The Classification of Symmetric Transversal Designs STD4[12; 3]’s

Abstract

In this article we prove that there is only one symmetric transversal design STD 4 [12; 3] up to isomorphism. We also show that the order of the full automorphism group of STD 4 [12; 3] is 2 5 · 3 3 and Aut STD 4 [12; 3] has a regular subgroup as a permutation group on the point set. We used a computer for our research.

A contraction of square transversal designs

“…(see [9]), , j 12). It follows that we can put A = (A ij ), where A ij (1 i, j 12) is a 6 × 6 matrix represented in the following form:…”

“…By Remark 3.8, the submatrix A ij of A corresponding to i and j (1 i, j 36) is equal to one of the following 2 × 2 matrices as the dual of D is also assumed to be a transversal design obtain a 36 × 36 incidence matrix M = (m ij ) of an STD4 [12, 3], where m ij = 0, if A ij = O, and 1, otherwise. On the other hand, there is a unique STD 4 [12, 3] by a result of[9]. Hence we may assume that M is represented in the following form using the three types of 3 × 3 matrices…”

On D. G. Higman's note on regular 3-graphs

“…We used classifications of generalized Hadamard matrices with suitable parameters (see [3], [9] and [14]) to construct all the corresponding non-isomorphic cyclic covers using the Klin-Pech method, which provide different regular 3-graphs. A summary of our new constructions of regular 3-graphs:…”

On ${\rm STD}_6[18,3]$'s and ${\rm STD}_7[21,3]$'s Admitting a Semiregular Automorphism Group of Order 9

“…We can easily check that n 1 = 1. We also checked that n 2 = 1 by a similar manner as in [13] without a computer, but we do not give the proof in this paper. Remark 7.4).…”

“…Let n λ be the number of nonisomorphic STD λ [3λ, 3]'s. It is known that n 1 = 1, n 2 = 1, n 3 = 4( [12]), n 4 = 1( [13]), and n 5 = 0 ( [5]). We can easily check that n 1 = 1.…”