COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

On property D neofields and some problems
concerning orthogonal latin squares
A. D. KEEDWELL
THE concept of the property D neofield arose from the attempt to find an
explanation for the non-existence of pairs of orthogonal latin squares of
order 6. The intention was to formulate a standard method of constructing
a pair of orthogonal squares of any sufficiently small order r distinct
from 6. The reason for failure when Y = 6 could then be observed.
The standard method devised (which is reported fully in [l]) is essentially
a modification of the Bose method for constructing a complete set of mutually
orthogonal latin squares from a field. This construction can be exhibited
as follows. One square L1 is the Cayley table of the addition group
of the field and its rows may be regarded as permutations SO E I, S1, SZ, . . . ,
S,-, of its first row. If 0, 1, X, . . . , xrH2, denote the elements of the field,
the remaining squares LT (i = 2, 3, . . . , Y- 1) are as follows
L-4 = O.&f-1s 0 1. kPISo Xr-2Mi--IS
O.M’-‘S 1 l.M’-?!?I : : . Xr-2Mi-lS
1
. . .
. . .
O.M'-lS,-l - 1 .M’-1s * ,-I . . .
Xr-2&.fi-lS
r-1
where M s (O)(l x x2 _ . . xr-2) and the first columns of all the squares are
the same.
Since the squares L1 = (So, S1, . . . ,S,-,} and L,* = {MSO, MS1,. . . ,MS,-,}
are orthogonal, it follows from a theorem due to H. B. Mann that the
permutations S,-l MSo, S,-lMSl, . . . , Sr?lMS,-l are a sharply transitive
set. Conversely, when these permutations form a sharply transitive set,
the squares Ll and L,* will be orthogonal. It is useful to observe that, again
by a result due to H. B. Mann, the squares Ll and LZ = {MSWT~-~,
M&M-l,. . . , MSr-lM-l} will also be orthogonal. Moreover, the permutations
MSiMel, being conjugate to the permutations Si, are easy to
compute, and the squares Ll and L2 have the same first row.
When we exhibit the permutations S,:lMSi as in Diagram 1, we observe
that the r x r matrix obtained is the Cayley table of the addition group of the
field, and that, if the first row and column are disregarded, the quotients
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0