3.1 ensures that N > ALGORITHM. If N = n pε, p prme, ε > 0, n 2 then t Z satsfyng n u t = 1 where u = N/p ε are found as follows. Step 0. Defne u (m := m j=1 pε j j p ε Recursvely we construct t (m for 1 m n. such that m u(m Step 1. Defne u (2 1 := p ε 2 2, u(2 2 := p ε 1 1. By the Eucldean Algorthm fnd t (2 1, t(2 2 Z such that u (2 1 t(2 1 +u(2 2 t(2 2 = 1. Step m < n. From the prevous step we have t (m such that m t (m = 1. By the Eucldean Algorthm fnd x, y such that p ε m+1 { m+1 defne t (m+1 := t (m.x for 1 m y for = m + 1. t (m = 1. u(m.x + u(m+1 m+1.y = 1, and then

11 Greg Gamble 11 Step n. After n 1 steps we take u := u (n and t := t (n. PROOF. At step 1 we have 2 u(2 t (2 have m Steps 1 through n 1 are valdated by nducton. = 1. By the nductve assumpton from step m 1 we t (m = 1. Observe u (m.p ε m+1 m+1 = u(m+1 for 1 m, so for the mth step u(m we have ( m 1 = u (m t (m ( m p ε m+1 m+1.x + u(m+1.y = (u (m m+1 (t(m.p ε m+1 ( m m+1 = (u (m+1 t (m+1 + u (m+1 m+1.y = u (m+1 t (m+1 as requred..x + u (m+1 m+1.y 3.3. REMARK. As noted above, Theorem 3.1 ( shows that for n = 1 (that s, for N = p ε 1 1 for some prme p 1 and ε 1 > 0 the set Ω s empty. As observed by Praeger [5, p. 196] ths stuaton occurs f, for example, p d 2 = p ε 1 for some prme p 1 and ε > 0 (as then d 2 ensures that p d 2 p 2 and thus that 1 N = p ε 1 1 for some ε 1 > 0. In partcular Praeger observed that for p = 3 and d {2,..., 6} that p d 2 s a prme or the square of a prme, and thus that Ω =, and so no non-trval block-transtve 3-desgns exst for these values of p and d. However for p = 3 and d = 7 we have p d 2 = = = N (as p 2 = 1; so that n = 3 and Ω contans (3 3 3/2 = 12 elements. Employng Algorthm 3.2 we fnd that t 1 = 33, t 2 = 132, t 3 = 8 where p 1 = 5, p 2 = 19, p 3 = 23 so that u 1 = 19.23, u 2 = 5.23, u 3 = Thus we have for ths case Ω = { k = (α 1.33( α ( α 3.8(5.19 mod α {0, 1, 2}, and 3 k < 3 7 /2 } = {115, 116, 230, 437, 552, 646, 667, 760, 761, 875, 876, 990}. In the next secton we show that block-transtve 3-(3 7, k, λ desgns do exst for k = 115 and for k = 116, thus answerng the queston posed by Praeger [5, p. 196] affrmatvely. In the followng example we construct the t mentoned above EXAMPLE. We now apply Algorthm 3.2 to N = Let p 1 = 5, p 2 = 19, p 3 = 23. Then at step 1 we have, u (2 1 = p 2 = 19, u (2 2 = p 1 = 5 and by the Eucldean Algorthm = 1. Hence, t (2 1 = 1, t (2 2 = 4. At the second and last step we fnd x, y such that p 3 x+u (3 3 y = 1 where p 3 = 23 and u (3 3 = p 1 p 2 = 95. By the Eucldean Algorthm = 1

12 12 Greg Gamble and so x = 33, y = 8. Hence fnally 1 = 33( (5.19 = 33( ( (5.19. That s, t 1 = 33, u 1 = 19.23, t 2 = 132, u 2 = 5.23, t 3 = 8, u 3 = CONSTRUCTIONS In ths secton we show how to construct a block B of a block-transtve 3-(3 7, k, λ desgn admttng G = AGL 7 (3. We do ths by frst constructng a large set W wth no collnear trples. Then smply by addng 4 ponts to W we create a set Y wth a large number of collnear trples. It turns out that we have a lot of control over modfyng the set Y to get a block B of a block-transtve 3-(3 7, k, λ desgn for k = 115 or 116. Frst we ntroduce some notaton that wll be useful n descrbng these blocks B NOTATION. Let S (Z p d then denote by T S the set of collnear trples of S, that s T S = { {u, v, w} (u, v, w S 3 and {u, v, w} s a collnear trple }. In partcular, f p = 3 then T S = { {u, v, w} w = u 1.(v u, u v, (u, v, w S 3} = { {u, v, w} u + v + w = 0, u v, (u, v, w S 3}. Denote by T S (u, the subset of T S consstng of collnear trples that contan u, that s T S (u = { {u, v, w } T S u {u, v, w } }. Let e be the th standard bass vector of (Z p d, that s the vector wth th coordnate 1 and all other coordnates 0. Let W be the subset of (Z 3 7 consstng of vectors wth no coordnate zero, that s { 7 } W = ε e ε 0. Where t s mportant to stress that a unon of sets s a dsjont unon the symbol wll be used n place of. Below we wll see there are a varety of ways to construct a block B of a 3-desgn by modfyng the set W gven above. The queston s: how do we tell whether two dfferent blocks B and B belong to non-somorphc desgns? The followng lemma wll assst us n fndng a partal answer to ths queston.

13 Greg Gamble LEMMA. B G B G (X, B G = (X, B G. PROOF. Let D = (X, B for {1, 2} be block-transtve t-desgns admttng AGL d (p. Then G = AGL d (p = Aut D = {g Sym(X B g = B }. Suppose now D 1 = D2. Then D φ 1 = D 2 for some φ Sym(X, that s, B φ 1 = B 2, and thus G φ = (Aut D 1 φ = Aut D 2 = G. So that φ N Sym(X (AGL d (p = AGL d (p = Aut D 1. So n fact B φ 1 = B 1. That s B 1 = B 2. Thus for any two t-desgns (X, B 1 and (X, B 2 ether B 1 B 2 = or B 1 = B 2. The author s grateful to Praeger for the above proof. Now for k large, checkng whether B B G (and hence B G = B G may be dffcult. Thus below we defne a way of dvdng up a block B nto classes C j (B of ponts. The numbers of ponts n each class of B provdes a sgnature J B for a desgn (X, B G. We shall see that all blocks of (X, B G have the same number of ponts n each class; and moreover so do all blocks of a desgn somorphc to (X, B G. Ths suggests a way of coarsely parttonng the set of all desgns nto dvsons contanng desgns of the same sgnature. A lower bound on the number of dvsons then provdes a lower bound for the number of parwse non-somorphc desgns. Ths motvates the followng defntons DEFINITION. Denote by C j (S, the class of ponts u n S, that are contaned n j dstnct collnear trples of S. That s, for j 0, defne C j (S = {u S T S (u = j}. Also let J S consst of those ordered pars (j, C j (S for whch C j (S s a non-empty subset of S. That s, J S = {(j, C j (S = C j (S S}. The mportance of J S s that s nvarant, under the acton of G = AGL d (p. That s, J S g = J S, for g G. Ths s because {u, v, w} s a collnear trple of S f and only f {u g, v g, w g } s a collnear trple of S g = {x g x S}. The njectvty of g: S S g ensures that u s contaned by j collnear trples of S f and only f u g s contaned by j collnear trples of S g, and also that (j, C j (S J S f and only f (j, C j (S g J S g. Thus desgns (X, B G and (X, B G wth dfferent sets J B and J B are necessarly non-somorphc. The followng lemma shows us that the set W can have no collnear trples LEMMA. Let a, b, c Z 3 \ {0}. Then a + b + c = 0 a = b = c. PROOF. Suppose wthout loss of generalty that a + b + c = 0 and a b then a + b = 0 and so c = 0 (contradcton COROLLARY. T W =.

14 14 Greg Gamble PROOF. Suppose {u, v, w} s a collnear trple of W, then consderng the th component of each pont we have u + v + w = 0 and thus by Lemma 4.4 we have u = v = w. But then u = v = w, contradctng that {u, v, w} s a collnear trple. Thus T W s empty. Now we modfy W to produce constructons of blocks for desgns for k = 115 and k = CONSTRUCTION. We may construct a block B, wth B = k = 115 such that for G = AGL 7 (3 and X = (Z 3 7, (X, B G s a block-transtve 3-desgn. Let Y = W {e 1, e 2, e 3, e 4 } then ( Nether W nor {e 1, e 2, e 3, e 4 } has collnear trples. That s, both the sets T W and T {e1,e 2,e 3,e 4 } are empty. Ths follows mmedately from Lemma 4.4. ( For each collnear trple of W {e }, the th coordnate of each pont s 1 (follows from Lemma 4.4, and u j = v j for j. That s, T W {e } = {e, 7 ε j e j, 2e j=1 7 ε j e j } ε j 0, ε = 1. Thus T W {e } = 2 6 /2 = 32, or puttng t another way C 32 (Y = {e 1, e 2, e 3, e 4 }. Observe that for any pont u of Y, other than e 1, e 2, e 3, e 4, that the number of collnear trples of T Y contanng u s determned by how many of the frst four coordnates are 1. Thus Y contans another fve classes, namely j=1 7 C j = ε e j elements of {ε 1, ε 2, ε 3, ε 4 } are +1,, 4 j elements of {ε 1, ε 2, ε 3, ε 4 } are 1, ε 0 for 5 7 for 0 j 4. So Y conssts of sx classes. Notce that any collnear trple of S conssts of one element from C j (S, one from C 32 (S and one from C 5 j (S for some j satsfyng 1 j 4. Also we have C 0 = C 4 = 8, C 1 = C 3 = 32, C 2 = 48, and thus J Y = {(0, 8, (1, 32, (2, 48, (3, 32, (4, 8, (32, 4}. ( The collnear trples of Y s a unon of the four dsjont sets, T W {e }, for 1 4. That s, T Y = 4 T W {e } and T Y = 4.32 = 128. (v Y = W + 4 = = 132. The dea now s to modfy the set Y to get a block B of a 3-desgn. To get a block B wth B = k = 115 we need T B = q 1 (k = 113. Such a block B can be obtaned from the set Y (whch has Y = 132 ponts and T Y = 128 collnear trples by deletng 17 ponts from Y n such a way as to reduce the number of collnear trples by 15. Ths can be done, for example, f we fnd a soluton of 4 n j = 17, j=0 4 j.n j = 15 ( j=0

15 Greg Gamble 15 where n j s the number of ponts n class C j (Y to be deleted (assumng that no two ponts of the 17 ponts are elements of a common collnear trple ths turns out to be easly arranged. Each lne of the followng table s a soluton of ( subject to 0 n j C j (Y. n 0 n 1 n 2 n 3 n n 0 n 1 n 2 n 3 n n 0 n 1 n 2 n 3 n Each of these 23 possbltes lead to desgn constructons. Consder, for example n 0 = 8, n 1 = 7, n 2 = n 3 = 0, n 4 = 2. Each pont n C 4 (Y les n 4 collnear trples of Y the other two ponts of each collnear trple beng a pont of C 32 (Y = {e 1, e 2, e 3, e 4 } together wth a pont from C 1 (Y. Thus after selectng 2 ponts x, y from C 4 (Y we avod 8 of the 32 ponts of C 1 (Y, namely those w that satsfy u + v + w = 0 where u {x, y} and v {e 1, e 2, e 3, e 4 }. Ths gves ( ( ( ways of obtanng a block B wth the desred characterstcs usng ths soluton of (. Of course, the 8 ponts of C 1 (Y that were avoded above no longer belong to a collnear trple of the set Y \ {x, y} so one can delete any of these 8 ponts nstead of the ponts of C 0 (Y to obtan the block B. That s C 0 (Y \ {x, y} = 16 and so the above number of ways can be multpled by a factor of ( Now we consder the queston as to how many of the desgns constructed n ths fashon are parwse non-somorphc. Certanly, (X, B G and (X, B G are non-somorphc f J B J B. For example blocks B and B, defned below are generated n the manner mentoned above, from the frst lne of the table. B = Y \ {(1, 1, 1, 1, 1, 1, 1, (1, 1, 1, 1, 1, 1, 1, (1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, 1, 1, 1, (1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, ε 5, ε 6, ε 7 ε 0 for 5 7}, B = (B {( 1, 1, 1, 1, 1, 1, 1} \ {( 1, 1, 1, 1, 1, 1, 1}. They dffer n that the nnth pont deleted from Y (a pont of C 1 (Y s the pont ( 1, 1, 1, 1, 1, 1, 1 for B, and ( 1, 1, 1, 1, 1, 1, 1 for B. Checkng we fnd that J B = {(0, 9, (1, 18, (2, 48, (3, 32, (4, 4, (28, 3, (29, 1}, J B = {(0, 9, (1, 17, (2, 49, (3, 33, (4, 3, (28, 3, (29, 1}.

16 16 Greg Gamble So, n fact (X, B G and (X, B G are non-somorphc. In ths way, each lne of the table may gve rse to several parwse non-somorphc desgns. Note that as ponts are deleted from Y the pont classes change (for example, suppose x C j (Y then j ponts of C 5 j (Y move nto C 5 j 1 (Y \ {x}, and all other ponts of Y \ {x} reman n ther prevous class. An example of ths was gven above. Wth ths n mnd we may relax the restrcton n j C j (Y. Ths opens up a further 21 possbltes for (n 0,..., n 4 all of whch are feasble. Consderng, these possbltes as well as those already mentoned a prelmnary computer nvestgaton turned up 37 somorphcally dstnct block-transtve 3-(3 7, 115, λ desgns CONSTRUCTION. In a smlar way to the prevous constructon we may construct a block B, wth B = k = 116 such that for G = AGL 7 (3 and X = (Z 3 7, (X, B G s a block-transtve 3-desgn. Ths tme we need T B = q 1 (k = 116. Thus from the set Y (whch has Y = 132 ponts and T Y = 128 collnear trples we must delete 12 ponts n such a way as to reduce the number of collnear trples by 16. Ths can be done, for example, f we fnd a soluton of 4 n j = 16, j=0 4 j.n j = 12. ( j=0 Each lne of the followng table s a soluton of ( subject to 0 n j C j. n 0 n 1 n 2 n 3 n n 0 n 1 n 2 n 3 n n 0 n 1 n 2 n 3 n From these 11 possbltes take n 0 = 8, n 1 = 6, n 2 = 1, n 3 = 0, n 4 = 1. One soluton generated from ths soluton of ( s the followng. B = Y \ {(1, 1, 1, 1, 1, 1, 1, (1, 1, 1, 1, 1, 1, 1, (1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, 1, 1, 1, (1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, 1, 1, 1, ( 1, 1, 1, 1, ε 5, ε 6, ε 7 ε 0 for 5 7}. Relaxng the restrcton that n j C j (Y yelds 21 more possbltes for (n 0,..., n 4. However, 4 of these appear to be nfeasble. A prelmnary computer nvestgaton found the number of parwse non-somorphc block-transtve 3-(3 7, 116, λ desgns to be at least 34. The usefulness of our set W fnshes wth these two constructons. It was not at all obvous how to generalse the above deas. Thus we fnsh wth the followng questons.

17 Greg Gamble QUESTION. What constructons are possble for blocks B for whch (X, B G s a 3- desgn admttng AGL 7 (3 and B = k {230, 437, 552, 646, 667, 760, 761, 875, 876, 990}? 4.9. QUESTION. Can the above technques be appled n the constructon of 3-desgns for other values of (p, d, k? REFERENCES [1] ALLTOP, W. O. : Some 3-desgns and a 4-desgn. J. Algebra (2 11 (1971, [2] BLOCK, R. E. : On the orbts of collneaton groups. Math. Zet 96 (1967, [3] CAMERON, P. J. and PRAEGER, C. E. : Block-transtve t desgns. I: pont-mprmtve desgns. Dscrete Math. 118 (1993, [4] HUGHES, D. R. and PIPER, F. C. : Desgn Theory (Cambrdge Unversty Press, 1985 [5] PRAEGER, C. E. : Block-transtve desgns and maxmal subgroups of fnte symmetrc groups. Australas. J. Combnatorcs 1 (1990, [6] PRAEGER, C. E. : prvate communcaton (1990 Greg Gamble Department of Mathematcs Unversty of Western Australa Nedlands WA 6009 Australa