I was reading about the Fano plane, the smallest possible projective plane. After playing around with it, it seems that any projective plane of 7 points will be isomorphic to the Fano plane.
However, I've always been troubled with showing isomorphisms between sets of points and lines, because it seems like any function between two sets is really just an assignment of points to other points that preserves lines, but there is no nice fixed function that maps elements to others based on some hard and fast rule, like say $x\mapsto x^2$ or $(x,y)\mapsto x-y$, etc.
Is there an elegant way to show that any projective plane of 7 points is necessarily isomorphic to the Fano plane? I could only think of exhausting all permutations of points and lines and saying "Look, if these points $A$, $B$, $C$ are on a line $l$ here, then $f(A), f(B), f(C)$ are on a line $f(l)$ here." But this seems like it's brute forcing the matter, and not very efficient at all, considering that there are many ways to connect the points. What is the best way to go about something like this? Thanks.
Best Answer
It seems you can use the enormous symmetry of the Fano plane for this. You can pick a line $ABC$ and a point off the line $D$ and pick $f(A)$ and $f(B)$ arbitrarily, $f(C)$ is the third point on the line $AB$, and pick $f(D)$ arbitrarily different from the first three. Then if the lines through $D$ are $ADE$, $BDF$, and $CDG$ you can define $f(E)$, $f(F)$ and $f(G)$ in the obvious way. Now you just need to prove that $f(CEF)$, $f(AGF)$, and $f(BEG)$ are lines. But there has to be a meet between the lines $f(CE)$ and $f(AG)$ and it can't be $f(B)$ or $f(D)$, so must be $f(F)$.
Added: support for this comes from that fact that any isomorphism should be able to be composed with all the automorphisms of the destination plane to make another. This isomorphism has $7*6*4$ possibilities as the first, second, and fourth points can be chosen at random and $7*6*4=168$ is the number of automorphisms of the Fano plane.