254 V Hadamard’s Determinant Problem
the same second triple of elements as the 4-th row, and that the 11-th and 12-th rows
have the same third triple of elements as the 4-th row.
In any row after the 6-th we have, in addition to the relations displayed above,
ξ 11 =1,ξ 10 +ξ 12 =0and
ξ 1 −ξ 4 −ξ 7 =ξ 2 −ξ 5 −ξ 8 =ξ 3 −ξ 6 −ξ 9 = 1.In the 7-th and 8-th rows we haveξ 1 =ξ 3 = 1 ,ξ 2 =−1, and henceξ 5 =ξ 8 =−1,
ξ 4 =−ξ 6 =−ξ 7 =ξ 9. Since the first six rows are still unaltered by an operation (ii),
and also by interchanging the first and third columns of the last block, we may assume
thatα 74 =−1,α 7 , 10 =1. The 7-th and 8-th rows are now uniquely determined and
are the same as the 7-th and 8-th rows of the array (∗).
In any row after the 8-th we have
ξ 2 −ξ 6 −ξ 7 +ξ 12 = 2 =ξ 2 −ξ 4 −ξ 9 +ξ 10.In the 9-th and 10-th rows we haveξ 5 = ξ 11 =1andξ 4 = ξ 6 =−1. Hence
ξ 2 =−ξ 8 =1,ξ 1 =ξ 7 =−ξ 3 =−ξ 9 , and finallyξ 9 =ξ 10 =−ξ 12. Thus the
9-th and 10-th rows are together uniquely determined and may be ordered so as to
coincide with the corresponding rows of the array (∗). Similarly the 11-th and 12-th
rows are together uniquely determined and may be ordered so as to coincide with the
corresponding rows of the displayed array.
It follows from Proposition 22 that, for any five distinct rows of a Hadamard ma-
trix of order 12, there exists exactly one pair of columns which either agree in all these
rows or disagree in all these rows. Indeed, by permuting the rows we may arrange that
the five given rows are the first five rows. Now, by Proposition 22, we may assume that
the matrix has the form (∗). But it is evident that in this case there is exactly one pair
of columns which either agree or disagree in all the first five rows, namely the 10-th
and 12-th columns.
Hence a 5-( 12 , 6 , 1 )design is obtained by taking the points to be elements of the
setP={ 1 ,..., 12 }and the blocks to be the 12·11 subsetsBjk,B′jkwithj,k∈P
andj=k,where
Bjk={i∈P:αij=αik}, B′jk={i∈P:αij=αik}.The Mathieu groupM 12 may be defined as the automorphism group of this design or
as the reduced automorphism group of any Hadamard matrix of order 12.
It is certainly not true in general that all Hadamard matrices of the same ordern
are equivalent. For example, there are 60 equivalence classes of Hadamard matrices of
order 24. The Mathieu groupM 24 is connected with the Hadamard matrix of order 24
which is constructed by Paley’s method, described in§2. The connection is not as
immediate as forM 12 , but the ideas involved are of general significance, as we now
explain.
A sequencex=(ξ 1 ,...,ξn)ofn0’s and 1’s may be regarded as a vector in the
n-dimensional vector spaceV =Fn 2 over the field of two elements. If we define the
weight|x|of the vectorxto be the number of nonzero coordinatesξk,then
(i)|x|≥0 with equality if and only ifx=0,