Steiner Triple Systems Without Parallel Classes

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For a prime p, let ordp(x) denote the multiplicative order of x in Zp. Let P be the set of odd primes given by p ∈ P if and only if ordp(−2) ≡ 0 (mod 4). Observe that P contains all of the inﬁnitely many primes congruent to 5 (mod 8), no primes congruent to 3 or 7 (mod 8), and some primes congruent to 1 (mod 8). Clearly, if p ≡ 3 (mod 4) is prime, then ordp(−2) 6≡ 0(mod 4). If p ≡ 5 (mod 8) is prime, then −2 is a quadratic non-residue in Zp by the law of quadratic reciprocity and it follows that ordp(−2) ≡ 0 (mod 4). Our theorem states that there is a Steiner triple system of order v with no parallel class whenever v ≡ 27(mod 30) and p ∈ P for each prime divisor p of v − 2. Thus, for any list p1,...,pt of (not necessarily distinct) primes from P in which the number of primes congruent to

5 (mod 6) is odd, we obtain a Steiner triple system of order v =5p1 ··· pt + 2 with no parallel class.

Theorem 1. Let V be the set of positive integers given by v ∈ V if and only if v ≡ 27 (mod 30) and ordp(−2) ≡ 0(mod 4) for each prime divisor p of v − 2. Then V is inﬁnite and for each v ∈ V there exists a Steiner triple system of order v having no parallel class.

Proof. It is clear from the remarks preceding the theorem that V is infinite. We now show that if v ∈V, then there exists a Steiner triple system of order v having no parallel class. Let v ∈V, define n by v = 5n + 2 (note that n is an odd integer because v ≡ 27(mod 30)), and let n = p1p2 ··· pt be the prime factorisation of n. By our hypotheses ordpi (−2) ≡ 0(mod 4) for i = 1, 2,...,t. Let p0 = 5 and Z Z Z Z let H = p1 × p2 ×···× pt and G = p0 × H be additive groups with identities 0H = (0, 0,..., 0) a and 0G = (0, 0H ). If m = b where a and b are integers such that gcd(b, pi) = 1 for i = 0,...,t, and −1 −1 −1 −1 g = (g0,g1,g2,...,gt) ∈ G, then we define mg by mg = (ab g0, ab g1,...,ab gt) where ab gi is Z evaluated in the field pi for i =0,...,t. Let A0 = {{g,g′,g′′} : g,g′,g′′ ∈ G, |{g,g′,g′′}| =3,g + g′ + g′′ =0}.

Observe that each pair of distinct elements of G occurs in at most one triple of A0, and that the pairs that occur in no triple of A0 are exactly the edges of the graph X with vertex set V (X)= G and edge

2 set

E(X)= {{g, −2g} : g ∈ G \{0G}}.

Thus, 0G is an isolated vertex in X and each vertex g ∈ G \{0G} has exactly 2 neighbours in X, namely 1 − 2 g and −2g. That is, X is a collection of disjoint cycles together with an isolated vertex. To each edge of X assign a weight equal to the sum in G of its endpoints. It follows that for each g ∈ G \{0G} there is a unique edge in X having weight g (namely the edge {−g, 2g}). Moreover, the edge of X having weight g has an adjacent edge of weight −2g. Henceforth we denote the edges of X by their weights and describe the cycles of X by listing their edges in cyclically ordered tuples. So the cycle containing 1 1 an arbitrary edge g of X has vertices −g, 2g, . . . , 2 g, edges {−g, 2g}, {2g, −4g},..., { 2 g, −g}, and is denoted by

1 (g, −2g, . . . , − 2 g).

Let θ : H \{0H } → H \{0H } be given by θ(h) = −2h, and let Y be the set of orbits of θ. Since ordpi (−2) ≡ 0(mod 4) for i =1, 2,...,t, we have |Y | ≡ 0(mod 4) for each Y ∈Y. Thus, each cycle of X except

C = ((1, 0H), (3, 0H), (4, 0H), (2, 0H)) is of the form 2 |Y |−1 (a, h), (3a, θ(h)), (4a, θ (h)),..., (2a, θ (h)) for some a ∈ Z5, h ∈ Y and Y ∈Y. Note that E(X) \ E(C)= Z5 × (H \{0H }). ′ We claim that there is a function γ : Z5 ×(H \{0H }) →{1, 2} such that for all g,g ∈ Z5 ×(H \{0H })

(1) γ(g) =6 γ(−2g); and

′ ′ (2) if g + g ∈ Z5 ×{0H }, then γ(g)= γ(g ).

For each Y ∈Y, let −Y = {−h : h ∈ Y }. It is clear that if Y ∈Y, then −Y ∈Y. Thus, either −Y = Y or Y and −Y are distinct orbits in Y. For each Y ∈ Y, choose a representative element hY ensuring that if Y and −Y are distinct, then h−Y = −hY . Write H \{0H } as a union of disjoint sets H1 and H2 given by

i i H1 = [ {θ (hY ): i =0, 2,..., |Y | − 2} and H2 = [ {θ (hY ): i =1, 3,..., |Y | − 1}. Y ∈Y Y ∈Y

Since |Y | is even for each Y ∈Y, H1 and H2 are well defined. For each a ∈ Z5 and each h ∈ H \{0H } define the function γ : Z5 × (H \{0H }) →{1, 2} by γ((a, h)) = 1 if h ∈ H1 and γ((a, h)) = 2 if h ∈ H2. It is clear that γ satisfies (1). It also satisfies (2) because |Y | ≡ 0(mod 4) for each Y ∈Y, and because we have h−Y = −hY when Y and −Y are distinct.

3 We now construct a Steiner triple system of order v = |G| + 2 which we will prove has no parallel class. Let ∞1 and ∞2 be two new points (not in G) and let

∞ B = {{−g, 2g, ∞γ(g)} : g ∈ Z5 × (H \{0H})}.

∗ ∗ Also, let ((Z5×{0H })∪{∞1, ∞2}, B ) be any Steiner triple system of order 7 such that {0G, ∞1, ∞2} ∈/ B 0 0 and let B = A \ {{0G, (1, 0H), (4, 0H)}, {0G, (2, 0H), (3, 0H)}}. It is easily seen that

0 ∞ ∗ S =(G ∪ {∞1, ∞2}, B ∪ B ∪ B ) is a Steiner triple system of order v. We now show that S has no parallel class. For a contradiction, suppose C is a parallel class of S. Deﬁne the weight of each triple of S to be the sum in G of its points, where the points ∞1 and ∞2 are treated as 0G. Since the sum of the elements of G is 0G, the sum of the weights of the triples in C is also 0G. Now note the following properties concerning 0 weights of triples which we will use in the arguments that follow. Every triple in B has weight 0G, ∗ ∞ every triple in B has weight in Z5 ×{0H }, and every triple in B has weight in Z5 × (H \{0H }). The weight of the triple that contains both ∞1 and ∞2 is not 0G. ∗ If the triple that contains both ∞1 and ∞2 is in C, then no other triple of B is in C (because any two triples of a Steiner triple system of order 7 intersect) and no triple of B∞ is in C (because every triple

∞ 0 of B contains either ∞1 or ∞2). This means that the remaining triples of C are from B . But this is a 0 contradiction because every triple in B has weight 0G, and the weight of the triple that contains both

∞1 and ∞2 is not 0G. We conclude that C contains distinct triples T1 and T2 such that ∞1 ∈ T1 and 0 ∗ ∞2 ∈ T2, and that the remaining triples of C are all from B ∪ B . 0 ∗ Since every triple in B ∪ B has weight in Z5 ×{0H }, the sum of the weights of T1 and T2 is also in ∗ ∞ Z5 ×{0H }. This means that either both T1 and T2 are in B or both are in B (because every triple in ∗ ∞ B has weight in Z5 ×{0H }, and every triple in B has weight in Z5 × (H \{0H })). The triples T1 and ∗ ∗ T2 cannot both be in B because any two triples of B intersect. On the other hand, if T1 and T2 are ∞ both in B , then the fact that the sum of the weights of T1 and T2 is in Z5 ×{0H } contradicts Property (2) of γ. We conclude that S has no parallel class.

Acknowledgements: The authors acknowledge the support of the Australian Research Council via grants DE120100040, DP120100790, DP120103067, DP150100530 and DP150100506.

4 References

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