Further results on transitive Kirkman triple systems

Journal of Statistical Planning and Inference 51 (1996) 189-194

ELSEVIER

journal of statistical planning and inference

Further results on transitive Kirkman triple systems 1 J i a n - g u o Lei Mathematical Institute, Hebei Normal College, Shijiazhuang 050091, China

Abstract

In this paper, we give a new construction for transitive Kirkman triple systems. As a consequence, it is shown that, to prove the existence ofa TKTS(v) for each admissible v, it is sufficient to prove the existence of a TKTS(3p) where p is a prime, p = 5(mod 6).

I. Introduction

F o r given positive integers t, k, and v, a Steiner system S(t, k, v) is a pair (X, ~ ) , where X is a v-set and ~ is a set of k-subsets (called block) of X, such that each t-subset of X is contained in a unique block. An S(2, 3, v), ( X , ~ ) , is called Kirkman triple system (briefly KTS(v)), if ~ can be partitioned into ~ S(1, 3, v)s. It is well known that there exists a K TS (v) if and only if v = 3 (mod 6). A permutation group G on X is called sharply transitive if IG[ = IX 1. A K i r k m a n triple system ( X , ~ ) is called transitive (briefly TKTS), if there exists a sharply transitive group G on X under which ~ is invariant, i.e. for any {x, y, z} ~ and any ct~ G, ct({x, y, z}) = {ct(x), ~(y), ~(z)} ~ . G is called the adjoint group of the TKTS(v). Transitive K i r k m a n triple systems have i m p o r t a n t applications e.g., in constructing large sets of disjoint K i r k m a n triple systems. We have the following results. L e m m a 1 (Denniston, 1979). There exists a TKTS(3m), i f m ~ {5, 11, 17, 25, 35}. L e m m a 2 (Wu, 1990). There exists a TKTS(3q), if q is a prime power, q - 1 (mod 6). L e m m a 3 (Lei and Chang, 1990). I f there exists a T K T S ( v l ) and a TKTS(v2), then there exists a TKTS(vl v2).

Research supported by NSFH Grant No. 193013. 0378-3758/96/$15.00 © 1996--Elsevier Science B.V. All rights reserved SSDI 0 3 7 8 - 3 7 5 8 ( 9 5 ) 0 0 0 8 3 - 6

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Jian-guo Lei / Journal of Statistical Planning and Inference 51 (1996) 189 194

Lemma 4 (Denniston, 1979; Lei and Chang, 1990). I f there exists a TKTS(v), then there exist a TKTS(3v) and a TKTS(qv), where q is a prime power, q = 1 (mod 6).

2. Property of TKTS(v) L e m m a 5. Let G be a sharply transitive group on X. (i) For any ct, fle G and any x e X , if ct(x) = fl(x) then ct = ft. (ii) For any ct e G, all the cycles in a cycle decomposition of ct are of the same length and which is a divisor of IX[. (iii) I f ~ e G, ct is not the identity element of G, and ~ is a product of t disjoint m-cycles, t > 1, then there exists fl • G, and fl ~ ( ~}, such that either flJ = ~ for some j, or the elements of every cycle of ~ belong to different cycles of fl, respectively. (iv) I f IX[ = pq, p < q, p and q are primes, and q ~ 1 (mod p), then G is a cyclic group. Proof. It is easy to prove the lemma by group theory. Here we only point out that to prove (iv) the assumption of sharply transitive of G is not necessary. []

Corollary 1. Let group on X. I f

IXI = 3p, p is a prime, p = 5 (mod 6), let G be a sharply transitive

O~-~- (XlYl21 ) (x2Y2Z 2 )'"(xvypZp) • G,

fl = (xlx2""xp) (Yl"") ( Z l " ' ) • G then fl = (xxx2...xp) (YlY2""Yp) (z122"" "2"p) and G = (ctfl>.

Proof. Since IXF = 3p, by L e m m a 5 and Sylow Theorem, G is a cyclic group, and there exists an element of order 3 and an element of order p. Obviously, ~t and fl are such elements, and ~fl = fla. We have fl(Yi) = flot(xi)= o~fl(xi)= a ( x i + x ) = Yi+ 1,

and similarly, fl(zi) = zi+l. Thus, fl = (xlx2...Xp) (YlY2""Yp)(zlz2...Zp). Obviously, 6 = < ~ > . []

Corollary 2. Let p be a prime, p - 5 ( m o d 6 ) , X={a,b,c}xlp,~=(abc), fl= (0 1 ...p - 1), then the sharply transitive group on X is isomorphic to G = ( ~ ) x (fl). Lemma 6. Let ( X , ~ ) be a T K T S ( v ) with adjoint group G, then (i) There exists an S(1, 3, v) c ~ , which is f i x e d by G. (ii) Let p be a prime, p - 5 (mod 6), v = 3p, i f ~ o is a f i x e d S(1, 3, v) in (i), then for any B e ~ \ ~ o and any ~ ~ f l e G , we have ~(B) ¢fl(B).

Jian-guo Lei / Journal of Statistical Planning and Inference 51 (1996) 189-194

Proof. Let ~ be partitioned into orbits ~o, ~ ..... ~ obvious that

191

under the action of G. It is

1___~.) = I~1 = ~ I~il and I~l ~< v.

1)(t)

6

i=O

Since v = 3 (rood 6), then ~

~ 0 ( m o d v), it follows that there exists an orbit ~'o with

I~ol < v. If T = {x, y,z} ~ o , then ~ o = {g(T); gEG}. Since IGI = v, I~ol < v, there exist O~ CgEeG, such that gl(T) = 92(T). Let Gr = {g~G; g(T) = T}, then Gr is a subgroup of G, and I G r l > l (e6Gr and g~g~X~Gr). For any g6GT, g # e , then Ü({x, y, z}) = {x, y, z}, by Lemma 5(i). It follows that o(x) = y,

o(y) = z,

o(x) = x,

or

g(x)= z, g(z)= y, g(y)= x. and g3(x) = x, thus g3 = e, and the cycle (x, y, z) is contained in 9 or Oz. By Lemma 5(ii), 9 consists of ~ 3-cycles, thus Gr = ( g ) and [Grl = 3. Furthermore, we can write T = {x, g(x), 92(x)}. Let {~i; 1 ~< i ~< ~} be all representative of left coset of Gr in G, then

is a S(1, 3, v) in ~ and fixed by G. (ii) For v = 3p, p is a prime, p - 5 (mod 6), G is a cyclic group by Lemma 5(iv), and Gr is the unique 3-subgroup of G by Sylow Theorem, thus Gr is independent of the choices of T(Te.~o). Let Gr = G3, then

~o =

{~i(x), o ( ~ ( x ) ) , 0~(~,(x))}; 1 <. i <~ 5 x e X

where 9 # e. Obviously, for any T ~ o 9 ~ G3 and 9 # e, then

,

and any g~G3, we have o(T)= T. Thus if

3

a = ~ (~,(x) o(~(x)) o2(oci(x))). i=1

In other words, if (x, y, z) is a 3-cycle of 9, then {x, y, z} e ~o. For any B ~ \ ~ o and • ¢: fl~G, we have ~(B) ¢ fl(B). Otherwise, if ~(B)= fl(B), then 0~- ~fl(B) = B, furthermore, ~- lfl is an element of order 3 in G, since G3 is unique, then ~-xfl ~ G3 and ~-~fl ¢ e. As the cycle decomposition of the elements of G3 has been determined, it can be easily seen that B ~ ~o, a contradiction. []

Jian-guo Lei/Journal of Statistical Planning and Inference 51 (1996) 189-194

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3. A recursive construction Let X be a v-set with an o r d e r " < ", (X,N) be a TKTS(v) and its adjoint group be G. If N can be partitioned into orbits No, ~a ..... ~ under G, such that [N'ol = ], [~1 = v, 1 ~< i ~< s, then it is called a good TK TS (v). By Lemma 6, we immediately have L e m m a 7. For p is a prime, p = 5(mod6), any TKTS(3p) must be good. Theorem 1. Let p be a prime, p -

5(mod6), /f there exist both a TKTS(v) and a TKTS(3p), then there exists a TKTS(pv). Construction. Let ( X , ~ ) be a good TKTS(v). For each ~i (1 ~< i ~< s), we can choose a fixed element {xi, Yi, Zi} E~i, such that ,~i = {{~(xl), ~(yi), ~(Zl)};~G}. Let

~ i = {(~(x3, ~(y3 ~(z3); ~ e G , x~ < y~ < z~},

i=1

then each element of ~ \ ~ o is ordered in ~ . If (I3 x Iv, ~ ) is a TKTS(3p), where p is a prime, p - 5(rood6), by Corollary 2, its adjoint goup is (c~)x (fl), where ( e ) = ((0 1 2)), (fl) = ((0 1 ...p - 1)). We can form the block set s~ o f a TKTS(pv) on X x Ip as follows: (i) For each B = {x, y, z} e N o , we construct a rKTS(3p) (B x lp, s4B) on B x I~, whose adjoint group is ((x y z ) ) x (fl), and ~ 8 can be partitioned into a~-i S(1, 3, 3p)s, s ~ , 0 <~j ~ 3~- 3. (ii) For any {x, y, z} e ~ \ ~ o , if B = (x, y, z) e ~ , let s~B = {{(x, t), (y, m), (z, n)}; t + m -= 2n(modp)}, s ~ = {{(x, t), (y, m), (z, n)}; t + m - 2n(mod p), t - m = k(modp)}, where 0 ~< k ~< p - 1. It can be seen that every ~ 0

~< k ~< p - 1 is a S(1, 3, 3p) on BxI r Let d = ~ { d B ; B e ~ o u ~ } . It is easy to see that I~1 - pv(P6-1)

Proof. It is easy to show that (X × Ip, d ) is a S(2, 3, pv). As ( X , ~ ) is a TKTS(v), we may let ~ be partitioned into ~ S(1, 3, v)s, C1, C2 ..... C~- ~. N o w let 2

~k=

U dk,

O<~k~p-1,

BeCk

~J=

U ~J, Be~o

v-1 l<~h<~--, 2

P~J<~3p--3 2

Jian-guo Lei/ Journal of Statistical Planning and Inference 51 (1996) 189 194

Since for each B e ~ , ~,k is a S(1, 3, 3p) on B x Ip, and C~,

C 2 .....

193

Cv- ~ and ~ o are 2

S(1,3, v ) s o n X , t h e n e a c h ~ k , O < ~ k < < . p - 1, l<<.h<~ -~2 andeachdJp<<.j<~ 3p-32 is a S(1, 3, pv) on set X x Ip. These S(l, 3, pv)s form a partition of ~¢. Since the total pv1 n u m b e r of S(1, 3, pv)s is ~v21~P+ p_l _ ~- , then (X x 1p, d ) is a KTS(pv). Obviously, the group G x (fl) is sharply transitive on set X x Ip. Now, we show that d is invariant under G x (fl). Let T = {(x, t), (y, m), (z, n)} e d . (1) If T is contained in part (i) of sO, then there exists a B e ~ o , such that T e riB, for any ~ e G , ~ ( B ) ~ o (as ~ o is fixed by G), and the T K T S ( 3 p ) = (~(B) x Ip, d~¢m) is contained in part (i) of d , its adjoint group is ((~(x) ~(y) ~(z))) x (fl). Thus for any (a, t/)eG x (fl), (a, r/) ( T ) ~ l ~ t m . (2) If T = {(x, t), (y, m), (z, n)} is contained in part (ii) of s¢, then B = {x, y, z ) ~ i for some i, 1 ~< i ~ s , and there exists a a e G , such that B = {a(xi),a(yi),a(zi)}. Suppose a(xl) = x, a(yi) = y, a(zi) = z. Since (a(xl), a(yl), a(zl))~ ~i, then (x, y, z)~ ~i. Further, we have t + m m 2n(modp). Since for any t / e ( f l ) , t / = fir for some r, 0 ~< r ~< p - 1, but for any selp, fl(s) = s + 1, therefore r/(t) = t + r,t/(m) = m + r, r/(n) = n + r, and r/(t) + r/(m) _= 2t/(n) (mod p). Since for any ct ~ G, (a(x), a(y), a(z)) E ~i, thus for any (a, t/) ~ G x (fl), (c~, r/) ( r ) = {(~(x), t/(t)), (c~(y),it(m)), (~(z), it(n))} is contained in part (ii) of ~¢. Thus (X x lp, s¢) is a TKTS(pv) and its adjoint group is G x (fl). Lastly, we show that this TKTS(pv) is good. Let B = {x, y, z} e~'o, (B x Ip, sJtj) be a TKTS(3p) on B x lp, with adjoint group GB = ((x y z)) x ((0 1... p -- 1)), and ~ o be invariable S(1, 3, 3p) under Gn, then the elements of order 3 in Gn are ((x y z), e) and ((xzy), e). By the proof of L e m m a 6(ii), ~1o = {{(x, i),(y, i),(z, i)}; ielp}. It is obvious that ~1o c ~1. Let

~¢o= U ~ffo= {{(x,i),(y,i),(z,i)}; ielp, { x , y , z } e ~ o } , Be~to

then ~¢o is a S(1, 3, pv) fixed by G × (fl). It is easy to see that the remaining orbits in ~¢ under G x (fl) are of length pv. This completes the proof. []

4. Main results Theorem 2. I f there exists a TKTS(3p) for any prime p v = 3 (mod 6), there exists a TKTS(v).

5(mod6), then for any

Proof. F o r v = 3 (mod 6), we can write v = 3'ql ... q,pl ... Pk, where r is a positive integer, qi are prime powers, qi -- 1 (mod 6), 1 ~< i ~< t, and pj are primes, pj = 5 (mod 6) 1 ~
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a T K T S ( 3 p l . . . p k ) by L e m m a 7 and T h e o r e m 1. Thus by L e m m a 4, there exists a T K T S ( 3 " q l ... q,Pt ""Pk) = T K T S ( v ) .

[]

Thus, the complete solution for T K T S ( v ) will only depend on the existence of T K T S ( 3 p ) for every prime p = 5 (mod6). But for prime p = 5 (mod 6), we only have k n o w n that there exists TKTS(3"5), T K T S ( 3 . 1 1 ) and TKTS(3.17). The existence of T K T S ( 3 p ) for p ~> 23 is still open. But we can still get some new T K T S s by Therem 1.

Corollary 3. There exists a TKTS(3"I5r~ll'317r'ql ""qt), where r2, rs, r4~{0, 1}, rt > 1, qi are prime powers, qi -= 1 (mod 6), 1 ~< i ~< t.

References Denniston, R.H.F. (1979), Four doubly resolvable complete 3odesign. Ars Combin., 7, 256-272. Denniston, R.H.F. (1979). Further case of double resolvability. J. Combin. Theory, Ser. A, 26, 298-303. Lei, J. and Y. Chang (1990). On the transitive Kirkman triple systems, J. Hebei Academy of Sciences, 2, 1-7. Wu, L. (1990). On large set of KTS(v), Combinatorial Designs and Application, Lecture Notes in Pure and Applied Mathematics, Vol. 126. Marcel Dekker, New York, 175-178.