III. LATIN SQUARES REVISITED

We return now to the subject of Latin squares, this time concentrating on construction techniques, in particular, how to construct orthogonal squares (which is hard) as opposed to constructing single squares (which is easy). The last section will deal with a construction which leads to the disproof of the Euler conjecture.

III.1 FIELD CONSTRUCTION

In section I.2.2 we mentioned that complete sets of MOLS were known only in the case of the order being a prime power. We will now prove the existence of these MOLS by construction. The restriction to prime powers comes from the fact that the construction uses finite fields (and works only for them) which only exist for prime power orders.

Construction III.1.1 - Let b1 ,b2 ,...,bn be the elements of a finite field GF(n) where n = pk . Let b1 be the multiplicative identity ( = 1) and bn be the additive identity ( = 0) of GF(n). For e = 1,2, ..., n-1 define the n × n array L(e) = (aij(e) ) by taking

aij(e) = (be x bi ) + bj ,
where the operations on the right are in the field. The resulting squares form a complete set of MOLS of order n.

For example, over the field GF(3), this construction gives the two squares,

L (1) =
201
012
120
      L (2) =
012
201
120
which are easily seen to be orthogonal mates.

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 aij(e) = aik(e) . So, we would have

be x bi + bj = be x bi + bk
so, by using field properties we have bj = bk , and therefore j = k. Thus, all the elements in the same row are distinct. Similarly, (fill in the details) we can show that all the elements in a column are distinct. Thus, L(e) is a Latin square for any value of e.

Now consider two squares L(e) and L(f). Suppose that

(aij(e) ,aij(f) ) = (akm(e) ,akm(f) )
which implies,
aij(e) = akm(e) and aij(f) = akm(f).
This leads to two equations,
be x bi + bj = be x bk + bm and bf x bi + bj = bf x bk + bm.
Subtracting the second from the first and then using the distributive laws gives,
(be - bf ) x bi = (be - bf ) x bk
and since ef, be - bf0 and we may conclude that bi = bk , so i = k. Using this in the first of our original equations, permits us to conclude that bj = bm and so j = m. We have shown that the ordered pairs are unique and so the squares are orthogonal.

Notice that in this construction the first square has ij-th entry equal to bi + bj , 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 p-group).

Research question: What is the largest number of MOLS in a set based on a cyclic group of non-prime order? (For prime order the cyclic group is the additive group of a field and so a complete set exists, but for non-prime orders in general the answer is not known.)

III.2 OTHER CONSTRUCTION METHODS

As we will see later in the term, there are complete sets of MOLS which do not arise from the construction of the last section. Also, there are methods of constructing sets of MOLS which do not guarantee complete sets as that construction does. We shall examine a few of these constructions in this section and one in the next.

III.2.1 - Generalized Bose Construction

The construction of section III.1 is due to Bose, but a means to generalize it was first observed by H.B. Mann. Before we state the method, we need some terminology from the theory of permutation groups.

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 onei 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;

(1 2 3), (1 3 2), (2 1 3), and (3 2 1)
and the last three permutations form a sharply transitive set.

Construction III.2.1.1 - Let S1 = I, S2 ,S3 ,...,Sr be the permutations representing the rows of an r x r Latin square L1 as permutations of its first row and M1 = I, M2 ,M3 ,...,Mh , hr , be permutations keeping one symbol of L1 fixed. Then the squares Li* whose rows are represented by the permutations Mi S1, Mi S2, ..., Mi Sr for i = 1,2,...,h are all Latin and will be mutually orthogonal if, for every choice of i,jh, the set of permutations

S1-1Mi-1Mj S1 , S2-1Mi-1Mj S2 ,..., Sr-1Mi-1Mj Sr
is sharply transitive on the symbols of L1 .

Let us illustrate this construction. Let L1 be the 4x4 Latin square

1 2 3 4
2 1 4 3
3 4 1 2
4 3 2 1
The rows of this square, considered as permutations, provide the S's, so S1 = (1 2 3 4), S2 = (2 1 4 3), S3 = (3 4 1 2), S4 = (4 3 2 1). Now, let M1 = (1 2 3 4), M2 = (1 4 2 3) and M3 = (1 3 4 2), all of which fix the symbol 1. Direct calculation now gives us the following:
M1S1 = (1 2 3 4)M2S1 = (1 4 2 3)M3 S1 = (1 3 4 2)
M1 S2 = (2 1 4 3)M2 S2 = (2 3 1 4)M3 S2 = (2 4 3 1)
M1 S3 = (3 4 1 2)M2 S3 = (3 2 4 1)M3 S3 = (3 1 2 4)
M1 S4 = (4 3 2 1)M2 S4 = (4 1 3 2)M3 S4 = (4 2 1 3)
So, the three latin squares produced are:
        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 3
and 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,

S1-1M2-1M3 S1 = (1 4 2 3)
S2-1M2-1M3 S2 = (3 2 4 1)
S3-1M2-1M3 S3 = (4 1 3 2)
S4-1M2-1M3 S4 = (2 3 1 4)

we see that it is a sharply transitive set of permutations on {1,2,3,4}. The same is true for any other choices of i and j.

Proof: First, notice that since L1 contains each symbol exactly once in each column, the permutations S1 ,S2 ,...,Sr 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 Li will contain each symbol exactly once, so Li will be Latin.

Secondly, if U1 ,U2 ,...,Ur are permutations representing the rows of one Latin square Li and if W1 ,W2 ,...,Wr are the permutations representing the rows of another square Lj , then the permutations U1-1W1 , U2-1W2 ,...,Ur-1Wr map the first, second, .... , r-th row of Li respectively to the first, second, ..., r-th row of Lj . If, and only if these squares are orthogonal, each symbol of Li must map exactly once onto each of the symbols of Lj since each symbol of Li occurs in positions corresponding to those of a transversal of Lj . Thus, Li and Lj are orthogonal iff the permutations U1-1W1 ,U2-1W2 ,...,Ur-1Wr form a sharply transitive set.

Of course, the one thing that the construction does not specify is how to find the permutations Mi with the required properties. It is really used as the starting point of a number of constructions, each differing by how the Mi 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 Mi are defined by (x)Mi = xgi (i.e., each Mi 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.

III.2.2 - Mann's Automorphism Construction

The generalized Bose construction of the last section does not give MOLS in standard form. This can easily be remedied. If Li* is a square given by the previous construction, define Li = Li*Mi-1. That is to say, the rows of Li are given by Mi Sj Mi-1. If the set of Li* are mutually orthogonal, then so are the set of Li and, moreover, the first row of each Li is the identity permutation, and so this set is in standard form. [Prove these statements].

Let L1 be the addition table of a group G (written additively, but not necessarily abelian), and let the mappings Mi-1 represent automorphisms i of G. If the elements of G are represented by a,b,c,... then the rows of L1 are given by the permutations S0 = I, Sa , Sb, ... where (x)Si = x + i for all x, iG. Because G is a group we also have (x)Sa Sb = (x + a)Sb = (x + a) + b = x + (a + b) = (x)Sa+b, so Sa Sb = Sa+b a,bG. Now the r-th row of Li is represented by the permutation, Mi Sr Mi-1, but

(x)Mi Sr Mi-1 = (xi-1)Sr Mi-1 = ( xi-1 + r )Mi-1 = ( xi-1 + r )i = x + ri = (x)Sri
since i is an automorphism. The squares Li and Lj will thus be orthogonal (see the proof of the last section), if {I, Sai-1Saj, Sbi-1Sbj, ... } is a sharply transitive set of permutations. But we can calculate that
Sai-1Saj = S-(ai)Saj = S-ai + aj.
Thus, the squares will be orthogonal provided that -si + sj-ti + tj for distinct elements s and t of G. This can be rewritten as,
ti - sitj - sj
and if we write t - s = u, we have that the squares Li and Lj will be orthogonal provided that the automorphisms i and j have the property that uiuj for any element u other than the identity in G. We have just proved the following construction.

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 uiuj 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 = x3m is an automorphism of the additive group, since (x + y)m = (x + y)3m = (x3 + y3)m = x3m + y3m = (x)m + (y)m.

These maps are not field automorphisms, but the cubing map is, in this field. The eight maps corresponding to the non-zero 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:

S0 = (0 1 2 3 4 5 6 7 8)
S1 = (1 5 8 4 6 0 3 2 7)
S2 = (2 8 6 1 5 7 0 4 3)
S3 = (3 4 1 7 2 6 8 0 5)
S4 = (4 6 5 2 8 3 7 1 0)
S5 = (5 0 7 6 3 1 4 8 2)
S6 = (6 3 0 8 7 4 2 5 1)
S7 = (7 2 4 0 1 8 5 3 6)
S8 = (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 M2 = 2-1 = (0 6 1 4 7 2 5 8 3) and M5 = 5-1 = (0 5 8 3 6 1 4 7 2). Thus, square L2, given by M2Si M2-1 is:

(0 6 1 4 7 2 5 8 3) Si (0 2 5 8 3 6 1 4 7) =
0 1 2 3 4 5 6 7 8
2 8 6 1 5 7 0 4 3
5 0 7 6 3 1 4 8 2
8 7 3 5 0 2 1 6 4
3 4 1 7 2 6 8 0 5
6 3 0 8 7 4 2 5 1
1 5 8 4 6 0 3 2 7
4 6 52 8 3 7 1 0
7 2 4 0 1 8 5 3 6
While the square L5 , given by M5 Si M5-1 is:
(0 5 8 3 6 1 4 7 2) Si (0 5 8 3 6 1 4 7 2) =
0 1 2 3 4 5 6 7 8
50 7 6 3 1 4 8 2
8 7 3 5 0 2 1 6 4
3 4 1 7 2 6 80 5
6 3 0 8 7 4 2 5 1
1 5 8 4 6 0 3 2 7
4 6 5 2 8 3 7 1 0
7 2 4 0 1 8 5 3 6
2 8 6 1 5 7 0 4 3
Direct comparison shows that these squares are mutually orthogonal to each other and to the square given by the S's.

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?

II.2.3 - The Direct Product Constructions

The constructions we have seen so far all give sets of MOLS which are just row rearrangements of a base square. Other methods are based on the idea of taking two or more squares and building larger squares from them. We have already seen one such construction in Thm. I.2.2.3, where MacNeish constructed what is called the Direct Product of Latin squares. This construction can easily be translated into the language of quasigroups. We say that two quasigroups (G,®) and (G,©) defined on the same set G, are orthogonal if the pair of equations x®y = a and x©y = b (where a and b are any given elements of G) are satisfied simultaneously by a unique pair of elements x and y from G. It is easily seen that orthogonal quasigroups have orthogonal multiplication tables when they are bordered in the same way (convince yourself of this). We can now restate the direct product construction for two pairs of orthogonal mates in terms of quasigroups.

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:

(x,a) @ (y,b) = (x®y, a×b) and (x,a) & (y,b) = (x©y, a¤b)
where x,yG and a,bH. Then (F,@) and (F,&) are orthogonal quasigroups.

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 a2 = 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 sub-quasigroup (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 =
1 2 3 4 5
2 1 4 5 3
3 5 1 2 4
4 3 5 1 2
5 4 2 3 1
Q =
1 2
2 1
R =
3 4 5
4 5 3
5 3 4
H =
a c b
c b a
b a c
Notice that G of order 5 = 2 + 3, contains the order 2 square Q as a subsquare in the upper left corner. R is an order 3 square on the elements of G which are not in Q and H is an idempotent square. T consists of the elements 1,2 and the ordered pairs (3,a),(3,b),..., (5,c), which we will write as 3a,3b,... and so on. The order 11 square produced by this construction is,
1  
2  
3a4a5a
4a5a3a
3b4b5b
4b5b3b
3c4c5c
4c5c3c
3a5a
4a3a
5a4a
124a
5a12
23a1
3c4c5c
4c5c3c
5c3c4c
3b4b5b
4b5b3b
5b3b4b
3b5b
4b3b
5b4b
3c4c5c
4c5c3c
5c3c4c
124b
5b12
23b1
3a4a5a
4a5a3a
5a3a4a
3c5c
4c3c
5c4c
3b4b5b
4b5b3b
5b3b4b
3a4a5a
4a5a3a
5a3a4a
124c
5c12
23c1

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.