Bitranslations and Symmetric Nets

Abstract

It is known that every class-regular symmetric (𝜇,𝑚)-net is tactical. Also it is known that all (𝜇,𝑚)-nets with 𝑚=2 or 𝜇=1 are tactical. In the work of Al-Kenani and Mavron (2003), it is proved that every symmetric net with 𝑚=3 is tactical if and only if it is class regular. In this paper, we construct (2,4)-net and show that it is class regular and therefore tactical. New necessary and sufficient conditions are given for a symmetric net to admit a nonidentity bitranslation.

1. Introduction

A 𝑡−(𝑣,𝑘,𝜆) design Π is an incidence structure with 𝑣 points, 𝑘 points on a block, and any subset of 𝑡 points is contained in exactly 𝜆 blocks, where 𝑣>𝑘,𝜆>0. The number of blocks is 𝑏 and the number of blocks on a point is 𝑟.

The design Π is resolvable if its blocks can be partitioned into 𝑟  parallel classes, such that each parallel class partitions the point set of Π. Blocks in the same parallel class are parallel. Clearly each parallel class has 𝑚=𝑣/𝑘 blocks. Π is affine resolvable, or simply affine, if it can be resolved so that any two nonparallel blocks meet in 𝜇 points, where 𝜇=𝑘/𝑚=𝑘²/𝑣 is constant. Affine 1-designs are also called nets. The dual design of a design Π is denoted by Π^⋆. If Π and Π^⋆ are both affine, we call Π a symmetric net. We use the terminology of Jungnickel [1] (see also [2]). In this case, if 𝑟>1, then 𝑣=𝑏=𝜇𝑚² and 𝑘=𝑟=𝜇𝑚. That is, Π is an affine 1−(𝜇𝑚²,𝜇𝑚,𝜇𝑚) design whose dual Π^⋆ is also affine with the same parameters. For short we call such a symmetric net a (𝜇,𝑚)-net.

If Π is a symmetric net, we shall refer to the parallel classes of Π as block classes of Π and to the parallel classes of Π^⋆ as point classes of Π.

A bitranslation of Π is an automorphism fixing every point and block class which is fixed-point-free or is the identity. It is well known that the bitranslations form a group of order of a factor of 𝑚. The bitranslation group has order 𝑚 if and only if it acts regularly on each point and block class. In which case we say that Π is class regular.

In this paper we find necessary and sufficient conditions for a permutation in 𝑆_𝑚 to induce a bitranslation, by considering regular subsets of 𝑆_𝑚.

Let Π be a symmetric (𝜇,𝑚)-net and let 𝐴,𝐵 be block classes. If 𝜃∶𝐴→𝐵 is a bijection, then the point subset 𝑆=⋃𝑎∩𝜃(𝑎), where 𝑎 ranges over all elements of 𝐴, is called an (𝐴,𝐵)-transversal of Π. If 𝑆 is a union of point classes of Π, then 𝑆 is said to be a regular transversal of Π, and 𝜃 is called an (𝐴,𝐵)-syntax of Π. We will denote the set of all (𝐴,𝐵)-syntaxes by Σ(𝐴,𝐵).

Π is defined to be tactical if and only if |Σ(𝐴,𝐵)|=𝑚 for all pairs of distinct block classes 𝐴 and 𝐵 of Π. Equivalently, the intersection of any two nonparallel blocks is contained in a (unique) transversal. See [3] for more details.

Label the 𝑚 blocks in each block class of Π by {1,2,…,𝑚} and similarly for point classes of Π. Then a bijection between point or block classes of Π may be regarded as an element of the symmetric group 𝑆_𝑚. If 𝐴,𝐵 are any two distinct block classes of Π, then we may regard Σ(𝐴,𝐵) as a semiregular subset of 𝑆_𝑚; that is, if 𝜃₁,𝜃₂∈Σ(𝐴,𝐵) and 𝜃₁(𝑎)=𝜃₂(𝑎) for some 𝑎∈𝐴, then 𝜃₁=𝜃₂ (see [3]).

Let 𝑌 be a given block class of Π which we will call the base block class, and let its blocks be labelled {𝑌₁,𝑌₂,…,𝑌_𝑚} arbitrarily. We call 𝑌_𝑖 the 𝑖th block of 𝑌. Label any point class 𝑃 of Π with integers {1,2,…,𝑚} such that its 𝑖th point 𝑝_𝑖 is on 𝑌_𝑖 for 𝑖=1,2,…,𝑚.

Now choose a fixed point class 𝑥={𝑥₁,𝑥₂,…,𝑥_𝑚}, called the base point class, and label any block class 𝐴 of Π with {1,2,…,𝑚} such that the 𝑖th block 𝐴_𝑖 is on 𝑥_𝑖,𝑖=1,2,…,𝑚. We can therefore refer to the 𝑖th block of a block class, and dually.

We call this labelling the standard labelling, relative to the given base block class 𝑌 and point class 𝑥.

Result 1.1 (see [3]). If Π is a tactical (𝜇,𝑚)-net with standard labelling for its block and point classes, then the identity bijection 1∈Σ(𝐴,𝐵), for any two block classes 𝐴 and 𝐵 of Π.

2. Bitranslation Groups

Let Π be a symmetric (𝜇,𝑚)-net.

Let 𝑌={𝑌₁,𝑌₂,…,𝑌_𝑚} be the base block class of Π and 𝑥={𝑥₁,𝑥₂,…,𝑥_𝑚} the base point class of Π in the standard labelling as above. If Φ is a bitranslation of Π, then Φ induces the permutation 𝜎∈𝑆_𝑚 defined by: Φ(𝑌_𝑖)=𝑌_𝜎(𝑖),𝑖=1,2,…,𝑚.

Then by definition of standard labelling it follows that Φ∶𝐾_𝑖→𝐾_𝜎(𝑖) for any point or block class 𝐾 of Π.

Notation 2. If 𝑋 is a subset of a group 𝐺, then 𝐶_𝐺(𝑋) denotes the centralizer subgroup {𝑔∈𝐺∣𝑔𝑥=𝑥𝑔,forall𝑥∈𝐺} of 𝑋 in 𝐺.

Theorem 2.1. Let Π be a tactical symmetric (𝜇,𝑚)-net with standard labelling.
Let 𝜎∈𝑆_𝑚 and define a mapping Φ∶Π→Π by Φ(𝑢_𝑖)=𝑢_𝜎(𝑖) and Φ(𝐴_𝑖)=𝐴_𝜎(𝑖) for all point classes 𝑢 and block classes 𝐴 (𝑖=1,2,…,𝑚).
Then Φ is a bitranslation of Π if and only if 𝜎∈𝐻, where 𝐻 is the subgroup _(𝐴,𝐵)𝐶_𝑆𝑚(Σ(𝐴,𝐵)). (2.1) Here (𝐴,𝐵) runs over all pairs of distinct block classes of Π.

Proof. Let 𝐴,𝐵 be distinct block classes of Π. Then 𝑆=⋃^𝑚_𝑖=1𝐴_𝑖∩𝐵_𝜃(𝑖) is a union of point classes for all 𝜃∈Σ(𝐴,𝐵), by definition of syntaxes.
Assume first that Φ is a bitranslation.
Since Φ is a bitranslation, it fixes every point and block classes. Hence Φ fixes 𝑆 since 𝑆 is a union of point classes.
Therefore, 𝑆=Φ(𝑆)=⋃^𝑚_𝑖=1𝐴_𝜎(𝑖)∩𝐵_{𝜎𝜃(𝑖)}=⋃^𝑚_𝑖=1𝐴_𝑖∩𝐵_{𝜎𝜃𝜎−1(𝑖)}.
It follows that 𝜎𝜃𝜎⁻¹∈Σ(𝐴,𝐵) for all 𝜃∈Σ(𝐴,𝐵) and all 𝐴,𝐵. Hence 𝜎∈𝐻.
Conversely, suppose 𝜎∈𝐻. Define Φ as in the statement of the theorem. Clearly Φ is bijective and fixes every point and block classes. Let 𝑝_𝑗 be any point and 𝐴_𝑖 any block of Π. Suppose 𝑝_𝑗∈𝐴_𝑖. To show Φ is a bitranslation, it is enough to show that 𝑝_𝜎(𝑗)∈𝐴_𝜎(𝑖) (i.e., that Φ is an automorphism).
By definition of standard labelling, 𝑝_𝑗∈𝑌_𝑗, where 𝑌 is the base block class. Therefore, 𝑝_𝑗∈𝑌_𝑗∩𝐴_𝑖. Since Π is tactical, then there is a unique 𝜃∈Σ(𝑌,𝐴) such that 𝑖=𝜎(𝑗).
Note that 𝜃 depends only on 𝑝_𝑗 and 𝐴_𝑖.
Since 𝑝_𝑗 and 𝑝_𝜎(𝑗) are parallel, they are in the same transversal determined by 𝜃: that is, 𝑝_𝜎(𝑗)∈𝑌_𝜎(𝑗)∩𝐴_{𝜃𝜎(𝑗)}. So 𝑝_𝜎(𝑗)∈𝐴_{𝜃𝜎(𝑗)}=𝐴_{𝜎𝜃(𝑗)}, since 𝜃∈𝐻≤𝐶_𝑆𝑚(Σ(𝑌,𝐴)). But 𝜃(𝑗)=𝑖, therefore 𝑝_𝜎(𝑗)∈𝐴_𝜎(𝑖), as required.

With the notation and hypothesis of the theorem, we prove the following corollaries.

Corollary 2.2. (a)  𝐻 is isomorphic to the bitranslation group of Π.
(b)  Π is class regular if and only if |𝐻|=𝑚.

Proof. (a) The mapping 𝜎→Φ of the theorem is easily verified to be an isomorphism from 𝐻 onto the bitranslation group.
(b) This follows easily from the definition of class regular.

Corollary 2.3. If all syntax sets of Π are the same subgroup 𝐺 of 𝑆_𝑚, then the bitranslation group of Π is isomorphic to 𝐶_𝑆𝑀(𝐺)=𝐻 and 𝐺∩𝐻=𝑍(𝐺).

Proof. It is clear that 𝐻=𝐶_𝑆𝑚(𝐺). The rest follows from Lemma 3.2.

3. Regular Subsets

Let 𝑛≥2 be an integer and Ω a set of size 𝑛. Let 𝑆_Ω be the symmetric group on Ω.

A subset 𝑇 of 𝑆_Ω is a semiregular subset of 𝑆_Ω if for any 𝛼,𝛽∈Ω, there exists at most one element 𝑡∈𝑇 such that 𝛼𝑡=𝛽.

If there exists always exactly one such 𝑡∈𝑇, then 𝑇 is a regular subset of 𝑆_Ω.

Suppose 𝑇 is a regular subset of 𝑆_Ω.

Clearly |𝑇|=𝑛. Let 𝐶=𝐶_𝑆Ω(𝑇)={𝑥∈𝑆_Ω∣𝑥𝑡=𝑡𝑥forall𝑡∈𝑇}.

Let 𝐺=⟨𝑇⟩, the subgroup generated by 𝑇 in 𝑆_Ω. Then it is easy to see that(a)𝐺 is transitive on Ω;(b)𝐶=𝐶_𝑆Ω(𝐺).

Using this notation, we prove the following results.

Lemma 3.1. If 𝑇 is regular, then 𝐶 is semiregular on Ω and |𝐶| divides 𝑛.

Proof. Let 𝑥∈𝐶 and suppose 𝛼𝑥=𝛼 for some 𝛼∈Ω.
Let 𝛽 be any element of Ω and let 𝑡∈𝑇 be such that 𝛼𝑡=𝛽. Then 𝛽𝑥=𝛼𝑡𝑥=𝛼𝑥𝑡=𝛼𝑡=𝛽. Therefore, 𝑥=1. The rest of the proof is straightforward.
It is clear that 𝑇=𝐺(=⟨𝑇⟩) if and only if 𝑇 is a subgroup of 𝑆_Ω.

Lemma 3.2. (a) If 𝑇≠𝐺, then |𝐶|<𝑛.
(b) If 𝑇=𝐺, then |𝐶|=𝑛. Moreover, 𝐶≅𝐺 and 𝐶∩𝐺=𝑍(𝐺).

Proof. (a) 𝑇≠𝐺. Let 𝛼∈Ω. Since 𝐺≠𝑇, then |𝐺|>|𝑇|≥𝑛.
Since |𝐺∶𝐺_𝛼|≤|Ω|=𝑛, then |𝐺_𝛼|≥|𝐺|/𝑛>1.
∴ If Δ is the set of fixed points of 𝐺_𝛼, then 𝛼∈Δ and Δ≠Ω.
Hence |Δ|<𝑛.
Let 𝑐∈𝐶 and 𝛽=𝛼𝑐. For any 𝑔∈𝐺_𝛼, we have 𝛽𝑔=𝛼𝑐𝑔=𝛼𝑔𝑐=𝛼𝑐=𝛽.
Hence 𝛽∈Δ. So Δ contains the 𝐶-orbit 𝛼𝐶 of 𝛼.
Therefore, |𝛼𝐶|≤|Δ|<𝑛. We also have |𝛼𝐶|=|𝐶|, since 𝐶 is semiregular on Ω. Hence |𝐶|<𝑛.
(b) 𝑇=𝐺. Choose any 𝛼∈𝐺. For each 𝑡∈𝑇, define 𝑔_𝑡∈𝑆_Ω by (𝛼𝑢)𝑔_𝑡=𝛼𝑡⁻¹𝑢∀𝑢∈𝑇. (3.1)
Note that {𝛼𝑢∣𝑢∈𝑇}=Ω, since 𝑇 is regular on Ω.
Let 𝐻={𝑔_𝑡∣𝑡∈𝑇}. If 𝑡₁,𝑡₂∈𝑇 and 𝑔_𝑡1=𝑔_𝑡2, then 𝛼𝑡₁⁻¹=𝛼𝑡₂⁻¹ and hence 𝑡₁=𝑡₂ (since 𝑇 is regular). Therefore, |𝐻|=|𝑇|=𝑛.
If 𝑡₁,𝑡₂,𝑢∈𝑇, then (𝛼𝑢)𝑔_𝑡1𝑔_𝑡2=(𝛼𝑡₁⁻¹𝑢)𝑔_𝑡2=𝛼𝑡₂⁻¹𝑡₁⁻¹𝑢=𝛼𝑔_𝑡1𝑡2.
Hence 𝑔_𝑡1𝑔_𝑡2=𝑔_𝑡1𝑡2, which means that the mapping 𝑡→𝑔_𝑡 defines an isomorphism 𝑇→𝐻 (𝐻 is just the left regular representation of 𝑇). Therefore, |𝐻|=|𝑇|=𝑛.
If 𝑡₁,𝑡₂∈𝑇, then (𝛼𝑢)𝑔_𝑡1𝑡₂=𝛼𝑡₁⁻¹𝑢𝑡₂=(𝛼𝑢𝑡₂)𝑔_𝑡1=(𝛼𝑢)𝑡₂𝑔_𝑡1. So 𝑔_𝑡1𝑡₂=𝑡₂𝑔_𝑡1 for all 𝑡₂∈𝑇 and 𝑔_𝑡1∈𝐻. Hence 𝐻⊆𝐶.
Since |𝐻|=𝑛 and, by Lemma 3.1, |𝐶| divides 𝑛, then |𝐶|=𝑛 and 𝐶=𝐻.
Finally, 𝐶∩𝐺=𝐶_𝑆Ω(𝐺)∩𝐺=𝑍(𝐺).

Lemma 3.3. If 𝑔,ℎ∈𝑆_Ω, then 𝑔𝑇ℎ is also a regular set.

Proof. Let 𝛼,𝛽∈Ω. There is a unique 𝑡∈𝑇 such that (𝛼𝑔)𝑡=𝛽ℎ⁻¹; that is, 𝛼(𝑔𝑡ℎ)=𝛽.
Since 𝑔𝑡ℎ∈𝑔𝑇ℎ, the result follows immediately.

Definition 3.4. 𝑇 and 𝑔𝑇ℎ are congruent regular sets. If 𝑔=ℎ⁻¹, they are said to be similar.

Lemma 3.5. If 𝑇,𝑇^′ are regular sets in 𝑆_Ω and |𝑇∩𝑇^′|≥𝑛−1, then 𝑇=𝑇^′.

Proof. The result is obvious if |𝑇∩𝑇^′|=𝑛, since |𝑇|=|𝑇^′|=𝑛. So suppose |𝑇∩𝑇^′|=𝑛−1.
Let 𝑇={𝑡₁,𝑡₂,…,𝑡_𝑛−1,𝑡} and 𝑇^′={𝑡₁,𝑡₂,…,𝑡_𝑛−1,𝑡^′}. Then 𝑇∩𝑇^′={𝑡₁,𝑡₂,…,𝑡_𝑛−1} and 𝑡≠𝑡^′.
Let 𝛼∈Ω. Then Ω=𝛼𝑇 and {𝛼𝑡}=Ω⧵𝛼(𝑇∩𝑇^′)={𝛼𝑡^′}, since 𝑇 is a regular set. Hence 𝛼𝑡=𝛼𝑡^′ and so 𝑡=𝑡^′, since 𝑇 is regular. Therefore, 𝑇=𝑇^′.

The above lemma shows that distinct regular sets must differ in at least 2 elements.

4. Syntax Sets as Regular Subsets

The general theory of regular sets developed in the previous section is now applied syntax sets.

From the introduction to this paper, we know that syntax sets of a tactical symmetric (𝜇,𝑚)-net are semiregular subsets of 𝑆_𝑚.

Lemma 4.1. Let Π be a tactical symmetric (𝜇,𝑚)-net and Σ any of its syntax sets. If Σ is not a subgroup of 𝑆_𝑚, then (a)|𝐶_𝑆𝑚(Σ)| divides 𝑚 and is less than 𝑚; (b)the bitranslation group of Π has order at most 𝑚−1.

Proof. (a) Follows from Lemmas 3.1 and 3.2.
(b) It is clear from Theorem 2.1 that the bitranslation group 𝐺 of Π has order |𝐻|.
Since 𝐻 is a subgroup of 𝐶_𝑆𝑚(Σ), then |𝐻| divides 𝑚 and |𝐻|<𝑚 by part (a).

Now, consider the special case 𝑛=4. Let Ω={1,2,3,4}.

Theorem 4.2. If 𝑛=4 and 𝑇 is a regular set in 𝑆₄, then 𝑇 is congruent to a regular subgroup of 𝑆₄.
The regular subgroups of 𝑆₄ are the three cyclic groups generated by 4-cycles and the Klein 4-group {1,(12)(34),(13)(24),(14)(23)}.

Proof. From Lemma 3.3 we may assume that 1∈𝑇. The order of any of the 3 nonidentity elements of 𝑇 is therefore either 2 or 4.
Suppose 𝑇 has an element of order 4. Without loss of generality, we may assume that the 4-cycle (1234)=𝜔∈𝑇. If 𝑡,𝑢 are the remaining nonidentity elements of 𝑇, then, say, 1𝑡=3,1𝑢=4.
Therefore, 3𝑡≠3 or 4; so 3𝑡=1 or 2.
If 3𝑡=1, then 𝑡=(13)(24), 𝑢=(1432), and so 𝑇=⟨𝜔⟩.
Suppose 3𝑡=2. Then 2𝑡≠2 or 3. Therefore, 2𝑡=1 or 4.
If 2𝑡=1, then 4𝑡≠1,2,3,4, using the fact that 𝑡 is a permutation and 𝑇 is a regular set. This is impossible.
Similarly, 2𝑡=4 would imply 4𝑡≠1,2,3, or 4, which again is impossible.
Therefore, the case 3𝑡=2 is impossible and so 3𝑡=1 as above.
If no element of 𝑇 has order 4, then all nonidentity elements of 𝑇 have order 2. Then from the regularity of 𝑇 it follows that 𝑇 must be the Klein 4-group.

We continue with the notation and hypothesis of Theorem 2.1.

(1)Suppose all syntax sets of Π are the same subgroup 𝐺 of 𝑆_Ω. Then by Lemma 3.2, 𝐻=𝐺 and the bitranslation group of Π is isomorphic to 𝐶_𝑆Ω(𝐺)≅𝐺. Furthermore, 𝐶_𝑆Ω(𝐺)∩𝐺=𝑍(𝐺).

(2) Consider the special case 𝑚=4. By Theorem 4.2, we know that any syntax set of Π is congruent to a subgroup of 𝑆₄ of order 4. This must be either the Klein 4-group or one of the 3 cyclic subgroups of order 4. Since Π is tactical, all its syntax sets contain the identity. Therefore, we can say that any syntax set of Π is conjugate to a subgroup of order 4 in 𝑆₄.

Suppose all syntax sets of Π are the same subgroup 𝐺. Then by Corollary 2.3, we have 𝐻=𝐺 and the bitranslation group of Π is isomorphic to 𝐶_𝑆4(𝐺).

From (1), 𝐺≅𝐶_𝑆4(𝐺) and 𝐶_𝑆4∩𝐺=𝐺, since 𝐺 is abelian. Hence 𝐶_𝑆4(𝐺)=𝐺. It follows that the bitranslation group of Π has order 4 and hence Π is class regular.

Below is an example of a tactical symmetric (2,4)-net in which every syntax set Σ(𝐴,𝐵),𝐴≠𝐵, is the Klein 4-group: {1,(12)(34),(13)(24),(14)(23)}. (The full automorphism group has order 5376. The author is grateful to V. D. Tonchev for this information.)

The incidence matrix 𝑀 of this symmetric (2,4)-net is as follows: 𝑀=1111001000001000000110000000000000000100100100000010000100010101000010010100010000000100100010100000000000100011110000100110000011000000010000001010010001000001000110100010000001000000000101100010010000010111000100001000000000000001100010000000101100101000100100010001010011000001000000000010100000000001001011000011000001000110101000000000000011001000000000000100101000010010000001111000000000111001000001000000110000100010100001001010001000000010010001010100001001001000000100000001100000000000000100011110000110100000000000100000000101011010000100000111000000101010100000000100100110000001100100000000010000000110000011000100010000100001100000001000000101001000100000110011010001000000100000000010110001001000001011100010000100000000000000110001000000010110010100001000001100000010001000001010010000101000110110000100000001000000010100000000010100000010000110010000010000100000100111010000001010000000111001000001000000110000000101010000100100100000010000100110000000000000010001111000010000001010000100101000100000001001. (4.1)

Acknowledgment

The author would like to thank Professor V. C. Mavron for reading carefully the manuscript and suggesting several corrections and improvements. This paper is a part of a project no. 169/428 supported by DSR, KAU, Jeddah, Saudi Arabia.