Steiner triple systems without parallel classes

We construct Steiner triple systems without parallel classes for an infinite number of orders congruent to $3 \pmod{6}$. The only previously known examples have order $15$ or $21$.

The existence of almost parallel classes (sets of disjoint triples containing all the points except one) in Steiner triple systems of order congruent to 1 (mod 6) has also been investigated. It is conjectured that for all v ≡ 1 (mod 6) except v = 13 there exists a Steiner triple system of order v with no almost parallel class [9]. Two sparse infinite families of such Steiner triple systems have been found, one by Wilson (see [9]) and one by the authors [2]. Also see [6] for a recent related result.
For a prime p, let ord p (x) denote the multiplicative order of x in Z p . Let P be the set of odd primes given by p ∈ P if and only if ord p (−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 ord p (−2) ≡ 0 (mod 4). If p ≡ 5 (mod 8) is prime, then −2 is a quadratic non-residue in Z p by the law of quadratic reciprocity and it follows that ord p (−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 p 1 , . . . , p t 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 = 5p 1 · · · p t + 2 with 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 = p 1 p 2 · · · p t be the prime factorisation of n. By our hypotheses ord p i (−2) ≡ 0 (mod 4) for i = 1, 2, . . . , t. Let p 0 = 5 and let H = Z p 1 × Z p 2 × · · · × Z pt and G = Z p 0 × H be additive groups with identities 0 H = (0, 0, . . . , 0) and 0 G = (0, 0 H ). If m = a b where a and b are integers such that gcd(b, p i ) = 1 for i = 0, . . . , t, and g = (g 0 , g 1 , g 2 , . . . , g t ) ∈ G, then we define mg by mg Observe that each pair of distinct elements of G occurs in at most one triple of B 0 , and that the pairs that occur in no triple of B 0 are exactly the edges of the graph X with vertex set V (X) = G and edge set Thus, 0 G is an isolated vertex in X and each vertex g ∈ G \ {0 G } 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 \ {0 G } 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 an arbitrary edge g of X has vertices −g, 2g, . . . is of the form We claim that there is a function γ : Z 5 ×(H \{0 H }) → {1, 2} such that for all g, g ′ ∈ Z 5 ×(H \{0 H }) (1) γ(g) = γ(−2g); and It is clear that γ satisfies (1). It also satisfies (2) because |Y | ≡ 0 (mod 4) for each Y ∈ Y, and because 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 Also, let ({(0, 0), (1, 0), (2, 0), (3, 0), (4, 0), ∞ 1 , ∞ 2 }, B * ) be any Steiner triple system of order 7 such that It is easily seen that 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 0 G . Since the sum of the elements of G is 0 G , the sum of the weights of the triples in C is also 0 G . Now note the following properties concerning weights of triples which we will use in the arguments that follow. Every triple in B 0 has weight 0 G , every triple in B * has weight in Z 5 × {0 H }, and every triple in B ∞ has weight in Z 5 × (H \ {0 H }). The weight of the triple that contains both ∞ 1 and ∞ 2 is not 0 G .
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 of B ∞ contains either ∞ 1 or ∞ 2 ). This means that the remaining triples of C are from B 0 . But this is a contradiction because every triple in B 0 has weight 0 G , and the weight of the triple that contains both ∞ 1 and ∞ 2 is not 0 G . We conclude that C contains distinct triples T 1 and T 2 such that ∞ 1 ∈ T 1 and ∞ 2 ∈ T 2 , and that the remaining triples of C are all from B 0 ∪ B * .
Since every triple in B 0 ∪ B * has weight in Z 5 × {0 H }, the sum of the weights of T 1 and T 2 is also in (2) of γ. We conclude that S has no parallel class.