International Journal of Combinatorics

VolumeΒ 2010, Article IDΒ 502989, 6 pages

http://dx.doi.org/10.1155/2010/502989

## Bitranslations and Symmetric Nets

Department of Mathematics, King Abdulaziz University, P.O. Box 80219, Jeddah 21589, Saudi Arabia

Received 20 November 2009; Accepted 13 April 2010

Academic Editor: CharlesΒ Semple

Copyright Β© 2010 Ahmad N. Al-Kenani. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

It is known that every class-regular symmetric -net is tactical. Also it is known that all -nets with or are tactical. In the work of Al-Kenani and Mavron (2003), it is proved that every symmetric net with is tactical if and only if it is class regular. In this paper, we construct -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 design is an incidence structure with points, points on a block, and any subset of points is contained in exactly blocks, where . The number of blocks is and the number of blocks on a point is .

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

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

A *bitranslation* of 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 . The bitranslation group has order if and only if it acts regularly on each point and block class. In which case we say that is *class regular*.

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

Let be a symmetric -net and let be block classes. If is a bijection, then the point subset , where ranges over all elements of is called an *-transversal *of . If is a union of point classes of , then is said to be a *regular transversal *of , and is called an *-syntax* of . We will denote the set of all -syntaxes by .

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

Label the blocks in each block class of by and similarly for point classes of . Then a bijection between point or block classes of may be regarded as an element of the symmetric group . If are any two distinct block classes of , then we may regard as a semiregular subset of ; that is, if and for some , then (see [3]).

Let be a given block class of which we will call the *base block class*, and let its blocks be labelled arbitrarily. We call the th block of . Label any point class of with integers such that its th point is on for .

Now choose a fixed point class , called the *base point class*, and label any block class of with such that the th block is on . We can therefore refer to the th block of a block class, and dually.

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

*Result 1.1 (see [3]). *If is a tactical -net with standard labelling for its block and point classes, then the identity bijection , for any two block classes and of .

#### 2. Bitranslation Groups

Let be a symmetric -net.

Let be the base block class of and the base point class of in the standard labelling as above. If is a bitranslation of , then induces the permutation defined by: .

Then by definition of standard labelling it follows that for any point or block class of

*Notation 2. *If is a subset of a group , then denotes the centralizer subgroup of in .

Theorem 2.1. *Let be a tactical symmetric -net with standard labelling.**Let and define a mapping by and for all point classes and block classes ().**Then is a bitranslation of if and only if , where is the subgroup
**
Here runs over all pairs of distinct block classes of .*

*Proof. *Let be distinct block classes of . Then is a union of point classes for all , by definition of syntaxes.

Assume first that is a bitranslation.

Since is a bitranslation, it fixes every point and block classes. Hence fixes since is a union of point classes.

Therefore, .

It follows that for all and all . Hence .

Conversely, suppose . Define as in the statement of the theorem. Clearly is bijective and fixes every point and block classes. Let be any point and any block of . Suppose . To show is a bitranslation, it is enough to show that (i.e., that is an automorphism).

By definition of standard labelling, , where is the base block class. Therefore, . Since is tactical, then there is a unique such that .

Note that depends only on and .

Since and are parallel, they are in the same transversal determined by : that is, . So , since . But , therefore , as required.

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

Corollary 2.2. *
(a)ββ is isomorphic to the bitranslation group of .**
(b)ββ is class regular if and only if .*

*Proof. *(a) The mapping of the theorem is easily verified to be an isomorphism from onto the bitranslation group.

(b) This follows easily from the definition of class regular.

Corollary 2.3. *If all syntax sets of are the same subgroup of , then the bitranslation group of is isomorphic to and .*

*Proof. *It is clear that . The rest follows from Lemma 3.2.

#### 3. Regular Subsets

Let be an integer and a set of size . Let be the symmetric group on .

A subset of is a * semiregular subset* of if for any , there exists at most one element such that .

If there exists always exactly one such , then is a * regular subset* of .

Suppose is a regular subset of .

Clearly . Let .

Let , the subgroup generated by in . Then it is easy to see that(a) is transitive on ;(b).

Using this notation, we prove the following results.

Lemma 3.1. *If is regular, then is semiregular on and divides .*

*Proof. *Let and suppose for some .

Let be any element of and let be such that Then . Therefore, . The rest of the proof is straightforward.

It is clear that if and only if is a subgroup of .

Lemma 3.2. *
(a) If , then .**
(b) If , then . Moreover, and .*

*Proof. *(a) . Let . Since , then .

Since , then .

If is the set of fixed points of , then and .

Hence .

Let and . For any , we have .

Hence . So contains the -orbit of .

Therefore, . We also have , since is semiregular on . Hence .

(b) . Choose any . For each , define by

Note that , since is regular on .

Let . If and , then and hence (since is regular). Therefore, .

If , then

Hence which means that the mapping defines an isomorphism ( is just the left regular representation of ). Therefore, .

If , then . So for all and . Hence .

Since and, by Lemma 3.1, divides , then and .

Finally,

Lemma 3.3. *If , then is also a regular set.*

*Proof. *Let . There is a unique such that ; that is, .

Since , the result follows immediately.

*Definition 3.4. * and are *congruent* regular sets. If , they are said to be *similar*.

Lemma 3.5. *If are regular sets in and , then .*

*Proof. *The result is obvious if , since . So suppose .

Let and . Then and .

Let . Then and , since is a regular set. Hence and so , since is regular. Therefore, .

The above lemma shows that distinct regular sets must differ in at least 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 -net are semiregular subsets of .

Lemma 4.1. *Let be a tactical symmetric -net and any of its syntax sets. If is not a subgroup of , then *(a)* divides and is less than ; *(b)*the bitranslation group of has order at most .*

*Proof. *(a) Follows from Lemmas 3.1 and 3.2.

(b) It is clear from Theorem 2.1 that the bitranslation group of has order .

Since is a subgroup of , then divides and by part (a).

Now, consider the special case . Let .

Theorem 4.2. *If and is a regular set in , then is congruent to a regular subgroup of .**The regular subgroups of are the three cyclic groups generated by -cycles and the Klein -group .*

*Proof. *From Lemma 3.3 we may assume that . The order of any of the nonidentity elements of is therefore either or .

Suppose has an element of order . Without loss of generality, we may assume that the -cycle . If are the remaining nonidentity elements of , then, say, .

Therefore, or ; so or .

If , then , , and so .

Suppose . Then or . Therefore, or .

If , then , using the fact that is a permutation and is a regular set. This is impossible.

Similarly, would imply or , which again is impossible.

Therefore, the case is impossible and so as above.

If no element of has order , then all nonidentity elements of have order . Then from the regularity of it follows that must be the Klein -group.

We continue with the notation and hypothesis of Theorem 2.1.

Suppose all syntax sets of are the same subgroup of . Then by Lemma 3.2, and the bitranslation group of is isomorphic to Furthermore, .

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

Suppose all syntax sets of are the same subgroup . Then by Corollary 2.3, we have and the bitranslation group of is isomorphic to .

From , and , since is abelian. Hence . It follows that the bitranslation group of has order and hence is class regular.

Below is an example of a tactical symmetric -net in which every syntax set , is the Klein -group: (The full automorphism group has order . The author is grateful to V. D. Tonchev for this information.)

The incidence matrix of this symmetric -net is as follows:

#### 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.

#### References

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