Homework Assignment 2

pg.36 prob. 6

Let n be an odd natural number. Show that A = [aij] and B = [bij] are orthogonal Latin squares of order n where
aij =n i + j - 1
and
bij =n 2i + j - 1.


We first show that A and B are Latin squares. If A had two elements in the same row equal, then i + j - 1 =n i + j' - 1, which implies that j =n j', so two elements in different columns must be different. If A had two elements in the same column equal, then i + j - 1 =n i' + j - 1, which implies i =n i', and therefore A is Latin. For B, the corresponding equations are 2i + j -1 =n 2i + j' -1, which implies j =n j', and 2i + j - 1 =n 2i' + j -1 which implies 2i =n 2i' and this implies i =n i' (but only because n is odd). Thus, B is also Latin.

Two show that A and B are orthogonal, suppose that the ordered pairs (i+j-1, 2i+j=1) and (s+t-1, 2s + t -1) are equal. This implies that i+j-1 =n s+t-1 and 2i+j-1 =n 2s+t-1. The first equation gives i+j =n s+t. Substituting this in the second equation gives i =n s, and then substituting this back in the first equation gives j =n t. So, A and B are orthogonal.

pg.40 prob. 1

Construct 3 orthogonal Latin squares from the affine plane of order 4 given by:
             OIDF    HBCD    HFKP    OAJH    HEIM
             ECPA    EFGJ    BEOL    GIPB    BFAN
             KMJB    IAKL    JNCI    FCML    CGKO
             GHLN    MNOP    AGDM    EDKN    DJLP

The columns here correspond to parallel classes. The construction starts with an arbitrary assignment of the numbers 1-4 to each line of each parallel class, and then choosing two parallel classes to be r and c. The results will vary by making different choices. We make the following choices:
             1-OIDF    1-HBCD    1-HFKP    1-OAJH    1-HEIM
             2-ECPA    2-EFGJ    2-BEOL    2-GIPB    2-BFAN
             3-KMJB    3-IAKL    3-JNCI    3-FCML    3-CGKO
             4-GHLN    4-MNOP    4-AGDM    4-EDKN    4-DJLP
                 r         c         1        2          3
We now fill in the entries of the three orthogonal squares in the following way: The (i,j) entry of each square (a square corresponds to one of the numbered parallel classes) is obtained by finding the point of intersection of the i-th line of r and the j-th line of c and then recording the number of the line containing this point in each of the other parallel classes. So, for instance, the element in the (1,1) position of each square is obtained by finding the intersection of the 1-line in r and the 1-line in c (this is D) and then putting the number of the line containing D in the (1,1) position of each square (in this case a 4 goes into each position). For another example, the (1,2) position (point F) would be filled by a 1, 3 and 2 respectively in the three squares. The completed squares are:
      D F I O        4 1 3 2      4 3 2 1     4 2 1 3
      C E A P        3 2 4 1      3 4 1 2     3 1 2 4
      B J K M        2 3 1 4      2 1 4 3     2 4 3 1
      H G L N        1 4 2 3      1 2 3 4     1 3 4 2
where the first square gives the points of intersection corresponding to the positions in the squares.

pg.41 prob.3

Find a pair of orthogonal Latin squares of order 3, different from the pair given in the text, and use the squares to construct an affine plane of order 3.


Using the following pair of orthogonal squares of order 3 and the array of 9 letters:
         1 3 2     3 2 1      A B C
         2 1 3     2 1 3      D E F
         3 2 1     1 3 2      G H I
The points of the affine plane are A, B, ..., I. The lines are obtained by taking the rows and columns of the array, and also by superimposing the array on each Latin square and taking the letters corresponding to the same number of the Latin square. We obtain:
{A,B,C}, {D,E,F}, {G,H,I} from the rows;
{A,D,G}, {B,E,H}, {C,F,I} from the columns;
{A,E,I}, {C,D,H}, {B,F,G} from the first square, and
{C,E,G}, {B,D,I}, {A,F,H} from the second square.

pg.45 prob.7

a) Prove that in every projective plane, four points may be found, no three of which are collinear.
b) State the dual of the theorem in a).


Axiom 3 states that every line has at least 3 points and that there are at least 2 lines. Let l and m be two distinct lines which meet at point P. By Axiom 3, there are at least two other points on each of these lines, say, A and B on l and C and D on m. If any three of these four points are collinear, then one of these four points would have to be on both lines and not equal to P. This contradicts Axiom 1, so no three of these points are collinear.

The dual statement is that there exist four lines, no three of which are concurrent.

pg. 46 prob. 11

A difference set in Zm is a subset S = {a1, ..., an+1} such that for all non-zero k in Zm there is a unique pair, ai, aj in S such that k = ai - aj (mod m). In the following exercises assume that S = {a1, ..., an+1} is a difference set in Zm.
  1. Show that m = n2 + n + 1.
  2. Define P' = {P1,...,Pm} and L' = {l1,...,lm} and show that (P',L') is a projective plane of order n where Pi is on lj if and only if i + j is in S.
  3. Find a three element difference set in Z7, and use it to construct a projective plane of order 2. Show that the plane is isomorphic to the Fano plane.
  4. Construct a projective plane of order 4 from the difference set {0,1,4,14,16} in Z21.


  1. There are (n+1)(n) = n2 + n ordered pairs (corresponding to differences) of elements of S, and each of the m-1 non-zero elements of Zm is associated to exactly one of these ordered pairs, so m-1 = n2 + n.
  2. Let Pi and Pi' be two distinct points. We need to show that there is a unique j so that i+j and i'+j are in S (thus Pi and Pi' determine a unique line lj). Since i and i' are not equal, i-i' is not 0, so there exist as and at in S with i-i' = as - at. Thus, i - as = i'-at = -j is the unique solution we are looking for. Similarly, let lj and lj' be two distinct lines. We want to find an i so that i + j and i + j' are in S. There exist au and av in S so that j - j' = au - av. Thus, j - au = j' -av = -i gives the required solution. Finally, if n > 1(which we must assume), there are at least 3 points on each line since all lines have n+1 points (the size of S) and m = n2 + n + 1 > 3, so there are at least two lines. (P',L') is thus a projective plane.
  3. Using S = {0,1,3} gives the following lines: {0,1,3}, {1,2,4}, {2,3,5}, {3,4,6}, {4,5,0}, {5,6,1}, and {6,0,2}. Using the example of the Fano plane given on page 21 (Example 5), we set up the isomorphism A = 0, B = 1, C = 3, D = 2, E = 5, F = 4, G = 6.
  4. The points are 0,...,20 and the 21 lines are: {0,1,4,14,16}, {1,2,5,15,17}, {2,3,6,16,18}, {3,4,7,17,19}, {4,5,8,18,20}, {5,6,9,19,0}, {6,7,10,20,1}, {7,8,11,0,2}, {8,9,12,1,3}, {9,10,13,2,4}, {10,11,14,3,5}, {11,12,15,4,6}, {12,13,16,5,7},{13,14,17,6,8}, {14,15,18,7,9}, {15,16,19,8,10}, {16,17,20,9,11}, {17,18,0,10,12}, {18,19,1,11,13}, {19,20,2,12,14}, and {20,0,3,13,15}.