Title: Partial projective planes
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Title: Partial projective planes
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Creator: Dow, Stephen John, 1955-
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Thesis: Thesis (Ph. D.)--University of Florida, 1982.
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PARTIAL PROJECTIVE PLANES


BY


STEPHEN JOHN DOW





















A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL
OF THE UNIVERSITY OF FLORIDA IN
PARTIAL FULFILLMENT OF THE REQUIREMENTS
FOR THE DEGREE OF DOCTOR OF PHILOSOPHY



UNIVERSITY OF FLORIDA


1982
















ACKNOWLEDGEMENTS


The author would like to thank the chairman of his su-

pervisory committee, David A. Drake, not only for directing

the research which led to this dissertation, but also for

his constant assistance and valuable advice throughout the

author's career at the University of Florida. The author

also thanks the other members of his committee for their

participation and for the memorable courses all of them

taught.

















TABLE OF CONTENTS



CHAPTER PAGE


ACKNOWLEDGEMENTS............................... ii


ABSTRACT.............................................. iv


I INTRODUCTION....................................... 1


II PRELIMINARIES... ................................. 5


III RELATED STRUCTURES...............................8


IV EXAMPLES OF MAXIMAL PPP'S.......................18


V PPP'S OF ORDER LESS THAN SIX....................25


VI EXTENDIBILITY OF PPP'S HAVING NO POINT
OF SMALL VALENCE ...............................46


VII EXTENDIBILITY OF PPP'S HAVING A LARGE
NUMBER OF LINES .................................55


VIII OTHER EXTENDIBILITY RESULTS.....................69


BIBLIOGRAPHY ... ................ .................74


BIOGRAPHICAL SKETCH.............................76





iii
















Abstract of Dissertation Presented to the Graduate Council
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy



PARTIAL PROJECTIVE PLANES

By

Stephen John Dow

August, 1982



Chairman: David A. Drake
Major Department: Mathematics

A projective plane of order n may be defined as a col-

lection of n2 + n + 1 subsets (called lines) of a set X of

cardinality n2 + n + 1 such that each line has cardinality

n + 1 and any two lines intersect in a single element of X.

Although extensive research has been done on projective

planes, fundamental questions remain unanswered. We hope

to further our understanding of projective planes by study-

ing partial projective planes. A partial projective plane

of order n is a collection of lines satisfying the same

properties as those of a projective plane of order n, ex-

cept that the number of lines may be fewer than n2 + n + 1.

The basic question of interest is when can the collection

be extended to include an additional line. We give exam-

ples of partial projective planes which cannot be extended,









detailed information on when this situation can occur for

small values of n, and extendibility conditions which hold

for general n.
















CHAPTER I
INTRODUCTION



In this work we undertake a study of certain finite in-

cidence structures which will be called partial projective

planes. In this chapter we provide some background, define

partial projective plane, and state the questions dealt

with in the remaining chapters.



Definition 1.1 An incidence structure is a triple (X,G,I),

where X and G are sets and I is a subset of X x G. Ele-

ments of X are called points, elements of G are lines, and

I is the incidence relation. The dual of the incidence

structure (X,G,I) is the incidence structure (G,X,I'),

where I' = {(x,y): (y,x)eI}. An incidence structure

(X,G,I) is said to be finite if both X and G are finite

sets.



We will use convenient geometric language to convey that

(P,g)EI; e.g., point P lies on line g. Often we will treat

a line as the set of points with which it is incident. In

particular, if X is any set and G is a collection of sub-

sets of X, then by convention the pair (X,G) is considered

an incidence structure with implicit incidence relation









I = {(P,g): Peg}. In this setting G is allowed to be a

multiset.



Definition 1.2 A projective plane is an incidence struc-

ture which satisfies the following three axioms:

(Al) Any two points lie on a unique line.

(A2) Any two lines meet in a unique point.

(A3) There exist four points, no three of which are col-

linear.



The third axiom serves to eliminate degenerate struc-

tures. An extensive theory has been developed for projec-

tive planes. Among the basic results is the following.



Theorem 1.3 If is a finite projective plane, then there

is an integer n > 2 such that

(i) The number of lines equals the number of points, each

being n + n + 1.

(ii) Every line contains exactly n + 1 points and every

point lies on exactly n + 1 lines.



The integer n is called the order of . A projective

plane of order n is known to exist whenever n is a prime

power, and is known not to exist whenever n H 1,2(mod 4)

and n cannot be expressed as the sum of two integral

squares. For any other n the question of existence is open.









Suppose = (X,G) is a projective plane of order n and

G1 is a subset of G. Removal of the lines of not in G1

produces an incidence structure 1 = (X,G1). From Defini-

tion 1.2 and Theorem 1.3 it is clear that 51 satisfies the

following properties:

(Bl) The number of points is n + n + 1.

(B2) Every line contains exactly n + 1 points.

(B3) Any two lines meet in a unique point.

Of course we also know that the number of lines of 5l is at

most n2 + n + 1. In the next chapter we show that this

fact can be deduced directly from properties (Bl),(B2),

(B3), and furthermore that any incidence structure satisfy-

ing (Bl),(B2),(B3) and having n2 + n + 1 lines is a projec-

tive plane. Thus we may view the problem of constructing a

projective plane of order n in the following way. Start

with a set X of cardinality n2 + n + 1. Build a collection

of (n + 1)-subsets of X, one subset at a time, so that each

subset added to the collection meets every previous subset

in exactly one point. If the size of the collection reaches

n2 + n + 1, then one has a projective plane of order n. We

are interested here in the structures obtained at the inter-

mediate stages. With this motivation we now make the fol-

lowing definitions.









Definition 1.4 A partial projective plane (ppp) of order

n, n > 2, is any incidence structure which satisfies prop-

erties (Bl),(B2), and (B3).



Definition 1.5 A ppp (X,G) is said to be extendible if

there exists a subset g of X such that ggG and (X,Gf{g})

is a ppp. Otherwise (X,G) is said to be maximal.



Throughout this work Q will denote a ppp of order n and

b will denote the number of lines of 2. We will seek con-

ditions which guarantee extendibility of 0 and present ex-

amples of maximal ppp's.

In the next chapter we present some definitions and eas-

ily proven results which will be frequently used in later

chapters. Chapter III is devoted to presenting results

others have obtained about certain combinatorial structures

which are closely related to partial projective planes. In

chapter IV we present infinite classes of maximal ppp's.

In chapter V we determine those integers b for which there

exist maximal ppp's of order 2,3, or 4 having b lines, and

we partially solve the same problem for n = 5. In chapters

VI, VII, and VIII we prove extendibility theorems for ppp's

of arbitrary order. In particular, we prove in chapter VII

that a ppp having nearly n2 + n + 1 lines is extendible.















CHAPTER II
PRELIMINARIES



Let Q be a ppp of order n with b lines. We now prove a

result mentioned in chapter I.



Proposition 2.1 The inequality b < n2 + n + 1 is satisfied,

with equality if and only if n is a projective plane.



Proof. It follows from property (B3) of the definition of

ppp that a given pair of points is contained in at most one

line. Each line contains (n + l)n ordered pairs of points.

Hence there are exactly b(n2 + n) distinct ordered pairs of

points which are joined by lines of 2. Altogether there

are (n2 + n + 1)(n2 + n) ordered pairs of points. There-

fore we have b < n2 + n + 1, with equality if and only if Q

satisfies axiom (Al) of Definition 1.2. By definition of

ppp, n satisfies axiom (A2). It is easily seen that any

ppp satisfies axiom (A3). The proof is complete.



Definition 2.2 The valence of a point P, denoted val(P),

is the number of lines through P. The type of a line g is

the nonincreasing sequence (xl,X2,. .,xn+) whose terms

are the valences of the points on g.









Proposition 2.3 For every point P, we have val(P) < n + 1,

with equality if and only if P is joined to every other

point.



Proof. The number of points joined to P is val(P)n. There

are n + n points other than P altogether.



Proposition 2.4 If a line g has type (x1,x2,. .,Xn+l)'

then xl + x2 + . + xn+1 = b + n.



Proof. Count incident point-line pairs (P,h) such that Peg.

There are n + 1 such pairs with h = g and b 1 such pairs

with h 7 g.



Proposition 2.5 The following are equivalent:

(i) Q is extendible;

(ii) there exist n + 1 points Pl,P2,. rP'n+l such that

every line contains exactly one Pi;

(iii) there exist n + 1 mutually unjoined points the sum of

whose valences is b.



Since the proof is routine it is left to the reader.



Definition 2.6 The deficiency of a point P, denoted dp, is

defined by the equation dp = n + 1 val(P).









Proposition 2.7 Let V be the set of points unjoined to a

given point P. Then IVI = dpn, and each line not through P

meets V in exactly dp points.



The easy proof is left to the reader.



Corollary 2.8 If any point P has valence n, then 0 is ex-

tendible.



Proof. It follows from Proposition 2.7 that IV U{P}I = n + 1

and that every line meets VJU{P} in exactly one point. By

Proposition 2.5, 0 is extendible.















CHAPTER III
RELATED STRUCTURES



Partial projective planes are closely related to at

least three other combinatorial structures on which signif-

icant research has been done. These structures are

(i) sets of mutually orthogonal latin squares (mols),

(ii) regular pairwise balanced designs, or (r,X)-designs,

and (iii) equidistant codes. Each of these structures will

be examined in turn in this chapter.

An extensive theory has been developed for sets of mols,

which are equivalent to structures called nets by Bruck [1].

The reader is referred to the book [4] of D4nes and Keed-

well and to the paper [1] of Bruck for definitions. In or-

der to make precise the connection between these structures

and ppp's, we make the following definition.



Definition 3.1 A partial latin square L of order n and

weight w is an nxn array such that (i) each cell either is

empty or contains one of w symbols, and (ii) each of the w

symbols occurs exactly once in every row and every column.

Two partial latin squares L1,L2 of order n are said to be

orthogonal if, when L1 is superimposed on L2, each of the









w1 symbols of L1 occurs precisely once with each of the w2

symbols of L2.



The reader is warned that the term partial latin square

has been used differently by other authors. Note that a

partial latin square of order n and weight n is a latin

square and that the definition of orthogonality given above

agrees with the usual definition when the partial latin

squares are latin squares. Also note that a set of s mols

of order n always defines a set of n 1 mutually orthogo-

nal partial latin squares (mopls): one adds n 1 s empty

squares (squares of weight zero) to the set.

It is well known that a set of n 1 mols of order n de-

fines a projective plane of order n, and vice versa (see

[4: Theorem 5.2.2]). An analogous correspondence exists

between sets of n 1 mopls of order n and ppp's of order n

containing at least two points of valence n + 1. If the set

of mopls consists of s mols and n 1 s empty squares,

then the corresponding ppp is an (s + 2)-net together with

a "line at infinity" (see [4: Theorem 8.2.1]). It is easily

seen that the construction goes through in the general case

and yields the following theorem.



Theorem 3.2 Let t = (n + l,n + 1,x,x2, . ,xn-1) be a

nonincreasing sequence of positive integers for some n > 2.

There exists a ppp Q containing a line of type t if and









only if there exists a set of n 1 mopls L1,L2,. ..,Ln

of order n such that Li has weight x 1, i = 1,2,. .,

n 1.



Remark 3.3 Let L1,L2,. .,L be a set of s mols, and let

n be the associated ppp given by Theorem 3.2. It is easily

seen that Q is extendible if and only if L1,L2,. .,Ls

have a common transversal.



Let L ,L2,. .,Ls be a set of s mols of order n; let

d = n 1 s; let t be the number of common transversals

of L,L2,. .,Ls. In [1], Bruck proved the following two

theorems.



Theorem 3.4 (Bruck) If n > (d 1)2, then

(i) t < dn, and

(ii) t = dn if and only if {L1,L2,. .,Ls can be extended

to a set of n 1 mols of order n.



Theorem 3.5 (Bruck) If n > p(d 1) where

p(x) = x4/2 + x3 + x2 + 3x/2, then t = dn.



Corollary 3.6 Let Q be a ppp of order n in which some line

contains k points of valence greater than or equal to n.

If n > p(n k), then 0 can be embedded in a projective

plane of order n.









Proof. By Corollary 2.8, Q can be embedded in a ppp Q' of

order n in which some line contains k points of valence

n + 1. It follows from the assumption n > p(n k) that

k > 2, so by Theorem 3.2 there is an associated set of

n 1 mopls LI,. ,Ln-1 of order n. Here L1,. .,Lk-2

have weight n; i.e., L1,. .,Lk-2 are mols. Setting

d = n 1 (k 2) we see that the hypotheses of Theorems

3.4 and 3.5 are satisfied. Hence LI,. ,Lk-2 have pre-

cisely dn common transversals and can be extended to a set

of n 1 mols of order n. Observe that the cells contain-

ing a given symbol in one of the remaining partial latin

squares Lk-l,. .,Ln1 form a common transversal of

L1,. ,Lk-2. Also note that to extend L1,. ,Lk-2 to

a set of n 1 mols, every one of the dn common transver-

sals of L1,. ,Lk-2 must be used as the set of cells con-

taining a single symbol within one of the d new squares.

It follows that these d squares may be obtained by adding

symbols to the squares Lkl,. .,Ln_1. Therefore the

associated ppp Q' (and hence the original ppp 2) can be

embedded in a projective plane of order n.



We now turn to a discussion of the second related struc-

ture mentioned at the outset of this chapter.



Definition 3.7 An (r,X)-design D is a pair (V,F) where V

is a set of v elements (called varieties) and F is a


I








collection of nonempty subsets of V (called blocks) such

that (i) every variety occurs in precisely r blocks, and

(ii) every pair of distinct varieties occurs together in

precisely A blocks. Here F is allowed to be a multiset.

We say that D is trivial if F contains A complete blocks

(blocks of cardinality v); otherwise D is non-trivial. We

say that D is extendible if there exist blocks B1,. .,Bk'

k < r, such that every variety occurs precisely X times in

B1,. .,Bk. Set n = r A. We say that D is symmetrical-

ly bounded if the number of blocks is at most n2 + n + 1.



An (r,X)-design is an incidence structure for which the

words "variety," "block" are used in place of "point,"

"line"; here we are following the usage found in most

papers on (r,X)-designs. Note that if an (r,X)-design D on

v varieties is extendible, then D can be extended to an

(r,X)-design on v + 1 varieties. The proof of the follow-

ing result shows that ppp's are essentially a special case

of dual (r,A)-designs.



Proposition 3.8 There exists a symmetrically bounded

(n + 1,1)-design D on v varieties if and only if there

exists a ppp Q of order n with v lines. Furthermore, ? is

extendible if and only if D can be extended to a symmetri-

cally bounded (n + 1,1)-design on v + 1 varieties.









Proof. Given D, let be the dual of D as in Definition

1.1. Since D has at most n2 + n + 1 blocks, has at most

n + n + 1 points. If has n + n + 1 k points, we let

0 be the incidence structure obtained from by adjoining k

new points, each incident with no lines. Thus Q has

n2 + n + 1 points. In view of the fact that lines of 0

correspond to varieties of D, we see that properties (B2)

and (B3) of the definition of ppp follow directly from (i)

and (ii) of Definition 3.7. Hence 0 is a ppp. Similarly,

given 0 we obtain D by removing all points of valence zero

and taking the dual. The last assertion is verified by

comparing Definition 3.7 with Proposition 2.5.



As mentioned in chapter I, we prove in chapter VII that

a ppp of order n with nearly n2 + n + 1 lines is extendible.

Weaker results of this form have been obtained in the equiv-

alent setting of (n + 1,1)-designs. In particular, Vanstone

has proven the following result in [12].



Theorem 3.9 (Vanstone) If D is a non-trivial (n + 1,1)-

design on v varieties where n2 v < n + n, then D is ex-

tendible.



Theorem 3.9 is improved in [14] to include the case
2
v = n 1. Theorem 3.9 with this improvement was reproved

independently by Hall [5] in the setting of partial









projective planes. The following two theorems, which

appear in [8], further improve Theorem 3.9.



Theorem 3.10 (McCarthy-Vanstone) The only non-extendible

(n + 1,1)-design on v = n2 2 varieties (n > 2) is the

projective plane of order 2 with a complete set of single-

tons.



Theorem 3.11 (McCarthy-Vanstone) An (n + 1,1)-design on

v = n a varieties is extendible if n > 2a2 + 3a + 2.



Note that Theorem 3.11 is stronger than Theorem 3.10 for

all n > 30. Also note that the one non-extendible design

of Theorem 3.10 is not symmetrically bounded: it is a

(4,1)-design with 14 blocks. The following result is

proved in lemmas ([12: Lemmas 2.2,2.4,3.2],[8: Lemmas 2.6,

2.9]) leading to the proofs of Theorems 3.9 and 3.11.



Theorem 3.12 (McCarthy-Vanstone) A non-trivial (n + 1,1)-

design on v varieties is symmetrically bounded if v > n2 or

if v = n a and n > (a2 + a)/2.



In view of Proposition 3.8 we have the following theorem

as an immediate consequence of the above results.









Theorem 3.13 Let 9 be a ppp of order n with b lines. If

(i) n2 2 < b < n2 + n, or if (ii) b = n2 a and

n > 2a + 3a + 2, then 0 is extendible.



Chapter VII is devoted to improving Theorem 3.13 so that

the inequality in (ii) reads n > a /4 + 3a + 6.

We now turn to our third related structure, the equidis-

tant code.



Definition 3.14 The Hamming distance between two s-tuples

is the number of coordinates in which they differ. An

equidistant d-code is a collection C of binary s-tuples

(called codewords) any two of which have Hamming distance d.

We say that C is non-trivial if there is some coordinate

for which the number y of codewords with a 1 in that coor-

dinate satisfies 2 < y < ICI 2.



In [5], Hall has obtained a connection between equidis-

tant codes and partial projective planes. Hall uses the

term partial projective plane more generally than we use it

here; as mentioned in [5], Hall's partial projective planes

are precisely the duals of (n + 1,1)-designs. For this

reason we state his result ([5: Theorem 1]) in terms of

(n + 1,1)-designs.









Theorem 3.15 (Hall) Let x > (n + 2)2/2 with n > 2. There

exists a non-trivial equidistant 2n-code with x codewords

if and only if there exists a non-trivial (n + 1,1)-design

on x 1 varieties.



One direction of Theorem 3.15 is easy to prove; it is a

special case of Proposition 3.17 below. First we require a

definition.



Definition 3.16 An (r,X)-design on v varieties is said to

be near-trivial if, for every block B, IBIE{l, v 1, v}.



Proposition 3.17 Let D be an (r,X)-design on v varieties

which is not near-trivial. Set n = r X. If r > 2n, then

there exists a non-trivial equidistant 2n-code C with v

codewords. If r < 2n, then there exists a non-trivial

equidistant 2n-code C with v + 1 codewords.



Proof. Let M be an incidence matrix for D in which rows

correspond to varieties. From the definition of (r,A)-

design we see that every row of M has weight r and any two

rows of M have inner product A. It follows that any two

rows of M have distance 2n. If r > 2n then we take all

rows of M as the codewords of C. If r < 2n then we adjoin

2n r columns of l's to M. The rows of the resulting ma-

trix M' have weight 2n and distance 2n. We then take as

codewords of C all rows of M' together with the zero vector.









In view of Hall's Theorem 3.15, we now have the follow-

ing result, which shows the importance of the case X = 1.



Theorem 3.18 Let D be an (r,A)-design on v varieties which

is not near-trivial. Set n = r X. If r < 2n and

v > n2/2 + 2n + 1, then there exists a non-trivial

(n + 1,1)-design on v varieties. If r > 2n and

v > n /2 + 2n + 2, then there exists a non-trivial

(n + 1,1)-design on v 1 varieties.



Vanstone and McCarthy have obtained similar results in

[15], but their version of Theorem 3.15 (which is implicit

in the proof of [15: Theorem 3.1]) requires larger x. In

[8] they use an unstated version of Theorem 3.18 together

with Theorem 3.11 to show that the existence of an (r,X)-

design on a large number of varieties implies the existence

of a projective plane of order r A. In chapter VII we

will be able to improve their result by reducing the number

of varieties required.
















CHAPTER IV
EXAMPLES OF MAXIMAL PPP'S



It follows from Theorem 3.13 that any ppp of order 2 can

be embedded in the projective plane of order 2. This

statement becomes false when 2 is replaced by any integer

n > 3. In this chapter we construct examples of maximal

ppp's of every order n having relatively few lines. Then,

using results of Ostrom found in [10], we construct an in-

finite class of maximal ppp's with b = n2 n3/2 + n + 1.



Theorem 4.1 For any even integer n > 2, there is a maximal

ppp of order n with b = 3n + 1.



Proof. It is well known that a cyclic latin square of even

order n has no transversal (see [4: p.445]). By Remark 3.3

the associated ppp, for which b = 3n + 1, is maximal.



Note that for n > 2, we have 3n + 1 < n2 + n + 1, so the

ppp of Theorem 4.1 is not a projective plane.



Theorem 4.2 For any odd integer n > 3, there is a maximal

ppp with b = n + 2.









Proof. Let V be an (n + 2)-set, F be the collection of all

2-subsets of V. It is easily checked that (V,F) is a sym-

metrically bounded (n + 1,1)-design on n + 2 varieties.

Since n + 2 is odd, no collection of blocks partitions V;

hence (V,F) is not extendible. The desired ppp is now

given by Proposition 3.8.



Our next construction is related to derivation, a tech-

nique for constructing non-Desarguesian projective planes

developed by Ostrom in [9], [10], and [11]. Our work is

self-contained except for the definitions (which can be

found in [1]) of net and related concepts and the proof

(which can be found in [10]) of the following theorem of

Ostrom.



Theorem 4.3 (Ostrom) Let N be a net of order n, deficiency

d; let t be the number of transversals of N. If n = (d-1)2

then t < 2dn, with equality if and only if N can be embed-

ded in distinct affine planes l', 2 by adjoining nets N1,

N2, respectively. Furthermore, each line of N1 is an affine

subplane of N2 of order d 1, and vice versa.



Corollary 4.4 Let N be a net of order n, deficiency

d = n /2 + 1, which can be embedded in distinct affine

planes T1, 2 by adjoining nets N1, N2.

(i) If gi is a line of Ni, i = 1, 2, then Igl g21 = 0 or

d 1.









(ii) If N can be embedded in a net N* by adjoining a single

parallel class C, where C is not a parallel class of

N1 or N2, then N* has no transversal.



Proof. (i) By the last assertion of Theorem 4.3, the

points of g2 together with certain of the lines of N1 form

an affine plane of order d 1. Therefore (d 1) +

(d 1) lines of N1 intersect g2 in d 1 points, and the

remaining lines of N1 intersect g2 in at most one point.

We must show that these remaining lines actually miss g2.

Since N1 is a d-net, each point of g2 is on exactly d lines

of N1. Thus there are exactly Ig2|d = n3/2 + n incident

point-line pairs (P,g) with P on g2 and g a line of N1.

Since [(d 1)2 + (d 1)](d 1) = n3/2 + n, all of these

pairs are accounted for by the lines which meet g2 in d 1

points. This completes the proof of (i).

(ii) It follows from Theorem 4.3 that each of the 2dn

transversals of N is a line of N1 or N2. If N* has a

transversal h, then h is a transversal of N, hence is a

line of N1 or N2, say N1. Now C consists of n disjoint

transversals of N which do not form a parallel class of N1

or N2. It follows that C must contain at least one line

from each of N1 and N2. Let g be a line of N2 which is in

C. As a transversal of N*, h must intersect every line of

C in exactly one point. But by (i) above, Jgfhl = 0 or

d 1. The contradiction completes the proof of (ii).









Corollary 4.4(i) can be found within the proofs in [10].

The author believes that Corollary 4.4(ii) and its applica-

tion to Theorem 4.5 below may be new. Related results have

been obtained by Bruen in [2].



Theorem 4.5 For any prime power q there exists a

(q2 q + l)-net of order q2 with no transversal.



Proof. Let K = GF(q2). We construct nets of order q2

having point set K x K. We begin with a well-known con-

struction of the unique Desarguesian affine plane I1 of
2
order q Let

gm,b = {(x,y): x,yEK, y = mx + b}, m,beK,

g,a = {(a,y): ycK}, aEK,
Ci = {g ij: jeK}, iEK L/{}.

The Ci's are the parallel classes of i.

Let F be the subfield of K of cardinality q. Let N be

the (q2 q)-net of order q2 having parallel classes C.,

ieK-F, N1 be the (q + l)-net of order q2 having parallel

classes Ci, iEFf {=}. With n = q N has deficiency

d = n /2 + 1. By Theorem 4.3, N has at most 2dn transver-

sals. We will show that this bound is attained by exhib-

iting dn transversals of N other than the dn lines of N1.

This is not a new result; it is equivalent to the fact that

1 is derivable (see [6: chapter XI).









Let K* = K-{0}, F* = F-{0}; let al,a2,. .,aq+ be a

system of representatives of K*/F*, with a lF. For b,cEK

and i = 1,2,. .,q + 1 let h, denote the set

{(a.x + b, aiY + c): x,yEF}. First note that hi cl = q

for all choices of i,b,c. For each i, a.F is an additive

subgroup of K having q costs aiF + b. Thus for each i

there are exactly q2 distinct sets h Since hc he,f
b,c b,c e,f
if i X j, there are (q + l)q2 distinct sets h altogether.
b,c
We claim that all are transversals of N. Let i,b,c be

given. To show that hb, is a transversal of N we must

show that no line of N contains two points of hbc. Sup-

pose on the contrary that points (aix + b, aiy + c) and

(aix' + b, aiy' + c) lie on line gm,d where (x,y), (x',y')

are distinct pairs of elements of F and mEK-F. Then by

definition of gm,d we have aiy + c = m(aix + b) + d and

aiy' + c = m(aix' + b) + d. Since a. is nonzero, these two

equations yield y y' = m(x x'). It follows that mEF, a

contradiction.

It is easily seen that no hbc is a line of N1, so we

have proved that N has at least 2dn transversals. It now

follows from Theorem 4.3 that N can be embedded in an affine

plane 2 by adjoining a net N2 whose lines are the sets

hb,. Thus we are in the setting of Corollary 4.4. Let
b,c
1 2
C = {gl,b: beK-F}U{h1 : ceK}. Note that C consists of q

transversals of N: q q from N1 and q from N2. If we
show that these q transversals are disjoint, then it will
show that these q transversals are disjoint, then it will









follow that C can be adjoined to N as a new parallel class.

Clearly the sets gl,b bEK-F, are disjoint. Also, distinct

sets h CEK, are disjoint, since they are precisely the
c, c
sets (F + c) x (F + c). Let beK-F, ceK, and suppose that
1
(x,y)sglb~n h Then y = x + b and x,yeF + c. Now

x,ycF + c implies y xcF, whereas we assumed bhF. There-

fore the q2 transversals of C are disjoint. Hence N can be

embedded in a (q2 q + 1)-net N* by adjoining C as a par-

allel class. By part (ii) of Corollary 4.4, N* has no

transversal.



The net of Theorem 4.5 is equivalent to a set of

q q 1 mols of order q2 with no common transversal. By

Remark 3.3 we have the following corollary.



Corollary 4.6 For any n which is an even power of a prime

there exists a maximal ppp of order n with

b = n n3/2 + n + 1.



Let 1i' 2 be any affine planes of order n related by

derivation. Then 1i', 2 are obtained by adjoining d-nets

N1, N2 to a common net N of order n and deficiency

d = n /2 + 1. It may be the case that in this more general

setting the parallel class C of Corollary 4.4 exists. All

that we can prove is the following result.









Proposition 4.7 Corresponding to any derivable plane of

order n > 9 there is a maximal ppp of order n with

(4.1) n2 n3/2 + n n1/2 + 2 < b < n2 n3/2 + n + 1.



Proof. Let fl, I2, N, N1, N2 satisfy the hypotheses of

Corollary 4.4. Let g be any line of N1, C any parallel

class of N2. The lines of C partition the point set, and

each line of C which meets g contains exactly d 1 points

of g, by Corollary 4.4(i). Therefore exactly

Igl/(d 1) = n /2 lines of C meet g and n nl/2 lines of

C miss g. We take these n n1/2 lines together with g as

a partial parallel class to be adjoined to N. Adjoining a

line at infinity yields a ppp 0 of order n with

b = (n + 1 d)n + n n /2 + 2. This expression simpli-

fies to the lower bound for b in (4.1). Let P be the point

on the line at infinity corresponding to the partial par-

allel class. Since this class contains lines of both N1

and N2,it follows from Corollary 4.4(i) that if 0 is ex-

tendible, then the new line must contain P. Since at most

nl/2 1 lines may be added to 0 through P, there is a max-

imal ppp with b in the desired range.
















CHAPTER V
PPP'S OF ORDER LESS THAN SIX



In this chapter we are interested in determining for

small n those integers b for which there exists a maximal

ppp of order n with b lines. We denote by B0(n) the set of

all such integers b. As noted in chapter IV, any ppp of

order 2 can be embedded in the projective plane of order 2.

Thus B0(2) = {7}. In Theorems 5.2 and 5.10 we determine

B0(3) and B0(4). Then in Theorem 5.17 we show that the

largest ppp of order 5 which cannot be embedded in the pro-

jective plane of order 5 has 19 lines.



Lemma 5.1 Let Q be a ppp of order n in which every line

contains a point of valence n + 1. If some point P has

valence 1, then every point on the line through P has va-

lence 1 or n + 1.



Proof. Suppose on the contrary that there is a point Q

joined to P for which 1 < val(Q) < n + 1. Let g be a line

through Q other than PQ. By hypothesis some point R on g

has valence n + 1. By Proposition 2.3, some line joins R

and P. Now P is on at least two lines, a contradiction.









Theorem 5.2 We have B0(3) = {5, 13}.



Proof. The projective plane of order 3 has 13 lines; the

example given by Theorem 4.2 has 5 lines; hence 5,13EB0(3).

It is easily seen that any ppp with fewer than 5 lines is

extendible; by Theorem 3.13 any ppp of order 3 with

7 < b < 12 is extendible. Hence the only number left in

doubt is 6. Suppose 2 is a maximal ppp of order 3 with

b = 6. By Corollary 2.8, no point of 0 has valence 3. By

Proposition 2.4, every line has type (4,2,2,1). Thus every

line contains a point of valence 4. Lemma 5.1 now asserts

that no line has type (4,2,2,1). We conclude that 6iB0(3).



Proposition 5.3 The set B0(4) contains both 13 and 21 but

none of 14, 15,. ., 20.



Proof. The projective plane of order 4 has 21 lines; the

example given by Theorem 4.1 has 13 lines; hence

13,21BO0(4). By Theorem 3.13 any ppp of order 4 with

14 < b < 20 is extendible.



Proposition 5.4 Any ppp of order 4 with b = 12 is extend-

ible.



Proof. By Corollary 2.8, we may assume that no point of

has valence 4. Then the only line types allowed by









Proposition 2.4 are tl = (5,5,3,2,1)

t2 = (5,5,2,2,2)

t3 = (5,3,3,3,2).

Clearly some point has valence 2 and its two lines each

contain a point of valence 5. Thus at least two points

have valence 5. Since any two points of valence 5 are

joined, some line g has type tl or t2. Let Pieg satisfy

2 < val(P.) < 3, i = 1,2. Let gl,g2 be lines other than g

through Pl,P2, respectively. Let P be the point of inter-

section of gl,g2. By Proposition 2.3, P is joined to the

two points of valence 5 on g. Thus P is joined to at least

four points of g; since val(P) is not 4, P is joined to all

five points of g. Therefore g has type t2. By Theorem 3.2,

n gives rise to a set of three mopls of order 4, each square

having weight 1. Let the point P correspond to the (1,1)-

cell of these squares, and let the symbol used in each

square be 0. Then the (1,1)-cell of each square contains a

0. The reader may now verify that there are only two such

sets of mopls, and that in either set the squares can be

completed to mols. Therefore 0 is extendible.



Notation 5.5 For any ppp we denote by vi the number of

points of valence i.



Proposition 5.6 Any ppp 0 of order 4 with b = 11 is extend-

ible.









Proof. We may assume that v4 = 0. If v5 > 2 then some

line has type (5,5,2,2,1) or (5,5,3,1,1). By examining the

possible associated sets of mopls one sees that if a has a

line of either type, then n is extendible. Hence we assume

that v5 < 1. Now the only line types which may occur are

tl = (5,3,3,3,1)

t2 = (5,3,3,2,2)

t3 = (3,3,3,3,3).

If v5 = 0, then every point has valence 0 or 3; by counting

incident point-line pairs we find that 3v3 = 5b = 55, which

cannot be. Hence there is exactly one point P of valence

5. Clearly no line has type t2: a point of valence 2 would

necessarily be joined to two points of valence 5. Hence

the five lines through P have type tl, and consequently we

have v1 = 5, v3 = 15. Each point of valence 3 lies on one

line of type tl (joining it to P) and two lines of type t3.

Thus each point of valence 3 is joined to exactly one point

of valence 1.

We intend to show that there is a set of five mutually

unjoined points, three having valence 3 and two having va-

lence 1. It will then follow from Proposition 2.5 that 0

is extendible. It suffices to find three mutually unjoined

points of valence 3, since only three of the five points of

valence 1 will be joined to any of these and no two points

of valence 1 are joined. Clearly there exist two unjoined

points Q and R of valence 3. There are exactly nine points









joined to both Q and R; one of these is P and the remaining

eight have valence 3. There are exactly six points which

are joined to one of Q, R but not the other; two of these

have valence 1 and the remaining four have valence 3. In-

cluding Q and R we have accounted for 14 points of valence

3. Since v3 = 15 there is one point of valence 3 joined to

neither Q nor R, and the proof is complete.



Proposition 5.7 For any ppp of order n the following three

equations hold.
n+l 2
(i) E v. = n + n + 1
i=0

n+l
(ii) E iv. = b(n + 1)
i=l 1

n+l
(iii) E i(i l)v = b(b 1)
i=2


Proof. Count in two ways each of the following three ob-

jects: (i) points, (ii) incident point-line pairs, and

(iii) ordered pairs of lines.



Proposition 5.8 Any ppp Q of order 4 with b = 10 is extend-

ible.



Proof. We may assume that v4 = 0. By Proposition 2.4, the

only line types which may occur are









tI = (5,5,2,1,1)

t2 = (5,3,3,2,1)

t3 = (5,3,2,2,2)

t4 = (3,3,3,3,2).

If any line has type tl, then Q is easily seen to be extend-

ible. Hence we assume that no line has type tI. It follows

that v5 5 1. If v5 = 0, then every line has type t4 and we

obtain 3v3 = 4b = 40, which cannot be. Hence there is ex-

actly one point P of valence 5. Since every point is joined

to P we have v0 = 0. The remaining via's are now determined

by Proposition 5.7: v3 = 10, v2 = 5, v1 = 5. It follows

that all five lines through P have type t2. Each point of

valence 3 lies on one line of type t2 (joining it to P) and

two lines of type t4, and hence is joined to 7 points of va-

lence 3 and unjoined to 2 points of valence 3. Let P1,P2 be

unjoined points of valence 3; let glg2 be the lines of type

t2 through Pl,P2, respectively; let Q1,Q2 be the points of

valence 2 on gl,g2, respectively. Since Plis joined to

three points of g2, none of which is P2 or the point of va-

lence 1, we see that P1 is joined to Q2. Similarly, P2 is

joined to Q1. Now P1,P2 are each joined to exactly three

points of valence 2, and we have shown that at least two

points of valence 2 are joined to both. Since v2 = 5 it

follows that some point Q of valence 2 is unjoined to both

P1 and P2. Each of PI,P2,Q is joined to exactly one point

of valence 1. Since v1 = 5 there are two points R1,R2 of









valence 1 unjoined to P1, P2' and Q. Since no two points

of valence 1 are joined, {P1,P2,Q,RI,R2} is a set of five

mutually unjoined points whose valences sum to 10. By

Proposition 2.5, 0 is extendible.



Proposition 5.9 Any ppp 2 of order 4 with b < 8 is extend-

ible.



Proof. If b < 6 or if any point has valence 4 or 5, it is

easily seen that 2 is extendible. Hence we consider only

the cases b = 6, 7, 8 and we assume that the valence of

every point is at most 3. First assume b = 6. By Propo-

sition 2.4, the only line types which may occur are

tI = (3,3,2,1,1)

t2 = (3,2,2,2,1)

t3 = (2,2,2,2,2).

If every line has type t3, then every point has valence 0

or 2. In this case we find that there exist five mutually

unjoined points, three of valence 2 and two of valence 0,

so 0 is extendible by Proposition 2.5. Hence we assume

that some point P has valence 3. If P is the only point of

valence 3, then 0 must be as drawn in figure 5a; in this

case there exist three mutually unjoined points of valence

2 and two points of valence 0, so 0 is extendible. If

there is another point Q of valence 3, then we consider

separately the case that P and Q are joined and the case









that they are unjoined; in either case we find that 0 can

be extended by adding a new line through P.

Now let b = 7. It follows from Proposition 2.4 that

every line contains a point of valence 3. Let P be any

point of valence 3. Let V be the set of points unjoined to

P. By Proposition 2.7, IV| = 8 and each of the four lines

not through P meets V in two points. By considering the

possible configurations for V, we find that 0 can be extend-

ed by adding a new line through P unless six of the seven

lines are as in figure 5a. There are only two essentially

different ways to add the seventh line; in either case 0

is extendible.

Finally, let b = 8. The only possible line types are

tl = (3,3,3,2,1) and t2 = (3,3,2,2,2). If every line has

type t2, then we obtain 3v3 = 2b = 16, which cannot be.

Hence some line g has type tl. Let P be a point of valence

3 which is off g and is joined to the point of valence 2 on

g. Since P is joined to only three points of g, some point

Q on g has valence 3 and is unjoined to P. The six lines

through P and Q are as drawn in figure 5b. In that figure

we have labeled the three points Pl,P2,P3 which are joined

to Q but unjoined to P. Here P1 is the point of valence 1

on g. If neither of the two remaining lines joins P2 and

P3, then we find that Q can be extended by adding a new

line through P. If one of the remaining lines joins P2 and

P3' then without loss of generality the remaining lines are

as drawn in figure 5c. Again we find that 0 is extendible.

















S* *
>-0 0


Figure 5a


2 \


P3


Figure 5b


p


ii?















































Figure 5c




















































Figure 5d









Theorem 5.10 We have B0(4) = {9, 13, 21}.



Proof. Figure 5d is a drawing of a ppp n of order 4 with

b = 9. It is routinely checked that 2 is maximal, perhaps

most easily by checking that no five points of 0 satisfy

Proposition 2.5(iii). Hence 9eB0(4). The inclusion or ex-

clusion of every other number has been decided in one of

Propositions 2.1, 5.3, 5.4, 5.6, 5.8, 5.9.



Proposition 5.11 The set B0(5) contains 7 and 31 but none

of 23, 24,. ., 30.



Proof. The projective plane of order 5 has 31 lines; the

example given by Theorem 4.2 has 7 lines. By Theorem 3.13,

any ppp of order 5 with 23 < b < 30 is extendible.



Proposition 5.12 Any ppp 0 of order 5 with b = 22 contains

a point of valence 5.



Proof. Suppose on the contrary that v5 = 0. Then by Prop-

osition 2.4, every line contains at least two points of va-

lence 6 and any line containing only two has type

(6,6,4,4,4,3). If every line has this type, then we obtain

3v3 = b = 22, which cannot be. Therefore some line g has

type (6,6,6,x4,x5,x6). Note that x5 + x6 < 6, so

(x5 1)(x6 1) < 4. Let P, Q, R be points on g having









valences 6, x5, x6, respectively. Each of the five lines

other than g through P contains at least one point of va-

lence 6 off g. By Proposition 2.3, all of these points are

joined to both Q and R. Since exactly (x5 1)(x6 1)

points off g are joined to both Q and R, we have

(x5 1)(x6 1) > 5, a contradiction.



Corollary 5.13 Any ppp of order 5 with b = 22 is extend-

ible.



Proposition 5.14 Any ppp 0 of order 5 with b = 21 is ex-

tendible.



Proof. We may assume that v5 = 0. Then by Proposition 2.4

every line contains at least one point of valence 6. If

any point has valence 1, then by Lemma 5.1 the line through

that point must have type (6,6,6,6,1,1); in this case 0 is

extendible by Corollary 3.6. Hence we assume that v, = 0.

Now the only possible line types are

tl = (6,4,4,4,4,4)

t2 = (6,6,4,4,3,3)

t3 = (6,6,4,4,4,2)

t4 = (6,6,6,4,2,2)

t5 = (6,6,6,3,3,2).

Suppose some point P has valence 2. Let g,h be the

lines through P. By Proposition 2.3. every point of









valence 6 lies on g or h. Lines g and h have type t3, t4,

or t5. Suppose that one, say g, has type t4 or t5. Then g

contains a point Q 7 P of valence 2 or 3. Let k be a line

through Q other than g. Since every point of valence 6

lies on g or h, k contains at most one point of valence 6.

Hence k has type tl, but QEk has valence 2 or 3, a contra-

diction. We conclude that both g and h have type t3. Let

Q be a point of valence 4 on g. Each of the three lines

other than g through Q contains a point of valence 6.

These three points of valence 6 must lie on h, contradic-

ting the fact that h has type t3. We conclude that no

point has valence 2.

At this point we know that vi = 0 for i $ 3,4,6 and that

every line has type tl or t2. From Proposition 5.7 we ob-

tain v3 = 10, v4 = 15, v6 = 6. Clearly some line g has

type t2. By Theorem 3.2, Q gives rise to a set of four

mopls, two of weight 2 and two of weight 3. The remainder

of the proof consists of showing that the structure of

these squares is forced to be that of the squares in figure

5e, which can be completed to mols.

Since v6 = 6, there are four points of valence 6 off g.

These correspond to four cells which are nonempty in all

four partial latin squares. Since no line contains three

points of valence 6, these cells are in distinct rows and

columns. By rearranging the rows and columns, we may as-

sume that the four cells are the cells (i,i), i = 1,2,3,4,









and that these cells are filled as in figure 5f. There are

three ways to add the remaining l's to L1. For each of

these three ways there are two ways to add the remaining

2's to L2. The six resulting pairs of squares are shown in

figure 5g. Three of these pairs can be obtained from the

remaining three pairs by reflection through the main diago-

nal. Hence we need only consider three of the pairs in

figure 5g. These three pairs can be completed to orthogo-

nal partial latin squares of weight 2 in exactly five ways,

as shown in figure 5h. For four of these five ways we find

that the remaining l's cannot be inserted into L3. For the

fifth way we find that L3,L4 can be completed in only one

way, as shown in figure 5e. Since these squares can be

completed to mols, we conclude that 0 is extendible.



Proposition 5.15 Any ppp 0 of order 5 with b = 20 contains

a point of valence 5.



Proof. Suppose on the contrary that v5 = 0. It follows

from Proposition 2.4 that every line contains a point of

valence 6. If some point has valence 1, then by Lemma 5.1

the line through that point contains only points of valence

1 or n + 1. No such line exists, by Proposition 2.4.

Hence vl = 0, and the only line types which may occur are

tI = (6,4,4,4,4,3)

t2 = (6,6,4,3,3,3)









t3 = (6,6,4,4,3,2)

t4 = (6,6,6,3,2,2).

Suppose some line g has type t4. Let P,Qcg have valence 2

and let Reg have valence 3. Only one point off g is joined

to both P and Q. The two lines other than g through R each

contain a point of valence 6 off g. By Proposition 2.3,

these two points of valence 6 are joined to both P and Q, a

contradiction. Hence no line has type t4. A similar argu-

ment shows that no line has type t3. Letting bi denote the

number of lines of type tiwe now obtain bl + b2 = 20,

bl + 2b2 = 6v, and b1 + 3b2 = 3v3. The second and third

of these equations imply that bl and b2 are divisible by 3,

a conclusion which contradicts the first equation.



Corollary 5.16 Any ppp of order 5 with b = 20 is extend-

ible.



Theorem 5.17 We have B0(5) = {7, 19, 31}US, where

S {8, 9,. ., 18}.



Proof. In figure 5i a set of mopls of order 5 is shown.

The corresponding ppp 0 has a line of type (6,6,6,2,2,2),

so b = 19. One may check that 0 is maximal by hand, or ar-

gue as follows. By Theorem 3.13, Proposition 5.14, and

Corollaries 5.13, 5.16 every ppp of order 5 with

20 < b < 30 is extendible. It follows that if 0 is






41


extendible then the squares of figure 5i can be completed

to mols. This is impossible since Mann [7] (alternatively,

see [4: Theorem 12.3.2]) has shown that a latin square of

order 4k + 1 which has a subsquare of order 2k has no or-

thogonal mate. Therefore Q is maximal. It is easily seen

that any ppp of order 5 with b < 7 is extendible. The

proof is completed by reference to Propositions 5.11, 5.14

and Corollaries 5.13, 5.16.


















2 1 2 3 2 3 1
1 3 2 1 3 1 2
1 3 2 1 2 3
2 1 3 1 2 1 2 3
2 3 1 3 2 1



Figure 5e


Figure 5f


















































Figure 5g









1 2
2 1


2 1 2




1 2
2 1



2 1
2 1
1 2
1 2


1 2
2 1
2 1
1 2
2 1


Figure 5h






















1 2 3 4 5
2 1 4 5 3
3 5 1 2 4
4 3 5 1 2
5 42 3 1


Figure 5i
















CHAPTER VI
EXTENDIBILITY OF PPP'S HAVING NO POINT OF SMALL VALENCE



In this chapter we obtain two theorems asserting the ex-

tendibility of ppp's based on the assumption that the va-

lence of every point is nearly n + 1. In Theorem 6.7 we

assume that every valence is at least n 2; in Theorem

6.10 we greatly weaken this assumption, but we add an as-

sumption on the number of lines.

The proofs of the theorems of this chapter and the next

depend on extending n by adding lines through a point of

valence n 1. Note that if val(P*) = n 1, then by Prop-

osition 2.7 there are exactly 2n points unjoined to P*.



Definition 6.1 Let P* be a point of valence n 1 in a ppp

Q of order n; let V be the set of 2n points unjoined to P*.

The point graph of V is the graph G whose vertices are the

points of V, with two vertices joined by an edge if the two

points are joined in . We say that a subset A of V is in-

dependent if no two points of A are joined. We say that G

is (a,b)-bipartite if V is the disjoint union of two inde-

pendent sets of cardinalities a and b.









Remark 6.2 The graph G of Definition 6.1 has exactly

b n + 1 edges, since each line not through P* gives rise

to one edge (see Proposition 2.7).



Proposition 6.3 Let G be as in Definition 6.1. If G is

(n,n)-bipartite, then 2 is extendible.



Proof. By Definition 6.1, V is the disjoint union of inde-

pendent sets A, B of size n. Every line not through P*

must contain one point of A and one point of B. Hence each

of the sets AU{P*}, B V{P*} satisfies Proposition 2.5(ii),

so Q is extendible.



The proof of the following proposition is routine.



Proposition 6.4 If P and Q are joined, then there are ex-

actly dpdQ points of n joined to neither P nor Q.



Proposition 6.5 Let val(P*) = n 1, and let V be the set

of 2n points unjoined to P*. Let P,QcV be joined. There

are at most dpd 1 points of V joined to neither P nor Q

and at most (dp 1)(dQ 1) points of V joined to both P

and Q.



Proof. The first assertion follows from Proposition 6.4.

Each line through P meets V in one other point. Thus P is









joined to val(P) = n + 1 dp points of V. Similarly Q is

joined to n + 1 dQ points of V. Let x be the number of

points of V joined to both P and Q, y the number of points

of V joined to neither. Counting points of V-{P,Q} and

taking into account that P and Q are joined, we obtain

2n 2 = n dQ + n dp x + y. Since y < dpd 1, we

have x < (dp 1)(dQ 1).



Proposition 6.6 Let val(P*) = n 1, V be the set of 2n

points unjoined to P*, G be the point graph of V. If

points P, Q, and R form a triangle in G, then

n dpdQ + dpdR + dQdR dp d dR'


Proof. Let T be the set of n + 1 d = val(P) points of V

joined to P. Let x be the number of points of T other than

R which are joined to Q, y be the number of points of T

other than Q which are joined to R. Then at least

IT-{Q,R}J (x + y) = n 1 dp x y points of T are

joined to neither Q nor R. By Proposition 6.5,

n 1 dp x y dQdR 1. Also by Proposition 6.5,

x < (dp 1)(d 1) 1 and y < (dp 1)(dR 1) 1.

Combining these three inequalities gives the desired result.



Theorem 6.7 Let 0 be a ppp of order n > 19 such that

val(P) > n 2 for every point P. If 0 is not extendible,

then either (i) 2 is a projective plane, or (ii) 31n and

b = n2 n 2.









Proof. Assume that 0 is not extendible. Then by Corollary

2.8 no point has valence n, so the only valences which may

occur are n 2, n 1, and n + 1. First suppose that

every point has valence n + 1. By Proposition 2.3 any two

points are joined. It follows that 0 is a projective plane.

Hence we assume that some point P* has valence n 1 or

n 2.

First assume that val(P*) = n 1. Let V be the set of

points unjoined to P*, G be the point graph of V. By Prop-

osition 2.3, no point of V has valence n + 1. Thus we have

dp = 2 or 3 for every P in V. Since n > 18 it follows from

Proposition 6.6 that G contains no triangle.

We now show that, if V contains two disjoint independent

sets A, B of cardinality at least n 2, then G is bipar-

tite. To do this it suffices to show that, if P E V-(AUB),

then one of the sets AL/UP}, BU{P} is independent. If

this is not the case then P is joined to points P1eA, P2EB.

By Proposition 6.5, there are at most 8 points of V un-

joined to both P and P1. Therefore P is joined to at least

JAl 8 > n 10 points of A. Similarly P is joined to at

least n 10 points of B. Since P is joined to at most

n 1 points of V altogether, we obtain 2(n 10) < n 1,

which contradicts our hypothesis that n > 19.

Let P, Q be any joined pair in V. Let S be the set of

points of V joined to P, T be the set of points of V joined

to Q. Since G contains no triangle, S and T are disjoint,









independent sets. Also ISI, JTI > n 2. Hence by the

preceding paragraph G is bipartite. Let A, B be disjoint

independent sets whose union is V, and let JAl IBI. Sup-

pose that IAI > IBI. Then IAl > n + 1, IBI < n 1. Let x

denote the number of edges of G. Then

n2 n 2 < A(n 2) < x < IBI(n 1) < n 2n + 1.

This implies that n < 3, which contradicts the hypothesis.

We conclude that G is (n,n)-bipartite. By Proposition 6.3,

0 is extendible. Since we have assumed that 0 is not ex-

tendible, we conclude that no point has valence n 1.

We now have that val(P*) = n 2 and that every point

has valence n 2 or n + 1. Let V be the set of points un-

joined to P*. By Proposition 2.7, IVI = 3n and each line

not through P* meets V in exactly 3 points. Counting inci-

dent point-line pairs (P,g) with P in V yields

3(b n + 2) = 3n(n 2), or b = n n 2. Let g be any

line; let x denote the number of points of valence n + 1 on

g. By Proposition 2.4 we obtain

x(n + 1) + (n + 1 x) (n 2) = b + n. Substituting

b = n n 2 we obtain 3x = n. Hence 31n, and the proof

is complete.



Proposition 6.8 Assume that no point of has valence n

that b > n2 n. If dp > 0, then some point unjoined to P

has valence n 1.









Proof. Let V be the set of points unjoined to P. We must

show that for some Q in V, val(Q) = n 1. Assume that

there is no such Q. Since no point has valence n and no

point of V has valence n + 1, we then have val(Q) < n 2

for every Q in V. Count incident pairs (Q,g) with Q in V,

using Proposition 2.7. One obtains
2 2
b < n 2n + val(P) < n n 1, contradicting the hy-

pothesis.

2
Proposition 6.9 Let b > n n and val(P*) = n 1. Let V

be the set of points unjoined to P*, G be the point graph

of V. If G is bipartite, then 0 is extendible.



Proof. By Corollary 2.8 we may assume that no point has

valence n. Since G is bipartite there exist disjoint inde-

pendent sets A, B whose union is V. By Proposition 6.3 the

proof will be complete if we show that IAl = IBI. Suppose

that IAI > IBI. Then IBI < n 1. Since no point has va-

lence n, the valence of any point of B is at most n 1.

Hence the number of edges in G is at most (n 1)2. By Re-
2
mark 6.2 we obtain b n + 1 < (n 1) which contradicts

the hypothesis.



Theorem 6.10 Let 0 be a ppp of order n with

n2 n < b < n2 + n. If val(P) > n (n + 12)1/2 + 4 for

every point P, then 0 is extendible.










Proof. By Corollary 2.8, we may assume that no point has

valence n. Since b < n2 + n, some point has positive defi-

ciency. Proposition 6.8 then guarantees the existence of a

point P* of valence n 1. Let G be the point graph of the

set V of points unjoined to P*. By Proposition 6.9, we

need only show that G is bipartite. Recall that G has

b n + 1 edges and that the valence of a point in G is the

same as its valence in 2. Therefore

2(b n + 1) = Z val(P) = Z (n + 1 dp).
PeV PEV

Since IVI = 2n, this may be rewritten as

Z (d 2) = 2n(n 1) 2(b n + 1) = 2(n2 1 b).
PEV

Since b > n2 n + 1, we have

(6.1) Z (dp 2) < 2n 4.
PEV

Note that dp > 2 for P in V, so the terms of the sum above

are nonnegative.

Set B = (n + 12)1/2 3 so that the hypothesis reads

val(P) > n + 1 8; i.e., dp < 8. Solving for n gives

(6.2) n = 82 + 68 3.

As the next step in proving Theorem 6.10 we present the

following two claims.

Claim 6.11 If PEV and dp < 4, then P is contained in no

triangle of G.









Proof. Suppose that points P, Q, and R form a triangle in

G. Since dQ, dR < 8, Proposition 6.6 implies that

n < 82 + 6 4, which contradicts (6.2).

Claim 6.12 If G contains disjoint independent sets A and B

such that JAl > n 2 and IBI > n 2, then G is bipartite.

Proof. It suffices to show that, if Q E V-(AUB), then one

of AU{Q}, BU{Q} is independent. Since IV-(AUB) < 4 and

Q is joined to n + 1 d points of V, Q is joined to at

least n 2 d points of AUB. If all of these

n 2 dQ points have deficiency 5 or more, then inequal-

ity (6.1) implies that 3(n 2 d ) < 2n 4, or

n < 3d + 2 < 38 + 2. This contradicts (6.2) unless 8 < 2,

in which case the hypothesis dp 8 6 cannot hold. It fol-

lows that Q is joined to some point P in AUB with dp < 4.

Without loss of generality assume that P is in A. We will

show that BU{Q} is independent; i.e., that Q is joined to

no point of B. Suppose on the contrary that Q is joined to

some point R in B. By Proposition 6.5, there are at most

d dR 1 points of V joined to neither Q nor R. Since no

point of B is joined to R, it follows that at most dQd 1

points of B are unjoined to Q. Since dp ~ 4, Claim 6.11

implies that P is contained in no triangle of G. Since P

and Q are joined, this means that Q is unjoined to every

point of V which is joined to P. There are n + 1 dp

points of V joined to P, none of which is in A. Therefore

at least n 3 dp points of B are joined to P, hence









unjoined to Q. Hence we must have n 3 dp < d d 1,

or n dQdR + dp + 2 < 2 + 8 + 2. This contradicts (6.2),

so Claim 6.12 is proved.

To complete the proof of Theorem 6.10 we need only pro-

duce the sets A, B of Claim 6.12. By Proposition 6.8,

there is some point P' in V with val(P') = n 1. Let A be

the set of points of V joined to P'. Then IAI = n 1.

Claim 6.11 guarantees that P' is contained in no triangle

of G, so A is independent. If every point of A has defi-

ciency 4 or more, then (6.1) implies that 2(n 1) < 2n 4.

The contradiction shows that there is some point P" in A

with deficiency 3 or less; i.e., val(P") > n 2. Let B be

the set of points of V joined to P". Then IBI > n 2.

Since A is independent, A and B are disjoint. Finally, by

applying Claim 6.11 to P", we see that B is independent.















CHAPTER VII
EXTENDIBILITY OF PPP'S HAVING A LARGE NUMBER OF LINES



In this chapter we prove that a ppp having nearly

n + n + 1 lines can be embedded in a projective plane, and

that the embedding is unique (see Theorem 7.2 and Corollary

7.13 for precise statements). The existence of the embed-

ding is a corollary of Theorem 7.8,'which improves Theorem

3.12 by lowering the number of lines required to guarantee

extendibility of a ppp. The reader will recall that Theorem

3.12 is a restatement of results of McCarthy and Vanstone

on (n + 1,1)-designs. In Theorem 7.14 we give an applica-

tion of results from this chapter and chapter III to (r,X)-

designs.



Proposition 7.1 (McCarthy-Vanstone [8: p.70]) Let 0 be a

ppp of order n with b n2 + n. If no point has valence n,

then b < n2 1 and val(P) > b (n2 n) for every point P.



Proof. Since b < n + n we have d > 0 for some point Q.

Since no point has valence n it follows that val(Q) < n 1.

Let V be the set of points unjoined to Q. Since a point of

V is unjoined to Q, it cannot have valence n + 1, so each









point of V has valence at most n 1. Count incident

point-line pairs (P,g) with P in V, using Proposition 2.7.

We obtain (b val(Q))dQ < dQn(n 1). Since d > 0 and

val(Q) < n 1, it follows that b < n2 1.

Now let P be any point of 2. If val(P) = n + 1, then

clearly val(P) > b (n2 n), so we assume dp > 0. We may

repeat the argument above with P in place of Q to obtain

(b val(P))dp 5 dpn(n 1). Hence val(P) > b (n2 n).



Theorem 7.2 Let 0 = (X,G) be a ppp of order n with

b > n2 n + 1. Suppose (X,G UGl) and (X,G UG2) are pro-

jective planes, where GAGG1 and G G2 are empty. Then

G1 = G2'



Proof. The proof is by induction on the nonnegative inte-

ger d = n2 + n + 1 b, the number of lines that must be

added to 0 in order to get a projective plane (see Proposi-

tion 2.1). We have IG11 = d = G21. When d = 0, we have

G1 = 0 = G2. Assume that Theorem 7.2 is true for d < dl,

and let d = dl + 1. To show that G1 = G2 it suffices to

show that G G2 is not empty.

First suppose there is some point P having valence n.

By Proposition 2.7, the set V of points unjoined to P has

size n. Clearly then the set g = VU{P} must be in both G1

and G2. Hence we may assume that has no point of valence

n.










Since we are assuming that d > 0, there must be some

point P with dp > 0. Thus Proposition 6.8 guarantees the

existence of a point P* with val(P*) = n 1. Let V be the

set of 2n points unjoined to P*. Each line not through P*

meets V in exactly two points. Proposition 6.8 applied to

P* shows that there is a point Q* in V with val(Q*) = n 1.

Each of the n 1 lines through Q* meets V in one other

point; let T denote the set of n 1 points of V joined to

Q*.

Now each of G1, G2 contains two lines through P*. Let

gl, hl E G1 and g2, h2 E G2 be these lines, where gl' 92
are the lines which contain Q*. Since the points of T are

joined to Q* in 2, none of these points can be in gl or g2.

Hence T is contained in hl h2. If hi = h2, then we are

done; assume that there are distinct points Pl. P2 in V

such that h. = {P*,P.}UT, i = 1, 2. Since P1 is not in

h2, it must be in g2. Now P1 is joined to the n 1 points

of T by hi, so is joined to no point of T by lines of 2.

Note that g2f T is empty and that P1 is also unjoined to

the other n 1 points of g2t V by lines of 2. Thus the

only point of V which can be joined to P1 in Q is P2. It

follows that val(P ) < 1. But by Proposition 7.1 we have

val(P ) > b (n n) > 1. This contradiction completes

the proof.









In Lemmas 7.3-7.7 we assume that no point of Q has va-

lence n, and that b = n a where 1 < a < n. Proposition

7.1 asserts that val(P) > n a (equivalently dp < a + 1)

for every point P.



Lemma 7.3 Let Q be any point of 0; let V be the set of

points unjoined to Q. Then

E (dp 2) = d (a + 1 dQ).
PcV


Proof. Count incident point-line pairs (P,g) with P in V

to obtain

Z (n + 1 dp) = dQ(b val(Q)).
PEV

Using the fact that JVI = dQn, the result follows.



Lemma 7.4 There are at least dQ(n a 1 + dQ) points of

valence n 1 unjoined to a given point Q.



Proof. Note that dp 2 for every point P in V, so the

terms of the sum in Lemma 7.3 are nonnegative. Hence at

most dQ ( + 1 d ) of those terms are nonzero. Since the

number of terms is dQn, we find that dp = 2 for at least

dQ(n a 1 + d ) choices of P in V.



Lemma 7.5 If d + dR > a + 2, then some point of valence

n 1 is unjoined to both Q and R.









Proof. Again let V be the set of points unjoined to Q.

Any line meets V in at most d points, so R is joined to at

most d val(R) points of V. By Lemma 7.4, at least

d (n a 1 + d ) points of V have valence n 1. Thus

we are done if d (n a 1 + d ) > d val(R). This ine-

quality follows directly from d + dR > a + 2.



Lemma 7.6 Let val(P*) = n 1; let V be the set of 2n

points unjoined to P*. Let PEV. If n > 3a, then at least

two points in V are joined to P and have valence n 1.



Proof. There are exactly val(P) points of V which are

joined to P. Recall that val(P) > n a. By Lemma 7.4, V

contains at least 2(n a + 1) points of valence n 1, so

at most 2a 2 points of V have valence less than n 1.

Since n a (2a 2) > 2, the result follows.



Lemma 7.7 Let P be any point of 2. If n > 3a, then some

point joined to P has valence n 1.



Proof. Apply Proposition 6.8 and Lemma 7.6.


2
Theorem 7.8 Let be a ppp of order n with b = n a,

a > 1. If

(7.1) n > a2/4 + 3a + 6,


then 0 is extendible.









Proof. By Corollary 2.8, we may assume that no point has

valence n. From (7.1) it follows that n > 3a, so Lemmas

7.3-7.7 are in force. Our strategy is to use Lemmas 7.5

and 7.7 to make a careful choice of P*, and then to apply

Proposition 6.9. For the graph G of Definition 6.1 we in-

troduce the following notation: for P in V, denote by Np

the set of points of V joined to P. Thus INpI = val(P).

By Proposition 7.1, val(P) > n a for every point P.

There will be two cases to consider.

Case 1: There is at most one point P in Q with dp > a/2 + 2.

Let P be a point of minimum valence in 2. By Lemma 7.7

there is some point P* of valence n 1 joined to Pm. Let

V be the set of points unjoined to P*, G be the point graph

of V. By our choice of P*, we have d < a/2 + 2 for every

P in V. Also, dp > 2 for every P in V. We claim that if

points P, Q, and R form a triangle in G, then dp > 2: other-

wise we could apply Proposition 6.6 with dp = 2 and

dQ,dR < a/2 + 2 to obtain

n < dQdR + d + dR 2 < a /4 + 3a + 6, which contradicts
R Q R -
(7.1). Therefore if dp = 2, then Np is an independent set.

By Lemma 7.6, there is an edge {P', P"} in G, where

d, = dp,, = 2. Hence Np, and Np,, are disjoint independent

sets of size n 1. There are only two remaining points in

G, say Q and R.

By Lemma 7.6, NQ contains at least two points of valence

n 1. One of these points, say P, must lie in Np,U Np,,









say in Np,. We now intend to show that Q is joined to no

point of Np,. Suppose on the contrary that Q is joined to

Q' Np,,. Since dp = 2, Np is another independent set of

size n 1. Since (P, P'} is an edge, Np and Np, must be

disjoint. Hence INpnNp,,I > n 3. Also Q E Np and

Q' E Np,, so neither Q nor Q' is joined to any of these

n 3 points. Then Proposition 6.5 implies that

n 3 Qdd 1 (a/2 + 2)2 1, which contradicts (7.1).

We have shown that Q is joined to no point of Np,. Sim-

ilarly, R is unjoined to every point of one of the sets

Np,, Np". If R is unjoined to the points of Np, then every

edge of G except possibly one (joining Q and R) meets Np,.

Since the points of Np, have valence at most n 1, we

would have b n + 1 < INp, (n 1) + 1. This inequality

simplifies to a > n 1, which contradicts (7.1). We con-

clude that R is unjoined to every point of Np,. Hence G is

bipartite, and Proposition 6.9 completes the proof of

Theorem 7.8 for Case 1.

Case 2: At least two points of Q have deficiency greater

than a/2 + 2.

Let Q, R be two points such that dQ,dR > a/2 + 2. By

Lemma 7.5, there is a point P* of valence n 1 unjoined to

both Q and R. We let V, G be as in Case 1. Thus Q,REV.

From Lemma 7.3 we have


(7.2)


E (dp 2) = 2a 2.
PEV









Denote W = V {Q, R}. It follows from (7.2) and our

choice of Q and R that

(7.3) (dp 2) < a 3.
PeW

Hence at most a 3 points of W have valence less than

n 1, so at least 2n a + 1 points of W have valence

n 1. Since 2n a + 1 > val(Q) + val(R), some point P'

in W with valence n 1 is unjoined to both Q and R. De-

note Np, by T, and set S = W T. Thus IS = ITI = n 1.

Suppose points A, B in T are joined. Then P', A, and B

form a triangle in G. It follows from (7.3) that

dA + dg < a + 1, so dAdg B (a + 1) /4. Then Proposition

6.6 gives n < dAdB + dA + dB 2 < (a + 1)2/4 + a 1.

This contradicts (7.1), so we conclude that T is independ-

ent.

Claim 7.9 If A,BES then some point of T is joined to both

A and B.

Proof. Suppose on the contrary that no point of T is

joined to both A and B. Then at most ITI = n 1 edges

join a point of T to either A or B. At least

val(A) + val(B) 1 edges meet the set {A, B}. It follows

that at least n + 2 dA dB edges meet {A, B) but not T.

Since T is independent, the number of edges which meet T

equals the sum of the valences of the points of T. As

there are only n a n + 1 edges in G, we have

E (n + 1 dp) + n + 2 dA dB < n2 a n + 1,
PeT









E (n + 1 dp) < n a + 1,
PET U{A,B}

Z (dP 2) > a 2.
PET {A,B}

The last inequality contradicts (7.3), so Claim 7.9 is

proved.

We can now show that no pair of points in S is joined.

We first show that any point of valence n 1 in S is

joined to no other point of S. Suppose A,BES are joined

and dA = 2. By Claim 7.9, some point C in T is joined to

both A and B. From (7.3) we have dB + dC < a + 1. As be-
2
fore we get n < (a + 1) /4 + a 1, contradicting (7.1).

Now suppose A, B in S are an arbitrary joined pair. Both

A and B are unjoined to every point of S having valence

n 1. From (7.3) we know that there are at least

ISI (a 3) = n a + 2 such points. Proposition 6.5

implies that n a + 2 < dAdB 1 < (a + 1) /4 1, again

contradicting (7.1).

We have now shown that both S and T are independent sets

of size n 1. Though T was originally distinguished from

S by the fact that all of the points of T are joined to a

common point, we will not need this fact in the remainder

of the proof. Therefore we may assume without loss of gen-

erality that


(7.4)


S (dP 2) > Z (d 2).
PET PES









From (7.3) and (7.4) it follows that

(7.5) Z (dp 2) < (a 3)/2.
PeS

Claim 7.10 One of Q, R is joined to at least (n a 1)/2

points of T.

Proof. The number of edges which meet S is at most

SJ(n 1) = (n 1)2. There are exactly n2 a n + 1

edges in G. Thus at least n a edges miss S. Since T is

independent, this means at least n a 1 edges join a

point of T to Q or R. The claim follows.

Assume without loss of generality that Q is joined to at

least (n a 1)/2 points of T.

Claim 7.11 At most one point of S is joined to Q.

Proof. Suppose that Q is joined to both A,BeS. By (7.5),

dA + dB < (a + 5)/2. Hence either dA or dB, say dA, is at

most (a + 5)/4. Since A is joined to no point of S, it

follows that A is joined to at least

val(A) 2 > n (a + 9)/4 points of T. Hence the number

of points of T joined to both Q and A is at least

(n a 1)/2 + n (a + 9)/4 IT = n/2 (3a + 7)/4. It

follows from (7.1) that this number is greater than a 3.

It follows from (7.3) that at most a 3 points of T have

deficiency greater than 2. Hence some point C in T with

dC = 2 is joined to both Q and A. Then C, Q, and A form a

triangle in G, so Proposition 6.6 implies that









n < dA + dQ + dA 2. Since d < a + 1 and dA (a + 5)/4

we have a contradiction to (7.1), so the claim is proved.

Set V1 = S U{Q}, V2 = TL {R). We have proved that at

most one pair of points in V1 is joined. Therefore we have

Z val(P) < 2 + E val(P),
PEV1 PcV2

(7.6) Z (d 2) < 2 + (dP 2).
PeV2 Pev

Add (7.2) and (7.6), and divide by 2 to obtain

(7.7) E (dp 2) < a.
PEV2

Claim 7.12 The set V2 is independent.

Proof. We already know that T is independent, so we need

only show that R is joined to no point of T. Suppose on

the contrary that R is joined to BET. At most

ITI(n 1) = (n 1)2 edges meet T, and there are
2
n a n + 1 edges in all, so at least n a edges lie

within V1 {R}. At most one edge lies within V1 by Claim

7.11, so R is joined to at least n a 1 points of V1.

Thus R is joined to at most n + 1 dR (n a 1) < a/2

points of T. Hence R is unjoined to at least n 1 a/2

points of T. Since B is also unjoined to these points,

Proposition 6.5 implies that n 1 a/2 < dRdB 1. By

(7.7), dR + dB < a + 4, so dRdB < (a + 4) /4. Thus we have

n < (a + 4)2/4 + a/2, which contradicts (7.1).









In view of Proposition 6.9, the proof of Theorem 7.8

will be complete if we show that V1 = SU{Q} is also inde-

pendent. We showed earlier that S is independent and that

Q is joined to at most one point of S. Assume that Q is


joined to AES.

come qualities




It follows that

Neither Q nor A

Proposition 6.5

a contradiction



Corollary 7.13


Then the inequalities (7.6) and (7.7) be-

and, using (7.2), we have

S(dp 2) = a 2.
PEV1

dQ + dA a + 2, so d dA (a + 2)2/4.

is joined to any other point of V1, so

implies that n 1 < d dA. We again have

to (7.1) and the proof is complete.



Let 2 = (X,G) be a ppp of order n with


b > n2 2(n + 3)1/2 + 6. Then n can be embedded in a pro-

jective plane of order n; i.e., there exists a collection

G1 of subsets of X such that (X,GUG ) is a projective

plane.



Proof. If b < n2 then set a = n2 b. Then the inequality

in the hypothesis is equivalent to inequality (7.1). Hence
2 2
by Theorem 7.8 2 is extendible. If n < b < n + n, then

by Proposition 3.13 2 is extendible. Continued use of

Theorem 7.8 and Proposition 3.13 show that there is a col-

lection G1 such that (X,GUG ) is a ppp of order n with

n + n + 1 lines. By Proposition 2.1, (X,GLUG) is a pro-

jective plane.









We now give an application to (r,A)-designs. It has

been established in [3] that there exists a least integer

v0(r,A) such that for v > v (r,A) any (r,A)-design on v

varieties is trivial. Let n = r A. In [13] it is shown

that v0(r,X) < max(A + 2, n2 + n + 1), with equality if n

is the order of a finite projective plane. This bound has

been improved in [8], [15]. The following result is a fur-

ther improvement.



Theorem 7.14 For positive integers r and A such that

n = r A is not the order of a finite projective plane,

we have

(i) v0(r,X) < max(X + 2, n2 (2n + 1/4)1/2 1/2), r < 2n,

(ii) v0(r,A) < max(X + 2, n2 (2n + 1/4)1/2 + 1/2), r > 2n.



Proof. One can always obtain a near-trivial (r,A)-design

on v = A + 2 varieties by taking all (X + 1)-subsets of a

(A + 2)-set; furthermore, any near-trivial (r,X)-design has

at most A + 2 varieties. Hence we need only consider de-

signs which are not near-trivial. The reader may verify

that any (A + 1,A)-design is near-trivial, so we assume

n > 1. Since we are assuming that n is not the order of a

projective plane, we have n > 6.

Let D be an (r,A)-design which is not near-trivial. As-

sume that r < 2n and that


v > n2 (2n + 1/4)1/2 1/2.


(7.8)









Since n > 6, (7.8) implies that v > n2/2 + 2n + 1. By

Theorem 3.18, there exists a non-trivial (n + 1,1)-design
2 2
D1 on v varieties. If v < n then set a = n v; in this

case (7.8) is equivalent to the inequality n > (a2 + a)/2.

Hence in any case D1 satisfies the hypothesis of Theorem

3.12, so D1 is symmetrically bounded. By Proposition 3.8,

there exists a ppp 2 of order n with v lines. Inequality

(7.8) implies that satisfies the hypothesis of Corollary

7.13. Therefore 2 can be embedded in a projective plane of

order n, but by hypothesis no such plane exists. This com-

pletes the proof in the case r < 2n. The case r > 2n is

handled similarly.
















CHAPTER VIII
OTHER EXTENDIBILITY RESULTS



In this chapter we use techniques similar to those of

chapters VI and VII to obtain further extendibility re-

sults. It was established in Theorem 3.13 that a ppp of

order n with n2 2 < b < n2 + n is extendible. In Theorem

8.4 we improve this assertion to include the case
2
b = n 3 for all n other than 4, 8, and 12.



Proposition 8.1 Let 0 be a ppp of order n with b = n2 a,

1 < a < n. Assume that no point has valence n. If

val(Q) = n a, then every point unjoined to Q has valence

n 1.



Proof. By Proposition 2.7, exactly dQn points are unjoined

to Q. By Lemma 7.4, at least dQn points of valence n 1

are unjoined to Q.


2
Theorem 8.2 Let Q be a ppp of order n with b = n a,

a > 3. If there exist two points Q, R of valence n a and

n > 3a + 3, then 0 is extendible.









Proof. By Corollary 2.8, we may assume that no point has

valence n. By Proposition 8.1, every point unjoined to Q

has valence n 1. Since val(R) = n a < n I, Q and R

are joined. Let P* be any point unjoined to both Q and R

(the existence of P* is given by Proposition 6.4). Then

val(P*) = n 1. Let V be the set of points unjoined to

P*, G be the point graph of V. By Lemma 7.3, we have

Z (dp 2) = 2a 2.
PsV

Since d = dR = a + 1, it follows that the other 2n 2
Q R
points of V have valence n 1. At most 2(n a) points of

V are joined to either Q or R, so some point P in V is un-

joined to both Q and R. Let S be the set of n 1 points

of V joined to P; let T = V (S U{Q, R}). Since

n > 3a + 3, Proposition 6.6 implies that every triangle in

G consists of Q, R, and one other point. Hence S is inde-

pendent. Exactly (n 1)2 edges of G meet S; the remaining

n a edges are contained in TU {Q, R}.

Let M,NET. At least 2n 3 edges contain M or N, so at

least 2n 3 (n a) edges join a point of S to M or N.

Since a > 3, it follows that some point of S is joined to

both M and N. Therefore M and N are unjoined. Since M,NET

were arbitrary, we have shown that T is independent.

Suppose Q is joined to points AES and BET. By Proposi-

tion 6.5, at most dAdQ 1 = 2a + 1 points of V are joined

to neither A nor Q. Since S is independent, it follows









that Q is joined to at least ISI (2a + 1) = n 2a 2

points of S. Similarly Q is joined to at least n 2a 2

points of T. Since Q is joined to R as well, we have

n a = val(Q) > 2(n 2a 2) + 1, which contradicts the

inequality n > 3a + 3 of the hypothesis. We conclude that

Q is unjoined to every point of one of the sets S, T. Sim-

ilarly R is unjoined to every point of S or T. Earlier we

noted that n a edges were contained in TU{Q, R). Since

T is independent, there are also n a edges contained in

SU{Q, R}. Hence one of Q, R is unjoined to the points of

S and the other is unjoined to the points of T. Therefore

G is bipartite, and Proposition 6.9 completes the proof.


2
Corollary 8.3 Let 0 be a ppp of order n with b = n a,

a > 2. Assume that there exist two points Q, R of valence

n a. If n > 3a + 3, then 0 can be embedded in a projec-

tive plane of order n.



Proof. The proof is by induction on a. If a = 2, then the

conclusion follows from Theorem 3.13. Assume Corollary 8.3

holds for all a < al and let a = al + 1 > 3. By Theorem

8.2, 0 can be extended to a ppp S1 with
2 2
b = n a + 1 = n al. By Corollary 2.8, 2l can be em-

bedded in a ppp ?2 with b = n al + k, k > 0, such that

no point has valence n. If k > a 2, then Theorem 3.13

completes the proof. Assume k < a1 2 and set a2 = al k.









Then 2 has b = n a2. By Proposition 7.1,

val(P) > n a2 for every point P of 02. Since Q2 was ob-

tained from 0 by the addition of a a2 lines, the valence

of any point cannot have increased by more than a a2.

Consequently, val(Q), val(R) < n a2 in 02. Therefore

val(Q) = n a2 = val(R) in Q2. The induction hypothesis

now implies that 2 can be embedded in a projective plane

of order n. Hence can be embedded in a projective plane

of order n, and the proof is complete.


2
Theorem 8.4 Let be a ppp of order n with b = n 3. If

n X {4, 8, 121, then Q is extendible.



Proof. For n = 2 the theorem is trivial. For n = 3, 5,

Theorem 8.4 was proved in Theorems 5.2, 5.17. Hence we as-

sume n > 6 and n $ 8, 12. We use Notation 5.5. By Corol-

lary 2.8, we may assume that vn = 0. By Proposition 7.1,

we have v. = 0 for all i < n 3. The equations of Propo-
1
sition 5.7 reduce to the following three equations.

6v ++ 2v = 3n2 -n 7n + 2
n+l n-3
(8.1) 2v1 4vn3 = n + 3n 4

3vn + 8-3 = 2n + 8

If some line contains x. points of valence i, then by Prop-

osition 2.4 we have


2xn-l + 3x-2 + 4xn-3 = n + 4.
n-l n-2 n-3


(8.2)










Suppose n is odd. Then from (8.2) it follows that every

line contains an odd number of points of valence n 2. In

particular, every line contains at least one point of va-

lence n 2. It easily follows that vn_2 > n + 1. From

(8.1) we have 2n + 8 = 3vn-2 + 8v n 3n + 3, which im-
n2 n-3 -
plies n < 5. Since we have assumed n > 6, we are done in

the case that n is odd.

Now assume that n is even. Then (8.2) implies that ev-

ery line contains an even number of points of valence

n 2. We consider separately the cases vn-2 = 0,

vn-2 ? 0.

First suppose vn_2 / 0; let val(P) = n 2. Each of the

n 2 lines through P contains at least one additional

point of valence n 2. Consequently, vn2 > n 1. If

v -3 > 0, then from (8.1) we have

2n + 8 = 3n-2 + 8vn3 3(n 1) + 8, or n < 3. If

vn_3 = 0, then from (8.1) we have

2n + 8 = 3v n2 3(n 1), so n < 11 and n = 2(mod 3);
n-2-
since n is even we have in this case n = 2 or n = 8. Since

we have assumed that n is not 2, 3, or 8, we are done in

the case vn_2 0.

Assume vn2 = 0. From (8.1) we have vn-3 = n/4 + 1, so

41n. Since we have ruled out n = 4, 8, and 12, we have

n > 12 and vn-3 > 4. By Theorem 8.2, 2 is extendible.















BIBLIOGRAPHY


[1] R. H. Bruck, Finite nets II, uniqueness and embedding,
Pacific J. Math. 13 (1963), 421-457.

[2] A. Bruen, Unimbeddable nets of small deficiency, Pa-
cific J. Math. 43 (1972), 51-54.

[3] V. Chvatal, On finite A-systems of Erd6s and Rado,
Acta Math. Sci. Hung. 21 (1970) 341-355.

[4] J. D4nes and A. D. Keedwell, Latin squares and their
applications, English Universities Press (1974).

[5] J. I. Hall, Bounds for equidistant codes and partial
projective planes, Discrete Math. 17 (1977), 85-94.

[6] D. R. Hughes and F. C. Piper, Projective planes,
Springer-Verlag (1973).

[7] H. B. Mann, The construction of orthogonal latin
squares, Ann. Math. Statist. 13 (1942), 418-423.

[8] D. McCarthy and S. A. Vanstone, Embedding (r,l)-
designs in finite projective planes, Discrete Math.
19 (1977), 67-76.

[9] T. G. Ostrom, Semi-translation planes, Trans. Amer.
Math. Soc. 111 (1964), 1-18.

[10] T. G. Ostrom, Nets with critical deficiency, Pacific
J. Math. 14 (1964), 1381-1387.

[11] T. G. Ostrom, Derivable nets, Canad. Math. Bull. 8
(1965), 601-613.

[12] S. A. Vanstone, The extendibility of (r,l)-designs,
Proc. 3rd Manitoba Conf. Numerical Math., Winnepeg
(1973), 409-418.

[13] S. A. Vanstone, A bound for v0(r,X), Proc. 5th South-
eastern Conf. Combinatorics, Graph Theory, and Com-
puting, Boca Raton (1974), 661-673.






75


[14] S. A. Vanstone, The structure of regular pairwise
balanced designs, Ph. D. Thesis, University of Water-
loo (1974).

[15] S. A. Vanstone and D. McCarthy, On (r,X)-designs and
finite projective planes, Utilitas Math. 11 (1977),
57-71.
















BIOGRAPHICAL SKETCH


Stephen John Dow was born February 7, 1955, in Jackson-

ville, Florida. He grew up in southwest Jacksonville and

attended public schools there. Following a two-year period

working at his father's business, he moved to Gainesville

in September, 1975, to attend the University of Florida. A

year later he began a concentrated study of mathematics.

He has continued this study at U.F. since, receiving his

B.A. in 1978 and his M.S. in 1980.
















I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.



David A. Drake, Chairman
Professor of Mathematics

I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.



Nicolae Dinculeanu
Professor of Mathematics

I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.



Robert L. Long
Assistant Professor o
Mathematics
















I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.



David J. Ny V
Associate Professor of
Finance, Insurance, and
Real Estate

I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.



Neil L. White
Associate Professor of
Mathematics

This dissertation was submitted to the Graduate Faculty of
the Department of Mathematics in the College of Liberal
Arts and Sciences and to the Graduate Council, and was
accepted as partial fulfillment of the requirements for
the degree of Doctor of Philosophy.


August, 1982
Dean for Graduate Studies
and Research




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