The Fano plane is the finite projective plane of order 2, having the smallest possible number of points and lines, 7 each, with 3 points on every line and 3 lines through every point.

Assume that we want to make Fano planes with the numbers from $1$ to $35$.

How many Fano planes can we make in this way?

Notice that for every two Fano planes like $F_1$ and $F_2$ , $F_1$ and $F_2$ must not share any blocks.

Note : What i tried ...
I said we can make $35 \choose 3$ blocks at all ( Call that number $Z$).
Every time that we want to build a Fano plane, We choose $Z \choose 7$ blocks such that the union of these blocks has exactly $7$ members. But the problem is that I don't know how to control this process.

Thanks in advance.

  • $\begingroup$ Hint: the projective space $PG(3,2)$ has 35 points. $\endgroup$ – Chris Godsil May 13 '16 at 23:22
  • $\begingroup$ @ChrisGodsil Thank you sir :) that helped me a lot ! $\endgroup$ – Arman Malekzadeh May 14 '16 at 2:10
  • $\begingroup$ @ChrisGodsil Hmm.. But if we take all the Fano planes in this space, every pair shares a block. $\endgroup$ – Morgan Rodgers May 14 '16 at 16:20
  • $\begingroup$ @ChrisGodsil Sir... I thought about that... can u please explain it more as an answer? that was not clear to me $\endgroup$ – Arman Malekzadeh May 14 '16 at 17:15
  • $\begingroup$ Note that Chris Godsil mistyped above, the projective space $PG(3,2)$ has 35 lines, not points. If you identify the lines of $PG(3,2)$ with your pointset $\{1\ldots35\}$, it is not hard to use the existing points and planes in $PG(3,2)$ to define 15 Fano structures on this set of "points". $\endgroup$ – Jeremy Dover Jun 2 '16 at 12:47

Let $L$ be the set of $35$ lines of $PG(3,2)$. For each plane $\pi$ contained in $PG(3,2)$, define $P_\pi$ to be the set of $7$ lines contained in $\pi$, and for each point $p \in \pi$, define $L_p = \{\ell \in P_\pi:p \in \ell\}$. Then the incidence structure $(P_\pi,L_\pi = \{L_p:p \in \pi\})$ is a Fano plane (axioms follow from the dual statements for $\pi$). Varying this construction over the $15$ planes in $PG(3,2)$ yields $15$ block-disjoint Fano planes on a set of $35$ points.

This is almost definitely not the maximum, but it is a neat construction.

  • $\begingroup$ Nice attack to solve the problem ! $\endgroup$ – Arman Malekzadeh Jun 2 '16 at 13:14

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