This paper is concerned with fundamental questions lying at the boundary of combinatorics, geometry, and group theory. For example, let A be the point-line incidence matrix of "PG(3, q). This question turns out to be intimately related to geometrical properties of certain line sets in , and to studies of the orbit structure of the collineation group P ¸(n#1, q) of . Academic Press

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We introduce techniques to obtain some new results on blocking sets and to extend some known ones. CHARACTERIZATION THEOREMSNotation. As usual, for any set X, its cardinality is denoted by IX I. For any 2 sets A, B, A -B denotes the set of elements of A not in B. In this section, D denotes a design with parameters v, b, r, k, A, where A = 1. This implies that each line of D has exactly k points and, dually, that through each point of D there are exactly r lines of D. For notational convenience we put r = m + 1, k = n + 1. S will denote a blocking set in D.By definition, this means that S is a set of points of D such that each line of D contains at least one point in S and at least one point not in S. We put I S I = t. Associated with each blocking set S is the set J of certain triples of D. J is defined as follows: J= {(X~, Xj, P) I X~,Xj~S, X~ ~ Xy, P~D -Sand X~, Xj, P collinear}. The strength of a line l of D is I I c~ S [ = the number of points of S on/. An ~-line is a line of strength ~. LEMMA 1. Let x be any integer such that x >1 t-r+ 1. Then [J[ ~< (v -t)x(x -1). If equality occurs then x = g = t -r + 1, where g is the maximum strength of the lines of D. Moreover, all lines of D then have strength 1 or strength g. Proof. For each point P E D -S, we compute the contribution at P, I/PI to the set J. HereJp = {(X~, xj, e)~l X~, Xj~S, X~ ~ Xj and X~, Xj, Pcollinear}. Let g be the maximum strength of any line of D through P and let l be a line of strength g, with P ~ l. Suppose there exists another line 1' through P with strength g', with 2 ~< g' ~< g. Then a larger value of IJPI would be obtained if we 'removed' a point from I' and placed it on L We could continue this process, bearing in mind that each line of D must contain at least one point of S, since S is a blocking set. Thus the largest value that [JPI could have is (t -r + 1)(t -r), which could only occur when all the lines through P save for one contain precisely one point of S; the remaining line through P then contains (exactly) t -r + 1 points of S. In summary, IJp[ ~< (t -r + 1) x (t -r). Since [J[ = Y~p IJp[, the result follows. Geometriae Dedicata 6 (1977) 193-203. All Rights Reserved Copyright ~ 1977 by D. Reidel Publishing Company, Dordrecht-Holland

A blocking set S in a projective plane w is a subset of the points of 7T such that every line of ir contains at least one point of 5 and at least one point which is not in S. Denoting the number of points in S by \S\ our main result, obtained by purely combinatorial means, is the following: If ir is finite of square order, say m 2 then | S| ^m 2 +m + 1 and if | S\ = ra 2 +m + l then the points of S are the points of a subplane of ir of order m (a Baer subplane). In this connection we first of all prove the following THEOREM. Baer subplanes form blocking sets.PROOF. Suppose ir is a plane of order m 2 which contains a subplane S of order m. Since any line of ir contains at most m-\-l points of S we have that every line of ir contains at least one point which is not in 5. Let I be any line of w and P be any point of I which is not in S. Then there is at most one line of S through P, S being a subplane. Also since any two points of T are connected by a unique line, the m 2 +m-\-l points of 5 are contained in the ra 2 + l lines of ir through P. If I contained no point of 5, the lines of w through P would account for at most (m + l) + (w 2 -1) -1 = m 2 +m points of S. Thus / must contain at least one point of S establishing our theorem.We now proceed to the main result, ir denotes a plane of order n and 5 is a blocking set in ir. S-l denotes all those points P such that P is contained in 5 but not in Z, and | 5 -l\ means the number of such points P; similarly for / -5, \l -S\. LEMMA 1. No line of TT contains more than \S\-n points of S.PROOF. Let / be any line of ir and suppose / contains exactly / points of S. Since S is a blocking set there is at least one point R in l -S. There are n lines of ir through R besides /, each containing at least one point of 5. Thus always | S\ ^t+n.LEMMA 2. Let a objects be packed into b boxes such that each box contains at least one object, with b^a<2b. Define a function ƒ on the objects X as follows:A MS Subject Classifications. Primary 5070, 5060; Secondary 0527.

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