RAYMOND HILL

P A C K I N G P R O B L E M S I N G A L O I S G E O M E T R I E S

O V E R GF(3)

ABSTRACT. A set of kind s in the Galois space S~,q is a set of points such that any s + 1 are linearly independent but there is at least one subset ofs + 2 linearly dependent points. The packing problem is that of finding m[,~, the largest size of set of kind s in S,,q. The main result is the evaluation of m~,a for all s ~< r ~< 5. Some partial results bounding m~,s are also given.

1. INTRODUCTION

Let Sr.a denote the projective space of dimension r over GF(q), the Galois field o f q elements. The points o f Sr.q are writ ten as (r + 1)-tuples (al, a2, . . . , a t+l) where a, ~ GF(q) and not all the a~'s are zero; (al, a 2 , . . . , a t+l) = (bl, b2, . . . , b~+l) if and only ifa~ = Ab~ for all i, for some A ~ GF(q), ~ # O. A k-set of kind s, abbrevia ted to kr~q, in S~,a is a set o f k points such that any s + I distinct points are linearly independent but there is at least one subset o f s + 2 linearly dependent points. We denote by rn~.a the largest size of set o f kind s in S~,q. The packing p rob lem is tha t o f finding this value, which is impor t an t in the theory of factorial designs in statistics (see Bose [2]) and also in the theory of error-correct ing codes (see Section 2). Since rn~,q is just the number (q~+l _ 1)/q - 1 of points in St.q, we assume always that 2 ~< x < r. F o r s = 2, the following results o f Bose [2] are well known (see, e.g., [9]):

(1.1) m~.2 = 2 r (1.2) m 2 2,q = q + 1 f o r q o d d (1.3) m 2 2,q = q + 2 for q even (1.4) rn~,q = q2 + 1 for all q # 2 (proved by Qvist for q even). For s = r = 3 and 4, Segre [10] has shown that (1.5) 3 m3,q = q + 1 f o r q o d d />5 (1.6) m ~ , q = q + l f o r q o d d >5 .

The following result o f Tallini [11] gives the value of m~,q when q is small relative to r and when s is sufficiently close to r,

(1.7) m ~ , q = r + 2 f o r q - 1 ~< r a n d [ q ( r + 1 ) - 1]/q+ 1 <<.s<<.r.

The only other values of m~,q known to the au thor are:

(1.8) m~,3 = 20 (Pellegrino [7]) (1.9) m~.3 = 56 (Hill [4]).

For an al ternative p r o o f of (1.9) which also gives the uniqueness o f a 56~,a, see Hill [5].

Geometriae Dedieata 7 (1978) 363-373. All Rights Reserved Copyright © 1978 by D. Reidel Publishing Company, Dordrecht, Holland

364 RAYMOND HILL

In this paper (§4) we will determine the remaining values of m~,a for s ~< r ~ 5, thus obtaining the following table.

Values of m~.3

r 2 3 4 5

2 4 3 10 5 4 20 11 6 5 56 13 12

Fig. 1

The results for s = 2 and for s = r have been mentioned above. The values of m~,~ ~ are established in Theorem 4.1, and that of m~,3 in Theorem 4.7. Of importance in proving the latter is the classification of all sets of kind 3 in $4,3, which is given in Theorem 4.6. We also make strong use of combina- torial properties of the 11~,8 and the fact that its automorphism group is the 4-transitive Mathieu group Mll. Thus the methods are rather special to this case and are unlikely to generalize easily to other values of q. Finally, in §5, we summarize the known results about the packing problems in $6,8.

2. EQUIVALENCE OF k-SETS AND CODES OF SETS OF KIND S

Let K b e a k~,~. A matrix A of K is a k x (r + 1) matrix over GF(q) whose rows are the points of K. Clearly, any permutation of the rows, or multiplica- tion of a row by a non-zero scalar, yields a matrix of the same k~,q. Let PGL(r + 1, q) denote the group of non-singular linear transformations of Sr.q. We regard the elements of PGL(r + 1, q) as non-singular (r + 1) x (r + 1) matrices over GF(q) which transform (r + 1)-tuples in Sr.q via right multiplication. Given a subset K = (al, a2, • •., ak} of S~,q and g EPGL(r + 1, q), we denote by Kg the set (alg, a2g . . . . , akg). Since non-singular linear trans- formation preserves the kind s property, it follows that if K is a k~,q, and g ~ PGL(r + 1, q), then Kg is also a k~,q. We say that/£1 and K2 are equivalent if and only if Klg = K2 for some g ~ PGL(r + 1, q). Since the operation of g on K corresponds to a sequence of elementary column operations, of types (C1), (C2) and (C3) below, on a matrix of K, we have

THEOREM 2.1. Two matrices represent equivalent k~,~'s i f and only i f one can be obtained from the other via a sequence of operations of the following types:

(C1) multiplication of a column by a non-zero scalar,

P A C K I N G PR O B L E MS IN G A L O I S G E O M E T R I E S 365

(C2) interchange of two columns, (C3) addition of a scalar multiple of one column to another, (R1) multiplication of a row by a non-zero scalar, (R2) interchange of two rows.

I f K i s a set of points in ST,~, we define its automorphism group, denoted Aut K, to be the subgroup {g : Kg = K) of PGL(r + 1, q).

An (n, d)-code over GF(q) is just a d-dimensional subspace of the n- dimensional vector space V(n, q). I t is convenient in this paper to write the points of a code as column vectors. A generator matrix for an (n, d)-code C is an n x d matrix whose columns are a basis for C. I f K i s a k~,q with matrix A, we define a code C of K to be the (k, r + 1)-code with generator matrix A. Two codes are said to be equivalent if one can be obtained from the other by a permutation of the coordinate indices combined with multiplication of some (or no) coordinates by a non-zero scalar. Thus two matrices generate equiv- alent codes if and only if one can be obtained from the other via operations (C1), (C2), (C3), (R1) and (R2), giving

T H E O R E M 2.2. Suppose CI and C2 are codes of the k-sets K1 and K2 respectively. Then CI and C2 are equivalent i f and only i f K1 and K2 are equivalent.

Let C be a (k, r + 1)-code. The weight of a point x of C is defined to be the number of non-zero entries of x and is denoted w(x). We denote by z(x) the number, k - w(x), of zero entries of x. The minimum weight of Cis the smallest of the non-zero weights of all vectors of C. The dual code C z of C is the (k, k - r - 1)-code consisting of all vectors y in V(k, q) such that (x, y ) = 0 for all x in C, where (x, y ) denotes the natural inner product of x and y. Denoting by rl, r2 . . . . . r~ the rows of a generator matrix A of C, a vector

y = Y2 is i n C ± if and only if y~r~ + y 2 r 2 + . . . + y e r e = O .

Hence a vector of weight w in C j- corresponds to a set of w linearly dependent rows of A, and we have

T H E O R E M 2.3. A (k, r + 1)-code C is the code of a k~,q if and only i f C ± has minimum weight s + 2.

Thus m~.a is the maximum value of n for which an (n, n - r - 1)-code (for a channel capable of transmitting q symbols) will correct [(s + 1)/2] errors and detect s errors.

An important observation in our theory is that the code C of a k~,q itself has bounded minimum weight:

366 R A Y M O N D H I L L

T H E O R E M 2.4. Let K be a k~,q with code C. Then the minimum weight of C is at least k - m~_ 1,q.

Proof. Let x be any non-zero vector in C and A a generator matr ix of C having x as its first column. Since the rows of A form a k~,q, and since any subset o f a set o f kind s is o f kind at least s, it follows that those rows o f A having a zero as first coordinate fo rm a z~_ 1.~, where z = z(x) = k - w(x), and t 1> s. Hence

z ~< m,~_l,~

and so

w(x) >1 k - - m r _ l , q.

3. T w o UPPER BOUNDS ON m~.q

There are two useful general upper bounds on m~,q; the first is the well-known packing bound (see, e.g., Barlott i [1, Theorem 6.3.1]), and the second a recursive bound.

T H E O R E M 3.1. The number of points k in a k~,q satisfies the following inequalities.

(i) qr+t~> 1 + ( k ) ( q - 1)

(ii) qr+l>~ 1 + ( k ) ( q - 1)

+ . . . + ( k ) ( q - 1 ) ~ / f s = 2 u - 1

+ " ' + ( k ) ( q - 1 ) ~ + ( k - l ) ( q -

if s = 2u.

T H E O R E M 3.2. , ~< ~- i mr ,q m r _ t , q + 1. Proof. Let K be a k~,q. By Theorem 2.1, we may assume K has a matr ix

i00 0 i] A = B

where B is a (k - 1) x r matrix.

The rows of B form a (k - 1)~-~,q, for if some s rows of B were linearly dependent, then the corresponding rows of A together with the first row of A would be linearly dependent , cont rary to the hypothesis that K is a set o f kind s. Thus k - 1 ~< m~-~,q, and the result follows.

P A C K I N G PROBLEMS IN GALOIS GEOMETRIES 367

4. DETERMINATION

It follows from (1.7) that r m ~ , 8 = r + 2 for

Indeed, it is easy to see that any

1 0 0 . . . 0 0 0 1 0 . . . 0 0

0 0 0 . . . 1 0 0 0 0 . . . 0 1 1 1 1 . . . 1 1

and that this is a maximal set of kind r.

8 OF mr,a FOR S ~< r ~< 5

r~>2.

(r + 2)~,a is equivalent to

T HEOR EM 4.1. (i) m3,82 = 10 (ii) 3 = 11 m 4 , 3

(iii) m s = 12 5,3

Proof. (i) is immediate from (1.4). Hence Theorem 3.2 gives 11 and 12 as upper bounds on m~,3 and m~,8

respectively. But an l l~,a is given explicitly by Barlotti [1]. Moreover, this can be extended to a 12~,8, a set described geometrically by Coxeter [3] and Todd [12], having the Mathieu group 3112 as its automorphism group,

A matrix representing the 12~,s is given in Figure 2; an 11~,3 is obtained by omitting the first row and last column.

1 0 0 0 0 2 2 2 1 0 1

0 0 0 0 1

0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 2 2 1 0 1 1 2 0 1 2 2 0 2 2 2 0 2 1 2 2 2 1 1 1 2 2 2 2 1 0

Fig. 2

It remains only to find the value of mg,s and this case is much more difficult than the above since the value turns out to be 13, which is substantially lower than the upper bounds given by Theorems 3.1 and 3.2. We make strong use of the following Theorems 4.2 and 4.6, which classify all sets of kind 3 in $4,3.

368 RAYMOND HILL

THEOREM 4.2. There is just one 11~.8 up to equivalence. Proof. Let K be an 11~.8 and C its code. Then by Theorem 2.3, C l is an

(11, 6)-code with minimum weight 5. But Pless [8] has shown that the only (11, 6)-code over GF(3) with minimum weight 5 is the perfect Golay code. (For an alternative proof, see [5].) Thus C = (C±) l is unique, and so too is K by Theorem 2.2. Also Aut(ll~,3) is the Mathieu group Mll, which acts 4-transitively on the points of 11~,3.

Remark 4.3. Theorem 4.2 is used several times to shorten the proof of Theorem 4.6, but its use can be bypassed, so that the uniqueness of the 11~,8 follows alternatively from the fact that the unique 10~,3 is contained in just one 11~,3.

LEMMA 4.4. Suppose each of K1 and K2 is a k~,3 with k >1 7 and that each is contained in an 11~,3. Then KI and K2 are equivalent.

Proof. By the equivalence of all ll~,3's, we may suppose that K~ and K2 are each contained in the same 11~,3 which we will denote by K. Since k 1> 7 and Aut K is 4-transitive on K, there exists g ~ Aut K such that (K\K~)g = K\K2 and hence Klg = K2, as required.

LEMMA 4.5. Let K be a k~.8 which contains the 'fundamental points' (10000), (01000), (00100), (00010) and (00001). Then

O) any further point o f K has at most one zero entry, (ii) there is at most one point with all its entries non-zero,

(iii) no two further points can have a zero entry in common.

Proof (i) is immediate from the kind-3 property. (ii) and (iii). If a and b were points each having all entries non-zero, or

each having just one zero in common, then either a + b or a - b would be dependent on two of the fundamental points, giving a set of four dependent points.

THEOREM 4.6. For each k with 6 <~ k <<. 11, there is just one k~,3, up to equivalence.

Proof Since 3 m3,3 = 5, any k~,3 with k /> 6 contains some five linearly independent points and by Theorem 2.1 we may assume that the first five rows r~, r2 . . . . . r5 of K are the fundamental points. By Lemma 4.5(i) we may assume that r6 = (11111) or (11110), for given any other possibility for r6, there are obvious column operations of types (CI) and (C2) which send it into one of these two rows and fix the set (r~, r2 . . . . . rs}.

We consider the cases k = 6, 7, 8, 9 and 10 in turn. (i) k = 6. If r6 -- (11111), then Kis of kind 4, and so there is just one 6~,3

up to equivalence. (ii) k = 7. First we observe that there is just one 7~.3 which contains the

fundamental points and also a point with all its entries non-zero. For, taking

P A C K I N G P R O B L E M S IN G A L O I S G E O M E T R I E S 369

r6 = (11111), in order that neither 1"7 + r6 nor r7 - r6 be dependent on two or fewer of rl, r2, • •. , rs, we require that r7 has two l 's, two 2's and one zero. We may take r7 = (11220), for given any other choice, there is an obvious interchange of columns which sends it to (11220) and fixes (r~, r2, • . . , r6).

Suppose Kis a 7~.3 which contains the fundamental points but no point with all non-zero entries. We may take r6 = (11110) and consideration of r~ + !"6 and r7 - r6 shows that we may take r7 = (11201).

3 Each of the two 74,3 s above can be completed to an 11~,3, the first by adjoining (12201), (21021), (20211) and (02121), the second by adjoining (11021), (20112), (01212) and (21122), and so they must be equivalent by Lemma 4.4.

(iii) k = 8. S i n c e m~,3 = 6 , any 7-subset of an 8~,3 must be a 7~.s and so by the previous case, we may assume that

rl = 10000 r2 = 01000 r3 = 00100 r4 = 00010 r5 = 00001 r6 = 11111 r7 = 11220.

By Lemma 4.5, r8 has just one zero, and an obvious interchange of columns can be performed to put this zero in the 4th place, leaving the set {r~, r2 . . . . , rv} unchanged. The kind-3 property then determines that either r8 = (12201) or r8 = (21201). The corresponding 8~,3's are equivalent, one transformed into the other by interchange of the first two columns.

(iv) k = 9. We may assume r l , r2 . . . . , re are as in the previous case, taking r8 = 12201. The point r9 has just one zero which is one of the first three entries and the kind-3 property easily determines a unique point for each of the three possibilities, viz. r9 = (21021), (20211) or (02121). But for each of these three choices of rg, the 9~.3 can be completed to an 1 l~,s by adjoining the other two, and so again Lemma 4.4 gives the equivalence of all 9~.3's.

(v) k = 10. Taking r9 = 21021, as above we have r t o = 20211 or 02121. For either choice, the corresponding 10~.3 can be completed to an 11~,3 by adjoining the other, thus giving the uniqueness of the 10~,3 by Lemma 4.4.

T H E O R E M 4.7. m5,33 = 13. P r o o f . Let K be a k~,3 and let C be the corresponding (k, 6)-code over

GF(3). Let z denote the largest number of zeros appearing in a non-zero vector of C. By Theorem 2.4, z ~< m~,3 = 11. We consider separately the different possibilities for z.

(i) Suppose z = 7. We d~note the rows and columns of a matrix A of K by r l , rz . . . . . rk and cl , c2, • . . , c6 respectively. K is equivalent to a set with

370 R A Y M O N D H I L L

matrix having the first seven entries of cn zero, the remaining entries o f c6 all equal to 1, and r8 = (000001). N o w rx, r2 . . . . , r~ fo rm a 7~.3, and so by Theorem 4.6 we may assume that A has the fo rm

1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 1 1 1 0 1 1 2 2 0 0 0 0 0 0 0 1

1

Since c5, ca - c6 and c5 + c6 each has at mos t seven zeros we require that the remaining entries of c5 contain at most one zero, two l ' s and two 2's, thus giving k ~< 13.

(ii) Suppose z ~< 6. Then the a rgument of (i) gives even stronger results. Fo r example, if z = 6, then k ~< 12.

Fo r the cases 11 > / z >t 8, we require the following two lemmas.

L E M M A 4.8. Le t K be an 11~.3 with points al, a2, • • . , a11. Then the points in

the set R = {a~, a2, . . ., a ~ ; a~ + aj, a~ - as; 1 <<. j < i <~ 11} are precisely

the points o f $4.3. Proof. The given points o f R are certainly distinct since no four points o f

K are linearly dependent. Hence the number of points in R is 11 + 2(21 ~) = 121. But the total number of points in S~,8 is (3 ~ - 1)/2 = 121, and the result follows.

L E M M A 4.9. L e t L = {al, a2, • • . , a l ~ } b e a n 11~,3. S u p p o s e K i s a k 35,3 with

code C o f minimum weight k - z, where 7 <~ z <~ 11. Then K is equivalent to

a k~,3 o f the f o r m

an+ 1 0

aa+z 0

al l 0 0 0 0 0 0 1

bl 1 b2 1 :

bt 1

Fig. 3

P A C K I N G PR O B L E MS IN G A L O I S G E O M E T R I E S 371

where d = 11 - z, t = k - z - 1, and bl, b2 . . . . , bt sa t i s fy the fo l lowing

proper t ies ( P 1 ) , . . . , (P4). L e t Xa denote the se t {+a~, +(a~ + aj); i = 1, 2 . . . . d ; j = 1, 2 , . . . , 11}. Then

(P1) b~ ~ Xa

(P2) b~ - bj ~ Xa (P3) The sum o f any two or three dist inct b,'s is non-zero and not equal to any

o f aa+l, aa+2 . . . . , a11, or their negatives.

(P4) b~ 1 - bsl = b~2 - bj2 i f and only i f (il , j l ) = (i2,j2), il # Jl , i2 # j2.

(No te : In this l emma the at and bj are regarded as vectors in V(5, 3) ra ther than points o f $4.3; i.e. a~ and bj are not identified with their negatives.)

Proof . By Theorem 4.6, K is equivalent to a kg,3 having matr ix A as in Figure 3, with the first z + 1 rows uniquely specified. By L e m m a 4.8, the set Xa is precisely the set o f points o f V(5, 3) which are not linear combinat ions of two points o f the set {aa + 1, aa + 2 , . . . , a l l ) . Hence (P1) and (P2) are, respec- tively, the condit ions that (bd) - (000001) and (bd) - (bjl) are not dependent on two of the first z rows of A. I t is now easy to see that the matr ix A represents a set o f kind 3 if and only if bl, b 2 , . . . , bt satisfy ( P 1 ) , . . . , (P4).

We now return to the p r o o f of Theorem 4.7. (iii) Suppose z = 11. The -Y0, defined in the previous lemma, is empty and

so there are no vectors available for the b~. Thus k ~< 12. (iv) Suppose z = 10. The b~'s mus t be chosen f rom the set X1. We m a y

reduce the possibilities for bl to just two cases: either b~ = al or bl = al + a2. Fo r if bl = - a l , then mult ipl icat ion o f the row (bd) by - 1 and the last co lumn o f A by - 1 gives an equivalent set in which the first 11 rows are unchanged, and ( - a l l ) is replaced by (al l) . I f bl = + (a~ + a~), then by the 4-transitivity of M~I on L, there exists g in Au t L sending ax to + a~ and a~ to +a2, and so again we can get an equivalent set in which (+a~ + a~l) is replaced by (ax + a21) and the first 11 rows o f A ( though permuted) remain unchanged. In the case bl = al, no further poin t m a y be chosen, for if b2 =

al + as then (P2) is violated, while if b2 = - a ~ + a~ then (P3) is violated. So we m a y suppose tha t bl = al + a2. I fb2 = al + aj, thenbz - b~. ~. ___a 2 + a s and this cannot be in X1, or else four o f the a, 's would be dependent , cont ra ry to the hypothesis that L is o f kind 3. I f b2 = - a l +_ aj, then bl - b2 = - al + a2 + as, and this is in Xx only if aj = + a2 in which case b~ + b2 = 0 or + a2, violating (P3). We have thus shown tha t z = i0 implies k ~< 12.

(v) Suppose z = 9. Then we require bl, b2 . . . . in X2 satisfying (P1) . . . . . (P4). By the 4-transitivity of Aut L, we need consider only the cases bl = al, al + a2, and a~ _+ a3. I t is easy to show that if bx = a~ then k ~< 12. Suppose b~ = al + a2. By the 4-transitivity o f Aut L, we m a y assume that b2 = + a l + a3. We will now consider explicit values o f al, a2 . . . . , al~ taking L to be the 11~.3 given by Figure 2. Thus bl = (11000). Suppose first tha t b2 = a~ + a3 = (10100). I f b3 = - a l + a~ or - a 2 + at (i >i 3), then

372 RAYMOND HILL

ba - b~ cannot be in X2 (or the kind-3 property of L would be violated), while if ba = az + ai (i 1> 3), then ba - b2 is not in X2. Hence we may assume that ba = a2 + at for some i/> 3. There are now no choices of b4 compatible with (P2). Thus if b~ = (11000) and b2 = (10100) we have k ~< 13. We observe that ba = a2 + aa = (01100) gives a 135aa. The other cases for bl = a~ + a2 and b~ = az + aa are easily dealt with similarly.

(vi) z = 8. This is the most difficult case, for even with the help of the 4-transitivity of Aut L there are many situations to be considered. Using the same arguments as in the previous cases, it required several hours of calcula- tion by hand to dispose of this case and the details are too lengthy to be included here. A large number of 13],a's arise, and also a large number of 14-sets which only just fail to be of kind 3. For example (taking L to be the 11~,8 of Figure 2),

0 0 0 1 0 0 0 0 0 0 1 0 2 2 2 1 0 0 2 1 2 0 1 0 2 2 0 2 2 0 1 0 2 1 2 0 0 2 1 1 1 0 1 2 2 2 1 0 0 0 0 0 0 1 1 0 0 1 0 1 1 2 2 1 0 1 2 0 2 1 0 1 0 1 1 1 1 1 0 0 2 1 0 1

is a 14-set in which bz = al + a4, b2 = - a ~ + a6, ba = a2 + a6, b4 = - a 2 + a~0 and b5 = - a 3 + as. The first 13 points form a 13~,3, while the addition of(bs1) violates (P1 ) , . . . , (P4) only in that ba + b4 + b5 = aT. Thus we have a 14-set in which there is just one subset of four dependent points.

The proof of Theorem 4.7 is now completed.

5. SETS OF KIND ,~ IN S6,a

In the following theorem, we summarize the present state of knowledge of the 8 values and bounds on m6,a.

T H E O R E M 5.1. (i) mg,a = 8 (ii) m~,8 = 8

Off) 13 ~< m~.a ~< 14 (iv) 17 ~< 8 m6.a ~< 33 (v) 112 ~< m~,a <~ 164.

PACKING PROBLEMS IN GALOIS GEOMETRIES 373

Proof. (i) and (ii) are immediate from (1.7). (iii) The upper bound of 14 is given by Theorems 3.2 and 4.7, while a

matrix of a 131,8 may be obtained from that of a 12~,3 by adjoining a seventh column of zeros and the row (0000001). I t may be possible to determine the value of mI,8 by imitating the method of Theorem 4.7, making use of the properties of the 121.3.

(iv) The upper bound of 33 is given by Theorem 3.1, while a 176a,~ was constructed by ad hoc methods. Further improvement on both these bounds should not be difficult.

(v) The lower bound is a particular case of the general bound mr,8 i> 2m,_1,3 (see [6]). The upper bound follows from the general upper bound m~,q <~ q.m~_~,~ - q - 1 for r /> 4, q v~ 2, proved in [5]. In fact, particular consideration of the code associated with a 164~,3 shows that it cannot exist, giving 163 as an upper bound. There appears to be no obvious way of improving further on either bound in this case.

B I B L I O G R A P H Y

1. Barlotti, A., 'Some Topics in Finite Geometrical Structures', Institute of Statistics Mimeo Series no. 439, University of North Carolina, 1965.

2. Bose, R. C., ' Mathematical Theory of the Symmetrical Factorial Design', Sankhyff 8, 107-166 (1947).

3. Coxeter, H.S.M., 'Twelve Points in PG(5, 3) with 95040 Self-Transformations', Proc. Roy. Soc. (A) 247, 279-293 (1958).

4. Hill, R., 'On the Largest Size of Cap in $5.3 ', Rend. Accad. Naz. Lincei (8) 54, 378- 384 (1973).

5. Hill, R., 'Caps and Codes', Discrete Mathematics 22, 111-137 (1978). 6. Hill, R., 'Caps and Groups', Atti Con. Teorie Comb. Acc. Lincei, Rome, 1973, II,

389-394 (1976). 7. Pellegrino, G., ' Sul massimo ordine delle calotte in S4,a', Matematiche 25, 149-157

(1971). 8. Pless, V., 'On the Uniqueness of the Golay Codes', J. Comb. Theory 5, 215-228

(1968). 9. Segre, B., 'Introduction to Galois Geometries', Atti Accad. Naz. Lincei, Mem. ser.

VIII, 7 (1967). 10. Segre, B., Lectures on Modern Geometry, Cremonese, Rome, 1961. 11. Tallini, G., 'Le geometrie di Galois e le loro applicazioni alia statistica e alia teoria

delle informazione', Rend. Mat. Appl. (5) 19, 379-400 (1960). 12. Todd, J.A., 'On Representations of the Mathieu Groups as Collineation Groups',

J. London Math. Soc. 34, 406-416 (1959).

Author's address:

Raymond Hill, Dept. of Mathematics University of Salford, Salford M 5 4 W T , England.

(Received April 15, 1977)