On 2-closures of rank 3 groups

0 Abstract

The largest permutation group on $\Omega$ having the same orbits as $G$ on $\Omega \times \Omega$ is called the $2$-closure of $G$, denoted by $G^{(2)}$.

A description of $2$-closure of rank $3$ groups is given.

1 Introduction

The orbits of $G$ acting on $\Omega \times \Omega$ are called $2$-orbits. If a rank $3$ group has even order, then its non-diagonal $2$-orbit induces a strongly regular graph on $\Omega$, which is called a rank $3$ graph. The strongly regular graph is defined here.

It is mentioned here and I give a proof as following.

Transclude of 2.0-for-winter-holiday#^rd1lbz

\begin{proof} Claim: any adjacent points has a common neighborhood of a constant cardinality. Suppose $O$ is an $2$-orbit and $(b,c_1),(b,c_2)\in O$. Let $S$ be the neighborhood of $a$ and $b,c_1,c_2\in S$. Then the cardinality of common neighborhoods of $b,c_1$ and $b,c_2$ can be expressed as $|S^{g_1}\cap S^{g_1^2}|$ and $|S^{g_2}\cap S^{g_2^2}|$ respectively, where $g_1:(a,b)\to (b,c_1)$ and $g_2:(a,b)\to (b,c_2)$. Since $g_1{g}_2^{-1}\in G_a$, $S^{g_1{g}_2^{-1}}=S$. Then $$ |S^{g_1}\cap S^{g_1^2}|=|S\cap S^{g_1}|=|S\cap S^{g_2}|=|S^{g_2}\cap S^{g_2^2}|

Now we prove the claim for non-adjacent points. Suppose $(c_1,c_2)$ and $(c_3,c_4)$ are not in the set $S$. Then there is $g_1,g_2,g$ satisfying $g:(c_1,c_2)\to (c_3,c_4)$ and $g_i:a\to c_i,i=1,2$. It remains to show $|S^{g_1}\cap S^{g_2}|=|S^{g_1g}\cap S^{g_2g}|$, which is trivial. Since $G$ has even order, there is a $g\in G$ such that $g^2=1$. Then for $b:=a^g$, both $(a,b)$ and $(b,a)$ are in the set $S$ and so the graph is undirected graph. `\end{proof}` Note that an arc-transitive strongly regular graph need not be a rank $3$ graph, since its automorphism group might be intransitive on non-arcs. Note also that given a rank $3$ graph $\Gamma$ corresponding to the rank $3$ group $G$, we have $\mathrm{Aut}(\Gamma)=G^{(2)}$. The main theorem describe the rank $3$ permutation group on a set $\Omega$ and its $2$-closure. The key point is to know which rank $3$ groups give rise to isomorphic graphs. > [!theorem] > > ![[Pasted image 20240210015103.png]] "The proof can be divided into three parts": which i did't understand. Then it gives a standalone result: > [!theorem] > > ![[Pasted image 20240211020323.png]] # 2 Reduction to affine case