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3‐Designs From PSL(2,q) $\,\text{PSL}\,(2,q)$ With Cyclic Starter Blocks.

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Title: 3‐Designs From PSL(2,q) $\,\text{PSL}\,(2,q)$ With Cyclic Starter Blocks.
Authors: Hanaki, Akihide1 (AUTHOR) hanaki@shinshu-u.ac.jp, Kobayashi, Kenji2 (AUTHOR), Munemasa, Akihiro3 (AUTHOR)
Source: Journal of Combinatorial Designs. Nov2025, p1. 11p.
Subject Terms: *COMBINATORIAL designs & configurations, *CYCLIC groups, *FINITE fields, *PROJECTIVE geometry, *BLOCK designs, *MATHEMATICAL equivalence
Abstract: ABSTRACT We consider when the projective special linear group over a finite field defines a block‐transitive 3‐design with a starter block which is a multiplicative subgroup of the field. For a prime power q≡1(mod20) $q\equiv 1\,(\mathrm{mod}\,20)$, we will show that the multiplicative subgroup of order 5 is a starter block of a 3‐design if and only if the multiplicative subgroup of order 10 is a starter block of a 3‐design. The former is the family of 3‐(q+1,5,3) $(q+1,5,3)$ designs investigated by Li, Deng and Zhang, while the latter appear in a different context by Bonnecaze and Solé for the case q=41 $q=41$. We also show a similar equivalence for multiplicative subgroups of orders 13 and 26 for a prime power q≡1(mod52) $q\equiv 1\,(\mathrm{mod}\,52)$. [ABSTRACT FROM AUTHOR]
Database: Academic Search Index
Description
Abstract:ABSTRACT We consider when the projective special linear group over a finite field defines a block‐transitive 3‐design with a starter block which is a multiplicative subgroup of the field. For a prime power q≡1(mod20) $q\equiv 1\,(\mathrm{mod}\,20)$, we will show that the multiplicative subgroup of order 5 is a starter block of a 3‐design if and only if the multiplicative subgroup of order 10 is a starter block of a 3‐design. The former is the family of 3‐(q+1,5,3) $(q+1,5,3)$ designs investigated by Li, Deng and Zhang, while the latter appear in a different context by Bonnecaze and Solé for the case q=41 $q=41$. We also show a similar equivalence for multiplicative subgroups of orders 13 and 26 for a prime power q≡1(mod52) $q\equiv 1\,(\mathrm{mod}\,52)$. [ABSTRACT FROM AUTHOR]
ISSN:10638539
DOI:10.1002/jcd.22014