3-Designs with Block Size 6 from P SL(2, q) and Their Large Sets^{*}

G. R. Omidi, M. R. Pournaki, B. Tayfeh-Rezaie

Abstract We investigate the existence of 3-designs and uniform large sets of 3-designs with block size 6 admitting P SL(2, q) as an automorphism group. We show the existence of simple 3-(28, 6, 10m) designs for 1 ≤ m ≤ 230. Most of these designs were previously unknown.

Keywords: 3-Design, Large set of 3-designs, Uniform large set, Projective special linear group. 2000 Mathematics Subject Classification: Primary 05B05, 05B30; Secondary 20D06.

1 Introduction

Let t, k, v, and λ be integers such that 0 ≤ t ≤ k ≤ v and λ > 0. Let X be a v-set and P_{k}(X) denote the set of all k-subsets of X. A t-(v, k, λ) design is a pair D = (X, D) in which D is a collection of elements of P_{k}(X) (called blocks) such that every t-subset of X appears in exactly λ blocks. If D has no repeated blocks, then it is called simple. Here we are concerned only with simple designs. It is well-known that a set of necessary conditions for the existence of a t-(v, k, λ) design is

λ\binom{v − i}{t − i} ≡ 0 (mod \binom{k − i}{t − i}),  (1)

for 0 ≤ i ≤ t. An automorphism of D is a permutation σ on X such that σ(B) ∈ D for each B ∈ D. An automorphism group of D is a group whose elements are automorphisms of D. A large set of t-(v, k, λ) designs, denoted by LS[N](t, k, v), is a set of N disjoint t-(v, k, λ) designs (X, D_{i}) such that D_{i} partition P_{k}(X) and N = \binom{v−t}{k−t}/λ. A large set is said to be G-uniform if each of its designs admits G as an automorphism group.

Let G be a finite group acting on X. For x ∈ X, the orbit of x is G(x) = {gx| g ∈ G} and the stabilizer of x is G_{x} = {g ∈ G| gx = x}. It is well-known that |G| = |G(x)||G_{x}|.

*This research was in part supported by a grant from IPM.

If there is an x ∈ X such that G(x) = X, then G is called transitive. The action of G on X induces a natural action on P_{k}(X). If this latter action is transitive, then G is called k-homogeneous.

Let q be a prime power and let X = GF (q) ∪ {∞}. Then the set of all mappings

g : x ↦ \frac{ax + b}{cx + d}

on X such that a, b, c, d ∈ GF (q), ad−bc is a nonzero square and g(∞) = a/c, g(−d/c) = ∞ if c ≠ 0, and g(∞) = ∞ if c = 0, is a group under composition of mappings called projective special linear group and is denoted by P SL(2, q). It is well-known that P SL(2, q) is 3homogeneous if and only if q ≡ 3 (mod 4). Note that |P SL(2, q)| = (q^{3} − q)/2. The structure of the elements of P SL(2, q) is well-known (see for example [5,6]) and is given in Table 1 for q ≡ 3 (mod 4) where φ denotes Euler’s function.

Throughout this paper, we let q be a prime power congruent to 3 (mod 4), X = GF (q) ∪ {∞}, and G = P SL(2, q) acting on X. We also denote G_{{0,1,∞}} by H. It is easy to see that

H = \{x ↦ x, x ↦ \frac{x − 1}{x}, x ↦ \frac{1}{1 − x}\}.

Table 1 The structure of the elements of P SL(2, q), q = p^{n}, q ≡ 3 (mod 4)

Order  Order of the centralizer  Number of conjugacy classes  Type

1  \frac{q^{3}−q}{2}  1  1^{q+1}
2  q + 1  1  2^{\frac{q+1}{2}}
p  q  2  1^{1}p^{\frac{q}{d}}
d|\frac{q−1}{2}  \frac{q−1}{2}  \frac{φ(d)}{2}  1^{2}d^{\frac{q−1}{d}}
d|\frac{q+1}{2}, d ≠ 2  \frac{q+1}{2}  \frac{φ(d)}{2}  d^{\frac{q+1}{d}}

The group P SL(2, q) has been used for constructing t-designs by different authors, see for example [1,3,4,7–10]. In [3], all 3-designs and uniform large sets of 3-designs with block sizes 4 and 5 admitting P SL(2, q) as an automorphism group were completely determined. In this paper, we investigate the existence of 3-designs and uniform large sets of 3-designs with block size 6 from P SL(2, q). Since P SL(2, q) is 3-homogeneous, a 3-(q + 1, k, λ) design admits P SL(2, q) as an automorphism group if and only if its block set is the union of orbits of P SL(2, q) on P_{k}(X). We determine the number of orbits for all possible orbit sizes from the action of P SL(2, q) on P_{6}(X) and then use the results to construct 3-(q + 1, 6, λ) designs and large sets of these designs. Finally, as a result, we establish the existence of 3-(28, 6, 10m) designs for 1 ≤ m ≤ 230. These designs were mostly unknown prior to this research [9]. Note that the method used in this paper is similar to the one in [10].

2 Orbit Counting

In this section we consider the action of G on P_{6}(X) and determine the possible sizes of orbits and the number of orbits for any fixed size. For a 6-subset B of X, let

Λ_{B} = {{x, y, z}| {0, 1, ∞, x, y, z} ∈ G(B)}.

The cardinality of Λ_{B} (denoted by λ_{B}) is called the index of G(B) which is clearly welldefined. Note that λ_{B} > 0. We denote the number of orbits of index i by N_{i}.

Lemma 2.1 Let B ∈ P_{6}(X). Then λ_{B}|G_{B}| = 60 and

(i) if q ≡ 11 (mod 20), then λ_{B} = 10, 12, 20, 30, 60, (ii) if q ≢ 11 (mod 20), then λ_{B} = 10, 20, 30, 60.

Proof. Since G(B) is a 3-(q + 1, 6, λ_{B}) design, we have |G(B)| = λ_{B}\binom{q+1}{3}/\binom{6}{3}. Therefore, by |G| = |G_{B}||G(B)|, we find λ_{B}|G_{B}| = 60. By (1), 4|λ_{B}(q − 1) and so λ_{B} is even. Moreover, 5|λ_{B}q(q − 1) and therefore if q ≢ 11 (mod 20), then 5|λ_{B}. It follows that λ_{B} = 2, 4, 6, 10, 12, 20, 30, 60, if q ≡ 11 (mod 20) and λ_{B} = 10, 20, 30, 60, otherwise. We now show that λ_{B} ≠ 2, 4, 6 or equivalently |G_{B}| ≠ 10, 15, 30 when q ≡ 11 (mod 20).

First suppose that |G_{B}| = 10. Let K be a normal subgroup of G_{B} of order 5 and g ∈ G_{B} be an element of order 2. Then there are k_{1}, k_{2} ∈ K such that gk_{1} = k_{2}g. Note that k_{1} and k_{2} fix exactly one element x of B. Since g(x) = k_{2}(g(x)), we have g(x) = x which is a contradiction with the fact that g has no fixed point.

Now let |G_{B}| = 15. As there is a unique group of order 15 which is cyclic, G_{B} has an element of order 15. But such an element cannot fix B and therefore |G_{B}| ≠ 15.

Finally, let |G_{B}| = 30. Let P_{1} and P_{2} be 3-Sylow and 5-Sylow subgroups of G_{B}, respectively. Since P_{2} is always normal in G_{B}, P_{1}P_{2} is a subgroup of order 15 of G_{B} which is impossible as described above. □

Lemma 2.2 Let H act on P_{3}(GF (q) \ {0, 1}). Then the number of orbits of sizes 1 and 3 are equal to L and (\binom{q−2}{3} − L)/3, respectively, where

L = \{ \begin{matrix}\frac{q−3}{3} & if q ≡ 3 (mod 12), \\ \frac{q−4}{3} & if q ≡ 7 (mod 12), \\ \frac{q−2}{3} & if q ≡ 11 (mod 12). \end{matrix}  (2)

Furthermore, the set of orbits of size 1 is

\{\{α, \frac{α − 1}{α}, \frac{1}{1 − α}\}| α ∈ GF (q) \ \{0, 1\}, α^{2} − α + 1 ≠ 0\}.

Proof. Let s_{1} and s_{2} be the number of orbits of sizes 1 and 3, respectively. It is easy to see that

s_{1} + 3s_{2} = \binom{q − 2}{3}.

The total number of orbits can be found by the Cauchy–Frobenius lemma. We have

s_{1} + s_{2} = \frac{1}{|H|} ∑_{g∈H} Fix(g)
= \frac{1}{3}(\binom{q−2}{3} + 2L),

where L is the number of 3-subsets of GF (q) \ {0, 1} fixed by x ↦ (x − 1)/x (note that x ↦ (x − 1)/x and x ↦ 1/(1 − x) have the same order and so, by Table 1, fixed the same number of 3-subsets). Therefore, s_{1} = L and s_{2} = (\binom{q−2}{3} − L)/3. By Table 1, L is easily determined as shown in (2). Now suppose that B lies in an orbit of size 1 and let α ∈ B. Then, (α − 1)/α, 1/(1 − α) ∈ B. If α^{2} − α + 1 ≠ 0, then B = {α, (α − 1)/α, 1/(1 − α)}.

If α^{2} − α + 1 = 0, then it is easily seen that every element of B satisfies the equation x^{2} − x + 1 = 0. But this is an impossibility since this equation has at most two solutions in GF (q). Finally, every subset of the form {α, (α − 1)/α, 1/(1 − α)} clearly belongs to an orbit of size 1. □

Lemma 2.3 Let A_{α} := {0, 1, ∞, α, (α − 1)/α, 1/(1 − α)}, where α ∈ GF (q) \ {0, 1} and let B ∈ P_{6}(X). Then

(i) |G_{B}| = 3, 6 if and only if G(B) = G(A_{α}) for some α, (ii) if |G_{B}| = 6, then G(B) contains exactly one element of the form A_{α}, (iii) if |G_{B}| = 3, then G(B) contains exactly two elements of the form A_{α}, (iv) |G_{B}| = 6 if and only if G(B) = G(A_{α}) for some α such that −α^{2} + α − 1 is a nonzero square or, α = 2 and q ≢ 0 (mod 3).

Proof. (i) If G(B) = G(A_{α}) for some α, then H ≤ G_{A_{α}}. Therefore, 3 | |G_{B}|. Conversely, let 3 | |G_{B}|. Then λ_{B} = 10, 20. Since Λ_{B} is the union of orbits of H on P_{3}(GF (q) \ {0, 1}), it must contain some orbits of size 1 which are of the form {α, (α − 1)/α, 1/(1 − α)}. Therefore, A_{α} ∈ G(B) for some α.

(ii), (iii) Suppose that A_{α}, A_{β} ∈ G(B) where A_{α} ≠ A_{β}. Hence, there is a g ∈ G such that g(A_{α}) = A_{β}. Note that G_{A_{α}} and G_{A_{β}} contain H as a normal subgroup. So there are h_{1}, h_{2} ∈ H such that h_{1}g = gh_{2}. Hence,

g(\{0, 1, ∞\}) = \{β, \frac{β − 1}{β}, \frac{1}{1 − β} \}.  (3)

Now let A_{γ} ∈ G(B) be distinct from A_{α} and A_{β} and k ∈ G such that k(A_{β}) = A_{γ}. With a similar argument as for g we have k({0, 1, ∞}) = {γ, (γ − 1)/γ, 1/(1 − γ)}. So by (3), kg({0, 1, ∞}) = {0, 1, ∞} and therefore kg ∈ H. Now kg(A_{α}) = A_{α} and on the other hand k(g(A_{α})) = A_{γ} which is a contradiction. Therefore, G(B) contains at most two distinct blocks of the form A_{α}. Since λ_{B} = 10 or 20, the assertion follows.

(iv) Let |G_{B}| = 6. Then by (ii), A_{α} ∈ G(B) for some α. Assume that g ∈ G_{A_{α}} is of order 2. Then, similar to (3), g({0, 1, ∞}) = {α, (α − 1)/α, 1/(1 − α)}. With no loss of generality, suppose g(0) = α. If g(∞) = 1/(1 − α), then clearly g(x) = (x − α)/((1 − α)x − 1) and therefore −α^{2} + α − 1 is a nonzero square. If g(∞) = (α − 1)/α, then g(x) = ((α − 1)x − α^{2} + α)/(αx − α + 1). Since g(1) = 1/(1 − α), we obtain (1 − α)^{3} = −1 and therefore α = 2 and necessarily q ≢ 0 (mod 3). Conversely, if −α^{2} + α − 1 is a nonzero square, then g(x) = (x − α)/((1 − α)x − 1) is of order 2, and g(A_{α}) = A_{α}, and therefore |G_{A_{α}}| = |G_{B}| = 6. If α = 2 and q ≢ 0 (mod 3), then g(x) = (x − 2)/(2x − 1) or g(x) = (2x − 1)/(x − 2) belongs to G and since both of them are of order 2 and fix A_{α}, we have |G_{A_{α}}| = |G_{B}| = 6. □

Lemma 2.4 The value of N_{10} is given by

N_{10} = \{ \begin{matrix}0 & if q ≡ 3 (mod 12), \\ \frac{q−1}{6} & if q ≡ 7 (mod 12), \\ \frac{q+7}{6} & if q ≡ 11 (mod 12). \end{matrix}

Proof. Let S = {t ∈ GF (q) \ {0, 1}| − t^{2} + t − 1 is a nonzero square}. By Lemma 2.3, it suffices to find |S|. Let T = {t ∈ GF (q) \ {0, 1}| t^{2} − t + 1 is a square}. Note that |S| + |T | = q − 2 and S ∩ T = ∅. We have

|T | = |{t ∈ GF (q) \ {0, 1}| ∃x ∈ GF (q), x^{2} − 3 = (2t − 1)^{2}}|
= |{y ∈ GF (q) \ {−1, 1}| ∃x ∈ GF (q), −3 = (y − x)(y + x)}|
= |{y ∈ GF (q) \ {−1, 1}| ∃a, b ∈ GF (q), ab = −3, y = (a + b)/2}|
= |{{a, b}| a, b ∈ GF (q), ab = −3, a + b ≠ ±2}|
= |{{a, b}| a, b ∈ GF (q), ab = −3, {a, b} ≠ {−1, 3}, {−3, 1}}|.

Therefore,

|T | = \{ \begin{matrix}q − 2 & if q ≡ 3 (mod 12), \\ \frac{q−3}{2} & if q ≡ 7 (mod 12), \\ \frac{q−5}{2} & if q ≡ 11 (mod 12). \end{matrix}

Using the fact that −3 is nonsquare if and only if q ≡ 11 (mod 12) (see [2] or deduce it from Lemma 2.2), we have

|S| = \{ \begin{matrix}0 & if q ≡ 3 (mod 12), \\ \frac{q−1}{2} & if q ≡ 7 (mod 12), \\ \frac{q+1}{2} & if q ≡ 11 (mod 12). \end{matrix}

Now Lemma 2.3 implies that

N_{10} = \{ \begin{matrix}0 & if q ≡ 3 (mod 12), \\ \frac{q−1}{6} & if q ≡ 7 (mod 12), \\ \frac{q+7}{6} & if q ≡ 11 (mod 12). □ \end{matrix}

Lemma 2.5 If q ≡ 11 (mod 20), then N_{12} = 2.

Proof. The number of B ∈ P_{6}(X) such that |G_{B}| = 5 is 12\binom{q+1}{3}N_{12}/\binom{6}{3}. On the other hand, by Table 1, each element of order 5 of G fixes exactly 2(q − 1)/5 elements of P_{6}(X) and there are exactly 2q(q + 1) elements of order 5 in G. Therefore, (q + 1)q(q − 1)/5 distinct 6-subsets are fixed by the elements of order 5 of G. We now have 12\binom{q+1}{3}N_{12}/\binom{6}{3} = (q − 1)q(q + 1)/5 and hence N_{12} = 2. □

Theorem 2.6 The number of orbits of P SL(2, q) on P_{6}(X) for all possible orbit indices are given below.

\begin{matrix}q (mod 60) & N_{10} & N_{12} & N_{20} & N_{30} & N_{60} \\ 3, 27 & 0 & 0 & \frac{q−3}{6} & \frac{q^{2}−4q+3}{24} & \frac{2q^{3}−33q^{2}+72q+27}{720} \\ 7, 19, 43 & \frac{q−1}{6} & 0 & \frac{q−7}{12} & \frac{q^{2}−8q+7}{24} & \frac{2q^{3}−33q^{2}+132q+7}{720} \\ 23, 47, 59 & \frac{q+7}{6} & 0 & \frac{q−11}{12} & \frac{q^{2}−8q−9}{24} & \frac{2q^{3}−33q^{2}+132q+167}{720} \\ 11 & \frac{q+7}{6} & 2 & \frac{q−11}{12} & \frac{q^{2}−8q−9}{24} & \frac{2q^{3}−33q^{2}+132q−121}{720} \\ 31 & \frac{q−1}{6} & 2 & \frac{q−7}{12} & \frac{q^{2}−8q+7}{24} & \frac{2q^{3}−33q^{2}+132q−281}{720} \end{matrix}

Proof. We use the Cauchy–Frobenius lemma to count the total number M of orbits of G on P_{6}(X). We have

M = \frac{1}{|G|} ∑_{g∈G} |Fix(g)|,

where |Fix(g)| is the number of 6-subsets of X fixed by g. We refer to Table 1 to find |Fix(g)|. If |Fix(g)| ≠ 0, then o(g) = 1, 2, 3, 5, 6, where o(g) denotes the order of g. Let n_{i} be the number of elements of order i in G. If o(g) = 1, then |Fix(g)| = \binom{q+1}{6}. If o(g) = 2, then |Fix(g)| = \binom{(q+1)/2}{3} and n_{2} = |G|/(q + 1). If o(g) = 3, then

|Fix(g)| = \{ \begin{matrix}\binom{\frac{q}{3}}{2} & if q ≡ 3 (mod 12), \\ \binom{\frac{q−1}{3}}{2} & if q ≡ 7 (mod 12), \\ \binom{\frac{q+1}{3}}{2} & if q ≡ 11 (mod 12), \end{matrix}

and

n_{3} = \{ \begin{matrix}\frac{2|G|}{q} & if q ≡ 3 (mod 12), \\ \frac{2|G|}{q−1} & if q ≡ 7 (mod 12), \\ \frac{2|G|}{q+1} & if q ≡ 11 (mod 12). \end{matrix}

If o(g) = 5, then q ≡ 11 (mod 20), |Fix(g)| = 2(q − 1)/5, and n_{5} = 4|G|/(q − 1). If o(g) = 6, then q ≡ 11 (mod 12), |Fix(g)| = (q + 1)/6, and n_{6} = 2|G|/(q + 1). We now obtain

M = \{ \begin{matrix}\frac{2q^{3}−3q^{2}+72q−243}{720} & if q ≡ 3, 27 (mod 60), \\ \frac{2q^{3}−3q^{2}+72q−323}{720} & if q ≡ 7, 19, 43 (mod 60), \\ \frac{2q^{3}−3q^{2}+72q+77}{720} & if q ≡ 23, 47, 59 (mod 60), \\ \frac{2q^{3}−3q^{2}+72q+1229}{720} & if q ≡ 11 (mod 60), \\ \frac{2q^{3}−3q^{2}+72q+829}{720} & if q ≡ 31 (mod 60). \end{matrix}

The total number of orbits gives the equation

N_{10} + N_{12} + N_{20} + N_{30} + N_{60} = M.  (4)

By Lemmas 2.2 and 2.3, we also have

N_{10} + 2N_{20} = \{ \begin{matrix}\frac{q−3}{3} & if q ≡ 3 (mod 12), \\ \frac{q−4}{3} & if q ≡ 7 (mod 12), \\ \frac{q−2}{3} & if q ≡ 11 (mod 12). \end{matrix}  (5)

On the other hand, by counting the 6-subsets of X containing 0, 1, ∞, we have

10N_{10} + 12N_{12} + 20N_{20} + 30N_{30} + 60N_{60} = \binom{q − 2}{3}.  (6)

According to Lemmas 2.4 and 2.5, N_{10} and N_{12} are known. Therefore, the system of equations containing (4)–(6) are easily solved and the values of N_{20}, N_{30}, and N_{60} are determined. □

3 3-Designs and Large Sets

In this section we use the results of previous section to find 3-(q+1, 6, λ) designs with automorphism group P SL(2, q) and large sets of these designs. Recall that every 3-(q + 1, 6, λ) design with automorphism group G = P SL(2, q) is a union of distinct orbits of G on P_{6}(X).

Theorem 3.1 Let q ≡ 11 (mod 20). Then, there exist 3-(q + 1, 6, λ) designs with automorphism group P SL(2, q) if and only if λ ≡ 0, 2, 4 (mod 10), 1 ≤ λ ≤ \binom{q−2}{3}, and λ ≠ i, \binom{q−2}{3} − i for i = 2, 4, 14.

Proof. Let D denote a 3-(q + 1, 6, λ) design with automorphism group P SL(2, q). If D exists, then by (1), 2|λ and 1 ≤ λ ≤ \binom{q−2}{3}. By Theorem 2.6, the indices of orbits of G on P_{6}(X) are 10, 12, 20, 30, 60. Therefore, λ ≡ 0, 2, 4 (mod 10) and λ ≠ i, \binom{q−2}{3} − i for i = 2, 4, 14.

Conversely, note that there are designs for λ = 10, 12, 20, 22, 24. It is easy to see that if there exists D, then by replacing some suitable orbits of D by some unused orbits, one can obtain a 3-(q + 1, 6, λ + 10) design. Otherwise, there are no more unused orbits and D is the complete 3-(q + 1, 6, \binom{q−2}{3}) design. □

Theorem 3.2 Let q ≢ 11 (mod 20). Then, there are 3-(q + 1, 6, λ) designs with automorphism group P SL(2, q) if and only if 10|λ and 1 ≤ λ ≤ \binom{q−2}{3} except for λ = 10 when q ≡ 0 (mod 3).

Proof. Suppose that a 3-(q+1, 6, λ) design with automorphism group P SL(2, q) exists. By Theorem 2.6, the indices of orbits of G on P_{6}(X) are 10, 12, 20, 30, 60. Therefore, 10|λ.

Conversely, similar to the proof of Theorem 3.1, one can show that for any λ such that 10|λ and 1 ≤ λ ≤ \binom{q−2}{3}, a 3-(q + 1, 6, λ) design exists except for λ = 10 when q ≡ 0 (mod 3). □

Theorem 3.3 Let q ≡ 11 (mod 20). Then, there are P SL(2, q)-uniform LS[N](3, 6, q+1) if and only if N = 2 and q ≡ 11, 91 (mod 120).

Proof. Consider the action of G on P_{6}(X). Suppose that there is a G-uniform LS[N](3, 6, q + 1). Since N_{12} = 2 and the other orbits have indices which are multiple of 10, we have N = 2. On the other hand, by the necessary conditions (1), \binom{q−2}{3}(q − 1)/8 must be integer. Therefore, q ≡ 11 or 91 (mod 120).

Conversely, let N = 2 and q ≡ 11, 91 (mod 120). First let q ≡ 11 (mod 120). In this case, N_{10} and N_{30} are odd and N_{20} is even. Let D be an empty set. Consider 1, (N_{10} − 3)/2, N_{20}/2, and (N_{30} − 1)/2 orbits of indices 12, 10, 20, and 30, respectively, and add them to D. If N_{60} is odd, then also add [N_{60}/2], 3, and 1 orbits of indices 60, 10, and 30, respectively, to D. If N_{60} is even, then add [N_{60}/2] and 1 orbits of indices 60 and 30, respectively, to D. Now {(X, D), (X, P_{6}(X) \ D)} is a LS[2](3, 6, q + 1). Now let q ≡ 91 (mod 120). Note that N_{10}, N_{20}, and N_{30} are odd. Consider (N_{10}−1)/2, (N_{20}−1)/2, and (N_{30} − 1)/2 orbits of indices 10, 20 and 30, respectively, and add them to D. If N_{60} is odd, then also add [N_{60}/2], 1, 1, and 1 orbits of indices 60, 10, 20, and 30, respectively, to D. If N_{60} is even, then add [N_{60}/2] orbits of index 60 and one orbit of index 30 to D. Now {(X, D), (X, P_{6}(X) \ D)} is a LS[2](3, 6, q + 1). □

Theorem 3.4 Let q ≢ 11 (mod 20). Then, there are P SL(2, q)-uniform LS[N](3, 6, q+1) if and only if one of the following holds:

(i) \binom{q−2}{3}/N ≡ 0 (mod 60), (ii) \binom{q−2}{3}/N ≡ 10, 40 (mod 60) and N ≤ N_{10} + [N_{20}/2], (iii) \binom{q−2}{3}/N ≡ 20, 50 (mod 60) and N ≤ [N_{10}/2] + N_{20}, (iv) \binom{q−2}{3}/N ≡ 30 (mod 60) and N ≤ N_{20} + N_{30} + [(N_{10} − N_{20})/2].

Proof. Consider the action of G on P_{6}(X). Let \binom{q−2}{3} = N(60m + l) where 0 ≤ l ≤ 60. Suppose that there is a G-uniform large set of 3-(q + 1, 6, 60m + l) designs. Then by Theorem 2.6, l ≡ 0 (mod 10). If l = 10, 40, then some designs of the large set contain orbits of index 10 and each of the other designs contains necessarily at least two orbits of index 20. Therefore, N ≤ N_{10} + [N_{20}/2]. If l = 20, 50, then some designs in the large set contain orbits of index 20 and each of the other designs contains at least two orbits of index 10. Hence, N ≤ [N_{10}/2] + N_{20}. If l = 30, then some designs in the large set contain orbits of index 30 and each of the other designs contains one orbit of index 10 and one orbit of index 20 or at least three orbits of index 10. Therefore, N ≤ N_{20} + N_{30} + [(N_{10} − N_{20})/2].

Conversely, let one of (i)–(iv) hold. Let D_{f} (1 ≤ f ≤ N) be N empty sets. Here is a useful observation. If there are x_{1}, x_{2}, x_{3}, and x_{4} orbits of indices 10, 20, 30, and 60, respectively, such that 10x_{1} + 20x_{2} + 30x_{3} + 60x_{4} = 60x, then it is easy to see that by suitable combinations of these orbits we can find x disjoint 3-(q + 1, 6, 60) designs. If (i) holds, then by this observation we are done. Now let (ii) hold. Note that N_{30} ≥ [N_{20}/2] and N_{30} ≥ N_{10}. Choose x (0 ≤ x ≤ min{N, N_{10}}) orbits of index 10 and add to each of D_{i} (1 ≤ i ≤ x) one of them. Choose 2(N − x) orbits of index 20 and add to each of D_{j} (x < j ≤ N) two of them. If l = 10 (respectively, l = 40), then also add to each of D_{j} (respectively, D_{i}) one orbit of index 30. If l = 10 (respectively, l = 40), this leaves \binom{q−2}{3} − 10x − 70(N − x) = 60xm + 60(N − x)(m − l) (respectively, \binom{q−2}{3} − 40x − 40(N − x) = 60xm + 60(N − x)m) 6-subsets unused. Therefore, by the observation above, D_{f} (1 ≤ f ≤ N) can be filled with suitable unused orbits to obtain N sets with the same size. Now {(X, D_{f})| 1 ≤ f ≤ N} is the desired large set. Now suppose that (iii) holds. Choose x (0 ≤ x ≤ min{N, N_{20}}) orbits of index 20 and add to each of D_{i} (1 ≤ i ≤ x) one of them. Choose 2(N − x) orbits of index 10 and add to each of D_{j} (x < j ≤ N) two of them. If l = 50, then also add to each of D_{i} (1 ≤ i ≤ N), one orbit of index 30 (note that N_{30} ≥ [N_{10}/2] + N_{20}). The number of unused blocks is equal to \binom{q−2}{3} − lx − l(N − x) = 60mN. Therefore, by the observation above, the remaining orbits can be divided between to D_{f} (1 ≤ f ≤ N) to obtain N sets of the same size which results in large set {(X, D_{f})| 1 ≤ f ≤ N}. Finally, assume that (iv) holds. Choose x orbits (0 ≤ x ≤ min{N, N_{30}}) of index 30 and add to each of D_{i} (1 ≤ i ≤ x) one of them. Choose y orbits (0 ≤ y ≤ min{N_{20}, N − x}) of index 10 and y orbits of index 20 and add to each of D_{j} (x < j ≤ N − x − y) one orbit of index 10 and one orbit of index 20. There are totaly \binom{q−2}{3} − 30x − 30y − 30(N − x − y) = 60mN unused 6-subsets and therefore, by the observation above, the remaining orbits can be appended in a suitable way to D_{f} (1 ≤ f ≤ N) to obtain N sets of the same size. Now {(X, D_{f})| 1 ≤ f ≤ N} is the desired large set. □

4 3-(28, 6, 10m) Designs

The existence of 3-(28, 6, 10m) designs has been known only for a small number of some large values of m (see [9]). In the previous section we showed that these designs exist for all possible values of m, i.e., 1 ≤ m ≤ 230 except for m = 1, 229 (Theorem 3.2). Note that the existence of 3-(28, 6, 10m) designs for m = 1 implies the existence of such designs for m = 229. By [9], a 3-(28, 6, 10) design with repeated blocks exists, but no simple one is known. We now construct a simple 3-(28, 6, 10) design by prescribing a suitable subgroup of P SL(2, 27) as its automorphism group. Let P SL(2, 27) = ⟨a, b⟩ where

a = (1 2)(3 4)(5 7)(6 8)(9 12)(10 13)(11 15)(14 19)(16 20)(17 21)(18 22)(23 25)(24 26)(27 28),

b = (2 3 5)(4 6 9)(7 10 14)(8 11 16)(12 17 20)(13 18 23)(15 19 22)(21 24 27)(25 26 28).

Consider the subgroup L = ⟨b, c⟩ of P SL(2, 27) in which

c = (2 3 19 21 25 27 20 15 10 23 5 16 17)(4 26 13 22 18 8 28 7 24 9 12 11 6).

The order of L is 351. The union of orbits of L with representatives

{1, 2, 3, 4, 7, 19}, {2, 3, 4, 5, 6, 9},
{2, 3, 4, 9, 22, 23}, {2, 3, 5, 8, 11, 16},
{2, 3, 4, 8, 23, 24}, {2, 3, 4, 6, 13, 21},

identify the block set of a 3-(28, 6, 10) design. This design was found via DISCRETA, a program to construct t-designs with prescribed automorphism group written at the Mathematics Department, University of Bayreuth, Germany.

Acknowledgment: This work was done while the second author was a Postdoctoral Research Associate at the School of Mathematics, Institute for Studies in Theoretical Physics and Mathematics (IPM). He would like to thank IPM for the financial support.

References

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[3] C. A. Cusack, S. W. Graham, D. L. Kreher, Large Sets of 3-Designs from P SL(2, q), with Block Sizes 4 and 5, J. Combin. Des. 3 (1995), no. 2, 147-160.

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[5] L. E. Dickson, “Linear Groups with an Exposition of the Galois Field Theory”, Dover Publications, Inc., New York, 1958.

[6] D. Gorenstein, “Finite Groups”, Second edition, Chelsea Publishing Co., New York, 1980.

[7] D. R. Hughes, On t-Designs and Groups, Amer. J. Math. 87 (1965), 761-778.

[8] S. Iwasaki, Infinite Families of 2- and 3-Designs with Parameters v = p + 1, k = (p − 1)/2^{i} + 1, where p Odd Prime, 2^{e} ⊤ (p − 1), e ≥ 2, 1 ≤ i ≤ e, J. Combin. Des. 5 (1997), no. 2, 95-110.

[9] D. L. Kreher, t-Designs, t ≥ 3, in: The CRC Handbook of Combinatorial Designs (C. J. Colbourn and J. H. Dinitz, eds.), CRC Press Series on Discrete Mathematics and its Applications, CRC Press, Boca Raton (1996), pp. 47-66.

[10] R. Laue, S. S. Magliveras, A. Wassermann, New Large Sets of t-Designs, J. Combin. Des. 9 (2001), no. 1, 40-59.

The Authors’ Addresses

G. R. Omidi, Department of Mathematics and Computer Science, Faculty of Science, University of Tehran, Tehran, Iran, and School of Mathematics, Institute for Studies in Theoretical Physics and Mathematics, P.O. Box 19395-5746, Tehran, Iran. E-mail address: omidi@ipm.ir

M. R. Pournaki, School of Mathematics, Institute for Studies in Theoretical Physics and Mathematics, P.O. Box 19395-5746, Tehran, Iran.

E-mail address: pournaki@ipm.ir

B. Tayfeh-Rezaie, School of Mathematics, Institute for Studies in Theoretical Physics and Mathematics, P.O. Box 19395-5746, Tehran, Iran.

E-mail address: tayfeh-r@ipm.ir