So far, I have put up the following posts related to finite projective planes:

Primitive roots

Finite fields

Finite fields (2)

Latin squares, solutions, more solns

Desargues theorem

Projective planes, solutions

Pappus and Desargues

Bruck-Ryser (1)

Bruck-Ryser (2)

The objective is to give the apparently relevant background to the long-standing conjecture that there are no projective planes of order \(n\) unless \(n\) is a prime power.

The material still to come is:

(1) Pappus implies Desargues (this post);

(2) Wedderburn’s theorem

(3) Desargues implies the plane is over a division ring

(4) Pappus implies the plane is over a field

(5) Stinson’s proof that there are no orthogonal Latin squares of order 6

(6) the construction of non-Desarguesian planes

(7) Lee’s proof that there are two orthogonal Latin squares for any order \(n\gt 6\)

Showing that Pappus implies Desargues was set as a problem here.

To recap, the Pappus configuration is:

0 1 2 = \(L_0\)

0 3 6 = \(L_1\)

0 4 7 = \(L_2\)

1 3 8 = \(L_3\)

2 5 6 = \(L_4\)

4 6 8 = \(L_5\)

1 4 5 = \(L_6\)

2 7 8 = \(L_7\)

3 5 7 = \(L_8\)

If we remove any one of these lines from the list, then its existence can be recovered from the other 8. The Desargues configuration is:

0 1 2 = \(L_7\)

0 3 4 = \(L_8\)

0 5 6 = \(L_9\)

7 8 9 = \(L_0\)

3 5 7 = \(L_2\)

4 6 7 = \(L_1\)

1 5 8 = \(L_4\)

2 6 8 = \(L_3\)

1 3 9 = \(L_6\)

2 4 9 = \(L_5\)

Here it is important that the point \(i\) does not lie on any of the lines apart from those listed, except possibly \(L_i\). Again if the line \(L_i\) is removed, then it can be recovered from the other 9.

So suppose that \(L_0\) is removed from the Desargues list. That leaves us with:

0 1 2 = \(L_7\)

0 3 4 = \(L_8\)

0 5 6 = \(L_9\)

3 5 7 = \(L_2\)

4 6 7 = \(L_1\)

1 5 8 = \(L_4\)

2 6 8 = \(L_3\)

1 3 9 = \(L_6\)

2 4 9 = \(L_5\)

Take: point 10 to be the intersection of \(L_3,L_6\); line \(L_{10}\) to be the line through points 3,6; point 11 to be the intersection of \(L_7,L_{10}\); line \(L_{11}\) to be the line through points 0,10; point 12 to be the intersection of \(L_2,L_{11}\); and point 13 to be the intersection of \(L_1,L_{11}\). That takes us to:

\(L_7\) = 0 1 2 11

\(L_8\) = 0 3 4

\(L_9\) = 0 5 6

\(L_2\) = 3 5 7 12

\(L_1\) = 4 6 7 13

\(L_4\) = 1 5 8

\(L_3\) = 2 6 8 10

\(L_6\) = 1 3 9 10

\(L_5\) = 2 4 9

\(L_{10}\) = 3 6 11

\(L_{11}\) = 0 10 12 13

These points and lines are a Pappus configuration:

\(L_7\) = 0 1 11

\(L_9\) = 0 5 6

\(L_{11}\) = 0 10 12

\(L_6\) = 1 3 10

\(L_4\) = 1 5 8

\(L_2\) = 3 5 12

\(L_{10}\) = 3 6 11

\(L_3\) = 6 8 10

from which we deduce the existence of the line

\(L_{12}\) 8 11 12

We also have this Pappus configuration:

\(L_7\) = 0 2 11

\(L_8\) = 0 3 4

\(L_{11}\) = 0 10 13

\(L_5\) = 2 4 9

\(L_3\) = 2 6 10

\(L_{10}\) = 3 6 11

\(L_6\) = 3 9 10

\(L_1\) = 4 6 13

from which we deduce the existence of the line

\(L_{13}\) = 9 11 13

Using these two lines we have finally this Pappus configuration:

\(L_{10}\) = 3 6 11

\(L_2\) = 3 7 12

\(L_6\) = 3 9 10

\(L_1\) = 6 7 13

\(L_3\) = 6 8 10

\(L_{11}\) = 10 12 13

\(L_{12}\) = 8 11 12

\(L_{13}\) = 9 11 13

from which we deduce the existence of the line

\(L_0\) = 7 8 9

as required.

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