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SOME PROBLEMS IN BLOCKING SETS By CYRUS L. KITTO A DISSERTATION PRESENTED TO TIE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 1991 ACKNOWLEDGEMENTS I thank my advisor Dr. David Drake for his patience and excellent guidance. TABLE OF CONTENTS ACKNOWLEDGEMENTS ...................... ii ABSTRACT . . . iv CHAPTERS 1. INTRODUCTION ....... ................. 1 2. A BOUND FOR BLOCKING SETS OF REDEI TYPE, PART I .. 2 3. A BOUND FOR BLOCKING SETS OF REDEI TYPE, PART II 14 4. A BOUND FOR BLOCKING SETS OF REDEI TYPE, PART III 28 5. A BOUND FOR BLOCKING SETS OF REDEI TYPE, PART IV 32 6. SMALL BLOCKING SETS OF HERMITIAN DESIGNS, PART I 46 7. SMALL BLOCKING SETS OF HERMITIAN DESIGNS, PART II 58 8. FINAL REMARKS ...................... 65 REFERENCES . . . 66 BIOGRAPHICAL SKETCH ..................... 68  iii  Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy SOME PROBLEMS IN BLOCKING SETS By Cyrus L. Kitto December 1991 Chairman: Dr. David A. Drake Major Department: Mathematics In this dissertation some new results are obtained on the cardinality of blocking sets in block designs. Firstly, lower bounds are established on the cardinality of blocking sets of R6dei type in finite projective planes. Secondly, a block design E is formed using the Hermitian unitals of PG(2,q2), q an odd prime power, as blocks, and then a lower and upper bound are established on the cardinality of a committee of E; also, a characterization of the Desarguesian projective plane PG(2,9) is established.  iv  CHAPTER 1 INTRODUCTION An incidence structure is an ordered pair (A, B), where A is a set and B a collection of subsets of A. The elements of A are called points and those of B are called blocks. A t (v, k, A) design is an incidence structure (A, B) where the cardinality of A is v, each block of B contains exactly k points and each subset of t points of A is contained in exactly A blocks of B. For t = 2 one calls such an incidence structure a block design. A projective plane of order n, n > 2, is a block design with v = n2 + n + 1, k = n + 1 and A = 1. If II represents a projective plane of order n, then it is straightforward to show that each pair of distinct blocks of I intersect in a unique point, and that there are four distinct points of H no three of which are contained in the same block. A block of a projective plane is called a line. A hitting set H of an incidence structure I = (A, B) is a subset of A so that each block of B has nonempty intersection with H. If H is a hitting set of I so that no block is a subset of H, then H is called a blocking set of I. The purpose of this dissertation is to prove some original theorems concerning the cardinality of blocking sets of block designs. CHAPTER 2 A BOUND FOR BLOCKING SETS OF REDEI TYPE, PART I We begin Chapter 2 by introducing the notation and terminology that will be used throughout Chapters 2, 3, 4 and 5. Most of the lemmas proven in Chapter 2 will also find use in the succeeding three chapters. Let II represent a finite projective plane of order n, and let S represent a blocking set of I. For each line I of I, call the number I S I I the strength of I and denote it by st(l). If st(l)=i > 2, then call I a secant or an i line. If st(l)=1, then call 1 a tangent. LEMMA 2.1. There exists an integer 6 > 1 such that I S I= n + 6. PROOF: Let 1 be a line of I. One can choose a point X on I yet not in S. The n lines other than I through X are each incident with at least one point of S. Thus, IS 1> n+st(l). Q.E.D. LEMMA 2.2. Let S be such that I S = n + 6. Let g and h represent distinct lines of H. If st(g)+st(h)> 6 + 2, then the point g n h is in S. PROOF: Assume, by way of contradiction, that for distinct lines g and h one has st(g)+st(h)> 6+2 and (gnh) S. The n1 lines other than g and h through point g n h are each incident with at least one point of S. Thus, I S > n 1+st(g)+st(h) > n + 6 + 1, a contradiction. Q.E.D. 2 3 LEMMA 2.3. Let S be such that I S = n + 6. Then for any line I one has st(l)< 6. PROOF: Assume, by way of contradiction, that st(l)> 6 for some secant I. Choose point X E I \ S. The n lines other than I through X are each incident with at least one point of S. Thus, I S 1> n+st(l) > n + 6, a contradiction. Q.E.D. One says that a blocking set S is of Redei type if I S J= n + 6 implies the existence of a secant l of strength 6. Such a secant I will sometimes be called a secant of maximum strength. By Lemma 2.2, if S is of R6dei type, then every secant intersects every secant of maximum strength in a point of S. Let S be of R6dei type with I S I= n + 6. Let U represent a point of II. One says that U is of type a = (a, ..., am) if U lies on exactly m secants li,..., m with ai=st(li) and al < ... < am. For point U of type a, define the constant F(U)=aY + ... + am; arguments using the computation of F(U) will often be called a fan count on U. For point U and integer i > 2, let ri(U), or just ri if the point U is clear, equal the number of ilines through U, and define r(U) = r2 +... + r6. Also, for integer i > 2, let bi equal the total number of ilines and b the total number of secants. LEMMA 2.4. Let S be a blocking set of Redei type with I S J= n+6. Let P E S be a point not incident with every secant of maximum strength. Then F(P) = n + 26 1. PROOF: Let I be a 6line not incident with point P E S. Then F(P) =1 S I +(61) since every secant through P meets I at a point of S. Q.E.D. LEMMA 2.5. Let S be a blocking set of Redei type with I S I= n + 6. Let X S be a point of type a = (al,..., am). Then F(X) = 6 1 + m. 4 PROOF: This should be clear since I S 1= F(X) + (n + 1 m). Q.E.D. LEMMA 2.6. Let S be a blocking set of Ridei type with I S 1= n + 6. Assume that there is a unique secant I of maximum strength. Let j > 3 be some fixed integer. If every P E S \ I lies on a secant g with st(g)> j, then every point P E S \ 1 lies on a secant h with st(h)< + 1 j. PROOF: Assume, by way of contradiction, that every point in S\ lies on a secant g with st(g)2 j and point P E S\l is of type a with al > 6+2j. Since 6+2j > 3, one can choose a point Q E S \ I such that PQ = 11. By assumption, Q lies on a secant g with st(g)> j. Since al 2 6 + 2 j, Lemma 2.2 implies that secant g intersects the 6 secants through P at points of S. Thus, st(g)> 6, a contradiction since Q 1, the unique 6line. Q.E.D. LEMMA 2.7. Let S be a blocking set with I S I= n + 6. Let P,Q and R be three distinct points of S such that PQ $ PR and st(PR)=j. Then through Q there are at most j 1 secants g # PQ of strength 6 + 2 j or more. PROOF: By Lemma 2.2, a secant g i PQ through Q of strength 6 + 2 j or more intersects line PR at a point of S. There are only j 1 points of PR where g can intersect. Q.E.D. For secant g, define g* to be the set of points in g \ S, and define L(g*) to be the set of secants h 0 g such that the point g n h is not in S. 5 LEMMA 2.8. Let S be a blocking set with I S J= n + S. Let g be a secant of strength less than 6. Then n + 1 st(g) =1 g* 1 PROOF: This should be clear since Lemma 2.5 and the fact st(g) < 6 imply that each X E g \ S lies on at least two secants. Q.E.D. We are now ready to introduce the main objective of Chapters 2 and 3. It was proven by A. Bruen that for any blocking set S, I S 1> n + n1/2 + 1. [8, Theorem, and; 9, Theorem 3.8] Under the additional assumption that n is a non square greater than 5, J. Bierbrauer proved that I S I> n + n1/2 + 2. [1, Theorem] (If n is square, then Baer subplanes offer counterexamples.) If n = 10 and S is of R6dei type, then J. Bierbrauer showed that I S I> 10 + 101/2 + 3. [3, Theorem] The main objective of Chapters 2 and 3 is to extend this last result and prove the following theorem. THEOREM 2.9. Let H be a finite projective plane of nonsquare order n = t2 e t > 4, 1 < e < 2t 2 and n # 10. Let S be a blocking set of Redei type. Then IS > n+t + 3. An immediate consequence of Theorem 2.9 and an observation by Bierbrauer [1, Corollary] is the following. COROLLARY 2.10. A net of nonsquare order > 11 with more than n(nl/2+2) parallel classes can be completed in at most one way to a projective plane of order n. In view of J. Bierbrauer's work in [1, Theorem], Theorem 2.9 can be proven by showing that I S I/ (t2 e) + t + 2. Henceforth, assume I S 1= (t2 e) + t + 2. 6 CLAIM 2.11. ([2, Theorem]) There is exactly one (t + 2)line. Henceforth, let I denote the unique secant of strength t + 2. CLAIM 2.12. For every point P E S \ 1, t2 + 5 < F(P) < t2 + 2t + 2. PROOF: This follows immediately from Lemma 2.4. Q.E.D. CLAIM 2.13. Every point P E S \ I lies on a secant of strength t 1 or more. PROOF: If not, then t2 + 5 < F(P) < (t + 2)(t 2) = t2 4. Q.E.D. By Lemma 2.6, every point P E S \ I lies on a secant of strength 4 or less. CLAIM 2.14. Every point P E S \ I lies on a secant of strength t or t + 1. PROOF: Assume not and let P be of type a with at+2 = t 1. Since al < 4, one has t2 + 5 < F(P) < (t+ 1)(t 1) + 4 = t2 +3. Q.E.D. By Lemma 2.6, every point P E S \ I lies on a secant of strength 2 or 3. CLAIM 2.15. Let g be a secant of strength t+1. Every P E S\(lUg) lies on exactly one 2line in L(g*). PROOF: Let P E S \ (U g). Set A = I g and let Qi,...,Qt be the t points in (g n S) \ {A}. The t distinct secants PQi intersect I at t distinct points of S \ {A}. 7 There is thus one point B # A on I and in S such that B P PQi for 1 < i < t. The secant h = PB intersects g at a point not in S. By Lemma 2.2, h must be a 2line. Q.E.D. If there is a secant g of strength t + 1, then Claim 2.15 implies that the total number of secants b = r(A) + t(t + 1) + (n t), where A denotes the point l n g. CLAIM 2.16. No point A E S n I lies on n secants of strength 2. PROOF: If A lies on n secants of strength 2, then the set S \ {A} is a blocking set with cardinality n + t + 1, a contradiction of the Bierbrauer bound in [1, Theorem]. Q.E.D. For the remainder of Chapter 2, assume that n > 37. The arguments used in dealing with nonsquare 11 < n < 35 are of an ad hoc nature and will be handled in Chapter 3. Note that n > 37 implies that t > 7. CLAIM 2.17. For every P E S \ 1 of type a, one has al + a2 + a3 + a4 > 14. PROOF: If not, then t2 + 5 < F(P) < (t 2)(t + 1) + 13 = t2 t + 11, a contradiction for t > 7. Q.E.D. CLAIM 2.18. Every point P E S \ I lies on a 2line. PROOF: Assume that some P of type a does not lie on a 2line. Then by Claim 2.14 and Lemma 2.6 al = 3 and all secants of strength t+1 pass through P. Choose 8 Q E S \ I so that PQ = 11. Claim 2.14 and al = 3 imply that Q lies on a tline. It follows from Lemma 2.2 and Claim 2.17 that P lies on exactly three 3lines. A fan count on P now shows that there are (t + 1)lines through P and, hence, Claim 2.15 implies that Q lies on a 2line. A fan count on Q shows that Q is of type (2,3,t,..., t). Let R E S \ I be such that PR = 12. The secant g = QR must be of strength t. But, a4 > 4 implies that g intersects at least t + 1 secants through P at points of S, a contradiction. Q.E.D. CLAIM 2.19. None of the points P E S \ l lies on a 3line. PROOF: Assume that some P does lie on a 3line. Let P be of type a. Claim 2.17 implies that P lies on at most three secants of strength 3 or less. There are three cases to consider. Case 1. P lies on two 2lines and one 3line. Let Q E S \ I be such that st(PQ)=3. A fan count on P shows that a4 > t and a5 = t + 1. Since t + 2 > 9, there are at least five (t + 1)lines through P. Thus, PQ is the only 3line of S. Since P lies on two 2lines, Q lies on at most two (t + 1)lines. Since a4 > t, Q lies only on secants of strength 2, 3, t or t + 1. A fan count on Q and Claim 2.18 imply that Q lies on a 2line, a 3line and t secants of strength t or t + 1. Let QB be the 2line, and remember that PQ is the only 3line of S. Then every iline g through B with i > 3 must meet the t + 2 secants through Q at points of S. This contradiction of Claim 2.11 implies that B must lie on n secants of strength 2, a contradiction to Claim 2.16. 9 Case 2. P lies on one 2line and two 3lines. Let Q E S \ I of type P and B E S n I be such that PQ and PB are a 3line and a 2line, respectively. A fan count of P shows that a4 > t 1 and a6 = t + 1. Since t + 2 5 = t 3 > 3 for t > 7, all 3lines pass through P. Let g # PQ be a secant through Q with st(g) > 2. Since a6 = t + 1, secant g has strength at least t 2. Since 04 + (t 2) > t + 4 for t > 7, Lemma 2.2 implies that g has strength at least t. Thus, Q lies on one 2line, one 3line and t secants of strength at least t. It also follows that st(QB)=2. Let h # I be a secant through B with st(h) > 2. Since /3 > t, #2 > 2 and the fact that all 3lines pass through P, secant h intersects the t + 2 secants through Q at points of S, a contradiction. Hence, B lies on n secants of strength 2, a contradiction of Claim 2.16. Case 3. P lies on a 2line, a 3line and a3 > 4. Let Q E S\ 1 and B E Sn be such that PQ=12 and PB=11, respectively. Set /a = a3. As Case 1 and Case 2 do not occur, Q also lies on exactly one 2line and exactly one 3line. Subclaim 2.20. If / < t 1, then Q lies on / secants of strength t + 4 / or larger. Proof. Assume not. Then, t2+ 5 F(Q) <(/ )(t + 1) + (t + )(t+3 ) + 5. Equivalently, f(/0) := 2 3(t + 3) + 3t + 2 > 0; but, f(/) < 0 for 4 < / < t 1, a contradiction. q.e.d. 10 If / < t 1, then Subclaim 2.20 would contradict Lemma 2.7. Thus, /3 t, or P lies on t secants of strengths t or t + 1. The same conclusion applies to Q, and Lemma 2.7 then yields / t + 1. It is then immediate that B must lie on n secants of strength 2, a contradiction of Claim 2.16. Q.E.D. CLAIM 2.21. None of the points P E S \ 1 lies on a 4line. PROOF: Assume some P of type a does lie on a 4line. Since ai=2, a fan count on P shows that P lies on at most three 4lines. We consider four cases. Case 1. P lies on two 4lines and a4 > 5. Subclaim 2.22. For 1 < k < t 5, rt+lk ... + rt+l > k + 3. Proof. Assume not. Then, 2 +5 < F(P) (k + 2)(t + 1)+(t k 3)(t k) + 4 + 4 + 2. Equivalently, f(k) := k2 + k(4 t + 7 > 0; but, f(k) < 0 for 1 < k < t 5, a contradiction. q.e.d. Let B E S n I be such that PB=11, and let g l I be an iline through B with i > 2. By Claim 2.19, i > 4, so g intersects secants of strength t or t +1 at points of S. As g also contains B, Subclaim 2.22 implies that i > (1 + 3) + 1 = 5. Using this argument repeatedly, one gets that i > (t 5 + 3) + 1 = t 1. As a4 > 5, Lemma 2.2 implies that i > t. Thus, g intersects the t + 2 secants through P at points of S, a contradiction to Claim 2.11. Hence, there must be n secants of strength 2 through B, a contradiction to Claim 2.16. 11  Case 2. P lies on two 2lines and a 4line. Let Q E S\l be such that PQ = 13. A fan count on P shows that F(P) < t2+7, c4 > t 1 and a5 > t. Since P lies on two 2lines, Q lies on at most two (t+ 1)lines. A fan count on Q reveals that Q lies on a 2line, a 4line and t secants of strength t 3 or more. Using the bounds on a4 and a5, one sees that Q lies on t secants of strength t or t + 1. Let B E S n I be such that QB is a 2line. Remembering that there are no 3lines, it is immediate that B must lie on n secants of strength 2, a contradiction to Claim 2.16. Case 3. P lies on three 4lines. Let Q E S \ l and B E S n I be such that PQ = 12 and PB = 11, respectively. A fan count on P shows that t = 7 and a5 = 8. This implies that any secant g # I through B has strength 2 or 6, and that QB is also a 2line. Since a3 = 4 and a5 = 8, point Q lies on at most three 8lines and only on secants of strength 2, 4, 6 or 8. Thus, Q is of type (2,4,6,6,6,6,8,8,8). Lemma 2.2 then implies that B is incident with n 2lines, a contradiction to Claim 2.16. Case 4. a3 > 5. Let Q E S \ I be such that PQ = 12. A fan count on Q shows that if a3 < t 2, then Q lies on a3 secants of strength t +4 a3 or larger. Lemma 2.7 then yields the contradiction a3 > t 1. Because Cases 1, 2 and 3 do not occur, Q lies on exactly one 2line and exactly one 4line. Since a3 > t 1, the remaining t secants through Q are of strength t + 1. If B E S n l is such that QB is the 2line through Q, it is straightforward to show that B must lie on n secants of strength 2, a contradiction to Claim 2.16. Q.E.D. 12 CLAIM 2.23. Every P E S \ 1 lies on at least two 2lines. PROOF: Suppose P is of type a and lies on exactly one 2line. Let B E S I I be such that PB = 11. First assume either that t > 8, or that t = 7 and F(P) > t2 +6. Arguing as in Claim 2.21, Case 1, it can be proved that for 2 < k < t 5, one has rt+lk + ... + rt+1 k + 3, and, hence, that B lies on n secants of strength 2. This contradiction yields t = 7 and F(P) = t2 + 5. Consider any iline g 5 I through B with i > 2. Claims 2.19 and 2.21 imply that i > 5. If a5 > 6, then P lies on five secants of strength 6 or more. This fact and Lemma 2.2 imply that i > 6. But, as a2 > 5, Lemma 2.2 implies that i = t + 2. This contradiction implies that a5 = 5, and, hence, that P is of type (2,5,5,5,5,8,8,8,8). This implies that there are no 6lines or 7lines. Thus, any point Q E S \ I is either of type (2,5,5,5,5,8,8,8,8), (2,2,5,5,8,8,8,8,8) or (2,2,2,8,8,8,8,8,8). Assume that Q is of one of the latter two types, and let R E S \ 1 be any point other than Q. Since Q lies on two 2lines, R lies on at most three 8lines. This contradiction yields that all points P are of type (2,5,5,5,5,8,8,8,8). Counting flags (T,g), where T E S \ I and g is an 8line, produces the contradiction 7b8 = 148. Q.E.D. CLAIM 2.24. None of the points P E S \ I lies on three 2lines. PROOF: Suppose P lies on three 2lines. A fan count on P shows that P lies on a (t + 1)line. Let Q E S \ l be such that PQ is a (t + 1)line. Since P lies on three 2lines, Q lies on exactly one (t + 1)line and at most three tlines. Claim 2.23 yields t2 +5 < F(P) < (t + 1) +3t +(t 1) + 4 =5t +4, or, t < 4, a contradiction. Q.E.D. 13 Proof of Theorem 2.9 for n > 37. By Claims 2.23 and 2.24, every point P E S\l lies on exactly two 2lines. Thus, no point P E S \ I lies on more than three (t + 1) lines. If P is of type a, then a fan count on P reveals that P lies on a tline and a4 > t 1. Let Q E S \ I be of type / such that PQ is a tline. Since Q lies on two 2lines, P lies on at most two (t + 1)lines. Since 34 > t 1 and #3 > 5, P lies on t secants of strength t or t+l. A fan count on P now shows that t2+5 < F(P) < t2+6. If F(P) = t2 + 5, then all points P are of type (2,2, t, ..., t, t + 1). If F(P) = t2 + 6, then all points P are of type (2,2, t, ..., t, (t + 1), (t + 1)). In either case, a flag count of (T, g), where T E S \ I and g is a (t + 1)line, reveals a contradiction. CHAPTER 3 A BOUND FOR BLOCKING SETS OF REDEI TYPE, PART II As previously stated, the objective of Chapter 3 is to prove Theorem 2.9 for 11 < n < 35, n nonsquare. Note that this implies 4< t < 6, and; if P E S \ I, then F(P)> t2 + 5 for t = 5 or 6 and F(P)> t2 + 6 for t = 4. Again, the terminology and notation used in Chapter 2 will also be employed here. Let us also add that P, Q, R, T and U will always represent points in S \ 1, and A and B will always represent points in S n 1. CLAIM 3.1. Let P be of type a with at+2 = t. Then a1=2. PROOF: Assume, by way of contradiction, that P is of type a with at+2 = t and al # 2. By Lemma 2.6 and Claim 2.14, al = 3. Let Q be such that PQ = 11. By Claim 2.14, there is a secant g through Q such that st(g)=t or t + 1. Since (t1)t + 3 x 3 = t2 9 for t=5 or 6, and since t2 t + 9 < t2 + 6 for t=4, one has a3 > 4. Hence, Lemma 2.2 implies that st(g)=t+2, a contradiction. Q.E.D. Claim 3.1, Claim 2.14 and Lemma 2.6 imply that every P lies on a 2line. 14 15 CLAIM 3.2. Let P be of type a with at+2 = t. Set g = lt+2. Then every Q not on g lies on exactly two secants from L(g*). PROOF: This should be clear; see Claim 2.15. Q.E.D. CLAIM 3.3. Let P be of type a with at+2 = t. Then at+3i > t + 4 i for 4 PROOF: Assume, by way of contradiction, that at+3i < t +4 i for some i. Then F(P) < (i 1)t + (t + 2 i)(t + 3 i) + 2 = i2 i(t + 5) + t2 + 4t + 8. If 4 < i < t + 1, then the inequality yields the contradiction F(P) < t2 + 5. Q.E.D. CLAIM 3.4. Let P be of type a with at+2 = t. Then a2 0 3. PROOF: Assume, by way of contradiction, that there is a P of type a with at+2 = t and a2 = 3. A computation of F(P) reveals that P is of type (2,3,t,...,t) and, hence, F(P)= t2 + 5. Thus, t=5 or 6. Choose Q of type # such that PQ=12. By Claim 3.3, Q lies only on secants of strength 2, 3 or t+ 1. By Claim 3.1, /1 = 2. By Claim 3.2, 33 < 3. Since (t 2)(t +1)+3+ 8= t2+9 < t2+5 for t > 5, Q lies on t 1 secants of strength t + 1. Thus, F(Q) > (t 1)(t+ 1)+ 7=t2 +6, a contradiction. Q.E.D. 16 CLAIM 3.5. Every P lies on a (t + 1)line. PROOF: Assume, by way of contradiction, that P is of type a with at+2 = t. By Claim 3.1, al = 2. A computation of F(P) shows that a2 2 3. Claim 3.4 then implies a2 > 4. Choose distinct points Q and R on 12 not equal to P. By Claim 3.3, Q lies only on secants g 5 l2 of strength 2, 3 or t + 1. Let A be such that PA = l1. Claim 3.3 and the fact a2 > 4 imply that A lies only on secants g : I of strength 2 or 3. Let t = 4. Then P is of type (2,4,4,4,4,4) and F(P)=22. Point Q lies on exactly one 4line. Thus, as is easily seen, all points T 0 P are of type (2,3,3,4,5,5). If QA is a 3line, then the two 5lines through R can not intersect QA at points of S. Thus, A lies on eleven 2lines, a contradiction of Claim 2.16. Thus, t 5 4. Since 2(t+1) + 1 x 2 + a2 + (t 2)3 = 5t 2 + for t = 5 or 6, Q and R each lie on at least three (t + 1)lines. If A lies on a secant g of strength 3, then g passes through Q and R, a contradiction. Thus, A lies on n secants of strength 2, a contradiction of Claim 2.16. Q.E.D. CLAIM 3.6. Let P be of type a. Then a2 $ 6. PROOF: Assume, by way of contradiction, that P is of type a with a2 = 6. Since (t+1)6 + 2 = 6t + 8 > t2+2t + 2 for t = 4 or 5, t = 6 and F(P)2 45. Let Q and R be distinct points on 12 not equal to P. Since a2 = 6, Q and R lie only on secants g # 12 of strength 2, 3 or 7. Since 4x7+1x6+1x2+2x3=42<45, 17 Q and R each lie on at least five 7lines. Let A be such that PA=l1. As a2 = 6, A lies only on secants g A I of strength 2 or 3. If A lies on a 3line h, then either all the 7lines through Q or all the 7lines through R can not intersect with h at points of S, a contradiction. Thus, A lies on n secants of strength 2, a contradiction of Claim 2.16. Q.E.D. CLAIM 3.7. Let P be of type a. Then a2 Z 5. PROOF: Assume, by way of contradiction, that P is of type a with a2 = 5. Since F(P) > (t + 1)+ t x 5 + 2= 6t + 3 > t2 +2t+2 for t = 4, one has t = 5 or 6. Let Q and A be such that PQ = 12 and PA = II. Suppose t = 5. Then Q lies only on secants g # 12 of strength 2, 3 or 6. Since 2x6+1x5+1x2+3 x 3 28 < 30, Q lies on at least three 6lines. Point A lies only on secants g 6 I with st(g)=2 or 3. If A lies on a 3line h, then the 6lines through Q can not all intersect h at points of S, a contradiction, unless QA = h. As a2 > 3, there is a point R E 12 such that the 6lines through R do not all intersect h at points of S. Thus, A lies on n secants of strength 2, a contradiction of Claim 2.16. So, t = 6. Point Q lies only on secants g # 12 with st(g)=2, 3, 4 or 7. Since 3x7+1x5+1x2+3x4=40 <41, Q lies on at least four 7lines. Point A lies only on secants g Il with st(g)=2, 3 or 4. An argument similar to the one for t = 5 will show that A lies on n secants of strength 2, a contradiction. Q.E.D. 18 CLAIM 3.8. Let n 0 17 and P be of type a. Then a2 5 4. PROOF: Assume, by way of contradiction, that n 0 17 and P is of type a with a2 = 4. Let Q and R be distinct points on 12 not equal to P. Let A be such that PA = 11. Suppose t = 4. Then F(P)2 23, Q and R lie on exactly one 4line and A lies only on secants g 1 of strength 2 or 3. Since 2x5+1x4+1x2+2x3=22 < 23, Q and R lie on at least three 5lines. Let h be a 3line through A. Secant h must intersect each 5line through Q or R at a point of S. Hence, h passes through Q and R, a contradiction. Thus, A lies on n secants of strength 2, a contradiction of Claim 2.16. So, t = 5 or 6. Suppose a4 > 5. Then neither Q nor R lie on a secant g # 12 with st(g)= t 1 or t and A does not lie on a secant of strength t 1, t or t + 1. Since (t 3)(t +1)+ x4+1 x 2+3(t2)= t2+t3 for t = 5 or 6, points Q and R both lie on at least t 2 secants of strength t + 1. Any secant h through A such that 3 < st(h) < t 2 intersects each of the (t + 1)lines through Q or R at points of S. Thus, h passes through Q and R, a contradiction. Hence, A lies on n secants of strength 2, a contradiction. So, a4 = 4. Suppose t = 5. Remember that n 5 17. Since a4 = 4, F(P)=31 or 32 and point Q lies on at most three 6lines and no 5lines. As 2x6+1x4+1x2+3x4=30<31, 19 Q lies on exactly three 6lines and, hence, does not lie on a 3line. Thus, Q and P are both of type (2, 4,4,46,6,6) and F(P)=32. Let T be such that st(PT)=6. T lies on at most four 6lines and no 5lines. A computation of F(T) reveals that T lies on at least three 6lines. If T lies on four 6lines, then T is of type (2,3, 3,6,6,6,6) and some point in S \ 1 lies on at most two 6lines, a contradiction. Thus, all points in S \ I are of type (2,4,4,4,6,6,6). Counting flags (U, g), U E S \ I and g a 6line, yields the contradiction 19 x 3 = 5b6. So, t = 6. Since a4 = 4, P lies on at least three 7lines, a5 > 6 and F(P)=41 or 42. Thus, Q and R do not lie on secants of strength 3 or 6 and lie on exactly one 4line. A computation of F(Q) shows that Q and R are both of type (2,4,5,5,5,7,7,7). But, it is not possible that both Q and R lie on distinct 5lines and six secants of strength 5 or more. So, t 5 6. Q.E.D. CLAIM 3.9. Let t = 5 or 6 and P be of type a with 02 = 3. Then a3 # 4. PROOF: Assume, by way of contradiction, that t = 5 or 6 and P is of type a with a2 = 3 and a3 = 4. Let Q be of type / such that PQ = 12. Since a3 = 4, Q does not lie on a tline and lies on at most three (t + 1)lines. Suppose t=5. Since 2x6+1x3+1x2+3x4= 29 < 30, Q lies on exactly three 6lines. Since 3x6+1x3+2x2+1x4=29 <30, 20. /2 = 3. Thus, Q is either of type (2,3,3,4,6,6,6) or (2,3,4,4,6,6,6). Let R be of type 7 such that st(QR)=4. Since 35 = 6, R does not lie on a 3line. If 72 = 2, then Q lies on at most two 6lines. Hence, Claims 3.6, 3.7 and 3.8 imply that 72 = 4 and n = 17. So, Q is of type (2,3,3,4,6,6,6). Since #4 = 4, P lies on at most three secants of strength 5 or 6 and, hence, is of type (2,3,4,4,5,6,6). Since 72 = 4, all 5lines pass through R and, hence, R is of type (2,4,4,4,5, 5,6). Let T be a point on QR distinct from Q and R. T is of the same type as R. But, this is not possible since R and T both lie on 5lines and six secants of strength 4 or more. So, t = 6. A computation of F(Q) reveals that Q is of type (2,3,5,5,5,7,7,7). Thus, P does not lie on a secant of strength 5 or 6 and, hence, is of type (2, 3, 4, 4, 7, 7, 7, 7). Let A be such that PA = 11. Since P lies on four 7lines and N4 > 4, QA must also be a 2line. PA and QA being of strength 2 imply that A lies on n secants of strength 2, a contradiction of Claim 2.16. Q.E.D. Part 1. Assume 11 < n < 15. Let P be of type a. By Claims 3.7 and 3.8, c2 < 3 and, hence, F(P)< 25. So, n # 15. By the BruckRyser Theorem [12, Corollary 2.4], one need not consider n = 14. Assume n = 13. Every point P is either of type (2, 2, 5, 5, 5, 5) or (2, 3, 4, 5, 5, 5). If P is of type (2,3,4,5,5, 5) with PQ the 3line, then Q can not lie on a 4line, a contradiction. If P is of type (2,2, 5,5,5,5) with PQ a 5line, then Q is on at most three 5lines, a contradiction. Assume n = 12. Every point P is either of type (2,2,4,5, 5,5), (2,3,3,5,5,5) or (2,3,4,4,5,5). Denote by vl,v2 and v3 the number of points of each type, respectively. If P is of type (2,2,4,5,5,5) with PQ the 4line, then Q is on no 3lines and at most two 5lines. None of the point types satisfy these conditions. Thus, vi = 0. If P is of type (2,3,3,5,5,5) with PQ a 3line, then Q is of type 21  (2,3,4,4,5,5) since it is on at most one 3line. If P is of type (2,3,4,4,5,5) with PQ the 3line, then Q is of type (2,3,3,5,5,5) since it is on no 4lines. So, v2 > 1 if and only if v3 > 1. Now count: 12 = v2 + v3 4b5 = 3v2 + 2v3 3b4 = 2v3. The only solutions have either v2 = 0 or v3 = 0, a contradiction. Assume n = 11. Every P is either of type (2,2,3,5,5,5), (2,2,4,4,5,5), (2,3,3,4,5,5) or (2,3,4,4,4,5). Denote by vl, v2, v3 and v4 the number of points of each type, respectively. If P is of type (2,2,3,5,5,5) with PQ the 3line, then Q is on exactly one 3line and at most two 5lines. Thus, Q is of type (2,3,4,4,4,5). If P is of type (2,3,4,4,4,5) with Q the 3line, then Q is on no 4lines. Thus, Q is of type (2,2,3,5,5,5). So, vl = v4. Counting flags (R,g), g a 5line, yields the contradiction: 4b5 = 3vi + 2v2 + 2v3 + v4 = 2(v1 + v2 + v3 + v4) = 22. This completes Part 1. Part 2. Assume 17 < n < 24. Let P be of type a. By Claims 3.6, 3.7 and 3.8, F(P)< 35. Thus, n < 22. By BruckRyser [12, Corollary 2.4], one need not consider n = 21 or 22. Assume n = 20. Every P is of type (2,2,5,6,6,6,6) or (2,3,5,5,6,6,6). If P is of the latter type with PA a 2line, then Lemma 2.2 and Claim 2.11 imply that A lies on twenty 2lines, a contradiction to Claim 2.16. If P is of type (2,2,5,6,6,6,6) with PQ the 5line, then Q lies on at most two 6lines, a contradiction. 22  Assume n = 19. Every P must be one of four types: (2,2, 4,6,6,6,6), (2,3,3,6, 6,6,6), (2,2,5,5,6,6,6) or (2,3,5,5,5,6,6). If P is of type (2,2,4,6,6,6,6) with PQ the 4line, then Q is on exactly one 4line and at most two 6lines. None of the four point types satisfy these conditions. If P is of type (2,2,5,5,6,6,6) with PQ a 5line, then Q lies on no 3lines and at most two 6lines. None of the remaining three point types satisfy these conditions. Thus, P is either of type (2,3,3,6,6,6,6) or (2,3,5,5,5,6,6). Let v1 and v2 de note the number of points of each type, respectively. If P is of type (2, 3,3,6,6,6,6) with T # P, then T lies on at most one 3line. Hence, v1 < 1. Counting flags (R, g), g a 6line, one gets 5b6 = 4vl + 2v2 = 2vl + 38. Thus, vl = 1. Now counting flags (R, g), g a 5line, yields the contradiction 4b5 = 54. Assume n = 18. Every point P is one of five point types: (2, 2,3,6,6,6,6), (2,2,4,5,6,6,6), (2,3,3,5,6,6,6), (2,2,5,5,5,6,6) or (2,3,5,5,5,5,6). If P is of type (2,2,4,5,6, 6,6) with PQ the 4line, then Q is on exactly one 4line and at most two 6lines. None of the point types satisfy these conditions. If P is of type (2,3,3,5,6,6,6) with PQ a 3line, then Q is on exactly one 3line, at most four lines of strength 5 or 6 and at most two 6lines. None of the four remaining point types satisfy these conditions. Let P be of type (2,2, 3,6,6,6, 6) with PQ the 3line and PR a 6line. Q lies on exactly one 3line and at most two 6lines; Q must be of type (2,3,5, 5, 5,5,6). R is on no 3lines. Thus, b3 = 1. Reconsider Q of type (2,3,5,5,5,5,6) with QA the 2line. Since there is only one 3line and A does not lie on it, A lies on eighteen 2lines, a contradiction of Claim 2.16. 23  If P is of type (2,3,5,5,5,5,6) with PQ the 3line, then point Q lies on at most four lines of strength 4 or greater. Neither of the two remaining point types satisfy these conditions. Hence, all eighteen points P are of type (2, 2, 5, 5, 5, 6, 6). Counting flags (P, g), g a 6line, reveals the contradiction 5b6 = 36. Assume n = 17. Every P is of one of nine types: (2, 2, 2, 6, 6, 6, 6), (2, 2, 3, 5, 6, 6, 6), (2,2,4,4,6,6,6), (2,3,3,4,6,6,6), (2,2,4,5,5,6,6), (2,3,3,5,5,6,6), (2,4,4,4,4,6,6), (2, 2,5, 5, 5, 5, 6) or (2,4,4,4, 5, 5, 6). If P is of type (2,2,3,5,6,6,6) with PQ the 3line, then Q lies on exactly one 3line and at most two 6lines. No point type satisfies these conditions. If P is of type (2,2,2,6,6,6,6) with Q 5 P, then Q lies on exactly one 6line and at most four secants of strength 5 or 6; Q must be of type (2, 4, 4, 4, 5, 5, 6). Let R be such that st(QR)=4. R can not lie on a 5line and, hence, is neither of type (2,2,2,6,6,6,6) nor (2,4,4,4,5,5,6), a contradiction. If P is of type (2,3,3,4,6,6,6) with PQ a 3line, then Q lies on exactly one 3line. None of the remaining seven point types satisfy this condition. If P is of type (2,4,4,4,4,6,6) with Q # P, then Q does not lie on a 5line; Q is either of type (2,2,4,4,6,6,6) or (2,4,4,4,4,6,6). Denote by vj and v2 the number of points of each type, respectively. Now count: 17 = vl + v2 5b6 = 3vl + 2v2 = vj + 34 3b4 = 2v1 + 4v2 = 34 + 2v2. As is easily verified, either vi = 1 and v2 = 16, or vl = 16 and v2 = 1. In either case, v1 > 1. If Q is of type (2, 2, 4, 4, 6, 6, 6) with QR a 4line, then R lies on at most two 6lines. Thus, v2 > 4. Or, vl = 1 and v2 = 16. Thus, b6 = 7. Now, let 24  Q be a point of type (2,2,4,4,6,6,6) with QB a 4line. Without loss of generality, assume PQ is a 6line and PB a 4line. Let R and T be the points on PB not equal to P. Clearly, B is on no 6lines. Hence, each of the seven 6lines intersect PB at P, R or T. Thus, either P, R or T lies on three 6lines. Thus, vi > 2, a contradiction. Thus, there are no points of type (2,4,4,4,4,6,6). Let P be of type (2, 4, 4, 4, 5, 5, 6) with PQ a 5line. Q lies on exactly one 5line. None of the remaining five point types satisfy this condition. If P is of type (2,2,4,4,6,6,6) with PQ a 4line, then Q is on no 3lines, at most three lines of strength 5 or 6 and at most two 6lines. None of the remaining four point types satisfy these conditions. If P is of type (2,2,5,5,5,5,6) with Q 5 P, then Q does not lie on a 4line. Hence, Q is either of type (2,2,5,5,5,5,6) or (2,3,3,5,5,6,6). Denote by vi and v2 the number of points of each type, respectively. Now count: 17 = vi + v2 5b6 = 17 + v2 4b5 = 2vl + 34 b4 = 0 2b3 = 2v2 b2 =v + 17. As is easily verified, vi = 9 and v2 = 8. Thus, b6 = 5, b5 = 13, b4 = 0, b3 = 8 and b2 = 26. So, the total number of secants b = 53. Let A E S n I be such that A lies on a 6line. By Claim 2.15, r(A) = 11. As is easily checked, A is either on one 7line, one 6line, one 5line and eight 2lines, or on one 7line, one 6line, three 3lines and six 2lines. Since A is on exactly one 6line and b6 = 5, there are at least thirty 2lines, a contradiction. 25  So, P is either of type (2,2,4,5,5,6,6) or (2,3,3,5,5,6,6). Denote by v1 and v2 the number of points of each type, respectively. Counting flags (T,g), g a 6line, yields the final contradiction 5b6=2v1+2v2=34. Part 3. Assume 26 < n < 35. Let P be of type a. By Claims 3.6, 3.7 and 3.8, F(P)< 47. Thus, n < 32. By BruckRyser[12, Corollary 2.4], there is no need to consider n = 30. Assume n = 32. Then P is of type (2,3, 7, 7, 7, 7, 7, 7). Let PA be the 2line. Then by Lemma 2.2 and Claim 2.11, A must lie on thirtytwo 2lines, a contradiction of Claim 2.16. Assume n = 31. P is either of type (2,2, 7, 7, 7,7, 7, 7) or (2,3,6, 7, 7, 7,7,7). Let P be of type (2, 3,6, 7, 7, 7, 7, 7) with PA the 2line. Then A lies on thirtyone 2lines, a contradiction. Thus, all points P must be of type (2,2, 7, 7,7,7, 7, 7). But, if PQ is a 7line, then Q lies on at most three 7lines. Assume n = 29. Then P is one of four types: (2,3,5,6,7,7,7,7), (2,3,6,6,6,7,7,7), (2,2,5,7,7,7,7,7) or (2,2,6,6,7,7,7,7). If P is either of the first two types with PA the 2line, then A lies on twentynine 2lines, a contradiction of Claim 2.16. Thus, P is either of type (2,2,5, 7, 7, 7, 7,7) or (2,2,6,6,7, 7, 7, 7). But, if PQ is a 7line, then Q lies on at most three 7lines Assume n = 28. Then P is one of seven type: (2,3,5,5,7,7,7,7), (2,3,5,6,6,7,7,7), (2,3,3,7,7,7,7,7), (2,3,6,6,6,6,7,7), (2,2,4,7,7,7,7,7), (2,2,5,6,7,7,7,7) or (2,2,6,6,6,7,7,7). If P is either of the first two types with PA the 2line, then A lies on twentyeight 2lines, a contradiction of Claim 2.16. If P is of type (2,2,4, 7, 7, 7, 7,7) with PQ the 4line, then Q lies on at most two 7lines. None of the remaining five point types satisfy these conditions. 26  If P is either of type (2, 2, 5, 6, 7, 7, 7, 7) or (2, 2, 6, 6, 6, 7, 7, 7) with PQ a 6line, then Q lies on at most two 7lines and no 3lines. None of the remaining four point types satisfy these conditions. Thus, P is either of type (2, 3, 3, 7, 7, 7, 7, 7) or (2, 3, 6, 6, 6, 6, 7, 7). If P is of the former type with PQ a 7line, then Q can not lie on a 3line. Thus, all P must be of type (2, 3,6,6,6,6, 7, 7). Counting flags (T, g), g a 7line, yields the contradiction, 6b7 = 56. Assume n = 27. Then P is one of nine types: (2,3,5,5,6,7,7,7), (2,2,3,7,7,7,7,7), (2,2,4,6,7,7,7,7), (2,3,3,6,7,7,7,7), (2,2,5,5,7,7,7,7), (2,2,5,6,6,7,7,7), (2,2,6,6,6,6,7,7), (2,3,5,6,6,6,7,7) or (2,3,6,6,6,6,6,7). If P is the first type with PA the 2line, then A is on twentyseven 2lines, a contradiction. If P is either of type (2, 2, 5, 5, 7, 7, 7, 7) or (2, 2, 5, 6, 6, 7, 7, 7) with PQ a 7line, then Q is on at most three 7lines, no 5lines and at most five lines of strength 6 or 7. None of the remaining eight point types satisfy these conditions. If P is of type (2,2, 3, 7, 7, 7, 7, 7) with PQ a 7line, then Q is on no 3lines, at most three 7lines and at most five lines of strength 6 or 7. None of the remaining six point types satisfy these conditions. If P is of type (2,2,4,6, 7, 7, 7, 7) with PQ a 7line, then Q is on no 3lines and at most four lines of strength 6 or 7. None of the remaining five point types satisfy these conditions. If P is either of types (2,3,5,6, 6, 6, 7, 7) or (2,3,6,6,6,6, 7, 7) with PQ the 3 line, then Q does not lie on a 6line. None of the remaining four point types satisfy these conditions. Thus, P is either of type (2,2,6,6, 6,6, 7, 7) or (2, 3,3,6, 7, 7,7, 7). If P is of the later type with PQ a 3line, then Q is on exactly one 3line. Hence, all points P 27  are of type (2,2,6,6,6,6,7,7). Counting flags (T,g), g a 6line, yields the final contradiction, 5b6 = 108. Assume n = 26. P is one of twelve types: (2,2,2,7,7,7,7,7), (2,2,3,6,7,7,7,7), (2,2,4,5,7,7,7,7), (2,3,3,5,7,7,7,7), (2,2,4,6,6,7,7,7), (2,3,3,6,6,7,7,7), (2,2,5,5,6,7,7,7), (2,3,5,5,5,7,7,7), (2,2,5,6,6,6,7,7), (2,3,5,5,6,6,7,7), (2,2,6,6,6,6,6,7) or (2,3,5,6,6,6,6,7). If P is of type (2,2,2, 7, 7, 7, 7, 7) with PQ a 7line, then Q lies on exactly one 7line, no 3lines and at most three 6lines. None of the point types satisfy these conditions. If P is of type (2,2,3,6, 7, 7, 7, 7), (2, 3,3,5, 7, 7, 7, 7) or (2,3,3,6,6, 7, 7, 7) with PQ a 3line, then Q lies on exactly one 3line, at most two 7lines and no 5lines. None of the remaining eleven point types satisfy these conditions. If P is either of type (2, 2, 4, 5, 7, 7, 7, 7) or (2, 2, 4, 6, 6, 7, 7, 7) with PQ a 4line, then Q lies on exactly one 4line and at most two 7lines. None of remaining eight point types satisfy these conditions. If P is either of type (2, 2, 5, 6, 6, 6, 7, 7) or (2, 3, 5, 6, 6, 6, 6, 7) with PQ a 6line, then Q lies on at most two 7lines and at most five lines with strength 5 or more. None of the remaining six point types satisfy these conditions. If P is of type (2,2,5, 5,6, 7, 7, 7) with PQ a 5line, then Q lies on exactly one 5line and no 3lines. None of the remaining four point types satisfy these conditions. If P is either of type (2, 3, 5, 5, 6, 6, 7, 7) or (2, 3, 5, 5, 5, 7, 7, 7) with PQ a 5line, then Q lies on exactly one 5line. None of the remaining three point types satisfy these conditions. Hence, all twentysix points P are of type (2,2,6,6,6,6,6, 7). Counting flags (T, g), g a 7line, yields the final contradiction, 6b7 = 26. This completes the proof of Theorem 2.9. CHAPTER 4 A BOUND FOR BLOCKING SETS OF REDEI TYPE, PART III In Chapters 4 and 5, we extend by one the lower bound on the cardinality of blocking sets of R6dei type of finite projective planes of certain nonsquare orders. In Chapters 2 and 3, Claim 2.11 afforded us a certain luxury. We knew that there exists a unique iline of maximal strength. We begin by proving a result similar to Claim 2.11. The purpose of Chapter 4 is to prove the following theorem. THEOREM 4.1. Let 1 be a finite projective plane of nonsquare order n = t2 e, t > 9 and 1 < e < 2t 2. Let S be a blocking set of II of Ridei type with I S J= n + t + 3. Then there exists an unique line 1 of II such that I n S = t + 3. The proof of Theorem 4.1 will be by way of contradiction. So, assume the existence of at least two secants of strength t + 3. Remember that the notation and terminology used in Chapters 2 and 3 will also be used in Chapters 4 and 5. CLAIM 4.2. Let g be an iline with 2 < i < t + 2. If there exists a triangle of (t + 3) lines, or if all (t + 3) lines are incident with a common point on g, then 2 < i <6 and j L(g*) = (t + 2)(t+ 3 i). PROOF: Let I be an (t + 3)line and P = g n 1. Through each of the t + 2 points Q E (Sn l)\ {P} there are exactly t +3i secants h E L(g*). Every secant h E L(g*) in tersects 1 at a point Q E (Snl)\{P}. Thus, I L(g*) I = (t+2)(t+3i). Using Lemma 2.8 it is easy to complete the proof. Q.E.D.  28   29  PROOF OF THEOREM 4.1: Case 1. Assume there exists a triangle of (t + 3)lines with point P one of the vertices. By Lemmas 2.2 and 2.4, P lies on exactly t + 3 secants and F(P) < t2+2t+4. Since t2+2t+4 < (t+3)(t+3), P lies on an ilineg for some 2 < i < t+2. Claim 4.2 implies that i < 6. Let X be a point on g but not in S. Since t 3 > 6, there are no jlines through X with j > t 3. Lemma 2.5 implies that point X lies on at least three secants. Thus, (t + 2)(t + 3 i) =1 L(g*) 1> 2(n + 1 i). But, as is easily checked, this is false if t > 9 and i < 6. Therefore, there does not exist a triangle of (t + 3)lines. Case 2. All (t + 3)lines pass through a common point P. Let 6 equal the number of (t + 3)lines and let I represent one of them. Suppose the existence of an (t + $/)line h for some / with I 3 I< 2. Since t > 9, Claim 4.2 implies that the secant h does not pass through P. Let T represent the subset of points of S that are not on h or any of the (t + 3)lines; I T 1= n 6(t+ 1) + 2 p. Set Q = h n 1. Choose any one of the n +1 t f points X on h yet not in S. By Lemma 2.5, X is either of type (4 /, t + P) or lies on at least three secants. There are at most n b(t + 1) + 2 , :3 points X E h\S of type (4/, t+/) such that the (4/p)line through X also passes through P. Thus, there are at least (n+1 t3)' secants of L(h*) that intersect I at points in S\{P, Q}. There are exactly (t+1)(30) secants in L(h*) that intersect I at points in S \ {P, Q}. Thus, it must be that (t + 1)(3 8) > (n t + 1) 7. 30  Then the following inequalities must be valid: (t + 1)(3 /)2 > (n t + 1)(3 /) n + (t + 1) 2 + () (t + 1)(3 /)2 > n(2/3) (t +/ 1)(3 )+6(t +1) 2 + (t + 1)(3 3)2 > (t2 2t + 2)(2 ) (t + 1)(3 /) + 2t + / 0 > (t2 (7 #)t 1)(2 /). Thus, if 3 < 2, then 7 / > t2 1. This contradiction proves that there do not exist secants h with t 2 < st(h) < t + 1. If p = 2, then inequality (*) yields (2 6)(t + 1) > 0. Hence, 6 = 2 and I T = n 2t 2. Every point in (1 n S) \ {P, Q} lies on exactly one line of L(h*). Thus, I L(h*) I is the sum of t + 1 and the number e of lines of L(h*) through P. Every point X in h \ S is of type (2, t + 2), so n t 1 = h* =1 L(h*) I= t + 1 + . Hence, e = n 2t 2 =1 7T Thus, if there exists a (t + 2)line, then P lies on two (t + 3)lines, n t 2 secants of strength 2 and no other secants. Still under the supposition that there is a (t + 2)line h, let g represent an iline not through P with 2 < i < t 3. Let V be the subset of points of S that are not on g or the two (t + 3)lines; I V = n t i. If X E g \ S, then X is either on a 2line or a tangent through P. There are exactly n t i points X E g \ S each lying on a 2line through P and, since (t 3) + 2 2 < t + 2, Lemma 2.5 implies that each such X lies on at least three secants, one of which does not pass through P and is not g. The t + 1 points X E g \ S each lying on a tangent through P each lies on at least one secant not through P and not g. Thus, there are at least n + 1 i secants in 31  L(g*) not passing through P. There are exactly (t+1)(t+3i) secants in L(g*) not passing through P. It is straightforward to verify that n + 1 i > (t + 1)(t + 3 i) for i > 7. Hence, 2 < i < 6. If 3 < i < 6, then Lemma 2.5 implies that each of the t + 1 points X E g \ S lying on a tangent through P lies on at least two secants not through P which are not g. Again, it is straightforward to convince oneself that (n t i)+ 2(t + 1) > (t + 1)(t+ 3 i) for i > 5. So, 2 < i < 4. One last application of Lemma 2.5 shows that each of the t + 1 points X E g \ S lying on a tangent through P lies on at least three secants not through P which are not g, and each of the n t i points X E g \ S lying on a 2line through P lies on at least three secants not through P which are not g. But, 3(n + 1 i) > (t + 1)(t + 3 i) for i > 3. Therefore, if there exists a (t + 2)line, then all secants not through P are either 2lines or (t + 2)lines. Let P0 E I n S \ {P}. Then P0 lies on either t 2 or t 3 lines of strength t + 2. This implies the existence of a secant f not through P with 3 < st(f) < t + 1, a contradiction. Therefore, there are no (t + 2)lines. To complete the proof of the theorem, let g be an iline through P with 2 < i < 6. As there are no secants h such that t 2 < st(h) < t + 2, Lemma 2.5 implies that each of the n 1 + i points X E S \ I lies on at least three secants. Thus, we must have that (t + 2)(t + 3 i) =1 L(g*) 1> 2(n + 1 i). As is easily verified, this is not true for t > 9. Thus, P lies only on secants of strength t + 3. Thus, 6(t + 2) + 1 = n + t + 3. As is easily verified, 6 = t 2 or t 1. Since every secant intersects with every (t + 3)line at a point in S, there exist 6lines. But, as shown, there do not exist (t + /)lines with /3 1< 2. Q.E.D. CHAPTER 5 A BOUND FOR BLOCKING SETS OF REDEI TYPE, PART IV The purpose of Chapter 5 is to prove the following theorem. THEOREM 5.1. Let II be a projective plane of nonsquare order n = t2 e, t > 12 and 1 < e < t. Let S be a blocking set of I of Redei type. Then, I S I> n + t + 4. By Theorem 2.9, I S 1 n + t + 3. Hence, we will assume I S I= n + t + 3 and derive a contradiction. By Theorem 4.1, there exists an unique (t + 3)line which will be denoted as 1. Note that by Lemma 2.4, if P E S \ 1, then t2 + t + 5 < F(P)< t2 + 2t + 4. CLAIM 5.2. Let A be a point in S and on 1. Then A cannot lie on n secants of strength 2. PROOF: Assume, by way of contradiction, that A does lie on n secants of strength 2. Then the set S \ {A} is a blocking set of R6dei type of cardinality n + t + 2, a contradiction to Theorem 2.9. Q.E.D. CLAIM 5.3. Each point P E S \ 1 lies on a secant of strength t 1 or more. PROOF: Assume not. Then, t2 +t +5 < F(P) (t+ 3)(t 2) = t2 + t 6. Q.E.D. 33  Lemma 2.6 and Claim 5.3 imply that each point P E S \ I lies on a secant of strength 5 or less. CLAIM 5.4. Each point P E S \ 1 lies on a secant of strength t or more. PROOF: Assume not and let P be of type a with at+3 = t 1. Since al < 5, t2 +t +5 < F(P) < (t + 2)(t 1) + 5 = t2 +t 3. Q.E.D. Lemma 2.6 now implies that every P E S \ I lies on a secant of strength 4 or less. CLAIM 5.5. Let P E S \ 1 be a point of type a. If at+3 = t, then al = 2. PROOF: Assume, by way of contradiction, that P E S\ I is of type a with at+3 = t and al 2 3. Let Q E S \ 1 be such that PQ = l1. By Claim 5.4, there exists a secant g through Q of strength at least t. Since t2 + 3 x 4 a3 > 5. Since a3 > 5 and al > 3, secant g intersects with each of the t + 3 secants through P at a point in S, a contradiction. Q.E.D. CLAIM 5.6. Let P E S \ I be a point of type a. Then at+4i > t + 5 i for at+3 t + 5 < i < 2t at+3. PROOF: Assume, by way of contradiction, that at+4i < t+5i for some i. Define the constant 7 by at+3 = t + 7. By Claim 5.4, y=0, 1 or 2. Thus, F(P) < (i 1)(t + 7) + (t + 3 i)(t + 4 i) + al = = i2 i(t + 7 ) + t2 + 6t + 12 y + al. 34  If + 5 < i < t , this inequality yields the contradiction F(P) < t2 + t + 5. Q.E.D. CLAIM 5.7. Each point P E S \ l lies on a secant of strength t + 1 or t + 2. PROOF: Assume not and let P be of type a with at+3 = t. By Claim 5.5, ai=2. Since (t+ 1)t + 2 2 = t2 +t +4 one can choose point Q E S \ I of type / such that PQ = 12. Claim 5.6 implies that Q does not lie on a secant g 0 12 with 5 < st(g) < t. Since t2+2 x 3+2 =t2+8 < t2+t + 5, ac3 > 4 and, hence, Q does not lie on a secant g 5 12 of strength t + 1. That is, Q lies only on secants g # 12 with st(g) = 2, 3, 4 or t + 2. Since (t 3)(t + 2) + 5 x 4 + a2 = t2 t + 14 + a2 Q lies on at least t 2 secants of strength t + 2. This implies that a3 > t 1. Suppose a3 = t. If Q lies on fewer than t 1 secants of strength t + 2, then Q lies necessarily on at least rt[L] > 5 secants h 12 of strength at most 4. Hence, Q lies on at least (t 2) + 5 + 1 = t + 4 secants, a contradiction. Therefore, Q lies on exactly t 1 secants of strength t + 2. Computing one gets, F(P) = (t + 1)t + a2 + 2 = 2 + t + a + 2 and F(Q) = (t 1)(t +2)+ 4 + /3 + /2 +/ = t2 + t 2 + 34 + /3 + 32+ 81. 35  Or, a2 + 4 = /4 + /3 + ?2 + i1. Since a2 equals /j for some 1 < j < 4, we have a contradiction. So, a3 5 t. Suppose a3 = t1. It is straightforward to show that a2 > 4. Choose R E S\l such that PR=lt+3 and st(QR) = t + 2. As a3 = t 1, point R can lie on at most t 2 secants of strength t + 2. Since a2 > 4 and a3 = t 1, point R lies only on secants g # h1+3 with st(g) = 2, 3, 4, 5 or t + 2. Since Q lies on t 2 > 6 secants of strength t + 2, all secants h with st(h) = 3, 4 or 5 must pass through Q. Thus, R lies only on secants g 0 1t+3 with st(g) = 2 or t + 2. Thus, t2 +t + 5 < F(R) < (t2)(t +2)+t +4 x 2 t2+t +4, a contradiction. Q.E.D. Lemma 2.6 now implies that every point P E S \ I lies on a secant of strength 2 or 3. CLAIM 5.8. Let P E S \ 1 be of type a. Then c4 > 3. PROOF: Assume not and let P E S \ l be of type a with 4 = 2. Let Q E S \ l be of type # such that PQ = 16. It is easily shown that a5 = t + 1 or t + 2 and that P lies on at least t 2 secants of strength t + 2. Thus, Q lies only on secants g $ 16 with st(g)=2, t 1 or t. If Q lies on fewer than t + 1 secants g 0 16 with st(g) > 2, then t2 + t + 5 < F(Q) < 4t + (t 4)(t 1) + a +2 x 2= = t2 + 10. Thus, Q lies on one (t + 2)line, one 2line and t + 1 secants of strength t or t 1. Choose an R E S \ I different from Q such that PR = 16. Clearly, R must also lie 36  on t + 1 secants of strength t or t 1. But, #2 > t 1 and Lemma 2.2 imply that R can not lie on a secant of strength t or t 1, a contradiction. Q.E.D. CLAIM 5.9. Let P E S \ I be of type a. Then a3 > 3. PROOF: Assume, by way of contradiction, that there is a point P E S \ I of type a such that a3 = 2. By Claim 5.8, there is a point Q E S \ I of type 3 such that PQ = 14. Claim 5.6 implies that Q does not lie on a secant g 9 14 with 7 < st(g) < t 2. Since (t 2)(t+ 2)+2 x5 +3 x 2 =t2 +12 a5 > 6 and, hence, Q does not lie on a secant g 14 of strength t 1. Since 5(t + 2)+ (t 6)(t 1) + 4 + 3 x 2 = t2 2t + 22 +a4 < t2 t + 5, at2 2 t and, hence, Q does not lie on a secant g / 14 of strength 5 or 6. Thus, Q does not lie on any secants g # 14 with 5 < st(g) < t 1. Since a3 = 2, any (t + 2)line must pass through P. Thus, t2 + t + 5 < F(Q) < 3(t + 1) + (a5 4)t + 4 + (t + 3 a5)4 = = a5(t 4) + 3t + 15 + a4, so a5 > t + 1. Suppose a5 = t + 2. Then Q lies only on secants g 5 14 with st(g) = 2, t or t+1. Also, note that a5 = t +2 implies the nonexistence of 3lines or 4lines, except 37 for possibly 14. If Q were to lie on fewer than t + 1 secants g $ 14 with st(g) > t, then Q would necessarily lie on at least t 1 secants of strength 2 and, hence, as is easily shown, F(Q) < + t + 5. Thus, Q lies on t + 1 secants g 0 14 of strength t or t + 1, secant 14 and one 2line. Let A E S n I be such that st(QA) = 2. As A is not on 14, there are no 3lines or 4lines through A. As 82 > 3 and 83 > t, there are no secants g through A with 5 < st(g) < t + 2. Thus, A lies on n secants of strength 2, a contradiction of Claim 5.2. So, 05 $ t + 2. Thus, a5 = t + 1. Then Q lies only on secants g # 14 for which st(g) = 2, 3, t or t + 1, and there are no 4lines with the possible exception of 14. The equality a5 = t + 1 implies that there are at least two secants h : 14 through Q of strength 2 or 3. If Q lies on fewer that t secants g # 14 with st(g) > t, then Q must lie on at least [Ft1 > 6 secants h 0 14 of strength 2 or 3 and, hence, as is easily shown, F(Q) < t2 + t + 5. Thus, Q lies on t secants g 5 14 with st(g) = t or t + 1, secant 14 and two secants of strengths 2 or 3. Thus, t2 + t + 5 < F(Q) < 3(t + 1) + (t 3)t + 6 + 4= t2 + 9 + r4, so a4 > t 4 > 5. Let R E S\l be of type 7 such that R 14. Since a5 = t+1 and c4 > t4 > 5, point R lies only on secants g # PR such that st(g) = 2, 3, t or t + 1. If R lies on fewer than t secants g # PR with st(g) > t, then R necessarily lies on at least [rf > 5 secants h $ PR of strength 2 or 3 and, thus, F(R) < t2 + t + 5. By the contradiction, a4 = t + 1. Thus, Q lies on t + 1 secants of strength t or t + 1. As Q lies on at most four (t + l)lines, one has that /3 = t. Choose U E S \ 1 such that st(QU) = t. Since /3 = t, U lies only on secants g $ QU for which st(g) = 2, 3, 4, 38 t + 1 or t + 2. Since a3 = 2, U lies on at most one (t + 2)line and at most four secants of strength t + 1 or t + 2. Thus, F(U) < (t + 2) + 3(t + 1) + t + (t 3)4 + 3 = 9t 4 < t2 + t + 5. So, a5 t + 1. This contradiction completes the proof of Claim 5.9. Q.E.D. CLAIM 5.10. Let P E S \ l be of type a. Then al + a2 > 6. PROOF: By way of contradiction, assume the existence of a point P of type a with al + a2 < 5. By Claim 5.9, there exists a point Q E S \ I of type / such that PQ = 13. Claim 5.6 implies that Q does not lie on a secant g 13 with 7 < st(g) < t 2. Since (t 2)(t + 2) + 3 x 5 + a + 2 t2 + 16 < t2 + t + 5, a5 > 6 and, hence, all secants g 9 13 through Q of strength t 1 or more intersect with 15 at points of S. Thus, Q does not lie on a secant g 3 l3 of strength t 1. As 4(t + 2) + (t 3)(t 1) + 5 = t2 + 16 < t2 + t + 5 and 5(t + 2) + (t 4)(t 2) + 5= t2 t + 23 at_1 > t and at2 > t 1, respectively. Hence, Q also does not lie on secants g $ 13 of strengths 5 or 6. Thus, Q does not lie on a secant g 0 13 with 5 < st(g) < t 1. 39  Note that al +a2 5 implies that Q lies on at most two (t+2)lines. (If st(13) = t+2, then F(P) > t2 + 2t + 4.) Thus, t2 + t + 5 < F(Q) < 2(t + 2) + (a5 3)(t + 1) + (t + 3 a5)4 + 3 = = ag(t 3) + 3t + 13 + a3. So, a5 > t. Note that if a5 = t, then a3 > t 8 > 4, and since Q lies on a secant of strength less than 4, one has t2 +t + 5 < F(Q) < 2(t + 2) + (t 3)(t +1)+2 x 4+a3 + 3= = t2 + 12 + a3. So, if a5 = t, then a3 > t 7 > 5. Suppose a5 = t + 2. Except for possibly 12,13 and 14, there are no 3lines or 4lines. Thus, Q lies only on secants g # 13 with st(g) = 2, t, t + 1 or t + 2. If Q lies on fewer than t + 1 secants g # 13 with st(g) > t, then Q necessarily lies on at least t 1 secants of strength 2 and, hence, F(Q) < t2 + t + 5. So, Q lies on t + 1 secants g 4 13 with st(g) > t. If a4 > 5, then these t + 1 secants g # 13 would each intersect 14 at a point of S, implying that a4 = t + 2. But, a4 = t + 2 implies that F(P) > t2 + 2t + 4, a contradiction. So, if a5 = t + 2, then a4 < 4. Let R E S \ I be such that PR = 15. Point R lies only on secants of strengths 2, t 1, t, t + 1 or t + 2. If R does not lie on an (t 1)line, then since a4 < 4 one has that F(R) < 3(t + 2) + (t + 1)+t +(t 2)2= 7t + 3 40  Let g be a (t 1)line. There are n t 2 points X E g \ S such that X 0 lj, j = 1,..., t + 3. Since there are no 5lines and all 3lines and 4lines pass through P, Lemma 2.5 implies that each of the points X is of type (2,2, 2,2, t 1). Since 4(n t 2) > 2n, there is some point U E S \ I which lies on at least three 2lines. This contradiction of Claim 5.8 implies that a5 : t + 2. Suppose a5 = t. Remember that a3 > 5. Let R E S \ 1 be of type y with PR = 14. Together, a5 = t and a3 > 5 imply that R does not lie on a secant g # 14 with 5 < st(g) < t. If R were to lie on fewer than t 1 secants g # 14 with st(g) > t + 1, then F(R) < 2(t + 2) + (t 4)(t + 1) +4 +3 x 4+3 = = t2 t + 15 + a4 <2 +t+5. Thus, R lies on exactly t 1 secants g : 14 with st(g) 2 t + 1. Thus, a3 = t. Note that R must lie on three secants h that are of strength 4 or less. Let U E S \ 1 be such that PU = 14, yet U Z R. Then U as R lies on exactly t 1 secants g = 14 with st(g) > t + 1. Thus, (73 1) + (72 1) + (71 1) > t 1, a contradiction since y3 < 4. Thus, a5 7 t and, so a5 = t + 1. Except for possibly 13 and 14, there are no 4lines. Thus, Q lies only on secants g # 13 with strengths 2, 3, t, t +1 or t + 2. This fact and the equality a5 = t +1 imply that Q lies on at least two secants h 5 13 with strengths 2 or 3. If Q lies on fewer than t secants g $ 13 with st(g) > t, then Q necessarily lies on at least FtL] 6 secants with strength 3 or less and, hence, as is easily shown, F(Q) < t2 + t + 5. So, 41 Q lies on exactly t secants g 5 13 with st(g) > t. If a4 < 4, then Q lies on at most three secants of strength t + 1 or more and, hence, F(Q) < 2(t + 2) + (t + 1) + (t 3)t + 10 = t2 + 15 Thus, a4 > 5. By Lemma 2.2, the t secants g 13 through Q of strength t or more intersect with 14 in points of S. Thus, a4 = t + 1 and Q lies only on secants g 0 13 of strengths 2, 3, t + 1 or t + 2. Assume, by way of contradiction, the existence of a secant g 0 13 of strength t. Choose a point R E S \ of type 7 such that R (12 U 13 U g). As a4 = t + 1, R lies only on secants of strengths 2, 3, t, t + 1 or t + 2. As with Q, point R lies on t secants h 0 PR of strength t or more. Since st(PR) t + 1, y3 > t. This inequality yields the contradiction that g meets t + 1 of the secants through R in points of S. Hence, there are no tlines, except for possibly 13. Thus, any point R E S \ I such that R 12 U 13 lies on t +1 secants of strengths t + 1 or t + 2. This yields the contradiction t2 + 2t +4 > F(R) > (t+ 1)(t + 1) + 2 x 2= t2 + 2t +5. Thus, a5 5 t + 1. Therefore, al + a2 > 6. Q.E.D. CLAIM 5.11. There are no 3lines and no (t + 1)lines. PROOF: Assume, by way of contradiction, that there does exist a secant of strength 3. By Claim 5.10, there is a point P E S \ I of type a with c1 = 3. Let Q E S \ I be such that PQ = 11. Point Q lies on a secant g of strength t + 1 or t + 2. Since 42  ac = 3, all secants of strength t + 2 pass through P. It is clear that a3 = 3, for else st(g) > t + 3 by Lemma 2.2. If at+3 = t + 2, then the n t 1 points X E t+3 \ S are each of type (2, t + 2). Thus, one of the n t 1 points in S \ (It+3 U 1) lies either on two 2lines or a 2line and a 3line, both not possible by Claim 5.10. Thus, at+3 = t + 1 and there are no secants of strength t + 2. Claim 5.6 now implies that Q does not lie on secants g l1 such that 6 (t 2)(t + 2) + 2 x 4 + 9 =t2 + 13 4(t+2) + (t1)(t+2)+9= t2 t + 21 < t2 + 5 and 3(t + 2) + (t 3)t 9 = t2 +15 a5 5, at_ > t and at 2 t + 1. Hence, Q also does not lie on a secant g : 11 of strength t, 5 or 4. Thus, Q lies only on secants of strength 3 or t + 1. Thus, Q must lie on t secants of strength t + 1 and three secants of strength 3. Thus, a4 = t + 1. Let R E S \ I be of type 7 with PR = 14. Point R lies only on secants of strength 2, 3, t or t + 1. Also, R lies on at least two secants of strengths 3 or less since at+3 = t + 1. If R lies on fewer than t secants g # 14 with st(g) > t, then R must lie on at least [rt1] 6 secants of strength 2 or 3, which is easily shown to imply F(R) < t2 + t + 5. Thus, R lies on t + 1 secants of strength t or more. Thus, either y1 + 72 < 5 or 71 = 72 = 3 and y3 > 3, both of which have been eliminated as possibilities. Thus, there are no 3lines. Since there are no 3lines, there are no (t + 1)lines. For otherwise, if g is a (t+ 1)line, then the n t points X E g\S are each of type (2, 2, t+1) and, thus, one of the n t points P E S\ (U g) lies on at least two 2lines, a contradiction of Claim 5.10. Q.E.D. 43  CLAIM 5.12. Let P E S \ I be of type a. Then, a2 > 7. PROOF: Assume, by way of contradiction, that there is a point P E S \ I of type a with a2 = 4, 5 or 6. By Claims 5.7 and 5.11, al = 2. Let A E S n 1 be such that PA = 11. Let Q E S \ I be of type / such that PQ = 12. Claim 5.6 implies that Q does not lie on a secant g # 12 with 7 < st(g) < t 2. Since 5(t + 2) + (t 4)(t 2) + 8 = t2 t + 26 < t2 + t + 5, at2 > t 1 and, hence, Q does not lie on a 6line g l12. Thus, Q lies only on secants g : 12 of strength 2,4,5, t 1, t or t + 2. (Remember, there are no 3lines or (t + 1)lines.) Computations of F(P) show that arg 7 and a5 > 5. Thus, t2 + t + 5 < F(Q) < (a6 1)(t + 2) + (t + 2 r6)5 + 2 + 2 = = a6(t 3) + 4t + 10 + a2. So ag > t 1. Suppose ac = t 1. Then Q lies on at most t 2 secants of strength t + 2. If Q lies on fewer than t 2 secants of strength t + 2, then F(Q) <(t3)(t +2) + +3 x5+a2+2=t2t + 11 + +a 2, where x = t if r2 = 4, x = t 1 if a2 = 5 and x = 5 if a2 = 6. In any case, F(Q) < t2 + t + 5. Thus, Q lies on exactly t 2 secants of strength t + 2. A computation of F(Q) shows that Q lies on at least three 5lines. Let g = 12 be a 5line through Q. As a2 > 4, there exists R E (S \ 1) U 12 distinct from P and Q. The t 2 > 10 secants of strength t + 2 through R must intersect with g at points 44 of S, clearly a contradiction since st(g)=5. Since there are no (t + 1)lines, a6 = t or t + 2. Suppose a6 = t + 2. Then A does not lie on a secant g with 3 < st(g) < t 2. Since a2 > 2, A does not lie on an (t + 2)line. There are no (t + 1)lines. Assume the existence of a secant g through A of strength t 1 or t. Since a6 = t + 2, all 4lines and 5lines, if any, pass through P. There are no 3lines. Thus, each of the points X E g \ S is either of type (2, 2, 2, 2, t 1) or (2, 2, 2, t), respectively. Thus, P lies on at least four 2lines, a contradiction of Claim 5.8. Thus, point A lies on n secants of strength 2, a contradiction of Claim 5.2. Thus, a6 Z t + 2, so a6 = t. Point A lies only on secants g 5 l1 of strength 2, 4, t 1 or t. Assume the existence of a tline g through A. Since 05 > 5, a2 = a3 = a4 = 4. Since 4(t + 2) + (t 5)t + 3 x 4 + 2 = 2 t + 22 there are at least five (t + 2)lines through P. Thus, all 4lines pass through P and each point X g \ S is of type (2,2,2,t). Hence, P lies on four 2lines, a contradiction of Claim 5.8. Assume the existence of a (t 1)line g through A. Then a5 = 5 and as is easily shown, there are at least five (t + 2)lines through P. Thus, all 4lines pass through P. Since ag = t, all 5lines also pass through P. Hence, each of the points X E g \ S is of type (2, 2, 2, 2, t 1). Hence, P lies on five 2lines, a contradiction of Claim 5.8. 45  Assume the existence of a 4line g through A. Then there are at most three (t + 2)lines through P. A computation of F(P) shows that a4 > 6. Let R E S \ I be such that PR = 13. Since a4 > 6, t2 + t + 5 < F(R) < (a2 1)(t + 2) + (t + 2 a2)4 + 03 + 2= = a2(t 2) + 3t + 8 + a3, a contradiction since a2 < 6. Thus, A lies on n secants of strength 2, a contradiction of Claim 5.2. Therefore, a2 > 7. Q.E.D. Therefore, there are no secants of strength 3, 4, 5 or 6. PROOF OF THEOREM 5.1: Let P E S \ I be of type a. Then aI = 2 and a2 > 7. Let A E S n I be such that PA = 11. Claim 5.6 and a2 7 imply that A lies on n secants of strength 2, a contradiction of Claim 5.2. Q.E.D. CHAPTER 6 SMALL BLOCKING SETS OF HERMITIAN DESIGNS, PART I Motivation for the next two chapters began with a paper by D. Jungnickel [15]. In his work Jungnickel introduces the idea of a self blocking block design. Given any block design 2 that admits blocking sets, form the collection C(2) of all committees of E. (A blocking set of minimum cardinality is called a committee.) If C(Z) is itself a block design, where the points of C(E) are the points of E and the blocks of C(E) are the committees of 27, then E is said to be self blocking when the committees of the block design C(E) are the blocks of E. That is, E is selfblocking if C(C(E)) = E. Let q be a prime power, K = GF(q) the finite field of q elements and V a 3dimensional vector space over K. One can form a finite projective plane H in the following manner: Let the 1dimensional subspaces of V be the points of H and the 2dimensional subspaces of V be the lines of H. As is well known, H is of order q, is denoted by PG(2, q) and is called the Desarguesian projective plane of order q. The points of H = PG(2, q) can be represented by homogeneous triples (x, y, z); that is, (x, y, z) can be thought of as a 1 x 3 matrix where (x, y, z) = (kx, ky, kz) for all nonzero k E K and x, y and z are not simultaneously zero. The lines of 7 can be represented by the transpose (a, b, c)t of a homogeneous triple. Then the point P = (x, y, z) lies on line 1 = (a, b, c)t if and only if ax + by + cz = 0; that is, the "dot product" is zero. [12, Chapter II]  46  47  In Jungnickel's paper [15] it is shown that for q a prime power, the projective planes H = PG(2, q2) are selfblocking; the committees of HI are its Baer subplanes. Inside of H are substructures called Hermitian unitals which are block designs and, though not committees, are blocking sets. The collection of all Hermitian unitals forms a block design, call it E. The original motivating question for Chapters 6 and 7 was for which values of q, if any, are the committees of Z the lines of the plane. In order to address that question we will need the following definitions. A collineation a of the plane H = PG(2, q), q a prime power, is a map of the set of points of H onto itself such that point P lies on line 1 if and only if point a(P) lies on line a(l); that is a preserves incidence. (Here, a(l) denotes the set of points a(Q), Q a point on 1. It is easily verified that a(l) is a line of I.) Denote by AutH the group of all collineations of H. The projective subgroup PGL(3, q) of AutH is the set of all collineations that can be represented by nonsingular 3 x 3 matrices A with entries from the Galois field GF(q). That is, if P = (x, y, z) is a point of 7, then the collineation a represented by A is defined by a(P) = (x, y, z)A. (It is easy to convince oneself that for nonzero k E K, kA and A represent the same collineation.) [12, Chapter II] For positive integer n, a unital is a 2(n3 + 1, n + 1,1) block design. For odd prime power q, set H = PG(2, q2). The plane H possesses substructures which are 2(q3 + 1, q + 1, 1) block designs; that is, there are unitals inside H. Let U represent a unital of H. Each line of H is either tangent to U, intersecting with U in exactly one point, or is a secant intersecting with U in exactly q+ 1 points [13, Chapter 6.3]. Thus, every unital of H is a blocking set of H. Denote by U(q) the (nonempty) set of all unitals of H = PG(2, q2), q an odd prime power. The plane 7 admits the doubly transitive automorphism group G = PGL(3, q2), and G preserves unitals, that is, elements of G map unitals to unitals. Thus, U(q) is itself a block design 48  where its points are the points of H and its blocks are the unitals of H. Call U(q) the unitals design of H. [12, Theorem 2.49] A correlation a of a (finite) projective plane f is a map of the set of points of 7 onto the set of lines of f, and vice versa, such that point P lies on line 1 if and only if point a(l) lies on line a(P). A polarity a is a correlation of order 2; that is, a2 is the identity collineation. [12, Chapter 11.6] Let H = PG(2,q2), q an odd prime power. A unitary polarity a of 17 is a polarity that can be represented by a 3 x 3 nonsingular Hermitian matrix H with entries from F = GF(q2) in the following manner. (The matrix H = (hi), hij E F, is Hermitian if and only if hij = hq. That is, H = (Hq)t.) The point P = (x, y, z) is sent by a to the line H(xq, yq, zq)t and the line I = (a, b, c)t is sent to the point (aq, b, cq)Hl. Point P = (x, y, z) of H is said to be an absolute point of a if and only if P lies on line a(P); that is, if and only if (x, y, z)H(xq, yq, zq)t = 0. Similarly, line I = (a, b, c)t is said to be an absolute line of a if and only if (aq bcq)Hl(a, b, c)t = 0. The absolute points and nonabsolute lines of a unitary polarity form a unital U, a 2(q3 + 1, q + 1, 1) block design. (The absolute lines of a unitary polarity a are called tangents, because they intersect with U in exactly one point.) Call such a unital a Hermitian unital and denote by H(q) the set of all Hermitian unitals of H. Since the projective group G = PGL(3, q3) preserves (Hermitian) unitals and its unitary subgroup, the subgroup of G that fixes some Hermitian unital U, has order (q3 + 1)q3(q2 1), it is routine to verify that H(q) is a block design with parameters v = q4 + q2 + 1, k = q3 + 1 and A = q4(q2 1). Call H(q) the Hermitian design of H. [12, Chapter II.8] For the remainder of this chapter, let q denote an odd prime power. Then H will denote PG(2, q2) and C will denote a committee of H(q). Also, let F be the 49  Galois field GF(q2) and K the Galois field GF(q). For real number r, denote the smallest integer greater than or equal to r by fr]. In Chapter 7 it is shown that for q = 3 the committees of the Hermitian design H(q) are the lines of H; that is, C(H(3))=7H. Initially it was hoped to extend this result and show that C(H(q))=nH for all values of q. No progress was made in this direction. After contacted by my advisor D. Drake, A. Blokhuis demonstrated that this extension was unattainable. [4] (An outline of Blokhuis' argument will be given later in this chapter.) So, are there any values of q other than 3 for which C(H(q))=H? For which values of q is it true that C(H(q)) is not equal to H? In attempting to answer these questions, a lower and upper bound are found on the cardinality of a committee C of H(q). The following definitions and Lemmas 6.1 through 6.5 will be needed to es tablish a lower bound on the cardinality of C and to show in Chapter 7 that C(H(3)) =PG(2,9). Let V be a vector space over K=GF(q) of dimension 2d. A spread of V is a set of qd + 1 ddimensional subspaces V1,..., Vqd+l of V such that V n Vj = {0} for i # j. [12, Exercise 7.7] The field F=GF(q2) is a 2dimensional vector space over K. Since there are exactly q + 1 1dimensional subspaces of F over K, there is a unique spread El of F over K. Let L[a, b] denote the set {x E F I (ax) + ax + b= 0,b E K and 0 : a E F}. LEMMA 6.1. The collection Z1 := { L[a,0]j a E F,a # 0 } is a spread of F over K. PROOF: Since (x + y)q = xq + yq for all x,y E F, and since L[a,0] $ F, L[a,0] is a 1dimensional subspace of F over K. The polynomial Xq + X has q 1 non zero roots in F. If r is one such root, then given any nonzero c E F, one has 50  c E L[rc1,O]. Clearly, 0 E L[a,0] for all nonzero a E F. Thus, all vectors c E F lie in some L[a, 0]. As 1dimensional subspaces are either equal or intersect only at 0, Z1 is a spread of F. Q.E.D. The Desarguesian affine plane of order q can be represented as the collection of costs of the (unique) spread Z1 of F over K. Henceforth, E will denote the affine plane of order q formed from the costs of 1E. LEMMA 6.2. Every set L[a, b] is a coset of the subspace L[a, 0] and, hence, a line of the afine plane Z, and; every coset of L[a, 0] is of the form L[a, b]. Further, lines L[a,b] and L[c,d] are parallel if and only if ac1 EK; in particular, L[a,0]=L[c,0] if and only if ac1 EK. PROOF: For any nonzero a E F, the mapping fa from F into K defined by fa(x) = (ax)q + ax is linear and onto. Thus, there exists c E F such that (ac)q + ac = b. (Remember, b E K.) For x E L[a, 0], one has (a(x + c))q + a(x + c) + b= (ax)q + ax + (ac)q + ac + b = 0+0 = 0. Thus, c + L[a,0] C L[a,b]. If x E L[a,b], then (a(x c))q + a(x c) = (ax)q + ax (ac)q ac = b (b) = 0. Thus, (x c) E L[a, 0], or L[a, b] C c + L[a, 0]. Hence, L[a, b] = c + L[a, 0], or L[a, b] is a coset of L[a, 0]. If d + L[a, 0] is a coset of L[a, 0], then it is straightforward to check that d + L[a, 0] = L[a, b], where b = (ad)q + ad. For all nonzero a, the line L[a, 0] contains the point 0. The line L[a, b] is parallel to the line L[a, 0] for all b. Thus, L[a, b] is parallel to L[c, d] if and only if L[a, 0] = L[c, 0]. Assume ac1 E K. Then, (ac1)q = ac1. Thus, (cx)q + cx = 0 if and only if acl((cx)q + cx) = 0 if and only if (ax)q + ax = 0. So, L[a, 0] = L[c, 0]. 51  Assume L[a,0] = L[c,0]. If 0 $ x E L[c,0], then L[c,0] = { ax a E K}. It is straightforward to verify that ac1x E L[c, 0], so ac1 E K. Q.E.D. For c and e elements of F, denote by H[c, e] the Hermitian unital represented by the following matrix. 0 0 c A= 0 1 e cq eq eq+1 Since A is nonsingular, c is nonzero. LEMMA 6.3. The point (r,s, 1) of the plane HI is in H[c,e] if and only if c is on the line L := L[r,f(e,s)], where f(e,s) := (e + sq)q+1; line L contains the point 0 if and only if e = sq. PROOF: The point (r, s, 1) is in H[c, e] if and only if 0 = (cr)q+cr+eq+l + (es)q + es + sq+l= (cr)q + cr + (e + sq)q+1 if and only if c EL[r, (e + sq)q+l]. Clearly, L contains the point 0 if and only if f(e, s) = 0, if and only if e = sq. Q.E.D. Henceforth, let f(e, s) := (e + sq)q+l LEMMA 6.4. The lines (0,0,1)t and (1,0,0)t of I are tangent to H[c,e] at the points (1,0,0) and (0, e, 1), respectively. PROOF: If the matrix A represents H[c, e], then the proof is simply a matter of matrix multiplication once A1 has been determined. Note that x+l1 + eqxq + ex + eq+1 equals (x + e)q+1 Q.E.D. 52  LEMMA 6.5. [14, Theorem 1', p.257] Let V be a vector space of dimension n over a finite field K with q elements. Then any covering of the nonzero elements of V with hyperplanes not containing zero must consist of at least n(q 1) hyperplanes. We are now in a position to establish a lower bound on the cardinality of a committee C of H(q). PROPOSITION 6.6. ICI > 2q + 2 for q > 3. PROOF: Assume, by way of contradiction, that ICI 5 2q+ 1. Choose distinct lines g and h from H that maximize ICn(g U h)l. Thus, ICn (g U h) > 4. Since 2q+ 1 < q2, there are distinct points P E g and T E h not in C so that the line I = PT has empty intersection with C. Coordinatize H so that P = (1,0, 0), Q = g n h = (0, 1, 0) and T = (0,0,1). Denote the i points in C\(gUh) by Rj = (r, sj, 1), 1 < j < i < 2q3. Note that for all j, the elements rj and sj are nonzero. Lemma 6.4 implies that for every c in F, the lines g and h are tangent to the unital H[c, 0] at the points P and T, respectively. Since sj 0 for all j, Lemma 6.3 implies that none of the j lines L[rj, f(0, sj)] contains the element 0 of F. Thus, by Lemma 6.5, the i < 2q3 lines L[rj, f(O, sj)] do not cover the nonzero elements of F. Hence, there exists a nonzero element c in F such that the unital H[c, 0] has empty intersection with C. This contradicts the fact that C is a blocking set of H(g). Q.E.D. There will now be presented three arguments for upper bounds on the cardinal ity of C. The bounds will be successively lower. The last argument is independent of the first two. The first argument is an outline of the Blokhuis' argument mentioned earlier. 53  LEMMA 6.7. Consider four points in standard position and the six lines joining them. (That is, there are four distinct points, no three collinear.) Then these six lines cannot all be tangents of the same Hermitian unital U. PROOF: Label the four points as (1,0,0), (0, 1,0), (0,0, 1) and (1, 1, 1). This com pletely determines the representations of the six lines. If U is represented by the Hermitian matrix A and 1 = (a, 6, c)t is any one of these six lines, then I is tangent to U if and only if (a, bq, cq)Al(a, b, c)t = 0. Simple matrix computations will show that the six lines can not all be tangents. Q.E.D. As is well known, a Baer subplane r of I intersects with each line I of I either in q + 1 points or in exactly one point.[12, Theorem 3.7] If F intersects with I in q + 1 points, then F n 1 is called a Baer subline of 1.[16, Definition 1] LEMMA 6.8. [15, Lemma 2] The lines of H that intersect with a Hermitian unital in q + 1 points do so in a Baer subline. Let I be an incidence structure and P a point of I. The internal structure Ip of I is defined as the set of points of I minus P and the blocks of I that contain P, where the incidence relation of Ip is that of I. A 3(q2 + 1, q +1, 1) design is called an inversive plane of order q. Denote this design by A(q). The blocks of such designs are called circles. An inversive plane can be characterized as those (finite) incidence structures I such that the internal structure Ip is an affine plane for every point P in I.[10, Chapter 6.1] LEMMA 6.9. [7, Lemma 3.1] The Baer sublines of a line I of H form an inversive plane. 54  Blokhuis then made the following count. Let A(q) be the inversive plane formed from the Baer sublines of line 1 of II. Let n be any positive integer and T any collection of n circles of A(q). In two ways count the ordered pair (T, d), where d represents a circle of A(q) that has disjoint intersection with every circle of T. It is straightforward to show, see [10, Chapter 6.1], that A(q) contains exactly q3+ q circles and each circle is disjoint from exactly q(q1(q2) circles. If a represents the average number of circles having disjoint intersection with every circle in T, then one gets the following. = (q3 + q) (3 3q2 + 2q ( 1)n Clearly, a < (q3 + q)2n. That is, if n > log2(q3 + q), then a < 1. Hence, there exists a hitting set of A(q) with at most ( := (q + 1)([log2(q3 + q)]) points. Lemma 6.7 now implies that there exists a blocking set of H(q) that contains at most 6/ points. Thus, 63 is an upper bound on the cardinality of C. Since 6/ is approximately 18q log2 q, and since it is clear that 18q log2(q) < q2 + 1 for large enough q, for at least these large enough values of q, C(H(q)) / II. This completes the outline of Blokhuis' argument. The second argument con sists of two minor improvements to Blokhuis' argument. LEMMA 6.10. Given that 6 is a primitive root of F = GF(q2), the four concurrent lines (1,0, 0)t, (0,1, 0)t, (1, 1, 0)t and (6, 1, 0)t can not all be tangent to the same Hermitian unital U. PRooF: Let U be represented by the Hermitian matrix A. The inverse A1 of A is also Hermitian. Set A1 = (aij). Assume by way of contradiction that the four lines are tangent to U. Simple computations show that all = 0, a22 = 0, a12 = a21 55  and a21(b 6q) = 0. If a21 = 0, then A1 is singular. If 6 8q = 0, then 6 is not primitive. In either case there is a contradiction. Q.E.D. For any line I of H, let A(q) be the inversive plane formed from the Baer sublines of 1. Call any set of q + 1 distinct points of A(q) a pseudo circle. Let n be positive integer and T be a collection of n pseudocircles. In two ways count (T, d), where d represents a circle having disjoint intersection with each of the pseudocircles of T. If a represents the average number of circles having disjoint intersection with every pseudocircle of T, then the following is true. a = (q3 +q) x +1 ( (2+1))n Define R as follows. R q+1 (q+1 Then there exists a hitting set of A(q) containing at most e := (q + 1)n + [aL points. (Here, [aj denotes the greatest integer less than or equal to a.) Lemma 6.10 now implies that there exists a blocking set of H(q) containing at most 4e points. Thus, 4e is an upper bound on the cardinality of C. It is easily seen that 4c < 63, so that this is a smaller upper bound. If one chooses integer n = [ logR(q3 + q)], then one has a < 1. It is now straightforward to verify that 4((q + 1)( [ logR(q3 + q)]) < q2 + 1 for q > 47. Thus, C(H(q)) can not equal H for q > 47. In fact, by carefully choosing n, it can be shown that C(H(q)) can not equal H for q > 25. This completes the second argument. 56  The third and sharpest argument begins with another personal communication from A. Blokhuis. In his communication Blokhuis credits T. Szonyi for pointing out the survey paper by Z. Furedi from which the following lemma by L. Lovasz was taken. [11] LEMMA 6.11. [11, Corollary 6.29] Let G be a hypergraph and denote its maximum degree by r. If 7 represents the covering number of G and r* represents the fractional covering number of G, then r < (1 + + ... + )r* < (1 + log(r))7*. A hypergraph G is just an incidence structure where the points are called vertices and the blocks are called edges. The hypergraph to which this lemma will be applied is the design H(q). The number r equals the maximum degree of a vertex (point) in G; that is, if r(p) denotes the number of edges (blocks) passing through vertex (point) p, then r := max r(p), the maximum taken over all vertices p of G. All points of H(q) have the same degree. The covering number 7 is the minimum cardinality of a hitting set of G. To define r*, the fractional covering number of G, start by letting t represent any real valued function defined on the set of vertices (points) of G such that t(p) > 0 for all points p in G and the sum Et(p), taken over all points in edge (block) E, is greater than or equal to 1 for all edges (blocks) E in G. If Itl := Et(p), the sum taken over all points p in G, then the fractional covering number 7* := minltJ, taken over all t. PROPOSITION 6.12. Cl < r((q4 + q2 + 1)(q3 + 1)1)(1 + log(q7 q3))] for q > 5. PROOF: Set 7 = (q4 + q2 + 1)(3 + 1)1). Recall that the block design H(q) has parameters v = q4 + 2 + 1, k = q3 + 1 and A = q4(q2 1). It is straightforward to verify that each point has degree r = q7 q3. Define the constant function t on the v points of H(q) by t(p) = where p is a point of H(q). Then 7* < Itl = = 7. 57  By Lemma 6.11, there is a hitting set S of H(q) with at most b := [7(1 + logr)] points. Since 6 < q3 + 1 for q 2 5, S is actually a blocking set of H(q) for q > 5. Q.E.D. It is easily verified that 6 < q2 + 1 for q 2 25. It is just as easy to verify that for q = 23, [(1 + +... + )7] < 232 + 1. Hence, by Lemma 6.11, there are blocking sets of the block design H(q) with fewer than q2 + 1 points for q 2 23. Thus, for q 2 23, C(H(q)) # II. CHAPTER 7 SMALL BLOCKING SETS OF HERMITIAN DESIGNS, PART II In Chapter 6 it was shown that C(H(q)) 0 PG(2, q2) for q an odd prime power greater than or equal to 23. In this chapter we prove the following. THEOREM 7.1. C(H(3))= PG(2,9). The notation and terminology employed in Chapter 6 will also be used here. Define F:=GF(9) as the polynomial ring GF(3)[X] modulo the ideal generated by the irreducible polynomial X2 + 1. The polynomial X + 1 is a primitive root of F. Set = X + 1. Let C be a committee of the block design H(3). Since every line 1 of H = PG(2, 9) intersects with every unital U of 7, and as 1 does not contain a unital U, it is clear that C contains at most 10 points. If C contained fewer than 10 points, then clearly there would exist a blocking set B of H(3) with exactly 10 points such that the 10 points are not linear. So, we will assume that B is a blocking set of H(3) with exactly 10 points, the 10 points not collinear. It will be shown that such a blocking set B can not exist. Hence, it will follow that the committees of H(3) are the lines of H. LEMMA 7.2. For any line g of H, g n AB < 8. PROOF: Assume, by way of contradiction, that Ig n BI = 9. Let P be the point on g not in B, Q any point on g in B and T the point in B not on g. Coordinatize 58 59 H such that P = (1,0,0), Q = (0,1,0) and T = (0,0,1). Then for any 0 e EF, Lemma 6.5 implies that the unital U[c, e] is disjoint from B, a contradiction. Q.E.D. LEMMA 7.3. Given distinct lines g and h of H, (g U h) n BJ < 5. PROOF: Assume, by way of contradiction, that there are two distinct lines g and h such that n := (g U h) n B\ > 6, where n is maximum in the sense that n > I(gl U hl) n 81 for any pair of distinct lines gl and hl. Without loss of generality assume Ig n Bi > Ih n BI and set Q = g n h. If n = 10, then by Lemma 7.2 one can choose points P on g and T on h not equal to Q and not in B. Coordinatize H so that P = (1,0,0), Q = (0,1,0) and T = (0, 0, 1). By Lemma 6.4, any unital U[c, 0] has disjoint intersection with B, a contradiction. Hence, n < 9. For 1 < i < 10 n, denote by Ri the points in B \ (g U h). Assume that n > 7. Let P be any point on g not in B. There are 9 lines other than g passing through P. Since 9 > Ih n BI + (10 n), one can choose a point T on h so that the line 1 := PT is disjoint from B. Coordinatize H so that P = (1, 0, 0), Q = (0,1,0), T = (0,0,1) and Ri = (ri, si, 1), 1 < i < 10 n. Since none of the points Ri lie on the line 1 = (0, 1, 0)t, not only is ri nonzero but so is si for all i. Thus, none of the lines Li := L[ri, f(O, si)] of the affine plane E contain the point 0. Since n > 7 implies that 10 n < 3 < 4, Lemma 6.5 implies that there exists a nonzero c E F not covered by the lines Li. Thus, none of the points Ri lie in the Hermitian unital H[c, 0]. Thus, by Lemma 6.4, H[c, 0] has disjoint intersection with B, a contradiction. Hence, n = 6. Since n = 6 implies that igl Bj < 4, and since we are assuming IhnBI < IgnBj, one can choose points P on g and T on h not in B and not equal to Q such that 60  P, T, and R4 are collinear; denote this line by I. Since n is maximum there is at least one Ri not on 1; without loss of generality assume i = 1. Coordinatize H so that P = (1,0,0), Q = (0,1,0), T = (0,0,1) and Ri = (ri,si, 1), i = 1,...,4. Note that sl 5 0 and s4 = 0. For any given e in F, represent the line L[ri, f(e, si)] by Li, i = 1, ...,4, and define Je = L1 U ... U L4. By Lemma 6.3, for each of the 5 values of e not equal to 2sM, 2s3, 2s or 0, the point 0 of E is not contained in any of the lines Li. Thus, for these 5 values of e, no set of three of these lines Li constitute a parallel class of the affine plane E. Assume that three or more of the lines Li are in the same parallel class A. Let e be any of the 5 values not equal to 2s 2s 2s or 0. Then the parallel lines constitute at most two lines of A since no line Li contains the point 0. Thus, for each of these 5 values of e, the set Je contains at most 7 points of E. Thus, for each of these 5 values of e, there is a nonzero value c(e) in F so that the unital U[c(e), e] does not contain any of the four points Ri. As IhnBJ < 5, Lemma 6.4 implies that there is a unital U[c, e] which has disjoint intersection with B, a contradiction. Hence, no three of the lines Li are parallel. Assume that the four lines Li are pairwise nonparallel. Again, let e be any of the 5 values not equal to 2s", 2s 2s or 0. If for any of these 5 values of e three of the four lines Li form a triangle, then the set Je contains at most 7 points of E. If for any of these 5 values of e the 4 lines are concurrent, then since none of the lines Li contains the point 0 two of the lines must be equal and, hence, Je again contains at most 7 points of E. In either case, for each of these 5 values of e there is a nonzero c(e) in F so that U[c(e), e] contains none of the points Ri. Since IhnBI < 5, Lemma 6.3 implies the existence of a unital which has disjoint intersection with B, a contradiction. 61  Thus, two of the lines Li are parallel but no third line is in the same parallel class as these two. Let e = 0. Since the point 0 of E is on L4 and no three of the lines Li are parallel, the set J0 contains at most 7 nonzero points of E. Thus, there is a nonzero c(0) in F such that U[c(0), 0] does not contain any of the points Ri. By Lemma 6.4, U[c(0), 0] has disjoint with B, a contradiction. Q.E.D. LEMMA 7.4. Given any line g of I, Ig n BI < 2. PROOF: Assume, by way of contradiction, that IgnBI > 3. By Lemma 7.3, IgnBI = 3. Label the points in B\g as Ri, i = 1,...,7. Also by Lemma 7.3, there are (7) = 21 distinct lines RiRj, i 0 j. Thus, since 21 > 2 x 10, there is a point QEg such that Q lies on 3 of these lines RiRj. Without loss of generality assume that R2R3, R4R5 and R6R7 are these lines. Again by Lemma 7.3, there is an integer 2 < i < 5 so that R1Ri intersects g and R6R7 at points P and T, respectively, not in B. Without loss of generality assume i = 2. Coordinatize H so that P=(1,0,0), Q=(0,1,0), T=(0, 0, 1), R5=(,1, 1) and Ri=(ri, si, 1) for i = 1,...,4. Then g=(0, 0, 1)t and I := R6R7 = (1,0,0)t. Note that sl = 0, s2 = 0, r2 = r3, s3 f 0, r4 = 1 and s4 # 0. Set r = r2. For any given e in F, define Li := L[r, f(e, s)] for i = 1,...,4, L5 := L[1, f(e, 1)] and Je=L1U...UL5. By Lemma 6.2, L4 and L5 are parallel, and L2 and L3 are parallel. Assume that r EK. Then by Lemma 6.2, the lines L2, L3, L4 and L5 are in the same parallel class A. By Lemma 6.3, for the 5 values of e not equal to 0, 2s3 2s4 or 2, the lines L2, L3, L4 and L5 do not contain the point 0 of E and, hence, constitute at most 2 lines of A. For these 5 values of e, the line L1 also does not contain the point 0 by Lemma 6.3. Thus, for each of these 5 values of e, the set Je contains at most 7 points of E. Hence, for each of these 5 values of e, there is a 62  nonzero c(e) in F so that the unital U[c(e), e] does not contain the points R1,...,R5. Since II n BI < 5, Lemma 6.4 implies the existence of a unital U[c(e), e] which has disjoint intersection with B, a contradiction. Thus, r 1K. Assume rliK. As is easily checked, the points a2 and a6 of the affine plane E are on the line L[1,0]. (Recall that a is a primitive root of F as defined at the beginning of this chapter.) Let e = 0. Since r and ri are not in K, Lemma 6.2 implies that neither L1 nor L2 contains the point a2 or a6. Since s4 # 0, neither does L4 nor L5. Since r K, L3 can not contain both a2 and a6. Thus, the set Jo does not contain the 8 nonzero points of E. Hence, there is a nonzero c(0) in F so that the unital U[c(0),0] has disjoint intersection with B, a contradiction. So, rl EK, and since rl # 0 or 1, must equal 2. This implies that L1, L4 and L5 are in the same parallel class A. If s4 = 1, then L1, L4 and L5 constitute exactly 2 lines of A and the point 0 is on L1. Since L3 is not in A and the point 0 is on L2, the set Jo contains at most 7 nonzero elements of F. As above, this implies the existence of a unital U[c(0),0] which has disjoint intersection with B. Hence, sA = 2. To summarize: rl EK, r K, s4 = 2, the 3 lines L1, L4 and L5 are in the same parallel class A and the 2 lines L2 and L3 are in a parallel class P distinct from A. We now want to show that there exists 3 values of e such that for each value the set Je does not contain the 8 nonzero points of E. This would imply that for these 3 values of e there exists three distinct nonzero values c(e) in F such that the unitals U[c(e),e] do not contain the points R1,...,R5. As l\{Q} contains at most 2 points of B, Lemma 6.4 would then imply the existence of a unital U[c(e), e] which has disjoint intersection with B, a contradiction. First, we want to prove the existence of 7 values of e such that the lines L1, L4 and L5 constitute at most 2 lines of A. For each of the 6 values of e not equal to 2s4, 2 or 0, none of these 3 lines contain the point 0 of the affine plane E. Hence, 63  for these 6 values of e, the 3 lines constitute at most 2 lines of A. If e = 2, then 2e4 = (e + s3)4 and L1=L4 unless (2 + s3)4 = 2. If (2 + s3)4 = 2, then for e = 2s, since s = 2, one has that 2e4 = 1. So, (e + 1)4 = (2s + 1)4 = (s + 2)4 = 1 and, hence, Li=L5. Thus, there are 7 values of e so that L1, L4 and L5 constitute at most 2 lines of A. Let H denote the set of these 7 distinct values of e. Second, we want to prove the existence of 5 values of e so that the lines L2 and L3 are either equal or one of them contains the point 0 of E. Clearly, L2 contains the point 0 if e = 0 and L3 contains the point 0 if e = 2sA. Since e4 = (e + s3)4 for values of e equal to As, as3 or a3s3, the lines L2 and L3 are equal for these 3 distinct values of e. Let G denote the set of these 5 distinct values of e. The set HnG contains at least 3 values. For each of these values of e, the set Je contains at most 7 nonzero points of E. Thus, we have proved Lemma 7.4. Q.E.D. Therefore, the 10 points of B constitute an oval of H=PG(2,9). The proof of the following lemma will therefore complete the proof of Theorem 7.1. LEMMA 7.5. For q an odd prime power, an oval e of I =PG(2,q2) is not a blocking set of H(q). PROOF: As is well known, an oval in T can be represented as a conic with coefficients from F=GF(q2). [17, Theorem 1] It is also well known that all conics in Q are projectively equivalent. [12, Theorem 2.36] Let w be a primitive root of K=GF(q) and define F as the polynomial ring K[T] modulo the ideal generated by T2 w. Let a = aT2 + b be a primitive root of F, a and b in K. 64 Case 1. Assume q is congruent to 3 modulo 4. Since all conics are projectively equivalent, represent 0 by the equation X2 + Y2 Z2 = 0; that is, a point P=(x, y, z) from IF is also in 0 if and only if x2 + y2 az2 = 0. Let U be the Hermitian unital represented by the equation Xq+1 + yq+l + bZq+1l 0; that is, point P=(x, y, z) is in U if and only if xq+l + yq+1 + bzq+l = 0. Assume that point P=(x, y, z) is in OnU. If z = 0, then x2 = 1 and x29q = 1. But, since q is congruent to 3 modulo 4, x2 = 1 implies xq+l = (x2) = 1. 2_1 So, z = 1. If y = 0, then x2 = a and x+1 = b imply a 2 = 1, contrary to a being primitive. So, y # 0. Similarly, x : 0. Set x = cT+d and y = eT+ f, where c, d, e and f are in K. Then 2 y2 = a implies d2f+(c2+e2)w = b, and xq+l +y+l = 6 implies d2+f2 (c2+e2)w = b. Adding one gets d2 + f2 = 0. As q is congruent to 3 modulo 4, d = 0 = f. Thus, x2 + y2 = a is an element of K, a contradiction. Case 2. Assume q is congruent to 1 modulo 4. Represent O by X2 Y2 aZ2 = 0 and U by Xq+l + yq+l bZq+l = 0. Suppose that the point P = (x, y, z) belongs to the set OnU. If z = 0, then x2 = 1 and xq+l = 1, clearly impossible. So, z = 0. If y = 0, then x2 = a and xq+l = b g21 imply a 2 = 1, contrary to a being primitive. Thus, y # 0. Similarly, x 5 0. As above, set x = cT+d and y = eT+f. Then x2y2 = a implies d2f2+(c2e2)w = b, and Xq+1 + yq+1 = b implies d2 + f2 (c2 + e2)w = b. Subtracting one gets that w = (1)2 is a square in K, a contradiction, unless c = 0 = f. If c = 0 = f, then 2 y2 = a is an element of K, another contradiction. Q.E.D. To summarize, in Chapters 6 and 7 it has been shown that C(H(3)) =PG(2,9) and C(H(q)) OPG(2,q2) for q > 23. CHAPTER 8 FINAL REMARKS As this dissertation does not lend itself to a conclusion or comprehensive sum mary, let the author end it by commenting upon possible future research related to the work done here. Because Theorem 4.1 is true for t > 9 and 1 < e < 2t 2, it would be desirable to have a theorem similar to Theorem 5.1 but with these weaker restrictions on t and e. The author believes that such a theorem does exist and is at present trying to show it. A. Blokhuis and A. E. Brouwer have shown that if q is odd, greater than 7 and not 27, then any blocking set of PG(2,q) has cardinality at least q + (2q)z + 1. [5] Is there such a result for all projective planes? As a first step, one might consider blocking sets of R6dei type. In Chapter 6 the large difference between the lower and upper bounds given on the cardinality of a committee of H(q) is unpleasant, but it appears that closing this gap is difficult. The author and others have worked on it with no success. The author has also tried to find something significant to say concerning the size of a committee in H(5), but also with no success. 65 REFERENCES 1. J. Bierbrauer, On minimal blocking sets, Arch. Math. 35 (1980), 394400. 2. J. Bierbrauer, Blocking sets of maximal type in finite projective planes, Rend. Sem. Mat. Univ. Padova 65 (1981), 85101. 3. J. Bierbrauer, On blocking sets of order 16 in projective planes of order 10., Preprint (1982). 4. A.Blokhuis, Personal Communication (1990). 5. A. Blokhuis and A. E. Brouwer, Blocking Sets in Desarguesian Projective Planes, Bull. London Math. Soc. 18 (1986), 132134. 6. A. Blokhuis and T. Szonyi, Personal Communication (1991). 7. R. H. Bruck, "A Survey of Combinatorial Theory," NorthHolland Publishing, Amsterdam, 1973. 8. A. Bruen, Baer subplanes and blocking sets, Bull. Amer. Math. Soc. 76 (1970), 342344. 9. A. Bruen, Blocking sets in finite projective planes, Siam J. Appl. Math 21 (Nov., 1971), 380392. 10. P. Dembowski, "Finite Geometries," SpringerVerlag, New York, 1968. 11. Z. Furedi, Matchings and covers in hypergraphs, Graphs and Combinatorics (1988), 115206. 12. D. R. Hughes and F. C. Piper, "Projective Planes," SpringerVerlag, New York, Heidelberg and Berlin, 1973. 13. D. R. Hughes and F.C. Piper, "Design Theory," Cambridge University Press, Cambridge, England, 1985.  66  67  14. R. E. Jamison, Covering finite fields with costs of subspaces, J. Comb. Theory, Series A 22 (1977), 253266. 15. D. Jungnickel, Some selfblocking block designs, Preprint (1987). 16. R. Metz, On a class of unitals, Geometriae Dedicata 8 (1979), 125126. 17. B. Segre, Ovals in a finite projective plane, Canad. J. Math. 7 (1955), 414416. BIOGRAPHICAL SKETCH Cyrus Kitto was born December 28, 1947, in Los Angeles, California. He re ceived an undergraduate degree in liberal arts from Rollins College in 1970. He received a master's degree in mathematics from the University of Florida in 1987.  68  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 thesis 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 thesis for the degree of Doctor of Philosophy. n Casagr e professor Linguistics 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 thesis for the degree of Doctor of Philosophy. Beverly L.Brechner 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 thesis for the degree of Doctor of Philosophy. Jean A. Larson 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 thesis for the degree of Doctor of/hil6phy, /Jorge Martinez Professor of Mathematics This thesis was submitted to the Graduate Faculty of the Department of Mathematics in the College of Liberal Arts and Sciences and to the Graduate School and was accepted as partial fulfillment of the requirements for the degree of Doctor of Philosophy. December, 1991 Dean, Graduate School UNIVERSITY OF FLORIDA 3 1262 08285 433 1 