Some progress on the existence of 1-rotational Steiner triple systems

A Steiner triple system of order v (briefly STS(v)) is 1-rotational under G if it admits G as an automorphism group acting sharply transitively on all but one point. The spectrum of values of v for which there exists a 1-rotational STS(v) under a cyclic, an abelian, or a dicyclic group, has been established in Phelps and Rosa (Discrete Math 33:57–66, 1981), Buratti (J Combin Des 9:215–226, 2001) and Mishima (Discrete Math 308:2617–2619, 2008), respectively. Nevertheless, the spectrum of values of v for which there exists a 1-rotational STS(v) under an arbitrary group has not been completely determined yet. This paper is a considerable step forward to the solution of this problem. In fact, we leave as uncertain cases only those for which we have v = (p3−p)n + 1 ≡ 1 (mod 96) with p a prime, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${n \not\equiv 0}$$\end{document} (mod 4), and the odd part of (p3 − p)n that is square-free and without prime factors congruent to 1 (mod 6).


Introduction
A Steiner 2-design of order v and block-size k, briefly S (2, k, v), is a pair D = (V, B) where V is a set of v points and B is a set of k-subsets of V (blocks) such that every 2-subset of V is contained in exactly one block. An automorphism group of such a design is a group G of bijections on V preserving B. The design is 1-rotational under G if G fixes one point ∞ and acts regularly (i.e., sharply transitively) on the others. In this case it is natural to identify the point-set V with G ∪ {∞} and the action of G on V with the multiplication (or addition) on the right with the rule that ∞ · g = ∞ (or ∞ + g = ∞) for every g ∈ G.
An S (2, 3, v) is usually called a Steiner triple system of order v, briefly STS(v), and it is very well known that an ST S(v) exists if and only if v ≡ 1 or 3 (mod 6) (see [6,10]). In this paper we carry on the investigation, begun in [8], regarding the existence of a 1-rotational STS of an assigned order v. The spectrum C 1r of values v for which there exists a 1-rotational ST S(v) under the cyclic group, namely under Z v−1 , was determined by Phelps and Rosa [16]: C 1r = {v | v ≡ 3 or 9 (mod 24)}.
The spectrum A 1r of values of v for which there exists a 1-rotational ST S(v) under a suitable abelian group was determined by the second author [8]: Finally, the spectrum Q 1r of values of v for which there exists a 1-rotational ST S(v) under the dicyclic group 1 was determined by Mishima [15]: Nevertheless, the spectrum G 1r of all values of v for which there exists a 1-rotational ST S(v) under an arbitrary group G has not been completely determined yet. As observed in [8] every 1-rotational STS is, in particular, a reverse STS, i.e., it admits an involutory automorphism. So G 1r is obviously contained in the set (determined by Rosa, Doyen and Teirlinck [17,13,19]) of values of v for which there exists a reverse STS(v): G 1r ⊂ {v | v ≡ 1 or 3 or 9 or 19 (mod 24)}.
Thus, in view of (1), the problem of determining G 1r reduces to that of establishing for which v ≡ 25 or 43 or 49 or 67 (mod 72) there exists a 1-rotational STS(v) under a suitable group of order u = v − 1. Some partial answers have been already got in [8]. For instance it was proved that the existence is ensured in the case that u has at least one prime factor p ≡ 1 (mod 6).
As a consequence of more general results concerning Steiner 2-designs, we briefly recall how 1-rotational STSs can be constructed. The list of differences of a given subset B of an additive (or multiplicative) group G is the multiset ∆B consisting of all possible differences x − y (or quotients xy −1 ) between two distinct elements x and y of B. A partial spread (PS) of a group G is a set Σ of subgroups of G intersecting each other trivially. A (G, Σ, k, λ) difference family (DF) is a set F = {B 1 , ..., B t } of k-subsets of G whose list of differences ∆F = ∆B 1 ∪ ... ∪ ∆B t is disjoint with every group of the partial spread Σ and covers exactly λ times all elements of G not lying in some group of Σ. In other words, every g ∈ G − ∪ S∈Σ S is representable in exactly λ ways in the form g = x − y (or g = xy −1 ) with (x, y) an ordered pair of distinct elements of some B ∈ F , while no element of ∪ S∈Σ S admits such a representation. Note that a (G, Σ, k, λ)-DF in which Σ contains only the trivial subgroup of G is an ordinary difference family; one usually says that it is a (v, k, λ)-DF in G where v denotes the order of G (see [1] or [6]). A k-PS of a group G is a partial spread of G all members of which have order k; a k * -PS of G is a partial spread of G with exactly one member of order k − 1 and all the others of order k. If k is a prime, every S (2, k, v) admitting an automorphism group G acting sharply transitively on the points is completely equivalent to a (G, Σ, k, 1)-DF where Σ is a k-PS (see [7]). Similarly, for k a prime, every S(2, k, v) design that is 1-rotational under G is completely equivalent to a (G, Σ, k, 1)-DF with Σ a k * -PS of G (see [9]). Thus, in particular, we can state: Let F be a difference family as in the above theorem and let D be the STS generated by it. A complete system of representatives for the G-orbits of the blocks of D is given by Observe that if F is a (G, Σ, 3, 1)-DF with Σ a 3 * -PS, then G has exactly one involution, namely the involution of the unique component of Σ of order 2. In the opposite case we would have another involution h representable as a difference x − y (or xy −1 ) of a block of F. It would follow that h = y − x (or h = yx −1 ) is another representation of h as a difference from F, that is absurd. A group G with exactly one involution g * is usually said binary and its unique subgroup of order 2, that is {0, g * }, will be denoted by Λ(G). In view of the above paragraph an STS may be 1-rotational under G only if G is binary. Throughout the paper, in most cases the group G will be denoted additively. Sometimes, however, we will also adopt the multiplicative notation according to the situations. Now observe that if F is a (G, Σ, 3, 1)-DF with Σ a 3 * -PS, then we have 6|F| = |G| − 2|Σ| since every triple of F produces exactly 6 differences and the union of all subgroups in Σ has size 2|Σ|. Thus, if G has order divisible by 3 the size of Σ is also divisible by 3, i.e., the number of its components of order 3 is congruent to 2 (mod 3) and hence G, besides being binary, must have at least two subgroups of order 3. For this reason, throughout the paper, any binary group with more than one subgroup of order 3 will be said admissible. Hence we can state Proposition 1.2. A necessary condition for the existence of a 1-rotational In the next section we will determine the set of all admissible orders v − 1. The third section will deal with even starters in a binary group, a concept allowing us, in the fourth section, to establish the existence of a 1-rotational STS(v) for every v ≡ 1 or 19 (mod 24) such that the odd part of v − 1 is not square-free. In the fifth section we will give an explicit construction for a 1-rotational STS(v) for any v ≡ 25 (mod 48) and any v ≡ 49 (mod 96). Putting together all these results with those obtained in [8], we conclude that the existence question for a 1-rotational STS(v) remains open only in the very special cases that the following conditions simultaneously hold: • v = (p 3 − p)n + 1 with n ̸ ≡ 0 (mod 4) and p a prime; • (p 3 − p)n = 2 ℓ 3p 1 p 2 ...p t with ℓ ≥ 5 and the p i 's pairwise distinct primes congruent to 5 modulo 6.
Note that the second of the above conditions can be equivalently formulated by saying that v − 1 is divisible by 96 and that its odd part is square-free and without prime factors congruent to 1 (mod 6).
2 On the existence of admissible groups of order 24n + 18 or 24n In this section we are able to establish the set of values of u ≡ 0 or 18 (mod 24) for which at least one admissible group of order u exists obtaining in this way some non-existence results for 1-rotational STS(24n + 1). First recall the following known result (see Example 3 at page 106 in [18]). As a consequence of the above lemma we get the following.

Lemma 2.2.
Let G be a group of order 3t with t a square-free integer whose prime factors are all congruent to 2 (mod 3). Then G has exactly one Sylow 3-subgroup.
Proof. By Sylow's theorem, the number n 3 of Sylow 3-subgroups of G is a divisor of t and by assumption on the prime factors of t it is obvious that there is no prime power divisor of t that is congruent to 1 (mod 6). Hence n 3 can be expressed as a product of integers each of which is the number of Sylow 3-subgroups of a composition factor of G by Theorem 2.1. On the other hand G is solvable since its order is not divisible by 4. Hence every composition factor of G has prime order so that the number of its Sylow 3-subgroups is 0 or 1. We conclude that n 3 = 1. 2 From now on, V q will denote the elementary abelian group of order q, namely the additive group of the field F q with q elements. The set of all u ≡ 18 (mod 24) which are orders of an admissible group was essentially established in [8]  ( * ) 4n+3 is square-free and all its prime factors are congruent to 2 (mod 3).
Proof. If ( * ) holds, the assertion follows from Lemma 2.2. If ( * ) does not hold we can write 4n + 3 = 3t or 4n + 3 = qt for a suitable prime power q ≡ 1 (mod 6). In the former case an admissible group of order u is given by Z 2t × V 9 while, in the latter, it is given by where ϵ is a cubic primitive root of unity of F q and where + ϵ is the semidirect product of Z 3 by V q defined by the rule (a, b) For convenience of the reader we recall the basic definitions about some classical groups. The general linear group of degree n over F q is the group GL n (q) of all n × n invertible matrices with elements in F q . The center of this group is the set of all scalar matrices, namely those of the form kI n with k ∈ F q − {0} and where I n is the n × n identity matrix.
The special linear group SL n (q) is the subgroup of GL n (q) consisting of all matrices with determinant 1. Its center has order gcd(n, q − 1) and consists of all scalar matrices kI n with k a n-th root of unity in F q . We note, in particular, that SL 2 (q) is an admissible group for every odd prime power q.
The projective linear group P GL n (q) and the projective special linear group P SL n (q) are the quotients of GL n (q) and SL n (q) by their centers, respectively. Finally, the general semilinear group ΓL n (q) and the projective semilinear group P ΓL n (q) are the semidirect products of Aut(F q ) by GL n (q) and by P GL n (q), respectively, where Aut(F q ) is the group of field automorphisms of F q . When q is a prime it is clear that we have ΓL n (q) = GL n (q) and P ΓL n (q) = P GL n (q).
A binary 2-group is well known to be either cyclic or dicyclic (in the second case it is also called a generalized quaternion group). Now note that if S is a cyclic or dicyclic group, then the quotient S/Λ(G) is cyclic or dihedral, respectively. Hence, if S is a Sylow 2-subgroup of a binary group G, then S/Λ(G) is to be either cyclic or dihedral. So the structure of binary groups is intimately related to the one of groups with cyclic or dihedral Sylow 2-subgroups, and it admits the following description which exploits the Burnside's Transfer Theorem and the Gorenstein-Walter Theorem [14].

Theorem 2.4. Let G be a binary group and let O(G) be the largest normal subgroup of G of odd order. Then G/(Λ(G) · O(G))
is either a 2-group or isomorphic to one of the following groups: (i) a subgroup of P ΓL 2 (q) containing P SL 2 (q) for a suitable odd prime power q; (ii) the alternating group A 7 .
We also need the following theorem. As already observed in [5], the result of Theorem 2.5 is known to some group theorists, but we are not aware of an original proof in the literature. We refer to [5] for a proof suggested by Glaubermann.
Here is the main result of this section.
Theorem 2.6. There exists an admissible group of a given order u ≡ 0 (mod 24) if and only if at least one of the following good conditions hold: (γ 1 ) u is divisible by a prime power (possibly a prime) q ≡ 1 (mod 6); (γ 2 ) u = (p 3 − p)n with p an odd prime and n ̸ ≡ 0 (mod 4).
Proof. (=⇒) Let G be a binary group of order u ≡ 0 (mod 24) and assume that none of the conditions (γ 0 ), (γ 1 ), (γ 2 ) hold. First observe that the order of G is not divisible by 7 otherwise condition (γ 1 ) would be met. Hence no quotient of G can be isomorphic to containing P SL 2 (q) for a suitable odd prime power q, say q = p α with p a prime and α a positive integer. In this case q is a divisor of |G| since the order of P SL 2 (q) is divisible by q. It follows that α = 1 otherwise p 2 would be a prime power divisor of u which contradicts the assumption that neither (γ 0 ) nor (γ 1 ) holds considering that we have p 2 = 9 or p 2 ≡ 1 (mod 6) according to whether p = 3 or not. Hence q = p, namely P ΓL 2 (q) = P GL 2 (p). Now consider that we have |P GL 2 (p)| = p 3 − p and |P SL 2 (p)| = p 3 −p 2 so that S is either P GL 2 (p) or P SL 2 (p). We have |G| = 2(p 3 − p)|O(G)| in the former case and |G| = (p 3 − p)|O(G)| in the latter contradicting the fact that condition (γ 2 ) does not hold. Then G/(Λ(G) · O(G)) is a 2-group by Theorem 2.4 and hence the order of O(G) coincides with the largest odd divisor of the order of G. This implies that every subgroup of G of odd order is necessarily contained in O(G). We have |O(G)| = 3t with t a square-free integer whose prime factors are all congruent to 2 (mod 3) since we are supposing that neither (γ 0 ) nor (γ 1 ) holds. Thus, by Lemma 2.2, O(G) has exactly one subgroup of order 3 that, consequently, is also the the unique subgroup of G of order 3. In conclusion, G is not admissible. (⇐=) Assume that u ≡ 0 (mod 24) and that at least one of the three good conditions (γ 0 ), (γ 1 ), (γ 2 ) hold. We have to show that there exists an admissible group of order u. If (γ 0 ) holds we have u = 72t for a suitable t and Z 8t × V 9 is an admissible group of order u. If (γ 1 ) holds we have u = 24tq for a suitable t and a suitable prime power q ≡ 1 (mod 6). In this case an admissible group of order u is given by m for a suitable odd prime p and a suitable odd integer m. In the first case an admissible group of order u is given by SL 2 (p) × Z m . In the second case, considering that the Sylow 2-subgroups of P GL 2 (p) are dihedral (see, e.g., [11]), there exists a (unique) binary group G such that G/Λ(G) ≃ P GL 2 (p) by Theorem 2.5. This group G obviously possesses more than one subgroup of order 3 since this is also true for its quotient P GL 2 (p). Hence it is clear that G × Z m is an admissible group of order 2(p 3 − p)m. 2 As a consequence of the above theorem we deduce the non-existence of a 1-rotational STS(v) for infinitely many values of v ≡ 1 (mod 24). For instance, it is an easy matter to see that the following result holds.

Some even starters with a prescribed missing element
An even starter of a binary group G is a set E of |G|/2 − 1 pairs partitioning G − {0, m E } for a suitable m E ∈ G − {0} and whose differences partition G − Λ(G) (see [12]). We will call m E the missing element of E. An earlier result on this concept was given by B.A. Anderson who proved that every symmetrically sequenceable binary group 2 admits an even starter (see [2], Theorem 2). In the same paper it is observed that the symmetric sequenceability is a sufficient but not necessary condition since, for instance, even though the quaternion group of order 8 (that is the multiplicative group 2 An additive group G is sequenceable if there is a permutation (a 0 = 0, a 1 , ..., a |G|−1 ) of its elements such that all the partial sums ∑ j i=0 a i are pairwise distinct. A binary additive group G is symmetrically sequenceable if there is a permutation as above with a |G| 2 −i = −a |G| 2 +i for 0 ≤ i ≤ |G| 2 . We have analogous definitions in multiplicative notation.
, ±i, ±j, ±k} with composition law defined by the rules (−1) 2 = 1 and i 2 = j 2 = k 2 = ijk = −1) does not admit a symmetric sequence, it is evident that is as an even starter of it with missing element −1.
We also recall that B.A. Anderson himself and E.C. Ihrig proved that every finite solvable binary group G ̸ = Q 8 is symmetrically sequenceable [3] and hence we can state: Theorem 3.1. Every finite solvable binary group admits an even starter.
Our aim is to establish under which conditions a given binary group G may admit an even starter with an assigned missing element m E .

Lemma 3.2.
Let G be a binary group admitting an even starter E 0 with missing element m E 0 = g * . Then, for any group H of odd order and for any even starter E 1 of Λ(G) × H, there exists an even starter Proof. Consider the set  {(a, b), (c, d)} of elements of Z 2 × H is of type t 00 or t 11 or t 01 according to whether we have a = c = 0 or a = c = 1 or a ̸ = c, namely according to whether the number of its even elements is 2 or 0 or 1, respectively.
Denote by x ij the number of pairs of E of type t ij . The size of E is n − 1 so that we have x 00 + x 11 + x 01 = n − 1. Now note that the two differences of any given pair of E are both even or both odd and that the latter case happens exactly when the given pair is of type t 01 . Thus 2x 01 gives the number of odd elements of (Z 2 × H) − {(0, 0), (1, 0)}, namely 2x 01 = n − 1. Finally, the number 2x 00 +x 01 of even elements covered by the pairs of E must be equal to the number of even elements of the set ( It is straightforward to see that the above equalities give The assertion immediately follows. 2 Given k, with 1 ≤ k ≤ 2n + 1, a k-extended Skolem sequence of order n is a sequence (s 1 , ..., s n ) of n integers such that The following Baker's theorem [4] holds. Now note that k-extended Skolem sequences also produce some even starters.
is an even starter of Z 2n with missing element k.

Proof. Straightforward. 2
Here is an immediate special consequence of the above lemma and the theorem of Baker.
Lemma 3.6. If G = Z 2n with n ≡ 0 or 1 (mod 4), then there exists an even starter of G with missing element g * = n.
The following two lemmas are crucial for proving the result of the next section.

Lemma 3.7. Let np α ≡ 3 (mod 4) with p a prime not dividing n.
There exists an even starter of Z 2n × V p α with missing element m for any prescribed Proof. It is enough to prove that there exists an even starter E of Z 2n ×V p α whose missing element m E lies in {0}×V p α . In this case in fact m E is mapped into m by a suitable automorphism ϕ of Z 2n × V p α and hence it is clear that ϕ(E) is an even starter of Z 2n × V p α with missing element m.
1 st case: n ≡ 1 (mod 4). Set G = Z 2n and H = V p α . By Lemma 3.6 there exists an even starter of G with missing element g * = n. Thus applying Lemma 3.2 we get an even starter of G × H with missing element m E where E is an even starter of Λ(G) × H whose existence is ensured by Theorem 3.1. Now note that n ≡ 1 (mod 4) implies that |H| = p α ≡ 3 (mod 4) and hence, by Lemma 3.3, we necessarily have m E ∈ {0} × H.
2 nd case: n ≡ p ≡ 3 (mod 4). Set G = Z 2np and H = V p α−1 . We have np ≡ 1 (mod 4) and hence, by Lemma 3.6, there exists an even starter of G with missing element g * = np. Applying Lemma 3.2 we get an even starter of G × H with missing element m E where E is an even starter of Λ(G) × H whose existence is ensured by Theorem 3.1. Also here we have |H| = p α−1 ≡ 3 (mod 4) and hence, by Lemma 3.3, we necessarily have m E ∈ {0} × H.
3 rd case: n ≡ 3 (mod 4) and p ≡ 1 (mod 4). Set G = Z 2 × V p α−1 and H = Z p × Z n . By induction on α and using Lemmas 3.2 and 3.6, it is easy to see that there exists an even starter of G with missing element g * = (1, 0). Now note that by Baker's theorem there exists a 2n-extended Skolem sequence of order pn − 1 which, by Lemma 3.5, yields an even starter of Z 2pn with missing element 2n. Of course such an even starter can be also seen as an even starter of Λ(G) × H with missing element ((0, 0), (2n, 0)) in view of the isomorphism Z 2pn ≃ Λ(G) × H. Thus, applying Lemma 3.2 we get an even starter of G × H with missing element ((0, 0), (2n, 0)) and the assertion very easily follows. Proof. Reasoning as in Lemma 3.7 it suffices to prove that there exists an even starter of Z 8n × V p α with missing element in {0} × V p α . By Baker's theorem there exists an 8n-extended Skolem sequence of order 4np − 1 which, by Lemma 3.5, yields an even starter of Z 8np with missing element 8n. So, identifying Z 8np with Z 8n × V p we have an even starter of Z 8n × V p with missing element in {0} × V p and the assertion is true for α = 1. Now assume α > 1, set 8np = 4r, and identify Z 8n ×V p α with G := Z 4r ×V p α−1 .
Consider the set E 0 consisting of the following 4r pairs of elements of G where i runs, in each case, from 1 to r − 1: It is not difficult to check that we have: with γ z = 3 for 2r + 1 ≤ z ≤ 3r and γ z = 1 otherwise. We also have: with δ z = 4 for any odd z ̸ = ±1 and δ z = 2 otherwise. Thus, denoting by S a complete system of representatives for the cosets of {1, −1} in the multiplicative group of F p α−1 , we have since we have supposed p > 3. This implies that is a set of pairs of G partitioning G − (Z 4r × {0}) and whose differences also partition G − (Z 4r × {0}).
As observed at the beginning of this proof, there exists an even starter E ′ of Z 4r = Z 8np with missing element 8n. So, if E 2 is the set of all pairs {(x, 0), (y, 0)} of G with {x, y} a pair of E ′ , it is obvious that E := E 1 ∪ E 2 is an even starter of Z 8np × V p α−1 with missing element (8n, 0). The natural isomorphism between Z 8np × V p α−1 and Z 8n × V p α maps this element into the pair (0, m) where m is the α-tuple (8n, 0, ..., 0). Thus E can be viewed as an even starter of Z 8n × V p α with missing element (0, m) and the assertion follows. 2

1-rotational STS(v) with the odd part of v − 1 non-square-free
We already mentioned that the existence of a 1-rotational STS(v) with v ≡ 1 or 19 (mod 24) whenever v − 1 has a prime factor p ≡ 1 (mod 6) has been established in [8]. Using a similar construction, now we show how our results about even starters allow to prove that the existence is also guaranteed in the weaker hypothesis that v − 1 is divisible by a prime power q ≡ 1 (mod 6).
We obviously have p 2 ≡ 1 (mod 6). Let ϵ be a primitive cubic root of unity of F p 2 and consider the unit w = (1, ϵ) of the ring with additive group H. Observe that ϵ 2 + ϵ + 1 = 0 so that we have w 2 + w + 1 = (3, 0). This implies that (w 2 +w+1)h = 0 for every h ∈ {0}×V p 2 . Let G = Z 3 + w H be the group with elements in the cartesian product Z 3 × H and composition law + w defined by the rule (a, h) + w (a ′ , h ′ ) = (a + a ′ , w a ′ h + h ′ ).
The hypothesis that v ≡ 1 or 19 (mod 24) implies that we have np α ≡ 0 or 3 (mod 4). By Lemma 3.8 in the former case and by Lemma 3.7 in the latter, there exists an even starter Consider the triples T 1 , ..., T np α −1 of elements of G defined as follows: Given h ∈ H, the opposites of (0, h), (1, h) and (2, h) in G are (0, −h), (2, −w 2 h) and (1, −wh), respectively. Taking account of this, we easily see that Hence, by the definition of an even starter, we see that we have where h * is the involution of H. Now note that the fact that m E ∈ {0} × V p 2 implies that (w 2 + w + 1)m E = 0 and hence that (w + 1) } is a subgroup of G of order 3. We conclude that {T 1 , ..., T np α −1 } is a (G, 3, Σ, 1)-DF where Σ is the following partial spread of G: The assertion follows. 2

Existence of a 1-rotational STS(24n+1) with n ̸ ≡ 0 (mod 4)
In this section we prove that the existence of a 1-rotational STS(v) with v ≡ 1 (mod 24) is guaranteed in the case that the largest power of 2 in v − 1 does not exceed 16. Hence we prove that there exists a 1-rotational STS(48n + 25) and a 1-rotational STS(96n + 49) for every n ≥ 0. We first need the following fundamental lemma.
Lemma 5.1. Assume that there exists a 1-rotational STS(6m + 1) under G. Also assume that for a given positive integer n there exist 2m triples for suitable elements δ g ∈ Z 2n+1 with gcd(δ g , 2n + 1) = 1 for every g ∈ G.
Proof. The assertion is true for n = 0 since the existence of a 1-rotational STS(25) under SL 2 (3) has been proved in [8]. It is unique up to isomorphism and it is 1-rotational under the unique admissible group of order 24, that is the special linear group ) .
Suppose that n ≥ 1 and consider the following eight triples of SL 2 (3)×Z 2n+1 : It is straightforward to check that we have where δ g = 2 or 1 according to whether g ∈ Q or not. Thus the assertion follows from Lemma 5.1. 2 At the moment, the existence of a 1-rotational ST S(96n + 49) is uncertain in the case that 2n + 1 is a square free product of primes ≡ 2 (mod 3) and hence, in particular, for n = 0. Also here we are able to give a direct construction covering these cases.
If q ̸ = 0, its inverse is given by q −1 = q ||q|| 2 . The multiplicative group O under examination consists of the following quaternions: ±1, ±i, ±j, ±k; 1 2 (±1 ± i ± j ± k) with all possible choices of the signs; and all possible choices of the signs.
We see that Q 8 is a subgroup of O and we have Λ(O) = {1, −1}. Also note that every element of O has norm 1 so that its inverse simply is its conjugate. Now consider the following seven triples of elements of O: Let us calculate their lists of differences: We see that Thus, observing that Then the assertion is true for n = 0. The assertion is also true when 2n + 1 > 0 is divisible by 3 in view of (1).
It is tedious but not difficult to check that we have For the reader who would like to check the above calculation, we point out that we have T 2i = ϕ(T 2i−1 ) for 3 ≤ i ≤ 8 where ϕ is the automorphism of O × Z 2n+1 defined by ϕ(g, z) = (g, −z) for every (g, z) ∈ O × Z 2n+1 ; hence we have ∆T 2i = ϕ(∆T 2i−1 ) for 3 ≤ i ≤ 8. We have gcd(δ g , 2n + 1) = 1 for every g since we are supposing that 3 does not divide 2n + 1 and hence, considering that a 1-rotational STS(49) under O has been proved to exist, the assertion follows from Lemma 5.1.

Conclusion
Putting together the results of [8] and those obtained in the previous sections we conclude that the existence of a 1-rotational STS(v) is uncertain only in the case of v = (p 3 − p)n + 1 ≡ 1 (mod 96) with p a prime, n ̸ ≡ 0 (mod 4), the odd part of v − 1 square-free and without prime factors ≡ 1 (mod 6). At the moment solving these open cases does not seem an easy matter to us. Maybe, the first step could be to find a solution when n = 1 or 2. Thus we propose the following problems.

Problem 1.
Given an odd prime p, does there exist an STS(2p 3 − 2p + 1) that is 1-rotational under an extension of P GL 2 (p) by Z 2 ?
Problem 2. Given an odd prime p, does there exist an STS(p 3 − p + 1) that is 1-rotational under SL 2 (p)?
For the time being, the above problems have been positively solved only in the smallest case of p = 3 (see the previous section). Also consider that the two solutions that we obtained in this special case allowed us, together with Lemmma 5.1, to prove the existence of a 1-rotational STS(v) for any v ≡ 25 (mod 48) and any v ≡ 49 (mod 96), namely the existence of a 1rotational STS((3 3 − 3)n + 1) for any n ̸ ≡ 0 (mod 4). Thus, it is maybe possible that if one positively solves Problems 1 and 2 for any odd prime p, then a clever use of Lemma 5.1 allows us to find a 1-rotational STS((p 3 − p)n + 1) for any odd prime p and any n ̸ ≡ 0 (mod 4). In this case our main question about the set of values of v for which a 1-rotational STS(v) exists would be completely solved. We point out, however, that even though Problems 1 and 2 are interesting in their own right, for our main purpose it is not necessary to solve them for all odd primes p. In Problem 1 it is enough to consider those primes p ≡ ±1 (mod 8) for which the odd part of p 3 − p is square-free and without prime factors ≡ 1 (mod 6). Analogously, in Problem 2 it is enough to consider those primes p ≡ ±1 (mod 16) for which, again, the odd part of p 3 − p is square-free and without prime factors ≡ 1 (mod 6