A construction of disjoint Steiner Triple systems

A construction of disjoint Steiner Triple systems

Citation for published version (APA):

Beenker, G. J. M., Gerards, A. M. H., & Penning, P. (1978). A construction of disjoint Steiner Triple systems. (EUT report. WSK, Dept. of Mathematics and Computing Science; Vol. 78-WSK-01). Eindhoven University of Technology.

Document status and date: Published: 01/01/1978

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TECHNISCHE HOGESCHOOL EINDHOVEN NEDERLAND

ONDERAFDELING DER WISKUNDE

TECHNOLOGICAL UNIVERSITY EINDHOVEN THE NETHERLANDS

DEPARTUENT OF MATHEMATICS

A construction of disjoint Steiner Triple Systems

by

G.F.}1. Beenker, A.H.H. Gerards, P. Penning

T.H.-Report 78-wSK-01 April 1978

Abstract

We show that there are at least 4t + 2 mutually disjoint, isomorphic Steiner triple systems on 6t + 3 points, if t ;?: 4.

- 1

-1. Introduction

2.

Given a finite, non-empty set S of v elements, a Steiner triple system of order v on S is a collection

S

of three-element subsets of S (called triples) such that each pair of distinct elements of S belortgs to exactly one triple of

S.

It is well known that there exists a Steiner triple system of order v if and only if v

=

1 or 3 mod 6.

Let

S1

and

S2

be two Steiner triple systems on the same set S;

Sl

and

S2

are called disjoint if

Sl

n

S2

0,

i.e. if they have no triple in common.

A Steiner triple system of order v is sometimes denoted simply by STS(v) • Let D(v) denote the maximum number of pairwise disjoint STS(v) that can be

*

constructed on a set S of v points. Furthermore denote by D (v) the maximum number of pairwise disjoint, isomorphic STS(v) that can be constructed on the set S.

Since for every 3 ~ i ~ v, the triple {1,2,i} occurs in at most one of the

*

pairwise disjoint STS(v), i t follows that 1 ~ D (v) ::; D(v) ::; v-2 for v;::: 3, v

=

1 or 3 mod 6.

A lower bound for D

*

(6t + 3)

In [lJ J. Doyen proved that for every nonnegative integer t

*

D (6t + 3) ;::: 4t + 1 if 2t +1

_t

0 mod 3 and

*

D (6t + 3) ;::: 4t - 1 if 2t + 1 - 0 mod 3

We shall give a construction which shows that

*

D (6t + 3) ;::: 4t + 2 for t ;::: 4 •

Let t be fixed, t;::: 4. Let G be the ring (Z mod 2t+1,+, ) and let S be the set {AO, ••• ,A2t,BO, ••• ,B2t,CO, ••• ,C2tL Note that, since gcd(2t+ 1,2) =

= gcd(2t+ 1,4) = gcd(2t+ 1,8) = 1,

~, ~

and

~

are well defined elements of G.

For every a E G we denote by

Sl(a)

the set consisting of

i)

all subsets {AJI, ,Ak ,Ba + (JI,+k) } ; {BJI, ,Bk ,C2a+ (JI,+k) }; {CJI, ,Ck,A -a-i-( Q,+k)} of S with JI"k E G, JI,

#

k.

ii) all subsets {AJI"Ba+2J1,'C4a+4J1,} of S with JI, E G.

For every 8 E G we denote by

S2

(13) the set consisting of

i) all subsets {AJI"Ak'C4f3+2(JI,+k)}i{CJI,'Ck,B_(f3+1)+~(JI,+k)}i{BJI"Bk,A~(1_f3)+~(JI,+k)} of S with JI"k c G, JI,

#

k.

2

-1 1 1 2 2 2

Lemma I .

The sets

S (O),S (1) f""S (2t),S (O),S (1), ••• ,S (2t),

obtained in

this

way~

are isomorphic under permutations of

s.

Proof. Let ~ be the permutation of S defined as follows:

for all t E G •

1

It is easy to see that ~ is a one-to-one mapping of the triples of S (a) on-to the triples of Sl (a + 1), for any a £ G.

1 1 1

So we have proved that the sets S (O),S (l), ••. ,S (2t) are isomorphic. Let a E G and let $a be the mapping of S into S defined by: $a(A

t) := At' $a(B t ) : B2t+2a , $a(Ct ) := C~~_(1+a) for all t E G.

Since gcd(2,2t+ 1) = gcd(~,2t+ 1) = 1, $ is a permutation of S. a

Let furthermore ~. be the transposition of S, which interchanges B. and C.

1 1 1

for any i E G.

With the help of the above mentioned permutations, we can define the permu-tation X of S by

a

Obviously Xa is a one-to-one mapping of Sl(a) into S2(a) I because

Xa

{CtICk,A_a~(t+k)}

+

{B~t-1-a,B~k-1-a'~(~t+~k-2-2a)+~(1-a)}

, Xa

{At,Ba+2~'C4a+4~}

+

{A~'C4a+4~,Ba+2t_1}

•

Since gcd(2,2t+ 1) = gcd(~,2t+ 1)

=

gcd(4,2t+ 1)

=

1, one knows that

( n + 2a), (~t - 1 - a) and (a + 2t - 1) run through G if t runs through G. So we can conclude that X is a one-to-one mapping of Sl(a) onto S2(a).

i

2 a

Thus

S

(a) and S (a) are isomorphic for any a E G.

1 1 2 2

Conclusion: The sets S (O)/ •••

'S

(2t),S (0) , •••

,S

(2t) are isomorphic.

n

1 1 1

Lemma II.

S

(0)

is a

STS (6t + 3),

and hence each of the sets

S (O), ••• ,S (2t),

- 3

-Proof.

i) The number of triples in Sl(O) is:

and this equals the number of triples in a STS (6t + 3) •

ii) We shall show that each pair of elements of S belongs to at least one triple of Sl (0) •

Trivially thepairs {A2,Ak},{B£,Bk}'{C2'Ck} (~,k E G, ~

#

k) occur once. The pair {A.,B.} belongs to at least one triple of Sl(O), since:

1. J

a) i f j 2i then {A.,B.} = {A.,B

2.} c {A.,B2

.,c

4.},

1. ] 1. 1. 1. 1. 1.

b) i f j

#

2i then {A. lB.} C {A. ,A . . ,B . . . }.

1. ] 1. J-1. 1.+J-1.

Also the pair {A.,C.} occurs at least once, since:

1. J

a) if j

=

4i then {A.,C.}

=

{A.,C

4.} C {A.,B2

.,c

4.},

1. J 1. 1. 1. 1. 1.

b) if j

#

4i then {A.,C,} C {C"C

S' "A,,(. 8' ')}.

1. J J 1.-J ~ J+ 1.-J

And finally the pair {B"C.} belongs t08at least one triple of Sl(O), 1. J since: a) if j

=

2i then {Bi,C j} b) if j

#

2i then {B.,C.} 1. J {B"C 2.} C {AL .,B.,C₂,}, 1. 1. ~1. 1. 1. c {B"B, "C, • • }. 1. J-1. 1.+)-1.

The combination of i) and ii) shows us, that each pair of distinct elements of S is contained in exactly one triple of Sl(D), and hence Sl(O) is a

STS (6t

+

3) •

IJ

*

Theorem. D (6t + 3) :;:> 4t + 2, j'or t ~ 4.

Proof. Because of the lemmas I and II, i t suffices to show that the 4t+ 2 STS(6t+ 3), Si(j), i ::= 1,2, j E G, are pairwise disjoint.

. \ 1 2

1) Suppose that

S

(a) n

S

(8)

#

~.

The only triples which S1(a) and S2(S) can have in common are the triples

{A.,B, ,C

k· }. Let {A"B. ,Ck} be such a triple. Then there exist elements

1. ] 1. J

~1 and £2 in G such that {A. IB, ,C, }

=

{An ,B 2~ ,C

4 42} and {A. IB, ,Ck} =

1. J K N1 a+ 1 a+ 1 1. ]

=

_{A2

IB2~

+6-1'C

_4S

_{+42 }.} 2 2 2 So we can conclude: i) A~ ::= A2 ' i.e. 21

=

2 2, 1 2 ii) _C4U+

21

=

C48+t2 ' so 4a = 46, which implies that a

= S

(since gcd ( 4 , 2t + 1) = 1),

B

4

-Conclusion: SI(a) n S2(8)

=

0

for any a,S E G. i i

2) Suppose now that S (a

l) n S (a2) ~

0,

i

=

1,2, al,a2 E G. Let (Xj,Xk,Y

t) E Si(al) n Si(a2), X,Y E {A,B,C}, X

':f

Y. Then there exist i

1,i2,i3,i4,a,b and c in G such that

and

Now we can conclude thati₁

=

i3 and i2

=

i4 (or il

=

i4 and i2

=

_{i 3 ,} but that gives the same result) and thus aa

1 + b(i1 + i2) + c

= aa2 + b(i

1 + i2) + c, which implies that aa1 = aa2•

As a E {1, 2,-1,4,4} and thus gcd (a, 2t + 1) = 1, we can conclude that

a₁ a_{2 •}

Finally assume that

then ~1

=

i2 and so again a

1

=

a2•

The combination of 1) and 2) shows us that siCa) n sj(S)

~

0

if and only if

i = j and a

=

S. [l

Reference

[1J J. Doyen: Constructions of disjoint Steiner Triple Systems, Proc. Am. Math. Soc. 32 (1972),409-416.