# Bitranslations and Symmetric Nets

Authors: Ahmad N. Al-Kenani
Journal: International Journal of Combinatorics
 (2010)

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## 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.


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## Body

## 1. Introduction

At-(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 designΠ 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.IfΠ 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 classesof Π.Abitranslation 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 inSm to induce a bitranslation, by considering regular subsets of Sm.LetΠ 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)-transversalof Π. If S is a union of point classes of Π, then S is said to be a regular transversalof 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 them 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 Sm. If A,B are any two distinct block classes of Π, then we may regard Σ(A,B) as a semiregular subset of Sm; that is, if θ1,θ2∈Σ(A,B) and θ1(a)=θ2(a) for some a∈A, then θ1=θ2 (see [3]).LetY be a given block class of Π which we will call the base block class, and let its blocks be labelled {Y1,Y2,…,Ym} arbitrarily. We call Yi the ith block of Y. Label any point class P of Π with integers {1,2,…,m} such that its ith point pi is on Yi for i=1,2,…,m.Now choose a fixed point classx={x1,x2,…,xm}, called the base point class, and label any block class A of Π with {1,2,…,m} such that the ith block Ai is on xi,i=1,2,…,m. We can therefore refer to the ith block of a block class, and dually.We call this labelling thestandard labelling, relative to the given base block class Y and point class x.Result 1.1 (see [3]).
IfΠ 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

LetΠ be a symmetric (μ,m)-net.LetY={Y1,Y2,…,Ym} be the base block class of Π and x={x1,x2,…,xm} 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∈Sm defined by: Φ(Yi)=Yσ(i),i=1,2,…,m.Then by definition of standard labelling it follows thatΦ:Ki→Kσ(i) for any point or block class K of Π.Notation 2.
IfX is a subset of a group G, then CG(X) denotes the centralizer subgroup {g∈G∣gx=xg,forallx∈G} of X in G.Theorem 2.1.
LetΠ be a tactical symmetric (μ,m)-net with standard labelling.
Letσ∈Sm and define a mapping Φ:Π→Π by Φ(ui)=uσ(i) and Φ(Ai)=Aσ(i) for all point classes u and block classes A (i=1,2,…,m).
ThenΦ is a bitranslation of Π if and only if σ∈H, where H is the subgroup
(2.1)⋂(A,B)‍CSm(Σ(A,B)).
Here (A,B) runs over all pairs of distinct block classes of Π.Proof.
LetA,B be distinct block classes of Π. Then S=⋃i=1mAi∩Bθ(i) is a union of point classes for all θ∈Σ(A,B), by definition of syntaxes.
Assume first thatΦ is a bitranslation.
SinceΦ 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)=⋃i=1mAσ(i)∩Bσθ(i)=⋃i=1mAi∩Bσθσ-1(i).
It follows thatσθσ-1∈Σ(A,B) for all θ∈Σ(A,B) and all A,B. Hence σ∈H.
Conversely, supposeσ∈H. Define Φ as in the statement of the theorem. Clearly Φ is bijective and fixes every point and block classes. Let pj be any point and Ai any block of Π. Suppose pj∈Ai. To show Φ is a bitranslation, it is enough to show that pσ(j)∈Aσ(i) (i.e., that Φ is an automorphism).
By definition of standard labelling,pj∈Yj, where Y is the base block class. Therefore, pj∈Yj∩Ai. Since Π is tactical, then there is a unique θ∈Σ(Y,A) such that i=σ(j).
Note thatθ depends only on pj and Ai.
Sincepj and pσ(j) are parallel, they are in the same transversal determined by θ: that is, pσ(j)∈Yσ(j)∩Aθσ(j). So pσ(j)∈Aθσ(j)=Aσθ(j), since θ∈H≤CSm(Σ(Y,A)). But θ(j)=i, therefore pσ(j)∈Aσ(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 mappingσ→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 ofΠ are the same subgroup G of Sm, then the bitranslation group of Π is isomorphic to CSM(G)=H and G∩H=Z(G).Proof.
It is clear thatH=CSm(G). The rest follows from Lemma 3.2.


## 3. Regular Subsets

Letn≥2 be an integer and Ω a set of size n. Let SΩ be the symmetric group on Ω.A subsetT of SΩ is a semiregular subset of SΩ if for any α,β∈Ω, there exists at most one element t∈T such that αt=β.If there exists always exactly one sucht∈T, then T is a regular subset of SΩ.SupposeT is a regular subset of SΩ.Clearly|T|=n. Let C=CSΩ(T)={x∈SΩ∣xt=txforallt∈T}.LetG=〈T〉, the subgroup generated by T in SΩ. Then it is easy to see that(a)
G is transitive on Ω;(b)
C=CSΩ(G).Using this notation, we prove the following results.Lemma 3.1.
IfT is regular, then C is semiregular on Ω and |C| divides n.Proof.
Letx∈C and suppose αx=α for some α∈Ω.
Letβ 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 thatT=G(=〈T〉) if and only if T is a subgroup of SΩ.Lemma 3.2.
(a) IfT≠G, then |C|<n.
(b) IfT=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:Gα|≤|Ω|=n, then |Gα|≥|G|/n>1.
∴ If Δ is the set of fixed points of Gα, then α∈Δ and Δ≠Ω.
Hence|Δ|<n.
Letc∈C and β=αc. For any g∈Gα, we have βg=αcg=αgc=αc=β.
Henceβ∈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 gt∈SΩ by
(3.1)(αu)gt=αt-1u∀u∈T.
Note that{αu∣u∈T}=Ω, since T is regular on Ω.
LetH={gt∣t∈T}. If t1,t2∈T and gt1=gt2, then αt1-1=αt2-1 and hence t1=t2 (since T is regular). Therefore, |H|=|T|=n.
Ift1,t2,u∈T, then (αu)gt1gt2=(αt1-1u)gt2=αt2-1t1-1u=αgt1t2.
Hencegt1gt2=gt1t2, which means that the mapping t→gt defines an isomorphism T→H (H is just the left regular representation of T). Therefore, |H|=|T|=n.
Ift1,t2∈T, then (αu)gt1t2=αt1-1ut2=(αut2)gt1=(αu)t2gt1. So gt1t2=t2gt1 for all t2∈T and gt1∈H. Hence H⊆C.
Since|H|=n and, by Lemma 3.1, |C| divides n, then |C|=n and C=H.
Finally,C∩G=CSΩ(G)∩G=Z(G).Lemma 3.3.
Ifg,h∈SΩ, then gTh is also a regular set.Proof.
Letα,β∈Ω. There is a unique t∈T such that (αg)t=βh-1; that is, α(gth)=β.
Sincegth∈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.
IfT,T′ are regular sets in SΩ 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.
LetT={t1,t2,…,tn-1,t} and T′={t1,t2,…,tn-1,t′}. Then T∩T′={t1,t2,…,tn-1} and t≠t′.
Letα∈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 least2 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 Sm.Lemma 4.1.
LetΠ be a tactical symmetric (μ,m)-net and Σ any of its syntax sets. If Σ is not a subgroup of Sm, then (a)
|CSm(Σ)| divides m and is less than m;(b)
the bitranslation group ofΠ has order at most m-1.Proof.
(a) Follows from Lemmas3.1 and 3.2.
(b) It is clear from Theorem2.1 that the bitranslation group G of Π has order |H|.
SinceH is a subgroup of CSm(Σ), then |H| divides m and |H|<m by part (a).Now, consider the special casen=4. Let Ω={1,2,3,4}.Theorem 4.2.
Ifn=4 and T is a regular set in S4, then T is congruent to a regular subgroup of S4.
The regular subgroups ofS4 are the three cyclic groups generated by 4-cycles and the Klein 4-group {1,(12)(34),(13)(24),(14)(23)}.Proof.
From Lemma3.3 we may assume that 1∈T. The order of any of the 3 nonidentity elements of T is therefore either 2 or 4.
SupposeT 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.
If3t=1, then t=(13)(24), u=(1432), and so T=〈ω〉.
Suppose3t=2. Then 2t≠2 or 3. Therefore, 2t=1 or 4.
If2t=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 case3t=2 is impossible and so 3t=1 as above.
If no element ofT 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 Theorem2.1.(1)Suppose all syntax sets of Π are the same subgroup G of SΩ. Then by Lemma 3.2, H=G and the bitranslation group of Π is isomorphic to CSΩ(G)≅G. Furthermore, CSΩ(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 ofΠ are the same subgroup G. Then by Corollary 2.3, we have H=G and the bitranslation group of Π is isomorphic to CS4(G).From(1), G≅CS4(G) and CS4∩G=G, since G is abelian. Hence CS4(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 matrixM of this symmetric (2,4)-net is as follows:
(4.1)M=(1111001000001000000110000000000000000100100100000010000100010101000010010100010000000100100010100000000000100011110000100110000011000000010000001010010001000001000110100010000001000000000101100010010000010111000100001000000000000001100010000000101100101000100100010001010011000001000000000010100000000001001011000011000001000110101000000000000011001000000000000100101000010010000001111000000000111001000001000000110000100010100001001010001000000010010001010100001001001000000100000001100000000000000100011110000110100000000000100000000101011010000100000111000000101010100000000100100110000001100100000000010000000110000011000100010000100001100000001000000101001000100000110011010001000000100000000010110001001000001011100010000100000000000000110001000000010110010100001000001100000010001000001010010000101000110110000100000001000000010100000000010100000010000110010000010000100000100111010000001010000000111001000001000000110000000101010000100100100000010000100110000000000000010001111000010000001010000100101000100000001001).



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*Source: 502989-2010-07-06.xml*