Construction III.1.1  Let b_{1} ,b_{2} ,...,b_{n} be the elements of a finite field GF(n) where n = p^{k} . Let b_{1} be the multiplicative identity ( = 1) and b_{n} be the additive identity ( = 0) of GF(n). For e = 1,2, ..., n1 define the n × n array L^{(e)} = (a_{ij}^{(e)} ) by taking
For example, over the field GF(3), this construction gives the two squares,
L^{ (1)} = 
 L^{ (2)} = 

Homework: Construct complete sets of MOLS of orders 8 and 9 and then put them into standard form
Proof: We must show that each L is in fact Latin and that any pair of them are orthogonal.
We first show that L^{(e)} is a Latin square. Suppose L^{(e)} had the same element appear in row i twice, then a_{ij}^{(e)} = a_{ik}^{(e)}. So, we would have
Now consider two squares L^{(e)} and L^{(f)}. Suppose that
Notice that in this construction the first square has ijth entry equal to b_{i} + b_{j} , so that this first square is an isotope of the addition table of the field. Also notice, in the examples you have worked out, that each of the other squares in the set is just a row permutation of the first square (try to prove this for homework). MOLS that are related in this way are said to be based on the original square. This construction thus gives a set of MOLS based on the additive group of a field (an elementary abelian pgroup).
Research question: What is the largest number of MOLS in a set based on a cyclic group of nonprime order? (For prime order the cyclic group is the additive group of a field and so a complete set exists, but for nonprime orders in general the answer is not known.)
As we will see later in the term, there are complete sets of MOLS which do not arise from this construction. Also, there are methods of constructing sets of MOLS which do not guarantee complete sets as this construction does.
Let L_{1}, L_{2}, ..., L_{q1} be a complete set of MOLS of order q and A a q × q matrix containing q^{2} distinct symbols. We define sets of size q (lines of the affine plane) on this set of q^{2} symbols (points of the affine plane) in the following way: There are q sets which are the rows of A, and q sets which are the columns of A. The remaining q^{2}  q sets are formed by taking each L_{i} in turn, superimposing it on A and taking as sets the elements of A which correspond to a single symbol in the L_{i}.
As an example consider the set of 3 MOLS of order 4:
1 2 3 4 1 2 3 4 1 2 3 4 2 1 4 3 3 4 1 2 4 3 2 1 3 4 1 2 4 3 2 1 2 1 4 3 4 3 2 1 2 1 4 3 3 4 1 2Now, let A be the matrix,
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16The lines of the affine plane are:
Exercise: Construct two affine planes from the 2 MOLS of order 3.
We can also reverse the construction and produce a complete set of MOLS from an affine plane in the following way:
Let be an affine plane of order n. For each parallel class of lines of , arbitrarily label with the digits 1,...,n each line of the parallel class. Now select two parallel classes. The lines contained in these parallel classes will be used to index rows and columns of the Latin squares, so label one of the classes R and the other C. The n^{2} points of intersection of the lines in R and C are associated with pairs of numbers, the number of the line in R and the number of the line in C. Now, for each parallel class in other than R or C, we will form a Latin square in the following way: If P is a parallel class, we have already labeled all the lines in P. The n^{2} points of intersection of the R and C lines must all lie on the n labeled lines in P. In the cell of the square corresponding to one of the intersection points we place the label of the line in P which passes through this point. It is easy to see that the square produced this way is a Latin square of order n. This procedure is repeated for each of the parallel classes other than R or C, giving n1 Latin squares.
To see that they are mutually orthogonal, consider two such squares and suppose that when superimposed there are two cells containing (a,b). Since the two cells of the first square received the label a, the two points which correspond to the cells must have been on the same line (labeled a) of the parallel class corresponding to the first square. Since these same cells have the label b in the second square, the two points must also be on the line labeled b contained in a different parallel class. This is impossible by axiom A1, so we see that these squares must be mutually orthogonal.
By Construction III.1.1 and the above construction we see that an affine plane of order n always exists if n is a prime or prime power, i.e., the order of a finite field. The only theorem that we have which rules out some affine planes is the following:
Theorem VIII.1.4  [BruckRyser Thm.] If n 1 or 2 mod 4 then an affine (projective) plane of order n does not exist unless n is the sum of two integral squares.
This theorem implies that there does not exist an affine (projective) plane of order 6, a result we could have inferred since no pair, let alone a complete set, of MOLS of order 6 exists. Notice that, since 10 = 9 + 1, this theorem says nothing about the existence of a projective plane of order 10. An extremely long computer search has finally proved that this plane does not exist (see note at the end of the chapter). The smallest open case is n = 12 which the BruckRyser theorem does not cover and then n = 18, which since 18 = 9 + 9, the theorem says nothing about. The only known orders are primes and prime powers, although alternate constructions exist which give projective planes which are not isomorphic to those produced by the Latin square construction. Instant fame and recognition will be the reward of the mathematician who settles the question of the existence of projective planes of composite order.
There are several interesting research questions concerning nets. For instance, what conditions on a net are sufficient and necessary for the net to be extended to an affine plane? What conditions are needed so that a net which can be extended to an affine plane can be extended in only one way?