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$7_3$ also know as Fano plane. How can I prove that only 1 configuration exists for 7 points and 7 lines with 3 points on every line and 3 lines at every point. I would think an incidence matrix would help, but I'm not sure how to begin. I know a lot of $10_3$ configurations exist, but why only 1 Fano plane configuration. Likewise, I need to figure out how to prove $8_3$ has a unique configuration as well. Any help?

Mark Bennet
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cakeyone
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  • What do you mean by $8_3$? Is it that there are three lines incident with each point and three points on each line? The Fano plane has the additional property that any two points define a unique line - so it is a finite projective plane. If what I suggest is correct the lines incident with $1$ are wlog $(1,2,3)(1,4,5)(1,6,7)$. There must be a line incident with $8$ which crosses two of these, wlog $(8,2,5)$. This forces the other lines through $8$ to be $(8,4,7)(8,6,3)$, then $(2,4,6)(3,5,7)$. You need to prove (or define) that there isn't a pair of lines with two points in common. – Mark Bennet Feb 09 '14 at 18:17

1 Answers1

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For the Fano plane it is easy just to be systematic.

Choose a line - it contains three points $P_1,P_2,P_3$

Choose a second line having a point in common with the first $P_1,P_4,P_5$

Choose a third line through $P_2$ and $P_4$ - and also therefore $P_6$ (noting that any pair of points defines a distinct line) $P_2,P_4,P_6$.

The fourth line is (necessary) the third line through $P_1$, which must be $P_1,P_6,P_7$ ($P_1$ is already joined to $P_2,P_3,P_4,P_5$)

The fifth line is the third line through $P_2$ which must be $P_2, P_5, P_7$

The sixth line is the third line through $P_4$ which must be $P_3,P_4,P_7$

The seventh line is necessarily $P_3, P_5, P_6$

Mark Bennet
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