Lecture Notes 4: Affine Planes and Mutually Orthogonal Latin Squares

Field Construction

In the previous section we mentioned that complete sets of MOLS are 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 × 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) =
      L (2) =
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 × bi + bj = be × 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 × bi + bj = be × bk + bm and bf × bi + bj = bf × bk + bm.
Subtracting the second from the first and then using the distributive laws gives,
(be - bf ) × bi = (be - bf ) × 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.)

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.

Affine Planes from MOLS

Complete sets of MOLS of order q can be used to construct an affine plane of order q.

Let L1, L2, ..., Lq-1 be a complete set of MOLS of order q and A a q × q matrix containing q2 distinct symbols. We define sets of size q (lines of the affine plane) on this set of q2 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 q2 - q sets are formed by taking each Li in turn, superimposing it on A and taking as sets the elements of A which correspond to a single symbol in the Li.

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  2
Now, let A be the matrix,
                          1   2   3   4
                          5   6   7   8
                          9  10  11  12
                          13 14  15  16
The lines of the affine plane are:
{1,2,3,4} {5,6,7,8} {9,10,11,12} {13,14,15,16} {1,5,9,13} {2,6,10,14} {3,7,11,15} {4,8,12,16} - from the rows and columns, and
{1,6,11,16} {2,5,12,15} {3,8,9,14} {4,7,10,13} {1,7,12,14} {2,8,11,13} {3,5,10,16} {4,6,9,15} {1,8,10,15} {2,7,9,16} {3,6,12,13} {4,5,11,14} - from the Latin squares.

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 n2 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 n2 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 n-1 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 - [Bruck-Ryser 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 Bruck-Ryser 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.


The construction of the last section will work even if the set of MOLS is not complete. You get, if there are s squares of order n in the set, n2 points and n(s+2) lines where the lines are partitioned into s+2 parallel classes of n lines apiece. The structure so obtained is called an (s+2)-net of order n (or as the book has it a partial affine plane.)

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?