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Pappus and Desargues solution

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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