7 Groups and Codes 253Ifρ 1 =±3, then all elements of the same triple of columns in ther-th row have the
same sign and orthogonality to thes-th row impliesσ 1 =0, which is impossible be-
causeσ 1 is odd. Henceρ 1 =±1. By changing the signs of some rows we may assume
thatρ 1 =1foreveryr>3. By permuting columns within each block of three we may
also normalize the 4-th row, so that the first four rows are now the same as the first four
rows of the array (∗).
In any row after the third, within a givenblock of three columns two elements have
the same sign and the third element the opposite sign. Moreover, these signs depend
only on the block and not on the row, sinceρ 1 =1. The scalar product of the triples
from two different rows belonging to the same block of columns is 3 if the exceptional
elements have the same position in the triple and is−1 otherwise. Since the two rows
are orthogonal, the exceptional elements must have the same position in exactly one
of the four blocks of columns. Thus if two rows after the 4-th have the same triple of
elements in thek-th block as the 4-th row, then they have no other triple in common
with the 4-th row or with one another. But this implies that if one of the two rows is
given, then the other is uniquely determined. Hence no other row besides these two
has the same triple of elements in thek-th block as the 4-th row. Since there are eight
rows after the 4-th, and since each has exactly one triple in common with the 4-th row,
it follows that, for eachk∈{ 1 , 2 , 3 , 4 }, exactly two of them have the same triple in the
k-th block as the 4-th row.
The first four rows are unaltered by the following operations:
(i) interchange of the first and last columns of any triple of columns,
(ii) interchange of the second and third triple of columns, and then interchange of the
second and third rows,
(iii) interchange of the first and fourth triple of columns, then interchange of the sec-
ond and third rows and change of sign of these two rows,
(iv) interchange of the second and fourth triple of columns and change of their signs,
then interchange of the first and third rows.
If we denote the elements of ther-th row(r> 4 )byξ 1 ,...,ξ 12 ,thenwehaveξ 1 +ξ 2 +ξ 3 = 1 =ξ 10 +ξ 11 +ξ 12 ,
ξ 4 +ξ 5 +ξ 6 =− 1 =ξ 7 +ξ 8 +ξ 9 ,
ξ 2 −ξ 5 −ξ 8 +ξ 11 = 2.In particular in the 5-th row we haveα 52 −α 55 −α 58 +α 5 , 11 =2. Thusα 52 andα 5 , 11
cannot both be−1 and by an operation (iii) we may assume thatα 52 =1. Similarly
α 55 andα 58 cannot both be 1 and by an operation (ii) we may assume thatα 58 =−1.
Thenα 55 =α 5 , 11 and by an operation (iv) we may assume thatα 55 =α 5 , 11 =−1. By
operations (i) we may finally assume that the 5-th row is the same as the 5-th row of
the array (∗).
As we have already shown, exactly one row after the 5-th row has the same triple
+−+in the last block of columns as the 4-th and 5-th rows and this row must be
the same as the 6-th row of the array (∗). By permuting the last seven rows we may
assume that this row is also the 6-th row of the given matrix, that the 7-th and 8-th rows
have the same first triple of elements as the 4-th row, that the 9-th and 10-th rows have