On the automorphism group of a putative Conway 99-graph

TL;DR

This paper refines the understanding of the automorphism group of a specific strongly regular graph called the Conway 99-graph, showing that divisibility by 7 constrains the group to be cyclic of order 7, and divisibility by 2 constrains it to be one of three small groups.

Contribution

It proves that if the automorphism group of the Conway 99-graph is divisible by 7, then it is cyclic of order 7, and if divisible by 2, then it is one of Z2, Z6, or S3.

Findings

Divisibility by 7 implies the automorphism group is Z7.

Divisibility by 2 implies the automorphism group divides 6.

Automorphism group is constrained to small, well-understood groups.

Abstract

Let $\Gamma$ be a {Conway 99-graph}, that is, a strongly regular graph with parameters $(99,14,1,2)$. In Makhnev and Minakova (On automorphisms of strongly regular graphs with parameters $\lambda =1$, $\mu= 2$, Discrete Math.\ Appl.\ 14 (2) (2004) 201-210), the authors prove that the automorphism group $G$ of $\Gamma$ must have order dividing $2\cdot 3^3\cdot 7\cdot 11$. They further show that if $|G|$ is divisible by $2$ then $|G|$ must divide $42$. In the present paper, we refine these results by proving that divisibility by $7$ implies $G \cong\mathbb Z_7$. As a consequence, divisibility by $2$ implies $|G|$ divides $6$, \ie $G$ is isomorphic to one of $\mathbb Z_2, \mathbb Z_6, S_3$.