The nonexistence of projective planes of order 12 with a collineation group of order 8

Abstract

Abstract:In this article, we prove that there does not exist a symmetric transversal design STD 2 [12; 6] which admits an automorphism group of order 4 acting semiregularly on the point set and the block set. We use an orbit theorem for symmetric transversal designs to prove our result. As a corollary of the result, we prove that there is no projective plane of order 12 admitting a collineation group of order 8.

Coloring and extremal problems in combinatorics

“…• Propositions 4.8, 4.9, 4.10, and Theorem 4.8 imply that f (n, ) ≥ ex(n, ) as well as f (n, 2 [2] ) < ex(n, 2 [2] ). As such, there is no inequality possible between f (n, H) and ex(n, H) for all n and H.…”

“…Unfortunately, there can be no similar stability result in the vein of Conjecture 4.1. We see this by examining ex(n, 2 [2] ). If true, Conjecture 4.1 gives ex(n, 2 [2]…”

“…For a pictorial representation of the projective plane of order 2, see Figure 2 examples of projective planes known are of prime power order. The smallest n for which it is not known whether there is a projective plane of that order is n = 12, although the work that has been done in [58], [71], [59], [1], and [2] suggests that such a structure does not exist. that is, the only colored triangles allowed are those which are monochromatic or totally multicolored.…”

“…N + 2 n ⌊n/2⌋ + n ⌊n/2⌋ + 1 Proposition 4.9. If n is a positive integer, N + 2 ≤ f (n, 2 [2] ) ≤ ex(n, V 4 ) + 2.…”

“…We construct a 5-coloring We are now ready to prove Proposition 4.9. There can be no rainbow copy of 2 [2] (induced or otherwise) under this coloring, since every 3-chain will contain at least 2 sets colored the same.…”