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 pointline 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, . ,xn1) 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,. ,Ln1 of order n. Here L1,. .,Lk2
have weight n; i.e., L1,. .,Lk2 are mols. Setting
d = n 1 (k 2) we see that the hypotheses of Theorems
3.4 and 3.5 are satisfied. Hence LI,. ,Lk2 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 Lkl,. .,Ln1 form a common transversal of
L1,. ,Lk2. Also note that to extend L1,. ,Lk2 to
a set of n 1 mols, every one of the dn common transver
sals of L1,. ,Lk2 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 nontrivial. 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 nontrivial (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 (McCarthyVanstone) The only nonextendible
(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 (McCarthyVanstone) 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 nonextendible 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 (McCarthyVanstone) A nontrivial (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 stuples
is the number of coordinates in which they differ. An
equidistant dcode is a collection C of binary stuples
(called codewords) any two of which have Hamming distance d.
We say that C is nontrivial 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 nontrivial equidistant 2ncode with x codewords
if and only if there exists a nontrivial (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 neartrivial 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 neartrivial. Set n = r X. If r > 2n, then
there exists a nontrivial equidistant 2ncode C with v
codewords. If r < 2n, then there exists a nontrivial
equidistant 2ncode 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 neartrivial. Set n = r X. If r < 2n and
v > n2/2 + 2n + 1, then there exists a nontrivial
(n + 1,1)design on v varieties. If r > 2n and
v > n /2 + 2n + 2, then there exists a nontrivial
(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
2subsets 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 nonDesarguesian projective planes
developed by Ostrom in [9], [10], and [11]. Our work is
selfcontained 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 = (d1)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 dnet, each point of g2 is on exactly d lines
of N1. Thus there are exactly Ig2d = n3/2 + n incident
pointline 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 wellknown 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.,
ieKF, 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 mEKF. 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: beKF}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 bEKF, are disjoint. Also, distinct
sets h CEK, are disjoint, since they are precisely the
c, c
sets (F + c) x (F + c). Let beKF, 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 dnets
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 pointline 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 pointline 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 pointline 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 (McCarthyVanstone [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
pointline 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 7.37.7 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 pointline 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
7.37.7 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 neartrivial (r,A)design
on v = A + 2 varieties by taking all (X + 1)subsets of a
(A + 2)set; furthermore, any neartrivial (r,X)design has
at most A + 2 varieties. Hence we need only consider de
signs which are not neartrivial. The reader may verify
that any (A + 1,A)design is neartrivial, 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 neartrivial. 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 nontrivial (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 n3
(8.1) 2v1 4vn3 = n + 3n 4
3vn + 83 = 2n + 8
If some line contains x. points of valence i, then by Prop
osition 2.4 we have
2xnl + 3x2 + 4xn3 = n + 4.
nl n2 n3
(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 = 3vn2 + 8v n 3n + 3, which im
n2 n3 
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 vn2 = 0,
vn2 ? 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 = 3n2 + 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);
n2
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 vn3 = n/4 + 1, so
41n. Since we have ruled out n = 4, 8, and 12, we have
n > 12 and vn3 > 4. By Theorem 8.2, 2 is extendible.
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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 twoyear 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
