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.)
Given a set of permutations {_{1}, _{2} ,..., _{k} } on a set S, we say that the set of permutations is transitive on S if for every ordered pair of elements a,bS, there exists at least one_{i} for which (a)_{i} = b. A permutation set for which there is exactly one _{i} which maps a to b is called sharply transitive (sometimes called exactly simply transitive).
For example, if on the set consisting of the three elements {1,2,3} we represent the permutation which maps 13, 22 and 31 by (3 2 1), then the following set of permutations is transitive;
Construction III.2.1.1  Let S_{1} = I, S_{2} ,S_{3} ,...,S_{r} be the permutations representing the rows of an r x r Latin square L_{1} as permutations of its first row and M_{1} = I, M_{2} ,M_{3} ,...,M_{h} , hr , be permutations keeping one symbol of L_{1} fixed. Then the squares L_{i}* whose rows are represented by the permutations M_{i} S_{1}, M_{i} S_{2}, ..., M_{i} S_{r} for i = 1,2,...,h are all Latin and will be mutually orthogonal if, for every choice of i,jh, the set of permutations
Let us illustrate this construction. Let L_{1} be the 4x4 Latin square
1  2  3  4 
2  1  4  3 
3  4  1  2 
4  3  2  1 
M_{1}S_{1} = (1 2 3 4)  M_{2}S_{1} = (1 4 2 3)  M_{3} S_{1} = (1 3 4 2) 
M_{1} S_{2} = (2 1 4 3)  M_{2} S_{2} = (2 3 1 4)  M_{3} S_{2} = (2 4 3 1) 
M_{1} S_{3} = (3 4 1 2)  M_{2} S_{3} = (3 2 4 1)  M_{3} S_{3} = (3 1 2 4) 
M_{1} S_{4} = (4 3 2 1)  M_{2} S_{4} = (4 1 3 2)  M_{3} S_{4} = (4 2 1 3) 
1 2 3 4 1 4 2 3 1 3 4 2 2 1 4 3 2 3 1 4 2 4 3 1 3 4 1 2 3 2 4 1 3 1 2 4 4 3 2 1 4 1 3 2 4 2 1 3and it is not too difficult to see that this is a complete set of MOLS. Examining, for the particular choice of i = 2 and j = 3, the set of permutations,
Proof: First, notice that since L_{1} contains each symbol exactly once in each column, the permutations S_{1} ,S_{2} ,...,S_{r} must form a sharply transitive set. If we multiply each of these by a fixed permutation, the new set of permutations is again sharply transitive (prove this). Consequently, the columns (and of course the rows) of L_{i} will contain each symbol exactly once, so L_{i} will be Latin.
Secondly, if U_{1} ,U_{2} ,...,U_{r} are permutations representing the rows of one Latin square L_{i} and if W_{1} ,W_{2} ,...,W_{r} are the permutations representing the rows of another square L_{j} , then the permutations U_{1}^{1}W_{1} , U_{2}^{1}W_{2} ,...,U_{r}^{1}W_{r} map the first, second, .... , rth row of L_{i} respectively to the first, second, ..., rth row of L_{j} . If, and only if these squares are orthogonal, each symbol of L_{i} must map exactly once onto each of the symbols of L_{j} since each symbol of L_{i} occurs in positions corresponding to those of a transversal of L_{j} . Thus, L_{i} and L_{j} are orthogonal iff the permutations U_{1}^{1}W_{1} ,U_{2}^{1}W_{2} ,...,U_{r}^{1}W_{r} form a sharply transitive set.
Of course, the one thing that the construction does not specify is how to find the permutations M_{i} with the required properties. It is really used as the starting point of a number of constructions, each differing by how the M_{i} are to be chosen and what type of square is used to initiate the construction. Details of these constructions can be found in Chapter 7 of Denes & Keedwell.
Prove that if the starting square is the addition table of a finite field, and the permutations M_{i} are defined by (x)M_{i} = xg_{i} (i.e., each M_{i} corresponds to multiplication by a different element of the field other than 0), then this construction gives the same result as the construction of section III.1.
Let L_{1} be the addition table of a group G (written additively, but not necessarily abelian), and let the mappings M_{i}^{1} represent automorphisms _{i} of G. If the elements of G are represented by a,b,c,... then the rows of L_{1} are given by the permutations S_{0} = I, S_{a} , S_{b}, ... where (x)S_{i} = x + i for all x, iG. Because G is a group we also have (x)S_{a} S_{b} = (x + a)S_{b} = (x + a) + b = x + (a + b) = (x)S_{a+b}, so S_{a} S_{b} = S_{a+b} a,bG. Now the rth row of L_{i} is represented by the permutation, M_{i} S_{r} M_{i}^{1}, but
Construction III.2.2.1  Let G be a group and suppose that there exist h automorphisms _{1} ,_{2} ,...,_{h} of G every pair of which possesses the property that u_{i}u_{j} for any element uG except the identity element, then we shall be able to construct h MOLS based on the group G.
As an example of this construction, consider the field GF(9). The mapping _{m} given by (x)_{m} = x^{3}m is an automorphism of the additive group, since (x + y)_{m} = (x + y)^{3}m = (x^{3} + y^{3})m = x^{3}m + y^{3}m = (x)_{m} + (y)_{m}.
These maps are not field automorphisms, but the cubing map is, in this field. The eight maps corresponding to the nonzero elements of the field form a set of sharply transitive automorphisms of the additive group of the field, as can easily be seen from their representations as permutations:
m = 1 : (0 1 4 7 2 5 8 3 6)
m = 2 : (0 2 5 8 3 6 1 4 7)
m = 3 : (0 3 6 1 4 7 2 5 8)
m = 4 : (0 4 7 2 5 8 3 6 1)
m = 5 : (0 5 8 3 6 1 4 7 2)
m = 6 : (0 6 1 4 7 2 5 8 3)
m = 7 : (0 7 2 5 8 3 6 1 4)
m = 8 : (0 8 3 6 1 4 7 2 5)
These automorphisms, acting on the base square (i.e., the additive group table of GF(9)) will give 8 mutually orthogonal latin squares of order 9. Representing the base square as permutations of the first row we have:
S_{0} = (0 1 2 3 4 5 6 7 8)
S_{1} = (1 5 8 4 6 0 3 2 7)
S_{2} = (2 8 6 1 5 7 0 4 3)
S_{3} = (3 4 1 7 2 6 8 0 5)
S_{4} = (4 6 5 2 8 3 7 1 0)
S_{5} = (5 0 7 6 3 1 4 8 2)
S_{6} = (6 3 0 8 7 4 2 5 1)
S_{7} = (7 2 4 0 1 8 5 3 6)
S_{8} = (8 7 3 5 0 2 1 6 4)
We will construct two squares of the set of MOLS, those corresponding to _{2} and _{5} . We have M_{2} = _{2}^{1} = (0 6 1 4 7 2 5 8 3) and M_{5} = _{5}^{1} = (0 5 8 3 6 1 4 7 2). Thus, square L_{2}, given by M_{2}S_{i} M_{2}^{1} is:
(0 6 1 4 7 2 5 8 3) S_{i} (0 2 5 8 3 6 1 4 7) = 

(0 5 8 3 6 1 4 7 2) S_{i} (0 5 8 3 6 1 4 7 2) = 

H.B. Mann, whose construction this is, was able to obtain an upper bound for h in terms of the number of conjugacy classes of G. What is the relationship of this construction to the research question asked in the last section?
Theorem III.2.3.1  Let (G,®) and (G,©), (H,×) and (H,¤) be pairs of orthogonal quasigroups and let us define the operations (@) and (&) on the cartesian product F = G×H by:
Proof: Left as an exercise (First prove that the construction gives a quasigroup and then show that the two quasigroups are orthogonal).
We will now discuss a construction by A. Sade called the singular direct product. This method requires the use of idempotent quasigroups. A quasigroup is idempotent if it satisfies a^{2} = a, a. The multiplication table of an idempotent quasigroup is a Latin square whose main left to right diagonal is a transversal with its elements in the natural order. Latin squares of this form will also be called idempotent.
Lemma III.2.3.2  Given a set of t MOLS of order n, we can form a set of t MOLS of which t  1 are idempotent.
Proof: Permute the rows of all the given squares so that the first square's main diagonal contains only one symbol. The main diagonals of each of the other squares are now transversals. Independently rename the elements in each of the other squares so that the main diagonal of each is in natural order. These operations preserve orthogonality.
Homework: Prove that every quasigroup with a complete mapping is isotopic to an idempotent quasigroup.
Sade's singular direct product construction is most easily expressed in terms of quasigroups (of which four are needed), but the example we will give will be in terms of Latin squares.
Construction III.2.3.3  Let (G,®) be a quasigroup of order n = m + l which contains a subquasigroup (Q,®) of order m, and let (R,©) be a quasigroup of order l defined on the set R = G\Q. Let (H,¤) be an idempotent quasigroup of order k. Finally, let the set T = Q(R × H). Define the operation @ on T by:
q @ q' = q®q' q,q' Q
q @ (r,h) = (q®r,h) qQ, rR and hH
(r,h) @ q = (r®q,h) qQ, rR and hH
(r,h) @ (r',h') = (r©r',h¤h') r,r'R and h,h'H, hh'
(r,h) @ (r',h) = (r®r',h) if r®r'R
(r,h) @ (r',h) = r®r' if r®r'Q
Then (T,@) is a quasigroup of order m + lk.
Proof: The proof of this construction, while not difficult, breaks up into a zillion cases due to the nature of @ and so will be omitted.
We will illustrate this construction by using the following squares.
G = 

Q = 

R = 

H = 




 














The usefulness of this construction comes from the following theorem whose proof is also omitted for much the same reason.
Theorem III.2.3.4  If in the construction of (T,@) above, each of the quasigroups G,Q,R and H are replaced by orthogonal mates, then the newly constructed (T',@) is orthogonal to (T,@).
We now come to a theorem which we need for the next section. While stated in the form we need it in, it can be generalized.
Theorem III.2.3.5  Suppose that there exists a pair of orthogonal mates of order p + q, with a pair of orthogonal subsquares of order q in their lower right hand corners, and a pair of orthogonal mates of order p, and three MOLS of order m. Then there exists a pair of orthogonal mates of order pm + q.
Proof: Given the pair of orthogonal mates of order p, we can rename their elements so that the symbols used are from the squares of order p + q but not in their subsquares. The 3 MOLS of order m provide, by Lemma III.2.3.2 a pair of orthogonal idempotent squares of order m. These are all the ingredients necessary for construction III.2.3.3 and so by Thm. III.2.3.4 a pair of orthogonal mates of order pm + q is produced.
Construct a pair of MOLS of order 22 = 5(4) + 2.