The first families of highly symmetric Kirkman Triple Systems whose orders fill a congruence class

Kirkman triple systems (KTSs) are among the most popular combinatorial designs and their existence has been settled a long time ago. Yet, in comparison with Steiner triple systems, little is known about their automorphism groups. In particular, there is no known congruence class representing the orders of a KTS with a number of automorphisms at least close to the number of points. We fill this gap by proving that whenever $v \equiv 39$ (mod 72), or $v \equiv 4^e48 + 3$ (mod $4^e96$) and $e \geq 0$, there exists a KTS on $v$ points having at least $v-3$ automorphisms. This is only one of the consequences of a careful investigation on the KTSs with an automorphism group $G$ acting sharply transitively on all but three points. Our methods are all constructive and yield KTSs which in many cases inherit some of the automorphisms of $G$, thus increasing the total number of symmetries. To obtain these results it was necessary to introduce new types of difference families (the doubly disjoint ones) and difference matrices (the splittable ones) which we believe are interesting by themselves.


Introduction
Steiner and Kirkman triple systems are undoubtedly amongst the most popular discrete structures. A Steiner triple system of order v, briefly STS(v), is a pair (V, B) where V is a set of v points and B is a set of 3-subsets (blocks) of V with the property that any two distinct points are contained in exactly one block. A Kirkman triple system of order v, briefly KTS(v), is an STS(v) together with a resolution R of its block-set B, that is, a partition of B into classes (parallel classes) each of which is, in its turn, a partition of the point-set V . It has been known since the mid-nineteenth century that a STS(v) exists if and only if v ≡ 1 or 3 (mod 6) [38]. The analogous result for KTSs has been instead obtained more than a century later [49]: a KTS(v) exists if and only if v ≡ 3 (mod 6).
An automorphism of a STS is a permutation of its points leaving the block-set invariant. Analogously, an automorphism of a KTS is a permutation of its points leaving the resolution invariant. Thus an automorphism of a KTS is automatically an automorphism of the underlying STS though the converse is not true in general.
For general background on these topics we refer to [18].
In the next section we will give a brief survey of the automorphism groups of STSs and KTSs. Looking at the results it will be evident how little we know about KTSs in comparison with STSs. For instance, it is has been known for close to a century that there is an STS(v) with an automorphism group of order v for any admissible v. Yet, we are still quite far from a similar result for KTSs. At the moment the existence of a KTS(v) with an automorphism group of order v or v − 1 is known only when v has a special prime factorization. In particular, there is no known congruence v ≡ k (mod n) which guarantees the existence of a KTS(v) with an automorphism group of order close to v.
Adopting the combinatorial-analog of the famous Erlangen program by Felix Klein [33], we believe that the interest of a discrete structure is proportional to the number of its automorphisms. Motivated by this and by the shortage of results mentioned above, in this paper we deeply investigate Kirkman triple systems which are 3-pyramidal, i.e., admitting an automorphism group acting sharply transitively on all but three points. Now we make a short digression to explain why adopting this "philosophy" also brings practical benefits. Given the definition of a discrete structure, the first natural target, which could be very difficult, is to determine under which constraints it exists. The second is to give an explicit construction for this structure; in some cases this could be even more difficult. For instance, thanks to the recent seminal work of P. Keevash [37], it is known that a Steiner t-design exists provided that the trivial necessary conditions are satisfied and that its order is sufficiently large (the case t = 2, due to R.M. Wilson [54,55,56], dates back to the 70's). This is an outstanding achievement which seemed completely out of reach only a decade ago; yet, the probabilistic methods used by Keevash are non-constructive, and do not provide an explicit lower bound on the order of a t-design that guarantees its existence. On the contrary, the request to have many symmetries, besides being in compliance with the Erlangen program, allows to develop constructive algebraic methods. This paper gives the complete recipe to construct, explicitly, a KTS(v) for any v as in (i), (ii), (iii) of our main result below. Theorem 1.1. A necessary condition for the existence of a 3-pyramidal KTS(v) is that v = 24n + 9 or v = 24n + 15 or v = 48n + 3 for some n which, in the last case, must be of the form 4 e m with m odd. This condition is also sufficient in each of the following cases: (i) v = 24n + 9 and 4n + 1 is a sum of two squares; (ii) v = 24n + 15 and either 2n + 1 ≡ 0 (mod 3) or the square-free part of 2n + 1 does not have any prime p ≡ 11 (mod 12); (iii) v = 48n + 3.
In particular, (ii) and (iii) allow us to reach our main target of getting some families of highly symmetric KTSs whose orders fill a congruence class. We point out that in many cases the constructed 3-pyramidal KTSs inherit some automorphisms of the group G acting sharply transitively on all but three points. This permits to increase their number of symmetries considerably (see Remarks 11.2,11.4 and 11.6).
After the brief survey of Section 2 concerning some results on the automorphism groups of Steiner and Kirkman triple systems, the article will be structured as follows. In Section 3 we provide the difference methods to construct a 3-pyramidal KTS and we prove that each group having a 3-pyramidal action on a KTS fixes one parallel class and acts transitively on the remaining ones. We also prove that such a group must have exactly three involutions, and these involutions are pairwise conjugate. Groups with this property will be called pertinent. Although in literature there is no lack of articles on groups with three involutions (see, for example, [34,39]), none of them allows us to determine the set of "relevant" orders. We prove (Theorem 3.9) that such orders are precisely those of the form 12n + 6 or 4 α (24n + 12), and from this we partially derive the necessary condition in Theorem 1.1. The proof of the "only if part" of Theorem 3.9 is purely group theoretical and for this reason the whole Section 4 is dedicated to it. However, its reading is not necessary for understanding the rest of the article, whose nature is purely combinatorial.
The constructive part of Theorem 1.1 will be proven in Section 11 and it is the result of numerous direct constructions (Sections 6, 7 and the appendix) and recursive constructions (Sections 8, 9, 10). These results are preceded by a brief section (Section 5) useful for understanding the notation and terminology used throughout the rest of the paper. The recursive constructions required the introduction of new concepts such as doubly disjoint difference family and splittable difference matrix which we believe may be important by themselves.
The article concludes (Section 12) with a short list of open problems.

A brief survey of the automorphism groups of Steiner and Kirkman triple systems
The literature on Steiner and Kirkman triple systems having an automorphism with a prescribed property or an automorphism group with a prescribed action is quite extensive.
For instance the set of values of v for which there exists an STS(v) with an involutory automorphism fixing exactly one point (reversed STS) has been established in [25,50,53]: it exists if and only if v ≡ 1, 3, 9 or 19 (mod 24). Results concerning the full automorphism group of a STS have been obtained by Mendelsohn [41] and Lovegrove [40].
Here, we just provide a brief survey of what is known on the existence of systems whose automorphisms are at least close to the number of points.
Adopting a terminology coined by E. Mendelsohn and A. Rosa [42], we say that a combinatorial design is f -pyramidal if it admits an automorphism group G fixing f points and acting sharply transitively on the others. If f = 0 one usually speaks of a regular design and, more specifically, of a cyclic design if the group G is cyclic. If f = 1 one usually speaks of a 1-rotational design.
It was proved a long time ago [47] that there exists a cyclic STS(v) for all admissible values of v except v = 9. On the other hand the unique STS (9), that is the point-line design associated with the affine plane of order 3, is clearly regular under the action of Z 2 3 . Thus there exists a regular STS(v) for all admissible values of v.
The analogous problem of determining the set of values of v for which there exists a regular KTS(v) is almost completely open and it appears to be very difficult. The few known results on this problem are the following. The parallel classes of the point-line design associated with the n-dimensional affine space over the field of order 3 clearly give a KTS(3 n ) that is regular under the action of Z n 3 . A necessary condition given in [43] for the existence of a cyclic KTS(6n + 3) is that 2n + 1 is not a prime power congruent to 5 (mod 6). This condition is also sufficient up to n = 32 [43] and when all prime factors of n are congruent to 1 (mod 6) [30].
The existence of a 1-rotational STS(v) has been thoroughly investigated in [3,6,44,48] leaving the problem open only for the orders v satisfying, simultaneously, the following conditions: v = (p 3 − p)n + 1 ≡ 1 (mod 96) with p a prime; n ≡ 0 (mod 4); the odd part of v − 1 is square-free and without prime factors ≡ 1 (mod 6).
The 1-rotational KTSs have a very nice structure. Indeed a group with a 1-rotational action on them (necessarily binary, i.e., admitting exactly one involution) is transitive on the parallel classes. On the other hand, as in the regular case, very little is known about their existence which has been proved only for orders v of the the following types: v is a power of 3 (the already mentioned regular KTS(3 n ) is also 1-rotational); all prime factors of v−1 2 are congruent to 1 (mod 12) [13]; v = 8n + 1 with all the prime factors of n congruent to 1 (mod 6) [14].
As an obvious consequence of the above result, a 3-pyramidal KTS(v) may exist only when v ≡ 9 (mod 24) or v ≡ 15 (mod 24) or v ≡ 3 (mod 48). The main result of this paper (Theorem 1.1) provides, in particular, a complete answer in the last case v ≡ 3 (mod 48).
Finally, a STS(v) is called 1-transrotational if it has an automorphism group G that fixes exactly one point, switches two points, and acts sharply transitively on the remaining v−3. This terminology was first used in [28] under the assumption that G is cyclic, though G just need to be binary. One cannot fail to notice a certain kinship between 3-pyramidal and 1-transrotational STSs, but apart from the fact that their groups are deeply different, the sets of orders for which they exist do not coincide. Indeed it was proved in [28] that a 1-transrotational STS(v) under the cyclic group exists if and only if v ≡ 1, 7, 9 or 15 (mod 24). It is easy to check that the same holds if we remove the assumption that the group be cyclic. As far as we are aware, nobody studied 1-transrotational KTS(v). Considering the above, they might exist only for v ≡ 9 or 15 (mod 24) but it is not difficult to exclude the case v ≡ 15 (mod 24). This will be shown in a paper in preparation [11] where we will deal with the case v ≡ 9 (mod 24).

Difference families and 3-pyramidal Kirkman triple systems
In this section we show that the existence of a 3-pyramidal KTS over a group G is equivalent to constructing a suitable difference family (DF) in G relative to a partial spread, a concept introduced by the second author in [6]. We point out that throughout the paper, except for Section 4, every group will be denoted additively.
A partial spread of a group G is a set Σ of subgroups of G whose mutual intersections are trivial. If τ = {d e 1 1 , . . . , d en n } is the multiset (written in "exponential" notation) of the orders of all subgroups belonging to Σ, we say that Σ is of type τ or a τ -partial spread. A spread (or partition) of G is a partial spread whose members between them cover the whole group G.
The list of differences of a triple B = {x, y, z} of elements of G is the multiset ∆B of size 6 defined by The list of differences of a family F of triples of G, denoted by ∆F , is the multiset union of the lists of differences of all its triples. Also, the flatten of F , denoted by Φ(F ), is the multiset union of all the triples of F .
A (G, Σ, 3, 1) difference family (DF) is a family F of triples of G (base blocks) whose list of differences is the set of all elements of G not belonging to any member of the partial spread Σ. If Σ = {H} we write (G, H, 3, 1)-DF or simply (G, 3, 1)-DF when H = {0}. If Σ is a partial spread of type τ , we also use the notation (G, τ, 3, 1)-DF.
If F is a (G, H, 3, 1)-DF, then its size is clearly equal to |G\H| 6 and then its flatten Φ(F ) has size |G\H| 2 . Thus, if J is a subgroup of H of order 2, one can ask whether Φ(F ) is a complete system of representatives for the left cosets of J that are not contained in H. In the affirmative case we say that F is J-resolvable.
Definition 3.1. Let F be a (G, H, 3, 1)-DF and let J be a subgroup of H of order 2. We say that F is J-resolvable if its flatten is a complete system of representatives for the left cosets of J in G that are not contained in H. So, equivalently, if we have Φ(F ) + J = G \ H.
We note that the development of a J-resolvable (G, H, 3, 1)-DF is a Kirkman frame [52] admitting G as a sharply point transitive automorphism group.
A multiplier of a J-resolvable (G, H, 3, 1)-DF, say F , is an automorphism µ of G leaving F invariant. We say that µ is a strong multiplier if it fixes H element-wise.
The following fact is straightforward.
Furthermore, F inherits the strong multipliers of F n−1 .
Difference families are a crucial topic in Design Theory [2,17]. In particular, as a special case of Theorem 2.1 in [12], it is possible to characterize the 3-pyramidal STS(6n+3) in terms of difference families as follows.
(ii) the action of G on V is the addition on the right with the rule that ∞ i + g = ∞ i for each g ∈ G; (iii) G has exactly three involutions, say j 1 , j 2 , j 3 ; (iv) a system of representatives for the G-orbits on B is of the form where: H is a set of e subgroups of G of order 3; F is a (G, Σ, 3, 1)-DF with Σ = {{0, j 1 }, {0, j 2 }, {0, j 3 }} ∪ H. Definition 3.5. Throughout this paper a finite group G will be called "pertinent" if it has precisely three involutions, and the three of them are pairwise conjugate in G.
Obviously, a pertinent group is necessarily non-abelian. Up to isomorphism, the smallest pertinent group is D 6 , i.e., the dihedral group of order 6. The next one is A 4 , the alternating group of degree 4.
As already said, we prefer to write every group in additive notation. So, differently from the mostly used representation, we prefer to see D 6 as the additive group D with underlying set Z 2 ×Z 3 and operation law+ defined by (a, b)+ (c, d) = (a+c, (−1) c b+d). Adopting this representation, it is easy to see that the difference− in D works as follows: The three involutions of D are (1, 0), (1,1) and (1,2). The fact that and confirms that D is pertinent. The alternating group A 4 will be also represented additively as the first term of an infinite series of pertinent additive groups that we will construct in the proof of the "if part" of Theorem 3.9.
Definition 3.6. Let F be a (G, {2 3 , 3}, 3, 1)-DF with G pertinent and let J be a subgroup of G of order 2. We say that F is J-resolvable if there exists a, b ∈ G such that • J, a + J − a and b + J − b are the three subgroups of order 2 of G; • Φ(F ) ∪ {0, a, b} is a complete system of representatives for the left cosets of J in G.
The following fact is straightforward.
Proposition 3.7. Let J be a subgroup of order 2 of a pertinent group G and let H be a pertintent subgroup of G containing J.
The following result gives a characterization of the 3-pyramidal KTSs and, more importantly, a way to construct them. Let P be the parallel class containing the block B ∞ . Obviously, we have P + g = P for each g ∈ G. It follows that the distinct right translates of the block H of P through 0 form a partition of G. This clearly implies that H is a subgroup of G and that the blocks of P are B ∞ and all the right cosets of H in G. Now let Q be the parallel class of R containing the block B 1 = {∞ 1 , 0, j 1 }. It is not difficult to see that the G-stabilizer of Q coincides with the G-stabilizer of B 1 that is J := {0, j 1 }. Thus, for i = 2, 3, the block of Q through ∞ i is of the form {∞ i , g i , g i + j 1 } for a suitable group element g i . This block necessarily belongs to the orbit of B i , hence we have {∞ i , 0, j i } + t i = {∞ i , g i , g i + j 1 } for a suitable t i . This equality easily implies that j i = g i + j 1 − g i , hence the three involutions j 1 , j 2 , j 3 are pairwise conjugate, i.e., G is pertinent.
The fact that the G-stabilizer of Q is J also implies that the 2n − 2 triples of Q not containing the "points at infinity" can be grouped into pairs {A i , A i + j 1 }, 1 ≤ i ≤ n − 1, and that the G-orbit of Q has length |G| 2 = 3n. Then, given that the resolution of a KTS(6n + 3) has size 3n + 1, we deduce that R = {P} ∪ Orb(Q). Also, if we set F = {A i | 1 ≤ i ≤ n − 1}, we can claim that a set of base blocks for B is given by Given that the blocks of Q partition V , we have (Φ(F ) ∪ {0, g 2 , g 3 }) + J = G. This means that Φ(F ) ∪ {0, g 2 , g 3 } is a complete system of representatives for the left cosets of J in G, i.e., F is J-resolvable.
(⇐=). Let G be a pertinent group of order 6n whose involutions are j 1 , j 2 , j 3 , and assume that F is a {0, j 1 }-resolvable (G, {2 3 , 3}, 3, 1)-DF. For i = 2, 3, there are suitable group elements g 2 , g 3 such that Let H be the subgroup of G of order 3 belonging to the partial spread associated with F . Two parallel classes of the STS(6n + 3) generated by F are clearly the following: Their G-stabilizers are, respectively, G and {0, j 1 } so that their G-orbits have size 1 and 3n. It easily follows that {P} ∪ Orb(Q) is a G-invariant resolution of the STS(6n + 3) generated by F , namely a 3-pyramidal KTS(6n + 3).
In view of the above theorem, it is important to determine the set of pertinent numbers, i.e., the set of orders of the pertinent groups. Proof. The proof of the "only if part" is purely group theoretical and for convenience it is postponed to Section 4. Here we prove the "if part".
If n ≡ 6 (mod 12), we have n = 6m for a suitable odd integer m. Then, recalling that D is pertinent, it is clear that D × H is a pertinent group of order 6m for every group H of order m. Now let n = 4 α m with α > 0 and m ≡ 3 (mod 6). Consider the matrix Θ =   1 0 0 0 0 1 0 −1 −1   and let G α be the group with underlying set Z 3 ×Z 2 α ×Z 2 α and operation + defined by the rule This is, up to isomorphism, the outer semidirect product of Z 2 2 α and Z 3 with respect to the group omomorphism θ : The difference of two triples (a, b, c) and (d, e, f ) of the group has a convenient form; it is the usual difference in the abelian group Z 3 × Z 2 2 α multiplied by the inverse of Θ d : More explicitly, we have It is a simple exercise to check that G α has exactly three involutions that are and that they are pairwise conjugate. Indeed we have: Thus G α is a pertinent group of order 4 α 3. Then, if H is any group of odd order m, it is clear that the direct product G α × H is a pertinent group of order 3 · 4 α m whose involutions are (0, 2 α−1 , 0, 0), (0, 0, 2 α−1 , 0) and (0, 2 α−1 , 2 α−1 , 0).
The alternating group A 4 can be seen, up to isomorphism, as the group G 1 .

Pertinent groups
Here we prove the "only if" part of Theorem 3.9. Considering that the arguments used are purely group theoretical, the reading of this section can be postponed to a later time without compromising the understanding of the rest of the article. Also, we point out that in this section, unlike the rest of the paper, we prefer to denote groups in multiplicative notation.
In the following, let G be a pertinent group and denote by K the subgroup of G generated by the three involutions i, j, k. Let C G (K) be the centralizer of K in G, that is and set C = C G (K). Clearly, K is a characteristic subgroup of G, so both K and C are normal in G. Since G acts on {i, j, k} by conjugation with kernel C, the quotient G/C is isomorphic to a transitive subgroup of S 3 , so either G/C ∼ = S 3 or G/C ∼ = A 3 (here A 3 denotes the alternating group of degree 3). Observe in particular that G has an element of order a power of 3 acting "cyclically" on the involutions (meaning that it sends i to j, j to k and k to i).
We will make use of the following well-known result in group theory.
We start providing sufficient conditions for a pertinent group to have subgroups or quotients that are pertinent.
3. If H has even order and contains a Sylow 3-subgroup of G, then H is pertinent.

Suppose the involutions of G commute pairwise. If H is normal in G and |H| is odd then G/H is pertinent.
Proof. (1) Since H has even order, it contains at least one involution. Considering that HC = G, it follows that the action of H on the involutions is transitive. Therefore, K ≤ H and H is pertinent.
(2) By Lemma 4.1, we have that HC = G. Since Q ≤ C, it follows that K centralizes Q, hence K ≤ H, therefore H has even order. By point (1), we obtain that H is pertinent.
(3) Since |H| is even, H contains at least one involution. Since |G/C| is a multiple of 3, a Sylow 3-subgroup S of G is not contained in C, hence S acts transitively on {i, j, k}. Therefore, H contains all three involutions and then it is pertinent.
(4) Suppose H is normal of odd order in G and the involutions commute pairwise. Denote by i, j, k the involutions of G. Since ij = ji, the element ij is an involution distinct from i and from j so ij = k and the elements iH, jH, kH ∈ G/H are involutions of G/H and G/H acts transitively by conjugation on them (because G acts transitively by conjugation on i, j, k), moreover they are pairwise distinct, for example iH = jH because i −1 j = ij = k ∈ H. We are left to show that G/H has precisely three involutions. If xH is an involution of G/H then x 2 ∈ H so x has order 2t with t odd (being |H| odd), x t is an involution of G, and xH = (xH) t = x t H. This means that the involutions of G/H 12Simona Bonvicini, Marco Buratti, Martino Garonzi, Gloria Rinaldi, Tommaso Traetta are of the form yH with y an involution in G, so they are precisely iH, jH, kH and we deduce that G/H is pertinent.
For a pertinent group of order 2 n · d with d odd, the following two lemmas give us sufficient or necessary conditions for n to be even or odd. Proof. G has an element g of order a power of 3 acting cyclically on the three involutions. We claim that g does not fix any non-trivial element of H. Indeed if 1 = h ∈ H is fixed by g then a suitable power of h is an involution fixed by g, but g does not fix any involution. This implies that the g -orbits of H distinct from {1} have size divisible by 3 so 2 m = |H| ≡ 1 mod 3 therefore m is even.
It is enough to show that two involutions, say i and j, commute, because then ij has order 2 so ij = k (it cannot be ij = i nor ij = j) therefore i and j commute with k. If this is not the case, then C must have odd order (otherwise an involution in C should commute with the others). But then the order of G/C is a multiple of 2 n ≥ 4 contradicting the fact that G/C is isomorphic to either A 3 or S 3 .
Write |G| = 2 n · d with d odd. We know that G/C is isomorphic to either A 3 or S 3 . Letting Q be a Sylow 2-subgroup of C, N = N G (Q) is pertinent by Lemma 4.2, so writing |Q| = 2 m , m is even by Lemma 4.3. If G/C ∼ = A 3 then Q is a Sylow 2-subgroup of G, so n = m is even. Conversely if G/C ∼ = S 3 then n = m + 1 is odd.
To prove the main result of this section, we need the following result. Proof. Let Q be a Sylow 2-subgroup of C and recall that |G/C| ∈ {3, 6}. Therefore, if |Q| = 2 m , then m = n or n − 1 according to whether |G/C| = 3 or 6. By Lemma 4.2, we have that N = N G (Q) is a pertinent group, and since Q is normal in N, by Lemma 4.3 we have that m is even, hence m = n − 1 (since n is odd by assumption) and |G/C| = 6. Considering that Q ≤ C ∩ N, and G = NC (by Lemma 4.1), it follows that |N/Q| is a multiple of |N/(C ∩ N)| = |NC/C| = |G/C| = 6, hence 2 n divides |N|.
Since |N/Q| ≡ 2 (mod 4), there exists a normal subgroup D/Q of N/Q of index 2. In particular, all the Sylow 3-subgroups of N/Q are contained in D/Q, so the number of Sylow 3-subgroups of N/Q is odd. Let P/Q be a subgroup of N/Q of order 2. Since P/Q acts by conjugation on the family consisting of the Sylow 3-subgroups of N/Q, there exists one of them, say H/Q, normalized by P/Q. This implies that P H/Q = (P/Q)(H/Q) ≤ N/Q hence L = P H ≤ N. Moreover |L| = |Q| · |L/Q| = |Q| · |P/Q| · |H/Q| = 2 n · 3 s where 3 s = |H/Q|. Now L has even order and contains a Sylow 3-subgroup of N, hence L is pertinent by Lemma 4.2.
We are now ready to prove the "only if part" of Theorem 3.9.
Theorem 4.6. If G is a pertinent group of order 2 n d with d odd and n ≥ 2, then n is even.
Proof. We prove the result by contradiction. Let G be a counterexample of minimal order, that is, assume that there exists a pertinent group G such that |G| = 2 n · d is minimal with respect to the property that both n and d are odd, and n ≥ 3. By Lemma 4.5, we have that G has a pertinent subgroup of order 2 n · 3 s for some positive integer s, so by the minimality of |G| we must have d = 3 s . Recall that K ∼ = C 2 × C 2 and G/C ∼ = S 3 by Lemma 4.4.
Let S be a Sylow 3-subgroup of C and note that |S| = 3 s−1 since |G/C| = 6. Now, by Lemma 4.2 we have that N = N G (S) is pertinent; in particular, K ≤ N hence 4 is a divisor of |N|. Also, since CN = G (by Lemma 4.1) and Also, by Lemma 4.4, it follows that m is odd. Finally, since S is a normal subgroup of N of odd order, Lemma 4.2 guarantees that N/S is pertinent. Since |N/S| = 2 m · 3 and m ≥ 3 is odd, by the minimality of |G| we must have s = 1, hence |G| = 2 n · 3.
We are reduced to the case |G| = 2 n · 3. Let H := C G (i) be the centralizer of i in G. We have |G : H| = 3 because i has 3 conjugates in G, in other words |H| = 2 n and H is a Sylow 2-subgroup of G. Let P be a Sylow 3-subgroup of G, then P = g is a cyclic group of order 3. Clearly g ∈ H because |H| = 2 n , so P acts transitively on the three involutions of G. We prove that N G (P ) = P . Suppose for a contradiction that N G (P ) = P , then there is an involution, say i, normalizing P , hence P, i is a group of order 6 containing all three involutions, hence K ≤ P, i , which is a contradiction since |K| = 4. So N G (P ) = P . We now prove that H is normal in G. Since N G (P ) = P , the subgroup P has precisely |G : P | = 2 n conjugates in G therefore G has precisely (|P | − 1) · 2 n = 2 · 2 n elements of order 3. Since |G| = 3 · 2 n we deduce that the number of elements of G of order not equal to 3 is 2 n hence there is room in G for only one Sylow 2-subgroup. We deduce that the Sylow 2-subgroups are normal hence H G, so n is even by Lemma 4.3. This is a contradiction and the result is proved. 14Simona Bonvicini, Marco Buratti, Martino Garonzi, Gloria Rinaldi, Tommaso Traetta

Notation and terminology
A maximal prime power divisor of any integer n will be called a component of n. As it is standard, given a prime power q, we denote by F q the field of order q. We extend this notation to any integer n > 1 denoting by F n the ring which is direct product of all the fields whose orders are the components of n. Thus, for instance, F 45 = F 5 × F 9 . The additive group of the ring F n will be denoted by V n and we set V * n := V n \ {0}. If n = 1, then V n is the trivial group with one element. If d is a divisor of n, any subgroup S of V n of order d is clearly isomorphic to V d and therefore, by abuse of notation, such a subgroup S will be often denoted by V d .
The group of units of F n will be denoted by U(F n ) and its order by ψ(n). Obviously, in the particular case that n = q is a prime power, U(F n ) is nothing but the multiplicative group F * q of the field F q and ψ(n) = q − 1. Otherwise, if n has more than one component, . The set of non-zero squares and of non-squares of the field F q will be denoted by F q and F q , respectively.
If A, B are non-empty subsets of F n , then AB will denote the multiset {ab | a ∈ A; b ∈ B}. If A = {a} or B = {b}, then AB will be written as aB or Ab, respectively.
Let q 1 , . . . , q ω be the components of an odd integer n. For every non-empty I belonging to the power-set 2 {1,...,ω} , choose an element c(I) ∈ I and consider the subset S(I) of F * n defined as follows: Then define S := I∈2 {1,...,ω} \{∅} S(I). Such a set S has size n−1 2 and then will be called a halving of F * n . It is easy to see that it has the following property: Given a group G and an element s of F n , the endomorphism of G × V n mapping (x, y) to (x, ys) will be denoted by µ s . It is evident that if s ∈ U(F n ), then µ s is an automorphism of G × V n .

The smallest examples
The five smallest pertinent values of n are 6, 12, 30, 36 and 48. In the following, for each of these values, a 3-pyramidal KTS(n + 3) will be given by means of a (G, {2 3 , 3}, 3, 1)-RDF with G = D, G 1 , D × V 5 , G 1 × V 3 and G 2 , respectively. By way of illustration, in the first two cases we follow the instructions of Theorem 3.8 and we concretely construct a 3-pyramidal KTS(9) and a 3-pyramidal KTS (15).
The realizations of these five small KTSs allow us to state the following. Proof. The first assertion follows immediately from Proposition 3.7. For the second assertion, it is enough to observe that the semidirect product G⋊M is a group of automophisms of the obtained KTS(n + 3).
The above proposition will enable us to construct infinite classes of 3-pyramidal KTSs in the subsequent sections.

A 3-pyramidal KTS(15)
The set of all non-trivial subgroups of G 1 is a spread of this group of type We see that B ∪ {0, a, b} is a system of representatives for the left cosets of J in G 1 , i.e., F is J-resolvable. Following the instructions given in the proof of the "if part" of Theorem 3.8, we obtain the following 3-pyramidal representation of a KTS (15) where, to save space, each element (a, b, c) ∈ G 1 is written as abc. It is known that, up to isomorphism, there exist exactly seven KTS(15), i.e., there are seven non-isomorphic solutions to the well-known Kirkman fifteen schoolgirls problem. It is possible to show, applying the proposition on page 894 of [27], that the solution above is necessarily isomorphic to the original solution given by Kirkman, that is the solution denoted by 1a in [17] (1, 1, 0) and (1, 2, 0).

A 3-pyramidal KTS(51)
One can check that the following six triples of G 2 :

Three direct constructions
The action of a group U on a set V is said to be semiregular if the non-identity elements of U do not fix any element of V . The following fact is straightforward. Proposition 7.1. If U is a group of units of F n whose action by multiplication on V * n is semiregular and S is a complete system of representatives for the orbits of U on V * n , then we have US = V * n .
We need the following lemma.
Lemma 7.2. Let n > 1 be an integer whose components are all congruent to 1 (mod λ).
Then there exist a unit u of F n of order λ and a subgroup T of U(F n ) such that 1. u j −1 is a unit for 1 ≤ j ≤ λ−1 and the group U generated by u acts semiregularly on V * n ; 2. the order of T is the greatest divisor of ψ(n) coprime with λ; 3. T leaves invariant a suitable complete system S of representatives for the orbits of U on V * n .
Proof. Let q 1 , . . . , q ω be the components of n. For 1 ≤ i ≤ ω, let u i be a generator of the subgroup U i of F * q i of order λ, and set u = (u 1 , . . . , u ω ). It is very easy to prove that u satisfies 1. (see Corollary 3.3 and Lemma 3.2 in [7], in this order). Let T i be the subgroup of F * q i whose order is the greatest divisor of q i − 1 coprime with λ, set T = T 1 × · · · × T ω , and let Σ i be a complete system of representatives for the cosets of T i U i in F * q i . Now, for every non-empty I belonging to the power-set 2 {1,...,ω} , choose an element c(I) ∈ I and consider the subset S(I) of F * n defined as follows: Then set S = I∈2 {1,...,ω} \{∅} S(I). It is not difficult to check that S is a complete system of representatives for the orbits of U on V * n and that T leaves S invariant.
In this section we give three direct constructions which can be understood without any further explanation. Anyway, to be more informative, we emphasize that, in each case, a suitable strong difference family satisfying a special resolvability property has been used. These concepts are defined as follows. Definition 7.3. Given a group G and an even integer λ, a (G, 3, λ) strong difference family (SDF) is a collection of triples of elements of G whose list of differences covers each element of G, 0 included, exactly λ times.
If J is a subgroup of G of order 2, then we say that a (G, 3, λ)-SDF is J-resolvable if its flatten contains exactly λ elements of each left coset of J in G.
Although the notion of a SDF was implicitly used in the literature for a long time, the formal definition has been given in [5]. Since then, SDFs have been crucial for the construction of various combinatorial designs in several papers such as [9,10,15,19,20,45,57]. As far as we are aware, the notion of a J-resolvable SDF is new.
Theorem 7.4. If all the components of 4n + 1 are congruent to 1 mod 4, then there exists a (D × V 4n+1 , D × V 1 , 3, 1)-RDF with a group of strong multipliers whose order is the greatest odd divisor of ψ(4n + 1).
Proof. Take u, T and S as in the statement of Lemma 7.2 applied with λ = 4. Thus u is a unit of order 4 such that u − 1 is a also a unit and U := u acts semiregularly on V * 4n+1 .
Note that we necessarily have u 2 = −1, hence U = {±1, ±u}. Also note that we have (u − 1)U = {±(u − 1), ±(u + 1)}. Let us consider the set B consisting of the following triples of D × V 4n+1 (recall that the underlying set of D is Z 2 × Z 3 ; see Section 3): It is straightforward to check that we have ∆B = g∈D {g} × ∆ g with if g = (0, 0) or g = (1, 1); It is also readily seen that Given that S is a complete system of representatives for the orbits of U on V * 4n+1 , we have US = V * 4n+1 by Proposition 7.1 and then, taking into account the previous identities, we easily obtain Finally, given that T is a subgroup of U(F n ) which leaves S invariant, we infer that {µ t | t ∈ T } is a group of strong multipliers of F . The assertion follows by observing that this group is clearly isomorphic to T and reminding that the order of T is the greatest odd divisor of ψ(4n + 1).
Observe that the set of initial base blocks B considered in the above theorem is a lifting of a J-resolvable (D, 3, 4)-SDF.
If we apply Theorem 7.4 with n = 1 we are forced to take u = 3 and one can see that the resultant RDF is exactly the one given in Subsection 6.3.
Theorem 7.5. If all the components of n are congruent to 1 mod 4, then there exists a (G 1 ×V n , G 1 ×V 1 , 3, 1)-RDF with a group of strong multipliers whose order is the greatest odd divisor of ψ(n).
Proof. Again, as in Theorem 7.4, take u, T and S as in the statement of Lemma 7.2 applied with λ = 4. Consider the set B consisting of the following triples of G 1 × V n (recall that the underlying set of G 1 is Z 3 × Z 2 × Z 2 ; see Section 3): One can check that ∆B = g∈G 1 {g}×∆ g with ∆ g = 2U or ∆ g = (u−1)U according to whether g belongs or does not belong to the Klein subgroup of G 1 , respectively. Also, we have B∈B B + J = g∈G 1 {g} × Φ g with J = {(0, 0, 0, 0), (0, 1, 1, 0)} and Φ g = U for every g ∈ G 1 .
Reasoning as in Theorem 7.4, we can see that } as a group of strong multipliers of order the greatest odd divisor of ψ(n).
Analogously to Theorem 7.4 the set of initial base blocks B considered above is a lifting of a J-resolvable (G 1 , 3, 4)-SDF.
Theorem 7.6. If all the components of n are congruent to 1 mod 6, then there exists a (G 1 ×V n , G 1 ×V 1 , 3, 1)-RDF with a group of strong multipliers whose order is the greatest divisor of ψ(n) coprime with 6.
Proof. Take u, T and S as in the statement of Lemma 7.2 applied with λ = 6. Thus u is a unit of U(F n ) of order 6 such that u i − 1 is a unit for 1 ≤ i ≤ 5 and U := u acts semiregularly on V * n . Note that we necessarily have u 3 = −1 so that U = {±1, ±u, ±u 2 }, and the identity u 2 − u + 1 = 0 holds 3  With a little bit of patience, taking into account the identity u 2 = u − 1, it is not difficult to check that we have: where J = {(0, 0, 0, 0), (0, 1, 1, 0)}. We can write u + 1 = −(u 4 − 1), hence u + 1 is a unit of F n by assumption on u. Now set Given that S is a complete system of representatives for the orbits of U on V * 4n+1 , we have US = V * n by Proposition 7.1 and then, taking into account (7.1), we easily obtain Finally, given that T is a subgroup of U(F n ) which leaves S invariant, we infer that {µ t | t ∈ T } is a group of strong multipliers of F . The assertion follows observing that this group is clearly isomorphic to T whose order is the greatest divisor of ψ(n) coprime with 6.
This time the set of initial base blocks B considered above is a lifting of a J-resolvable (G 1 , 3, 6)-SDF.

A composition construction via pseudo-resolvable difference families
Now we define a class of (G, {2 3 , 3}, 3, 1)-DFs that we call pseudo-resolvable and will be crucial for the construction of a 3-pyramidal KTS(v) with v = 36n + 3 or v = 48n + 3 or v = 108n + 3 with all the components of n congruent to 7 or 11 (mod 12).
Let F be a (G, Σ, 3, 1)-PRDF and let H be the union of the members of Σ. Thus H = {0, j 1 , j 2 , j 3 , x, −x} for a suitable element x of order 3. By definition, up to a reordering of {j 1 , j 2 , j 3 }, we have where L = {0, j 1 , j 2 , j 3 , x, x + j 1 }. Let n = q 1 . . . q ω be the prime power factorization of n. By assumption we have q i ≡ 3 (mod 4), hence −1 ∈ F q i for 1 ≤ i ≤ ω. Take any element σ i of F q i \{1} and set y i = σ i +1 σ i −1 . Set y = (y 1 , . . . , y ω ) and consider the following two triples of G × F n : We have y i +1 y i −1 = σ i ∈ F q i , thus y i − 1 and y i + 1 are both squares or both non-squares of F q i . Also, we have q i ≡ 3 (mod 4) for each i so that −1 ∈ F q i . On the basis of these facts it is evident that the projection of each ∆ h on F q i consists of a non-zero square and a non-square of F q i . As a consequence, if S is any halving of F n , we have S∆ h = F * n for every h ∈ H. Thus, the set of triples A =: {µ s (A i ) | i = 1, 2; s ∈ S} has list of differences ∆A = H × V * n , i.e., Now note that we have for each ℓ ∈ L and then Take a triple {u 1 , u 2 , u 3 } of units of F n with the property that the elements of its list of differences ∆{u 1 , u 2 , u 3 } are also units. For instance, one could take We note that the contribution of a single B ∈ F to ∆F + and Φ(F + ) is (∆B) × V * n and B × V * n , respectively. Thus that the two equalities in (8.1) imply that we have: Letting q 1 , . . . , q ω be the components of n, we finally note that is a group of strong multipliers of F + and its order is the greatest odd divisor of ψ(n) since each q i ≡ 3 (mod 4). The assertion follows.
Let us say that two difference families are strongly equivalent if each block of the first is a suitable translate of a block of the second 4 . Note how the previous definition reminds a bit of the notion introduced in [21] of a (v, k, λ) tiling of a group G, that is a set of mutually disjoint (v, k, λ) difference sets in G partitioning G \ {0}. Needless to say, however, that the members of a (v, k, λ) tiling of G are pairwise strongly inequivalent.
In the next section we will give a composition construction for RDFs where doubly disjoint difference families play a crucial role. For this reason, it is worth studying their possible existence. In particular, we are interested in (Z 3 × V 2n+1 , Z 3 × V 1 , 3, 1)-DFs. We are going to prove their existence in the case that all the components of 2n + 1 are congruent to 1 (mod 4). Proof. First observe that if q ≡ 1 (mod 4) is a prime power, then the set X = {x ∈ F q : x−2 ∈ F q } is not empty. For instance, using the cyclotomic numbers of order 2 (see, e.g., [24]) one can see that X has size q−1 4 . More elementarily and constructively, an element of X can be found as follows. Take any y of F q \ {2} and check that the set has at least one element in common with X. Thus, if n = q 1 . . . q ω with q i ≡ 1 (mod 4) is a prime power for 1 ≤ i ≤ ω, we can construct an element x = (x 1 , . . . , x ω ) ∈ F n with the property that x i ∈ F q i and x i − 2 ∈ F q i for 1 ≤ i ≤ ω. Consider the 3-subsets A and B of Z 3 × V n defined as follows: 4 We cannot simply say that they are equivalent since two difference families in a group G, say F = {B 1 , . . . , B n } and F ′ = {B ′ 1 , . . . , B ′ n }, are usually said to be equivalent if, up to the order, we have B ′ i = α(B i ) + t i for a suitable α ∈ Aut(G) and suitable t 1 , . . . , t n ∈ G.
We have Now take a halving S of F * n (see Section 5). In view of the choice of the element x, we see that the projection of each ∆ h and each Φ h on F q i consists of a square and a non-square. Thus, by (5.1), we have ∆ h S = Φ h S = F * n for h = 0, 1, 2. For each s ∈ S set A s = µ s (A), B s = µ s (B), and consider the family F = {A s , B s | s ∈ S}. We obviously have ∆A s = µ s (∆A) and ∆B s = µ s (∆B). Thus we have We also have: Note that the chosen halving S is symmetric, i.e., we have −S = S. Then there exists a subset T of S for which we have S = T ∪ (−T ) so that F is splittable in the two families Now note that A −t is a translate of A t and that B −t is a translate of B t for every t ∈ T . Indeed it is readily seen that we have: We deduce, in particular, that ∆A t = ∆A −t and ∆B t = ∆B −t for every t ∈ T . It follows that ∆F + = ∆F − . Thus, considering that ∆F is twice Z 3 × V * n , we necessarily have n . This means that both F + and F − are (Z 3 × V n , Z 3 × V 1 , 3, 1)-DFs.
Considering that each block of F − is a translate of a block of F + and that Φ(F + ∪ F − ) = Φ(F ) = (Z 3 × V n ) \ (Z 3 × V 1 ), we conclude that F + is doubly disjoint and the assertion follows.
We recall that a (h, k, 1) difference matrix in a group H of order h, briefly denoted by (H, k, 1)-DM, is a k × h matrix in which the difference of any two distinct rows is a permutation of H.
In particular, an (H, 3, 1)-DM is completely equivalent to a complete mapping of H. There is a large literature on complete mappings starting with the famous conjecture of Hall and Paige [31] according to which a group H of even order admits a complete mapping if and only if it is admissible, i.e., if and only if its 2-Sylow subgroups are not cyclic. The conjecture has been finally proved in [26]. Thus, as immediate consequence of the characterization of all abelian groups H for which there exists an (H, 4, 1)-DM (see [29,46]), we have the following. Difference matrices are also a crucial topic of Design Theory [2,17]. They have been used explicitly or implicitly in a lot of papers especially for the composition constructions of designs with a regular automorphism group starting from some early work by Jungnickel [35] and Colbourn [16]. Homogeneous difference matrices have been used later for the composition constructions of several kinds of resolvable designs (see, e.g., [1]). As far as we are aware the quite appropriate term homogeneous was coined in [36] when other authors choose other terms as good [13] about the same time.
We need to devise a new type of difference matrix that we call splittable. We give three examples of splittable difference matrices that will be crucial for the construction of some classes of resolvable difference families. The matrix of the third example is also homogeneous. The following theorem explains how doubly disjoint or resolvable difference families in a quotient group G/H can be combined with homogeneous or splittable difference matrices in H for the construction of resolvable difference families in G. Let j be an involution of G and assume that one of the following additional hypotheses holds.
Of course F M is a (G, H, k, 1)-DF which is strongly equivalent to F • M. Let us show that it is {0, j}-resolvable. We have: for suitable triples (i 1 , r 1 , c 1 ) and (i 2 , r 2 , c 2 ). We have to prove that these triples are necessarily equal. Without loss of generality we can assume that c 1 ≤ c 2 . So we have the following three possible cases.
On the other hand b i 2 ,r 2 and b i 1 ,r 1 + t i belong to Φ(F ) and Φ(F ′ ), respectively. This is absurd since Φ(F ) and Φ(F ′ ) are disjoint.
As an important consequence of the above theorem we get the following corollaries. Proof. Consider the group G = G 2 × V n and its subgroups L = G 1 × V n , and H = {0} × V n . We have G/H ≃ G 2 and L/H ≃ G 1 , hence there exists a doubly disjoint (G/H, L/H, 3, 1)-DF (see Subsection 6.5). There also exists a homogeneous (H, 3, 1)-DM by Theorem 10.2. Thus there exists a (G, L, 3, 1)-RDF, i.e., a (G 2 ×V n , G 1 ×V n , 3, 1)-RDF by Theorem 10.8(i). Now recall that there exists a (G 1 × V n , G 1 × V 1 , 3, 1)-RDF by Theorem 7.5. We get the assertion by applying Proposition 3.2 with the chain

Main results
We are finally able to prove the sufficient conditions given by the main Theorem 1.1.
Remark 11.2. Recall that Theorem 7.4 assures a group of strong multipliers of order the greatest odd divisor of ψ(4n + 1). Therefore, Proposition 6.1 guarantees that the number of symmetries of each KTS(24n + 9) obtainable via Theorem 11.1 is at least equal to (24n + 6)m, where m is the greatest odd divisor of ψ(4n + 1).
Case 2): the square-free part of 2n + 1 does not contain any prime congruent to 11 (mod 12).
We can assume that 2n + 1 is not divisible by 3 in view of Case 1). Thus, by assumption, we can write 2n + 1 = P Q where P is the product of all components of 2n + 1 that are congruent to 7 modulo 12 and Q is the product of all components of n that are congruent to 1 modulo 4. Of course it is understood that P and/or Q may be equal to 1 in the case that the components of the respective kinds do not exist. If P = 1 or Q = 1, we have a (G 1 × V 2n+1 , G 1 × V 1 , 3, 1)-RDF by Theorem 7.5 or Theorem 7.6, respectively. If both P and Q are greater than 1, consider the group G = G 1 × V 2n+1 and its subgroups L = G 1 × V P and H = {0}×V P . We have G/H ≃ G 1 ×V Q and L/H ≃ G 1 ×V 1 . Thus there exists a (G/H, L/H, 3, 1)-RDF by Theorem 7.5. Also, there exists a homogeneous (H, 3, 1)-DM by Theorem 10.2. It follows, by Theorem 10.8(i), that there exists a (G, L, 3, 1)-RDF, i.e. a (G 1 × V 2n+1 , G 1 × V P , 3, 1)-RDF. We also have a (G 1 × V P , G 1 × V 1 , 3, 1)-RDF by Theorem 7.6. Applying Proposition 3.2 with the chain Remark 11.4. We recall that Theorems 7.5 and 7.6, and Corollaries 8.4 and 8.5 show the existence of a group of strong multipliers. Therefore, it is not difficult to check that Propositions 3.2 and 6.1 guarantee that 1. the number of symmetries of each KTS(72n ′ + 39) obtainable via Theorem 11.3. (1) is at least equal to m(72n ′ + 39), where m is the greatest odd divisor of ψ(P ) and P > 1 is the product of all the components of 2n ′ + 1 congruent to 7 or 11 (mod 12); 2. the number of symmetries of each KTS(24n + 15) built in Theorem 11.3. (2) is at least equal to m(24n + 15) where m is defined as follows: (a) if Q > 1 is the product of all the components of 2n + 1 congruent to 1 (mod 4), then m is the greatest odd divisor of ψ(Q); (b) if all the components of 2n + 1 are congruent to 7 (mod 12), then m is the greatest odd divisor of ψ(2n + 1) coprime with 6.

3-pyramidal KT S(48n + 3)
In this subsection we will prove that the necessary condition for the existence of a KTS(v) is also sufficient when v ≡ 3 (mod 48), that is Theorem 1.1(iii). We recall that {G α : α ≥ 1} is the series of pertinent groups considered in Section 2. For 0 ≤ i ≤ α − 1, the subgroup of G α with underlying-set Z 3 × 2 i Z 2 α × 2 i Z 2 α is isomorphic to G α−i . Hence, by abuse of notation, this subgroup will be denoted by G α−i in the following.
Theorem 11.5. There exists a KTS(4 e 48n + 3) for every non-negative integer e and every positive odd integer n.
Proof. Set α = e + 2 and note that G α × V n is a pertinent group of order 4 e 48n. Hence, by Proposition 6.1, it is enough to prove the existence of a (G α × V n , G 1 × V i , 3, 1)-RDF with i = 1 or 3 for any α ≥ 2 and any odd n ≥ 1.
We distinguish five cases.
Remark 11.6. We recall that Corollary 8.6 shows the existence of a group of strong multipliers. Therefore, it is not difficult to check that Propositions 3.2 and 6.1 guarantee that the number of symmetries of each KTS(48n + 3) obtainable via Theorem 11.5, when n is not divisible by 3, is at least equal to m(48n + 3), where m is the greatest odd divisor of ψ(P ) and P > 1 is the product of all the components of n congruent to 3 (mod 4).

Open problems
The problem of classifying the 3-pyramidal KTS(v) remains open in the following cases.
• v − 3 = 24n + 6 and 4n + 1 is not a sum of two squares; • v − 3 = 72n ± 12 and its prime decomposition contains a prime factor p ≡ 11 (mod 12) raised to an odd power; The open cases above could be closed if one solves the following problems, respectively.
Our research naturally leads to consider also the following collateral problems which, in our opinion, are interesting on their own.