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2020, Volume 14, Issue 4: 603-611. Doi: 10.3934/amc.2020033

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Designs from maximal subgroups and conjugacy classes of Ree groups

1.
School of Mathematical Sciences, North-West University, (Mafikeng) 2754, South Africa
2.
School of Mathematics, Statistics and Computer Science University of KwaZulu-Natal, Durban 4000, South Africa

^* Corresponding author: Seiran Zandi
^* Corresponding author: Seiran Zandi

Received: August 2018

Revised: August 2019

Early access: November 2019

Published: November 2020

The first author acknowledges support of NRF and NWU (Mafikeng).
The second author acknowledges support of NRF through Grant Numbers 95725 and 106071.
The third author acknowledges support of NWU (Mafikeng) postdoctoral fellowship.
The fourth author acknowledges support of NRF postdoctoral fellowship through Grant Number 91495

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Abstract

In this paper, using a method of construction of $ 1 $-designs which are not necessarily symmetric, introduced by Key and Moori in [5], we determine a number of $ 1 $-designs with interesting parameters from the maximal subgroups and the conjugacy classes of the small Ree groups $ ^2G_2(q) $. The designs we obtain are invariant under the action of the groups $ ^2G_2(q) $.

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Mathematics Subject Classification: 20D05, 05E15, 05E20.

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Table 1. Non-trivial designs from $G = Ree(q)$ using construction Method 2

$Max$	$t =o(x)$	$v = \|x^G\|$	$k =\| M \cap x^G\|$	$ \lambda= \chi_{M_i}(x)$
$M_1$	$t=2$	$q^2(q^2-q+1)$	$q^2$	$q+1$
$M_1$	$t=3$	$(q^3+1)(q-1)$	$q-1$	1
$M_1$	$t=3$	$\frac{q(q^3+1)(q-1)}{2}$	$\frac{q(q-1)}{2}$	1
$M_1$	$t=9$	$\frac{q^2(q^3+1)(q-1)}{3}$	$\frac{q^2(q-1)}{3}$	1
$M_1$	$t=6$	$\frac{q^2(q^3+1)(q-1)}{2}$	$\frac{q^2 (q-1)}{2}$	1
$M_1$	$t \|(q-1)$, $t \ne 2$	${q^3}(q^3+1)$	$2q^3$	2
$M_2, M_3$	$ t=2$	$q^2(q^2-q+1) $	$ q^{\mp}$	$\frac{{{q(q^2-1)}}}{6}$
$M_2, M_3$	$t=3$	$\frac{q(q^3+1)(q-1)}{2}$	$ q^{\mp}$	$\frac{q^2}{3}$
$M_2, M_3$	$t=6$	$\frac{q^2(q^3+1)(q-1)}{2}$	$ q^{\mp}$	$\frac{q}{3}$
$M_2, M_3$	$t \| q^{\mp}$	${q^3(q^2-1)q^{\pm}}$	$ 6$	$1$
$M_4$	$ t=2$	$q^2(q^2-q+1) $	$ q^2-q+1$	$q^2-q+1$
$M_4$	$t=3$	$\frac{q(q^3+1)(q-1)}{2}$	$ \frac{q^2-1}{2}$	$q$
$M_4$	$t=6$	$\frac{q^2(q^3+1)(q-1)}{2} $	$ \frac{q^2-1}{2}$	$1$
$M_4$	$t \|(q-1)$, $t \ne 2$	$q^3(q^3+1)$	$ q(q+1)$	$1$
$M_4$	$t \|\frac{q+1}{2}$, $t \ne 2$	${q^3(q^2-q+1)(q-1)}$	$ 3q(q-1))$	$3$
$M_5$	$t=2$	$q^2(q^2-q+1) $	$ q+4$	$\frac{{{q(q-1)(q+4)}}}{6}$
$M_5$	$ t=3$	$\frac{{{{q}(q^3+1)(q-1)}}}{2}$	$ q+1$	$ \frac{{{q^2}}}{3}$
$M_5$	$ t=6$	$\frac{{{{q^2}(q^3+1)(q-1)}}}{2}$	$ q+1$	$ \frac{{{q}}}{3}$
$M_5$	$t \|\frac{q+1}{2}$, $t \ne 2$	${q^3(q^2-q+1)(q-1)}$	$ 6$	1
$M_6$	$t=2$	$q^2(q^2-q+1) $	$ q_0^2(q_0^2-q_0+1)$	$\frac{q(q^2-1)}{q_0(q_0^2-1)}$
$M_6$	$ t=3$	$(q^3+1)(q-1)$	$(q_0^3+1)(q_0-1)$	$ \frac{{{q^3}}}{q_0^3}$
$M_6$	$ t=3$	$\frac{q(q^3+1)(q-1)}{2}$	$\frac{q_0(q_0^3+1)(q_0-1)}{2}$	$ \frac{{{q^2}}}{q_0^2}$
$M_6$	$t= 9$	$\frac{q^2(q^3+1)(q-1)}{3}$	$\frac{q_0^2(q_0^3+1)(q_0-1)}{3}$	$ \frac{{{q}}}{q_0}$
$M_6$	$t= 6$	$\frac{q^2(q^3+1)(q-1)}{2}$	$\frac{q_0^2(q_0^3+1)(q_0-1)}{2}$	$ \frac{{{q}}}{q_0}$
$M_6$	$t \|(q_0-1)$, $t \ne 2$	${q^3(q^3+1)}$	${q_0^3(q_0^3+1)}$	$ \frac{q-1}{q_0-1} $
$M_6$	$t \|\frac{q_0+1}{2}$, $t \ne 2$	${q^3(q^2-q+1)(q-1)}$	${q_0^3(q_0^2-q_0+1)(q_0-1)}$	$ \frac{q+1}{q_0+1} $
$^*M_6$	$t\|q_0^{\pm}$	${q^3}\left( {q^2 - 1} \right){q^{\pm}} $	${q_0^3}\left( {q_0^2 - 1} \right){q_0^{\pm}} $	$ \frac{{q^{\mp}}}{{q_0^{\mp}}} $
$^{**}M_6$	$t\|q_0^{\pm}$	${q^3(q^3+1)}$	${q_0^3}\left( {q_0^2 - 1} \right){q_0^{\pm}} $	$\frac{q-1}{q_0^{\mp}}$

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