\documentclass[ALCO,Unicode]{cedram} \usepackage{mathrsfs, bm, latexsym, stmaryrd, array, bold-extra} \renewcommand{\a}{\alpha} \renewcommand{\b}{\beta} \newcommand{\g}{\gamma} \newcommand{\e}{\varepsilon} \renewcommand{\l}{\lambda} \newcommand{\p}{\varphi} \newcommand{\s}{\sigma} \renewcommand{\L}{\Lambda} \renewcommand{\O}{\Omega} \newcommand{\T}{\Theta} \newcommand{\normeq}{\trianglelefteqslant} \newcommand{\BB}{\mathbb{B}} \newcommand{\F}{\mathbb{F}_{q}} \newcommand{\Nat}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\z}{\mathbb{Z}} \newcommand{\mcC}{\mathcal{C}} \newcommand{\E}{\mathcal{E}} \newcommand{\ms}{\mathcal{S}} \newcommand{\M}{\mathcal{M}} \newcommand{\<}{\langle} \renewcommand{\>}{\rangle} \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\into}{\hookrightarrow} \renewcommand{\to}{\rightarrow} \newcommand{\too}{\longrightarrow} \newcommand{\leqs}{\leqslant} \newcommand{\geqs}{\geqslant} \renewcommand{\o}{\circ} \newcommand{\h}{H^{0}} \newcommand{\fpr}{\mbox{{\rm fpr}}} \newcommand{\ub}{\underline{b}} \newcommand{\perps}{\vdash} \newcommand{\wtilde}{\widetilde} \newcommand{\what}{\widehat} \newcommand{\Aut}{\operatorname{Aut}} \newcommand{\Inn}{\operatorname{Inn}} \newcommand{\Out}{\operatorname{Out}} \newcommand{\soc}{\operatorname{soc}} \newcommand{\PSL}{\operatorname{PSL}} \newcommand{\GL}{\operatorname{GL}} \newcommand{\PGL}{\operatorname{PGL}} \newcommand{\SL}{\operatorname{SL}} \newcommand{\PG}{\operatorname{PG}} \newcommand{\rank}{\operatorname{rank}} \newcommand{\Image}{\operatorname{Im}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\Fix}{\operatorname{Fix}} \newcommand{\PGammaL}{\operatorname{P\Gamma L}} \newcommand{\PSigmaL}{\operatorname{P\Sigma L}} \newcommand{\val}{\operatorname{val}} \newcommand{\PSp}{\operatorname{PSp}} \newcommand{\Sp}{\operatorname{Sp}} \newcommand{\PSU}{\operatorname{PSU}} \newcommand{\POmega}{\operatorname{P\Omega}} \newcommand{\LL}{\operatorname{L}} \newcommand{\UU}{\operatorname{U}} \newcommand{\SU}{\operatorname{SU}} \newcommand{\PGU}{\operatorname{PGU}} \newcommand{\PGammaU}{\operatorname{P\Gamma U}} \newcommand{\GU}{\operatorname{GU}} \newcommand{\suchthat}{\,:\,} \newcommand{\litspace}{\,} \def\L{{\rm L}} \def\Ma{{\rm M}} \makeatletter \newcommand{\imod}[1]{\allowbreak\mkern4mu({\operator@font mod}\,\,#1)} \makeatother \title[Saxl graphs of primitive groups]{On the Saxl graphs of primitive groups\\ with soluble stabilisers} \author[\initial{T.} \middlename{C.} Burness]{\firstname{Timothy} \middlename{C.} \lastname{Burness}} \address{School of Mathematics\\ University of Bristol\\ Bristol\\ BS8 1UG (UK)} \email{t.burness@bristol.ac.uk} \author[\initial{H.} \middlename{Y.} Huang]{\firstname{Hong Yi} \lastname{Huang}} \address{School of Mathematics\\ University of Bristol\\ Bristol\\ BS8 1UG (UK)} \email{hy.huang@bristol.ac.uk} \thanks{Both authors thank Eamonn O'Brien for his assistance with several computations in this paper. The second author thanks the China Scholarship Council for supporting his doctoral studies at the University of Bristol and he thanks the Southern University of Science and Technology (SUSTech) for their generous hospitality during a visit in 2021.} \keywords{Saxl graph, primitive group, base, soluble stabiliser} \subjclass{20B15, 20E32, 20E28, 05C40} \datereceived{2021-08-23} \daterevised{2022-03-14} \dateaccepted{2022-03-16} \begin{document} \begin{abstract} Let $G$ be a transitive permutation group on a finite set $\O$ and recall that a base for $G$ is a subset of $\O$ with trivial pointwise stabiliser. The base size of $G$, denoted $b(G)$, is the minimal size of a base. If $b(G)=2$ then we can study the Saxl graph $\Sigma(G)$ of $G$, which has vertex set $\O$ and two vertices are adjacent if and only if they form a base. This is a vertex-transitive graph, which is conjectured to be connected with diameter at most $2$ when $G$ is primitive. In this paper, we combine probabilistic and computational methods to prove a strong form of this conjecture for all almost simple primitive groups with soluble point stabilisers. In this setting, we also establish best possible lower bounds on the clique and independence numbers of $\Sigma(G)$ and we determine the groups with a unique regular suborbit, which can be interpreted in terms of the valency of $\Sigma(G)$. \end{abstract} \maketitle \section{Introduction}\label{s:intro} Let $G \leqs {\rm Sym}(\O)$ be a transitive permutation group on a finite set $\O$ with point stabiliser $H$ and recall that a \emph{base} for $G$ is a subset of $\O$ with trivial pointwise stabiliser. In turn, the \emph{base size} of $G$, denoted $b(G)$, is the minimal size of a base. This is a classical concept in permutation group theory, which arises naturally in a wide range of contexts. There is a long history of studying bases, which stretches all the way back to the nineteenth century, and there are many applications and connections to other areas of algebra and combinatorics. We refer the reader to the survey articles \cite{BC,LSh3} and \cite[Section 5]{Bur181} for further details. Historically, there has been a focus on studying bases for finite primitive permutation groups and there have been several major advances in this direction in recent years. For example, a number of highly influential conjectures due to Babai, Cameron, Kantor and Pyber from the 1990s have been resolved, which in turn has opened up new directions of research. For instance, Seress \cite{Seress} proved that $b(G) \leqs 4$ if $G$ is a finite soluble primitive group, which established a strong form of Pyber's base size conjecture in this setting (in \cite{Pyber}, Pyber conjectured that there exists an absolute constant $c$ such that $b(G)$ is at most $c \log_{n}|G|$ for every primitive group $G$ of degree $n$ and the proof was finally completed in \cite{DHM}). Seress' theorem has recently been extended in \cite{Burness2020base}, where the main result shows that $b(G) \leqs 5$ for every finite primitive group with soluble point stabilisers (it is worth noting that the bounds in \cite{Burness2020base} and \cite{Seress} are best possible). Moreover, the exact base size is computed in \cite{Burness2020base} for all almost simple primitive groups with soluble point stabilisers (recall that $G$ is \emph{almost simple} if there exists a non-abelian simple group $G_0$ such that $G_0 \normeq G \leqs {\rm Aut}(G_0)$; here $G_0$ is the \emph{socle} of $G$). Further motivation stems from an ambitious project initiated by Jan Saxl in the 1990s. Here the main aim is to classify all the base-two finite primitive permutation groups, where a group $G$ is said to be \emph{base-two} if $b(G)=2$. These groups arise naturally in many applications of bases to other problems. For example, see \cite{ExtremelyPrimitive} on the classification of extremely primitive groups and \cite{BurnessHarper} for applications concerning the $2$-generation properties of almost simple groups. Although a complete classification of the base-two primitive groups remains out of reach, there has been some significant progress. For instance, we refer the reader to \cite{faw2,faw1} for work of Fawcett on diagonal-type groups and twisted wreath products, respectively, and there are various partial results for affine-type groups (see \cite{aff1,aff2,Lee1,Lee2}, for example). Similarly, we refer the reader to \cite{Burness2020base,BGS0,BGS,BOW} for results towards a classification of the base-two almost simple groups. Let $G \leqs {\rm Sym}(\O)$ be a base-two finite transitive permutation group with point stabiliser $H$. In \cite{SaxlGraph}, Burness and Giudici define the \emph{Saxl graph} $\Sigma(G)$ of $G$ as follows: the vertex set is $\O$ and two vertices are joined by an edge if and only if they form a base for $G$. Clearly, $\Sigma(G)$ is vertex-transitive and it is easy to see that $\Sigma(G)$ is connected if $G$ is primitive (as discussed in \cite{SaxlGraph}, if $G$ is imprimitive then $\Sigma(G)$ can have arbitrarily many connected components). In addition, $\Sigma(G)$ is a complete graph if and only if $G$ is a Frobenius group. Various problems concerning the valency and connectedness properties of Saxl graphs are investigated in \cite{SaxlGraph} and we refer the reader to \cite{ChenHuang2020val} for further results on the valency of $\Sigma(G)$ when $G$ is primitive. A number of open problems concerning Saxl graphs are presented in \cite{SaxlGraph} and the aim of this paper is to address some of these questions for the permutation groups in the collection denoted $\mathcal{G}$, which is defined as follows: \[ \mathcal{G} = \{\mbox{\emph{finite almost simple primitive base-two groups with soluble point stabilisers}}\}. \] This is a natural collection to consider in this setting because one of the main results in \cite{Burness2020base} provides a complete classification of the groups in $\mathcal{G}$. We will also need to highlight the following subcollection, where $G_0$ denotes the socle of $G$ and $H$ is a point stabiliser: \[ \mathcal{L} = \{ G \in \mathcal{G} \suchthat\mbox{$G_0 = {\rm L}_{2}(q)$, $H$ is of type ${\rm GL}_1(q) \wr S_2$ or ${\rm GL}_{1}(q^2)$} \}. \] In terms of Aschbacher's subgroup structure theorem \cite{asch}, the subgroups $H$ in the definition of $\mathcal{L}$ comprise the collections denoted $\mathcal{C}_2$ and $\mathcal{C}_3$. Perhaps the most intriguing open problem in this area is \cite[Conjecture 4.5]{SaxlGraph}, which asserts that if $G$ is primitive, then any two vertices in $\Sigma(G)$ have a common neighbour. In particular, this implies that the diameter of $\Sigma(G)$ is at most $2$ for every base-two finite primitive group $G$. Evidence for the conjecture is presented in \cite[Sections 4-6]{SaxlGraph}, where it is verified in several special cases. Our first main result establishes \cite[Conjecture 4.5]{SaxlGraph} for the groups in $\mathcal{G}$. This extends recent work of Chen and Du \cite{ChenDu2020Saxl}, who have shown that $\Sigma(G)$ has diameter $2$ for every base-two almost simple primitive group with socle ${\rm L}_{2}(q)$. \begin{theorem}\label{t:main1} Let $G \leqs {\rm Sym}(\O)$ be a permutation group in $\mathcal{G}$. Then any two vertices in $\Sigma(G)$ have a common neighbour. In particular, $\Sigma(G)$ has diameter $2$. \end{theorem} \begin{rema} It is straightforward to extend our methods in order to establish \cite[Conjecture 4.5]{SaxlGraph} for all base-two almost simple primitive groups with socle ${\rm L}_{2}(q)$; we refer the reader to Theorem \ref{t:further}. \end{rema} Next we turn to the clique number of $\Sigma(G)$, which is denoted $\omega(G)$. Recall that this is the maximal size of a complete subgraph. By Theorem \ref{t:main1} we immediately deduce that $\omega(G) \geqs 3$ for every group $G$ in $\mathcal{G}$ and we can establish a stronger lower bound. \begin{theorem}\label{t:main11} Let $G \leqs {\rm Sym}(\O)$ be a permutation group in $\mathcal{G}$ with socle $G_0$ and assume that either $G \in \mathcal{G} \setminus \mathcal{L}$ or $G \leqs {\rm PGL}_{2}(q)$. Then $\omega(G) \geqs 4$, with equality if and only if $G = A_5$ and $\O$ is the set of $2$-element subsets of $\{1,\ldots, 5\}$. \end{theorem} \begin{rema} Let us record a couple of comments on the statement of Theorem \ref{t:main11}. \begin{enumerate}[label = (\alph*)] \item First note that if $G = A_5$ and $\O$ is the set of $2$-element subsets of $\{1,\ldots, 5\}$, then $\Sigma(G)$ is the complement of the Petersen graph, which also coincides with the Johnson graph $J(5,2)$. \item We expect that the bound $\omega(G) \geqs 5$ also holds for the groups $G \in \mathcal{L}$ with $G \not\leqs {\rm PGL}_{2}(q)$, but we have not been able to verify this in all cases. Here the main difficulty involves constructing an explicit clique of size $5$ when $G$ contains field automorphisms, working with a suitable geometric description of $\O$ (for example, if $H$ is of type ${\rm GL}_1(q) \wr S_2$, then we may identify $\O$ with the set of pairs of distinct $1$-dimensional subspaces of the natural module for $G_0$). With the aid of {\sc Magma} \cite{Magma}, we have verified the bound $\omega(G) \geqs 5$ for all groups in $\mathcal{L}$ with $5 |G| - |H|^2 \] and thus $r = |\{x \in S \suchthat |HxH| = |H|^2\}|$. It is straightforward to implement all of these methods in {\sc Magma}. \section{Random bases}\label{s:random} In this section we will prove the following theorem, which will be a key ingredient in the proofs of Theorems \ref{t:main1} and \ref{t:main11}. We adopt the notation $\mathcal{G}$ and $\mathcal{L}$ introduced above. In particular, we recall that the socle $G_0$ of a group in $\mathcal{G} \setminus \mathcal{L}$ satisfies the restrictions in \eqref{e:excl}. \begin{thm}\label{t:random} Let $G \leqs {\rm Sym}(\O)$ be a permutation group in $\mathcal{G} \setminus \mathcal{L}$ with socle $G_0$ and point stabiliser $H$. Then either $Q(G)<1/4$, or $(G,H)$ is one of the cases in Tables $\ref{tab:random}$ and $\ref{tab:random2}$. \end{thm} {\small \begin{table} \[ \begin{array}{llll} \hline G & H & r & Q(G)\\ \hline A_9 & {\rm ASL}_{2}(3) & 2 & 17/35\\ S_7 & {\rm AGL}_{1}(7) & 1 & 13/20\\ {\rm M}_{11} & 2.S_4 & 1 & 39/55\\ {\rm M}_{12} & A_4 \times S_3 & 13 & 16/55\\ {\rm M}_{12}.2 & S_4 \times S_3 & 4 & 31/55\\ {\rm J}_{1} & 19{:}6 & 9 & 257/770\\ & D_6 \times D_{10} & 34 & 443/1463\\ {\rm J}_{2} & 5^2{:}D_{12} & 2 & 59/84\\ {\rm J}_{2}.2 & 5^2{:}(4 \times S_3) & 1 & 59/84\\ {\rm J}_{3}.2 & 3^{2+1+2}{:}8.2 & 3 & 886/1615\\ & 2^{2+4}{:}(S_3 \times S_3) & 10 & 457/969\\ {\rm HS}.2 & 5^{1+2}.[2^5] & 3 & 106/231\\ {\rm McL}.2 & 2^{2+4}{:}(S_3 \times S_3) & 228 & 9419/28875\\ {\rm He}.2 & 2^{4+4}.(S_3 \times S_3).2 & 5 & 23011/29155\\ {\rm Suz} & 3^{2+4}{:}2.(A_4 \times 2^2).2 & 16 & 7529/25025\\ {\rm Suz}.2 & 3^{2+4}{:}2.(S_4 \times D_8) & 4 & 16277/25025\\ {\rm HN} & 5^{1+4}{:}2^{1+4}.5.4 & 47 & 332152/1066527\\ {\rm HN}.2 & 5^{1+4}{:}2^{1+4}.5.4.2 & 22 & 34457/96957\\ {}^2B_2(8) & 13{:}4 & 7 & 7/20\\ {}^2B_2(8).3 & 13{:}12 & 2 & 31/70\\ {}^2F_4(2)' & 5^2{:}4A_4 & 6 & 27/52\\ {}^2F_4(2) & 5^2{:}4S_4 & 3 & 27/52\\ G_2(3) & ({\rm SL}_{2}(3) \circ {\rm SL}_{2}(3)).2 & 4 & 563/819\\ G_2(3).2 & ({\rm SL}_{2}(3) \circ {\rm SL}_{2}(3)).2.2 & 1 & 691/819\\ \hline \end{array} \] \caption{The groups in $\mathcal{G} \setminus \mathcal{L}$ with $Q(G) \geqs 1/4$, part I} \label{tab:random} \end{table}} {\small \begin{table} \[ \begin{array}{llll} \hline G & \mbox{Type of $H$} & r & Q(G)\\ \hline {\rm L}_{4}(3) & {\rm O}_{4}^{+}(3) & 6 & 131/195\\ {\rm L}_{4}(3).2_3 & {\rm O}_{4}^{+}(3) & 3 & 131/195\\ {\rm L}_{4}(3).2_2 & {\rm O}_{4}^{+}(3) & 2 & 457/585\\ {\rm L}_{4}(3).2_1 = {\rm PGL}_{4}(3) & {\rm O}_{4}^{+}(3) & 1 & 521/585\\ {\rm L}_{3}(9).2^2 & \GL_1(9) \wr S_3 & 48 & 4093/12285\\ {\rm L}_{3}(5) & {\rm GL}_{1}(5) \wr S_3 & 30 & 199/775\\ {\rm L}_{3}(5).2 & {\rm GL}_{1}(5) \wr S_3 & 13 & 1379/3875\\ & {\rm GL}_{1}(5^3) & 13 & 791/2000\\ {\rm L}_{3}(4).2 \ne {\rm P\Sigma L}_{3}(4) & \GU_3(2) & 1 & 17/35\\ {\rm L}_{3}(4).6 & \GL_1(4^3) & 5 & 11/32\\ {\rm L}_{3}(4).S_3 = G_0.\la \delta,\gamma\ra & \GL_1(4^3) & 3 & 97/160\\ {\rm L}_{3}(4).D_{12} & \GL_1(4^3) & 1 & 59/80\\ {\rm L}_{3}(3) & \GL_1(3^3) & 2 & 11/24\\ & {\rm O}_{3}(3) & 5 & 19/39\\ {\rm L}_{3}(3).2 & {\rm O}_{3}(3) & 2 & 23/39\\ {\rm L}_{2}(17) & 2_{-}^{1+2}.{\rm O}_{2}^{-}(2) & 3 & 5/17\\ {\rm L}_{2}(13).2 & 2_{-}^{1+2}.{\rm O}_{2}^{-}(2) & 2 & 43/91\\ {\rm L}_{2}(11).2 & 2_{-}^{1+2}.{\rm O}_{2}^{-}(2) & 1 & 31/55\\ {\rm U}_4(5).2^2 & \GU_1(5) \wr S_4 & 409 & 361747/1421875\\ {\rm U}_4(4) & \GU_1(4) \wr S_4 & 80 & 259/884\\ {\rm U}_4(4).2 & \GU_1(4) \wr S_4 & 30 & 1661/3536\\ {\rm U}_4(4).4 & \GU_1(4) \wr S_4 & 15 & 1661/3536\\ {\rm U}_4(3) & \GU_2(3) \wr S_2 & 1 & 187/315\\ {\rm U}_4(3).2 = {\rm U}_4(3).\la \delta^2\phi \ra & \GU_1(3) \wr S_4 & 4 & 1811/2835\\ {\rm U}_4(3).2^2 \ne {\rm U}_{4}(3).\la \delta^2,\phi\ra & \GU_1(3) \wr S_4 & 1 & 2323/2835\\ {\rm U}_4(3).4 & \GU_1(3) \wr S_4 & 1 & 2323/2835\\ {\rm U}_3(9).2 & \GU_1(9) \wr S_3 & 40 & 1913/5913\\ {\rm U}_{3}(9).4 & \GU_1(9) \wr S_3 & 20 & 1913/5913\\ {\rm U}_3(8).2 & \GU_1(8) \wr S_3 & 78 & 1097/4256\\ {\rm U}_3(8).S_3 & \GU_1(8) \wr S_3 & 19 & 205/448\\ {\rm U}_3(8).6 & \GU_1(8) \wr S_3 & 25 & 2437/8512\\ {\rm U}_{3}(8).(3 \times S_3) & \GU_1(8) \wr S_3 & 6 & 2069/4256\\ {\rm U}_{3}(7) & \GU_1(7) \wr S_3 & 27 & 4381/14749\\ {\rm U}_{3}(7).2 & \GU_1(7) \wr S_3 & 10 & 7069/14749\\ {\rm U}_{3}(5).3 & \GU_1(5) \wr S_3 & 3 & 551/875\\ & 3^{1+2}.{\rm Sp}_{2}(3) & 5 & 67/175\\ {\rm U}_{3}(5).S_3 & \GU_1(5) \wr S_3 & 1 & 659/875\\ & 3^{1+2}.{\rm Sp}_{2}(3) & 2 & 443/875\\ {\rm U}_3(4) & \GU_1(4) \wr S_3 & 1 & 133/208\\ {\rm PSp}_6(3) & {\rm Sp}_{2}(3) \wr S_3 & 1 & 853/1365\\ {\rm Sp}_{4}(4).4 & {\rm O}_{2}^{-}(4) \wr S_2 & 2 & 103/153\\ {\rm P\O}_8^{+}(3) & {\rm O}_{4}^{+}(3) \wr S_2 & 12 & 45041/61425\\ {\rm P\O}_8^{+}(3).2 = {\rm PSO}_{8}^{+}(3) & {\rm O}_{4}^{+}(3) \wr S_2 & 4 & 151507/184275\\ {\rm P\O}_8^{+}(3).2 = {\rm P\Omega}_{8}^{+}(3).\la \gamma \ra & {\rm O}_{4}^{+}(3) \wr S_2 & 3 & 53233/61425\\ {\rm P\O}_8^{+}(3).3 & {\rm O}_{4}^{+}(3) \wr S_2 & 3 & 16379/20475\\ {\rm P\O}_8^{+}(3).2^2 & {\rm O}_{4}^{+}(3) \wr S_2 & 1 & 167891/184275\\ {\rm P\O}_8^{+}(3).4 & {\rm O}_{4}^{+}(3) \wr S_2 & 2 & 151507/184275\\ {\rm P\O}_8^{+}(3).S_4 & {\rm O}_{2}^{-}(3) \wr S_4 & 823 & 17810761/44778825\\ \O_8^{+}(2).3 & {\rm O}_{2}^{-}(2) \times \GU_3(2) & 1 & 2071/2800\\ \O_7(3) & {\rm O}_{4}^{+}(3) \perp {\rm O}_{3}(3) & 5 & 1945/2457\\ {\rm SO}_7(3) & {\rm O}_{4}^{+}(3) \perp {\rm O}_{3}(3) & 1 & 11261/12285\\ \hline \end{array} \] \caption{The groups in $\mathcal{G} \setminus \mathcal{L}$ with $Q(G) \geqs 1/4$, part II} \label{tab:random2} \end{table}} \begin{rema}\label{r:q14} We make several comments concerning Theorem \ref{t:random}. \begin{enumerate}[label = (\alph*)] \item In Table \ref{tab:random} we record the relevant cases where $G_0$ is non-classical (as previously noted, the condition $G \in \mathcal{G}\setminus \mathcal{L}$ implies that $G_0 \ne A_5,A_6$ or ${}^2G_2(3)'$). \item The cases with $G_0$ classical are listed in Table \ref{tab:random2}. As before, the \emph{type of $H$} provides an approximate description of the group theoretic structure of $H$ (the precise structure can be read off from \cite{KleidmanLiebeckClassicalGroups}). \item In both tables, the number of regular suborbits of $G$ is listed in the third column. \item We use the standard \textsc{Atlas} \cite{ATLAS} notation for describing the almost simple groups of the form ${\rm L}_{4}(3).2$. In particular, ${\rm L}_{4}(3).2_2$ and ${\rm L}_{4}(3).2_3$ contain involutory graph automorphisms $x$ with $C_{G_0}(x) = {\rm PSp}_{4}(3).2$ and ${\rm PSO}_{4}^{-}(3).2$, respectively. \item Suppose $G = G_0.S_3$, where $G_0 = {\rm L}_3(4)$ and $H$ is of type ${\rm GL}_{1}(3^4)$. There are two groups of this form, up to conjugacy in ${\rm Aut}(G_0)$, and we find that $r=6$ and $Q(G) = 17/80$ if $G = G_0.\la \delta, \phi \ra$, whereas $r=3$ and $Q(G) = 97/160$ if $G = G_0.\la \delta, \gamma\ra$. Here we are using the notation for automorphisms in \cite{Low-Dimensional}, where $\delta$, $\phi$ and $\gamma$ denote diagonal, field and graph automorphisms, respectively. We adopt similar notation to describe the relevant groups with $G_0 = {\rm U}_{4}(3)$ or ${\rm P\O}_{8}^{+}(3)$ (in the latter case, $\gamma$ is an involutory graph automorphism). \end{enumerate} \end{rema} \subsection{Alternating and sporadic groups}\label{s:altspor} \begin{prop}\label{p:alt} The conclusion to Theorem \ref{t:random} holds if $G_0$ is an alternating group. \end{prop} \begin{proof} Let $G_0 = A_m$ be the socle of $G$. If $m \leqs 12$ then the result is easily checked using {\sc Magma} \cite{Magma} (see Section \ref{ss:comp}), so let us assume $m \geqs 13$. By inspecting \cite[Table 14]{SolubleStabiliser} and \cite[Table 4]{Burness2020base} we deduce that $m$ is a prime and $H = {\rm AGL}_{1}(m) \cap G$, in which case \[ |H| \leqs m(m-1) = a, \;\; |x^G| \geqs \frac{m!}{((m-1)/2)!2^{(m-1)/2}} = b \] for all $x \in H$ of prime order (minimal if $x$ is an involution, noting that $x$ has at most one fixed point on $\{1, \ldots, m\}$). In view of Lemma \ref{l:calc}, this gives $\what{Q}(G) < a^2/b < 1/4$ and the result follows. \end{proof} \begin{prop}\label{p:spor} The conclusion to Theorem \ref{t:random} holds if $G_0$ is a sporadic group. \end{prop} \begin{proof} First recall that the maximal subgroups of $G$ have been classified up to conjugacy, with the exception of the Monster group $\mathbb{M}$, where the problem of determining all the almost simple maximal subgroups is still open. In particular, the possibilities for $(G,H)$ are known and \cite{Wilson} is a convenient reference. In addition, the groups with $b(G) \geqs 3$ are listed in \cite[Table 4]{Burness2020base}. First assume $G$ is not the Baby Monster $\mathbb{B}$ nor the Monster $\mathbb{M}$. Here we first use the \textsf{GAP} Character Table Library \cite{GAPCTL} to identify the relevant groups with $\what{Q}(G) \geqs 1/4$. Indeed, in each case the character table of $G$ is available in \cite{GAPCTL} and we can use the \texttt{Maxes} function to access the character table of the maximal subgroup $H$. Moreover, \cite{GAPCTL} also stores the fusion map from $H$-classes to $G$-classes and this allows us to compute precise fixed point ratios and subsequently determine the exact value of $\what{Q}(G)$. This reduces the problem to a small number of cases that require further attention. To handle these groups, we adopt the method described in Section \ref{ss:comp} to compute $Q(G)$ precisely. First we use the function \texttt{AutomorphismGroupSimpleGroup} to construct $G$ as a permutation group and we obtain $H$ via the \texttt{MaximalSubgroups} function (for $G_0 = {\rm HN}$ and $H \cap G_0 = 5^{1+4}.2^{1+4}.5.4$ we construct $H$ using the generators given in the Web Atlas \cite{WebAt}). Next we use \texttt{DoubleCosetRepresentatives} to construct a complete set $R$ of $(H,H)$ double coset representatives and this allows us to calculate $Q(G)$ via \eqref{e:QG} and \eqref{e:reg} (we thank Eamonn O'Brien for his assistance with this computation in the special case where $G_0 = {\rm HN}$ and $H \cap G_0 = 5^{1+4}.2^{1+4}.5.4$). In this way, we can read off the groups with $Q(G) \geqs 1/4$ and they are recorded in Table \ref{tab:random}. Finally, suppose $G = \mathbb{B}$ or $\mathbb{M}$. If $G = \mathbb{B}$ then $H = [3^{11}].(S_4 \times 2S_4)$ or $47{:}23$; in both cases we can use \cite{GAPCTL} and the \texttt{Maxes} function as above to show that $\what{Q}(G)<1/4$. Similarly, if $G = \mathbb{M}$ then $H = 13^{1+2}{:}(3 \times 4S_4)$ or $41{:}40$ and once again we can use \cite{GAPCTL} to verify the bound $\what{Q}(G)<1/4$ (here we use \texttt{NamesOfFusionSources} in place of \texttt{Maxes} since the latter is not available for $\mathbb{M}$). \end{proof} \subsection{Exceptional groups}\label{s:excep} Next let us turn to the groups in $\mathcal{G}\setminus \mathcal{L}$ where $G_0$ is an exceptional group of Lie type over $\mathbb{F}_q$ with $q=p^f$ for a prime $p$. As noted in the proof of \cite[Proposition 7.1]{Burness2020base}, the condition $b(G) = 2$ implies that $H$ is a maximal rank subgroup (that is, $H$ contains a maximal torus of $G$). More precisely, either $H = N_G(T)$ for some maximal torus $T$ of $G_0$ (see \cite[Table 5.2]{MaxExceptional}), or $(G,H)$ is one of the cases recorded in Table \ref{tab:ex}. Recall that $G_0 \ne {}^2G_2(3)' \cong {\rm L}_{2}(8)$ since $G \in \mathcal{G}\setminus \mathcal{L}$. {\small \begin{table} \[ \begin{array}{lll} \hline & G_0 & \mbox{Type of $H$}\\ \hline {\rm (a)} & G_2(3) & {\rm SL}_2(3)^2\\ {\rm (b)} & {}^3D_4(2) & 3\times {\rm SU}_3(2)\\ {\rm (c)} & {}^2F_4(2)' & {\rm SU}_3(2)\\ {\rm (d)} & F_4(2) & {\rm SU}_3(2)^2\\ {\rm (e)} & {}^2E_6(2) & {\rm SU}_3(2)^3\\ {\rm (f)} & E_8(2) & {\rm SU}_3(2)^4\\ \hline \end{array} \] \caption{The groups in $\mathcal{G} \setminus \mathcal{L}$, $G_0$ exceptional, $H \ne N_G(T)$} \label{tab:ex} \end{table}} \begin{lemm}\label{l:ex1} The conclusion to Theorem \ref{t:random} holds if $G_0$ is an exceptional group of Lie type and $H$ is the normaliser of a maximal torus. \end{lemm} \begin{proof} The possibilities for $H$ are recorded in \cite[Table 5.2]{MaxExceptional} and \cite[Proposition 4.2]{ExtremelyPrimitive} states that $b(G)=2$ whenever $H$ is the normaliser of a maximal torus (soluble or otherwise). We proceed by carefully inspecting the proof of \cite[Proposition 4.2]{ExtremelyPrimitive} in the relevant cases with $H$ soluble. If $G_0 = E_8(q)$ then one checks that the bound on $\what{Q}(G)$ in the proof of \cite[Lemma 4.3]{ExtremelyPrimitive} is sufficient and we note that $H$ is insoluble when $G_0 = E_7(q)$. Next assume $G_0 = E_6^{\e}(q)$. Here the proof of \cite[Lemma 4.11]{ExtremelyPrimitive} yields $\what{Q}(G)2^{31}=b_1$ and we calculate that $i_3(H) = 11438$ and $i_7(H) = 342$, so Lemma \ref{l:calc} implies that the contribution to $\what{Q}(G)$ from elements of odd prime order is less than $a_1^2/b_1$, where $a_1 = 11780$. Now assume $x \in H$ is an involution. We find that $H_0$ contains $a_2 = 441$ involutions, so the contribution from these elements is less than $a_2^2/b_2$, where $b_2 = 2^{21}$. Similarly, there are $a_3=406$ involutions $x \in H_0.2 \setminus H_0$, each of which acts on $G_0$ as a graph automorphism. Therefore $|x^G|>\frac{1}{3}2^{25}=b_3$ and we conclude that \begin{equation}\label{e:q14} \what{Q}(G) < \sum_{i=1}^3a_i^2/b_i < \frac{1}{4}. \end{equation} Next assume $G_0 = F_4(q)$, so $q$ is even and $G$ contains graph automorphisms (see \cite[Table 5.2]{MaxExceptional}). By applying the bounds on $\what{Q}(G)$ in the proof of \cite[Lemma 4.15]{ExtremelyPrimitive} we immediately reduce to the case $q=2$. Here $G = F_4(2).2$ and we may assume $H = 7^2{:}(3 \times 2.S_4)$ is the normaliser of a Sylow $7$-subgroup of $G$. The upper bound on $\what{Q}(G)$ in the proof of \cite[Lemma 4.15]{ExtremelyPrimitive} is larger than $1/2$, but we can use {\sc Magma} to construct $G$ and $H$ as permutation groups of degree $139776$ (more precisely, we use \texttt{AutomorphismGroupSimpleGroup} to construct $G$ and we find $H$ by taking the normaliser of a Sylow $7$-subgroup). Then by considering the fusion of $H$-classes in $G$ we calculate that \[ \what{Q}(G) = \frac{541861}{29328998400} \] and the result follows. Now suppose $G_0 = G_2(q)$, so $p=3$, $q \geqs 9$ and $G$ contains graph automorphisms (see \cite[Table 5.2]{MaxExceptional}). By arguing as in the proof of \cite[Lemma 4.21]{ExtremelyPrimitive} we reduce to the cases $q \in \{9,27\}$. Suppose $q=27$ and note that $|H_0| \leqs 12(q+1)^2 = a_1$. Let $x \in H$ be an element of prime order. If $x \in H_0$ then $|x^G| \geqs q^3(q^3-1)(q+1)=b_1$ (as noted in the proof of \cite[Lemma 4.21]{ExtremelyPrimitive}), whereas if $x$ is a field automorphism then $|x^G|>q^{28/3}=b_2$ and $H$ contains at most $a_2=24(q+1)^2$ such elements. Similarly, if $x$ is an involutory graph automorphism then $|x^G| = q^3(q^3-1)(q+1)=b_3$ and there are at most $a_3=12(q+1)^2$ such elements in $H$. It is straightforward to check that \eqref{e:q14} holds. Finally, suppose $q=9$. First we use {\sc Magma} to construct $G = G_2(9).4 = {\rm Aut}(G_0)$ as a permutation group of degree $132860$ and we note that $H = N_G(K)$, where $K$ is either a Sylow $\ell$-subgroup of $G_0$ (with $\ell \in \{5,13,73\}$) or $K = C_8 \times C_8$. In each case, it is straightforward to construct $H$ and verify the bound $\what{Q}(G)<1/4$. To complete the proof of the lemma, we may assume $G_0$ is one of the twisted groups ${}^3D_4(q)$, ${}^2F_4(q)'$, ${}^2G_2(q)$ ($q \geqs 27$) or ${}^2B_2(q)$. First assume $G_0 = {}^3D_4(q)$, in which case there are three possibilities for $H$ and one checks that the bound on $\what{Q}(G)$ in the proof of \cite[Lemma 4.24]{ExtremelyPrimitive} is sufficient if $q \geqs 9$. Suppose $q=8$ and let $x \in H$ be an element of prime order, which implies that $x \in H_0.3$. Then $|x^G|>8^{14}=b_1$ and $3|H_0| \leqs 383688 = a_1$, whence $\what{Q}(G)7^{14}$ for all $x \in H$ of prime order. The remaining groups with $q \leqs 5$ can be handled using {\sc Magma}. In each case, we can use \texttt{AutomorphismGroupSimpleGroup} to construct $G$ and we obtain $H$ as the normaliser in $G$ of an appropriate Sylow $\ell$-subgroup of $G_0$. For example, if $q=5$ then the three possibilities for $H$ correspond to the primes $\ell \in \{7,31,601\}$. In every case, it is straightforward to verify the bound $\what{Q}(G)<1/4$. Next assume $G_0= {}^2F_4(q)'$. The case $q=2$ can be checked using {\sc Magma} and we note that $Q(G)>1/4$ when $H \cap G_0 = 5^2{:}4A_4$ (as recorded in Table \ref{tab:random}). For $q \geqs 8$, the upper bound on $\what{Q}(G)$ in the proof of \cite[Lemma 4.26]{ExtremelyPrimitive} is sufficient (note that in the upper bound on $|H|$ given in the proof of this lemma, the $2\log q$ factor can be replaced by $|{\rm Out}(G_0)| = \log q$). The case $G_0 = {}^2G_2(q)$ is very similar. Indeed, if $q \geqs 3^5$ then the upper bound on $\what{Q}(G)$ in the proof of \cite[Lemma 4.37]{CameronConjecture} is good enough, while the case $q = 27$ can be handled using {\sc Magma}, noting that $H=N_G(K)$ with $K$ a Sylow $\ell$-subgroup of $G_0$ for $\ell \in \{7,19,37\}$. Finally, let us assume $G_0 = {}^2B_2(q)$. If $q \geqs 2^9$ then the bounds in the proof of \cite[Lemma 4.39]{CameronConjecture} are good enough. If $q=2^7$ and $x \in H$ is a field automorphism of order $7$ then $|x^G|>q^4$ and by arguing as in the proof of \cite[Lemma 4.39]{CameronConjecture} we deduce that $\what{Q}(G)<1/4$. The remaining cases with $q \in \{8,32\}$ can be checked using {\sc Magma} and we find that $Q(G)<1/4$ unless $q=8$ and $H \cap G_0 = 13{:}4$. The latter case is recorded in Table \ref{tab:random}. \end{proof} \begin{prop}\label{p:ex} The conclusion to Theorem \ref{t:random} holds if $G_0$ is an exceptional group of Lie type. \end{prop} \begin{proof} In view of the previous lemma, we may assume $G$ is one of the groups listed in Table \ref{tab:ex}. In cases (a), (b) and (c) we can use {\sc Magma} to prove the result (we get $Q(G)<1/4$ in cases (b) and (c), while $Q(G) \geqs 1/4$ in (a)). In (d), the upper bound on $\what{Q}(G)$ in the proof of \cite[Lemma 4.16]{ExtremelyPrimitive} is insufficient. But as explained in \cite[Example 1.4]{BTh_comp}, we can use {\sc Magma} to construct $G$ and $H$ and then it is straightforward to verify the bound $\what{Q}(G)<1/4$. Finally, let us consider cases (e) and (f). In (e) we observe that $2$ and $3$ are the only prime divisors of $|H|$ and we deduce that $\what{Q}(G)<1/4$ by applying the relevant bounds presented in Case 1 in the proof of \cite[Lemma 4.12]{ExtremelyPrimitive}. Similarly, in case (f) we note that the only prime divisors of $|H|$ are $2$ and $3$. If $x$ is a long root element, then \[ |x^G\cap H| = 4|y^L| = 36 = a_1, \;\; |x^G|>2^{58}=b_1, \] where $y$ is a long root element in $L = {\rm SU}_3(2)$. If not, then $|x^G| > 2^{92} = b_2$ and we note that $|H| = 104485552128 = a_2$. By applying Lemma \ref{l:calc} we deduce that \[ \what{Q}(G)< a_1^2/b_1 + a_2^2/b_2 < \frac{1}{4} \] and the result follows. \end{proof} In order to complete the proof of Theorem \ref{t:random}, we may assume $G_0$ is a classical group. It will be convenient to partition the proof into various subsections according to the socle $G_0$. The cases that we need to consider are recorded in the following result, which is an immediate consequence of \cite{SolubleStabiliser} and \cite[Theorem 2]{Burness2020base}. \begin{thm}\label{t:baseclass} Let $G \leqs {\rm Sym}(\O)$ be a permutation group in $\mathcal{G} \setminus \mathcal{L}$ with socle $G_0$ classical and point stabiliser $H$. Then $(G,H)$ is one of the cases in Table $\ref{tab:cases}$. \end{thm} \begin{table} \[ \begin{array}{lll} \hline G_0 & \mbox{Type of $H$} & \mbox{Conditions}\\ \hline {\rm L}_n(q) & {\rm GL}_1(q^n) & \mbox{$n \geqs 3$ prime, $G \ne {\rm L}_{3}(3).2$}\\ & {\rm GL}_{2}(q) \wr S_{n/2} & \mbox{$n \in \{6,8\}$, $q=3$}\\ & {\rm GL}_1(q)\wr S_n & n \in \{3,4\}, \litspace q \geqs 5\\ & {\rm O}_4^+(q) & \mbox{$(n,q)=(4,3)$, $G\ne {\rm Aut}(G_0)$}\\ & {\rm O}_3(q) & (n,q) = (3,3)\\ & 3^{1+2}.{\rm Sp}_2(3) & \mbox{$n=3$, $p=q\equiv 1 \imod{3}$}\\ & {\rm GL}_{2}(3) & \mbox{$n=2$, $q=3^f$, $f \geqs 3$ prime}\\ & 2_{-}^{1+2}.{\rm O}_{2}^{-}(2) & \mbox{$n=2$, $q=p \geqs 11$}\\ \UU_n(q) & \GU_1(q^n) & \mbox{$n \geqs 3$ prime}\\ & \GU_{1}(q) \wr S_{n} & \mbox{$n \in \{3,4\}$, $q \geqs 3$, $(n,q) \ne (3,3)$}\\ & \GU_{2}(q) \wr S_{n/2} & \mbox{$n \in \{4,6,8\}$, $q=3$}\\ & \GU_{3}(q) \wr S_{n/3} & \mbox{$n \in \{9,12\}$, $q=2$}\\ & 3^{1+2}.\Sp_2(3) & \mbox{$n=3$, $q=p\equiv 2\imod{3}$}\\ & \GU_3(2) & \mbox{$n=3$, $q=2^f$, $f\geqs 3$ prime}\\ \PSp_n(q) & \Sp_2(q)\wr S_{n/2} & \mbox{$n \in \{6,8\}$, $q=3$}\\ & {\rm O}_2^\epsilon(q)\wr S_2 & \mbox{$(n,p)=(4,2)$, $q \geqs 4$}\\ & {\rm O}_{2}^{-}(q^2) & \mbox{$(n,p)=(4,2)$, $q \geqs 4$}\\ \POmega_{n}^+(q) & {\rm O}_4^+(q)\wr S_{n/4} & \mbox{$n \in \{12,16\}$, $q=3$}\\ & {\rm O}_4^+(q) \wr S_2 & \mbox{$(n,q)=(8,3)$, $|G:G_0|<6$}\\ & {\rm O}_2^{\e}(q)\wr S_4 & \mbox{$n=8$, $q \geqs 3$}\\ & {\rm O}_2^{-}(q) \times \GU_3(q) & \mbox{$(n,q)=(8,2)$, $G=G_0.3$}\\ & {\rm O}_2^-(q^2) \times {\rm O}_2^-(q^2) & n = 8\\ \O_n(q) & {\rm O}_4^+(q) \perp {\rm O}_3(q) & (n,q) = (7,3)\\ \hline \end{array} \] \caption{The groups in $\mathcal{G}\setminus \mathcal{L}$ with $G_0$ classical}\label{tab:cases} \end{table} Note that in the final column of Table \ref{tab:cases} we list necessary conditions for the existence of a group in $\mathcal{G}\setminus \mathcal{L}$ with the given socle and point stabiliser. In general, these conditions are not sufficient and we refer the reader to \cite{Low-Dimensional,KleidmanLiebeckClassicalGroups} and \cite{Burness2020base} for a precise description of the conditions that are needed to ensure that $G$ has a maximal subgroup of the given type and a base of size $2$. We also refer the reader to \cite[Chapter 3]{BG_book} for detailed information on the conjugacy classes of elements of prime order in $G$. \subsection{Linear groups}\label{ss:lin} In this section we assume $G_0 = {\rm L}_{n}(q)$. Recall that the condition $G \not\in \mathcal{L}$ implies that $q \geqs 11$ if $n=2$, and $q \geqs 3$ if $n=3$. \begin{prop}\label{p:lin} The conclusion to Theorem \ref{t:random} holds if $G_0 = {\rm L}_{n}(q)$. \end{prop} \begin{proof} First assume $n$ is a prime and $H$ is of type ${\rm GL}_{1}(q^n)$. By applying the upper bound on $\what{Q}(G)$ in the proof of \cite[Lemma 6.4]{Burness2020base} we immediately reduce to the cases where $(n,q) = (7,2)$, or $n=5$ and $q \leqs 5$, or $n=3$ and $q \leqs 19$. With the aid of {\sc Magma}, it is straightforward to compute $Q(G)$ precisely in each of these cases and the result quickly follows (note that the condition $b(G)=2$ implies that $G \ne {\rm L}_{3}(3).2$). In particular, we find that $Q(G) \geqs 1/4$ only if $n = 3$ and $q \leqs 5$ (the precise exceptions are recorded in Table \ref{tab:random2}). Next assume $n \in \{3,4\}$, $q \geqs 5$ and $H$ is of type ${\rm GL}_1(q)\wr S_n$. First assume $n=3$. By inspecting the proof of \cite[Lemma 6.5]{Burness2020base} we deduce that $\what{Q}(G)<1/4$ if $q \geqs 43$. If $29 \leqs q \leqs 41$ then $G$ does not contain field automorphisms of order $2$ or $3$, nor graph-field automorphisms of order $2$, so we may set $a_8 = a_9=0$ in the bound on $\what{Q}(G)$ presented in the proof of \cite[Lemma 6.5]{Burness2020base}. One checks that this modified bound yields $\what{Q}(G)<1/4$. For $7 \leqs q \leqs 27$ we can use {\sc Magma} to show that $Q(G)<1/4$ in the usual manner, with the single exception of the case $G = \Aut(G_0)$ with $q=9$, where $Q(G) = 4093/12285$. Finally, for $q=5$ we calculate that $Q(G) = 199/775$ if $G = G_0$, otherwise $Q(G) = 1379/3875$; both cases are recorded in Table \ref{tab:random2}. Similarly, if $n=4$ then the result follows by combining explicit {\sc Magma} computations for $q \in \{5,7,8\}$ with the upper bound on $\what{Q}(G)$ presented in the proof of \cite[Lemma 6.6]{Burness2020base} for $q \geqs 9$ (in every case we get $Q(G)<1/4$). The case where $n=3$ and $H$ is of type $3^{1+2}.{\rm Sp}_2(3)$ is entirely similar, working with the bound on $\what{Q}(G)$ in the proof of \cite[Lemma 6.11]{Burness2020base}. Now let us turn to the relevant groups with $G_0 = {\rm L}_{2}(q)$. If $q=3^f$ with $f \geqs 3$ a prime and $H$ is a subfield subgroup of type ${\rm GL}_2(3)$, then the bound on $\what{Q}(G)$ in the proof of \cite[Lemma 4.9]{Burness2020base} is sufficient if $f \geqs 7$, while the cases $f \in \{3,5\}$ are easily checked using {\sc Magma}. Similarly, if $q=p \geqs 11$ and $H$ is of type $2^{1+2}_{-}.{\rm O}_{2}^{-}(2)$ then the bound in the proof of \cite[Lemma 4.10]{Burness2020base} is good enough if $q \geqs 71$ and we can use {\sc Magma} to handle the cases with $q<71$. There are four remaining cases to consider. If $G_0 = {\rm L}_3(3)$ with $H$ of type ${\rm O}_3(3)$ then we compute $Q(G) = 19/39$ if $G = G_0$, whereas $Q(G) = 23/39$ for $G = G_0.2$. Next suppose $G_0 = {\rm L}_{4}(3)$ and $H$ is of type ${\rm O}_{4}^{+}(3)$. Here the condition $b(G) = 2$ implies that $G \ne {\rm Aut}(G_0)$ and using {\sc Magma} one checks that $Q(G)>1/4$ in each case (the precise value of $Q(G)$ is recorded in Table \ref{tab:random2}). The case $G_0 = {\rm L}_{6}(3)$ with $H$ of type ${\rm GL}_{2}(3) \wr S_3$ can be handled using {\sc Magma}, working with the function \texttt{MaximalSubgroups} to construct $H$. Finally, suppose $G_0 = {\rm L}_{8}(3)$ and $H$ is of type ${\rm GL}_{2}(3) \wr S_4$. Here the \texttt{MaximalSubgroups} function is ineffective, but we can construct $H$ by observing that $H = N_G(K)$ with $|K|=2^{11}$, as noted in the proof of \cite[Proposition 6.3]{Burness2020base} (also see \cite[Example 2.4]{Burness2020base}). It is straightforward to check that $\what{Q}(G)<1/4$. \end{proof} \subsection{Unitary groups}\label{ss:unit} \begin{prop}\label{p:unit} The conclusion to Theorem \ref{t:random} holds if $G_0 = {\rm U}_{n}(q)$ with $n \geqs 3$. \end{prop} \begin{proof} First assume $n$ is a prime and $H$ is of type $\GU_1(q^n)$. For $n \geqs 5$, one checks that the upper bound on $\what{Q}(G)$ in the proof of \cite[Lemma 6.4]{Burness2020base} is sufficient unless $n=5$ and $q \leqs 5$. Suppose $n=5$, so $q \geqs 3$ by the maximality of $H$. If $q=5$ then it is easy to improve the given bound in \cite{Burness2020base} in order to show that $\what{Q}(G)<1/4$ (for example, we can use the fact that $|H| \leqs 10(5^5+1)/6$). For $q=4$ we observe that $H = N_G(K)$, where $K$ is a Sylow $41$-subgroup of $G$, so it is straightforward to construct $H$ in {\sc Magma} and verify the bound $\what{Q}(G)<1/4$ (note that it suffices to check this for $G = \Aut(G_0)$). The case $q=3$ can also be checked using {\sc Magma} (using \texttt{MaximalSubgroups} to construct $H$, or noting that $H$ is the normaliser of a Sylow $61$-subgroup). Similarly, if $n=3$ then $q \geqs 4$ and the bound in the proof of \cite[Lemma 6.4]{Burness2020base} is sufficient for $q \geqs 23$ (for $q=32$, we note that $|x^G| \geqs |G_0:{\rm U}_{3}(2)|$ if $x$ is a field automorphism of order $5$); the remaining cases with $q \leqs 19$ can be verified using {\sc Magma}. Next suppose $G_0 = {\rm U}_{3}(q)$ and $H$ is of type $\GU_1(q) \wr S_3$ with $q \geqs 4$. For $q \geqs 43$ it is easy to check that the upper bound on $\what{Q}(G)$ in the proof of \cite[Lemma 6.5]{Burness2020base} is sufficient. The same estimates are also good enough when $29 \leqs q \leqs 41$, noting that in each case $G$ does not contain any field or graph-field automorphisms of order $2$ or $3$. For $11 \leqs q \leqs 27$ we can use {\sc Magma} to verify the bound $\what{Q}(G)<1/4$. We find that there are examples with $Q(G) \geqs 1/4$ when $q \leqs 9$; they are easily identified using {\sc Magma} and they are recorded in Table \ref{tab:random2}. The case where $G_0 = {\rm U}_{4}(q)$ and $H$ is of type $\GU_1(q) \wr S_4$ is similar. Here $q \geqs 3$ and the bound on $\what{Q}(G)$ in the proof of \cite[Lemma 6.5]{Burness2020base} is good enough for $q \geqs 9$. If $q \in \{7,8\}$ then one can check that $Q(G)<1/4$ using {\sc Magma}. In the same way, we find that there are exceptions to this bound when $q \in \{3,4,5\}$ and each of these cases is listed in Table \ref{tab:random2}. Next assume $n \in \{4,6,8\}$, $q=3$ and $H$ is of type $\GU_2(q) \wr S_{n/2}$. If $n=4$ then $G = G_0$ is the only group with $b(G)=2$ (see \cite[Table 7]{Burness2020base}) and with the aid of {\sc Magma} we calculate that $Q(G) = 187/315$. Next assume $n=6$. Here $H = N_G(K)$ for some subgroup $K$ of $G_0$ of order $2^{10}$ and it is straightforward to check that $\what{Q}(G)<1/4$ (see \cite[Example 2.4]{Burness2020base} and the proof of \cite[Proposition 6.3]{Burness2020base}). Similarly, if $n=8$ then $H = N_G(K)$ with $|K|=2^{13}$ and once again one checks that $\what{Q}(G)<1/4$. (Note that in both cases, it suffices to check the bound for $G = \Aut(G_0)$.) Now assume $n \in \{9,12\}$, $q=2$ and $H$ is of type $\GU_3(q) \wr S_{n/3}$. As noted in the proof of \cite[Proposition 6.3]{Burness2020base}, if $n=9$ then $H = N_G(K)$ with $|K|=3^8$ and we can use {\sc Magma} to verify the bound $\what{Q}(G)<1/4$. For $n=12$ we find that the bound presented in the proof of \cite[Proposition 6.3]{Burness2020base} does not give $\what{Q}(G)<1/4$ and a more accurate estimate is required. To do this, it suffices to improve the upper bound on the contribution to $\what{Q}(G)$ from elements of order $3$. As in the proof of \cite[Proposition 6.3]{Burness2020base}, we may view $H$ as the stabiliser in $G$ of an orthogonal decomposition \[ V = V_1 \perp V_2 \perp V_3 \perp V_4 \] of the natural module, where each $V_i$ is a nondegenerate $3$-space. Suppose $x \in H$ has order $3$. If some conjugate of $x$ induces a nontrivial permutation of the $V_i$, then $|x^G|>2^{89}=b_1$ and we note that $|H|<2^{42}=a_1$. Following the argument in \cite{Burness2020base}, the contribution from the remaining elements of order $3$ in $H$ with $|x^G|>2^{69}=b_2$ is less than $a_2^2/b_2$, where $a_2 = 2^{31}$. As explained in the proof of \cite[Proposition 6.3]{Burness2020base}, the contribution from the elements with $|x^G| \leqs 3.2^{62}$ is less than $2\sum_{i=3}^7a_i^2/b_i$, where the integers $a_i$ and $b_i$ are defined as in the proof in \cite{Burness2020base}. Finally, if $3.2^{62} < |x^G| \leqs 2^{69}$ then one can check that $x$ is of the form $[I_8,\omega I_4]$, where $\omega \in \mathbb{F}_4$ is a primitive cube root of unity. Here we calculate \[ |x^G \cap H| \leqs 2\binom{4}{2}m + \binom{4}{2}m^2+2\binom{4}{2}m^3+m^4= 42480 = a_0 \] where $m = \frac{1}{3}|\GU_3(2):\GU_2(2)| = 12$. Therefore, the contribution to $\what{Q}(G)$ from elements of order $3$ is less than \[ a_1^2/b_1+a_2^2/b_2+2\left(a_0^2/b_0+\sum_{i=3}^7a_i^2/b_i\right)<\frac{1}{20} \] where $b_0 = 3.2^{62}$. Finally, the estimates in the proof of \cite[Proposition 6.3]{Burness2020base} imply that the contribution to $\what{Q}(G)$ from involutions is also less than $1/20$ and the result follows. To complete the proof of the proposition, we may assume $n=3$ and either $q=p \equiv 2 \imod{3}$ and $H$ is of type $3^{1+2}.{\rm Sp}_{2}(3)$, or $q=2^f$ with $f \geqs 3$ a prime and $H$ is a subfield subgroup of type $\GU_3(2)$. Suppose $H$ is of type $3^{1+2}.{\rm Sp}_{2}(3)$. Here the proof of \cite[Lemma 6.11]{Burness2020base} gives the result for $q > 29$ and we can use {\sc Magma} to handle the cases with $q \leqs 29$, noting that there are exceptions to the bound $Q(G)<1/4$ when $q=5$ (as recorded in Table \ref{tab:random2}). Finally, let us assume $H$ is of type $\GU_3(2)$, so $q=2^f$ with $f \geqs 3$ odd. If $f \geqs 7$ then the bound on $\what{Q}(G)$ in the proof of \cite[Lemma 6.10]{Burness2020base} is sufficient, while the cases with $f \in \{3,5\}$ can be handled using {\sc Magma}. \end{proof} \subsection{Symplectic groups}\label{ss:symp} Next assume $G_0 = {\rm PSp}_{n}(q)$ with $n \geqs 4$. Recall that $(n,q) \ne (4,2)$ since $G \not\in \mathcal{L}$. \begin{prop}\label{p:symp} The conclusion to Theorem \ref{t:random} holds if $G_0 = {\rm PSp}_{n}(q)$ with $n \geqs 4$. \end{prop} \begin{proof} First assume $n \in \{6,8\}$, $q=3$ and $H$ is of type $\Sp_2(q)\wr S_{n/2}$. If $n=6$ then the condition $b(G)=2$ implies that $G = G_0$ and using {\sc Magma} we calculate that $Q(G) = 853/1365$, so this case is listed in Table \ref{tab:random2}. For $n=8$ we use the functions \texttt{AutomorphismGroupSimpleGroup} and \texttt{MaximalSubgroups} to construct $G$ and $H$, and we apply \texttt{DoubleCosetCanonical} to establish the existence of sufficiently many regular $H$-orbits in order to force $Q(G)<1/4$ (see \eqref{e:QG}). Indeed, for $G = G_0$ we get $r \geqs 3113$, while $r \geqs 1557$ for $G = G_0.2$. Finally let us assume $G_0 = \Sp_4(q)$ with $q \geqs 4$ even and $H$ of type ${\rm O}_2^\epsilon(q)\wr S_2$ or ${\rm O}_{2}^{-}(q^2)$. Here $H$ is maximal only if $G$ contains graph automorphisms and with the aid of {\sc Magma} one checks that if $q \leqs 2^5$ then either $\what{Q}(G)<1/4$ or $q=4$, $G = \Aut(G_0)$ and $H$ is of type ${\rm O}_2^{-}(q)\wr S_2$. In the latter case we have $Q(G) = 103/153$ as recorded in Table \ref{tab:random2}. For the remainder, we may assume $q \geqs 2^6$. Suppose $H$ is of type ${\rm O}_2^\epsilon(q)\wr S_2$, so $H_0 = (C_{q-\e})^2{:}D_8$. By applying the upper bound in the proof of \cite[Lemma 6.9]{Burness2020base}, we deduce that $\what{Q}(G)<1/4$ if $q \ne 2^7$. So let us assume $q=2^7$ and write $\what{Q}(G) = \a_1+\a_2$, where $\a_1$ is the contribution from involutory graph automorphisms. The proof of \cite[Lemma 6.9]{Burness2020base} gives $\a_2<2^{-6}$, so it remains for us to estimate $\a_1$. If $\e=-$ then $H \leqs (C_{129})^2{:}(SD_{16} \times C_7)$ and it follows that every involution in $H$ is contained in $H \cap G_0 = (C_{129})^2{:}D_{8}$, whence $\a_1=0$ and the result follows. Now assume $\e=+$, so $H \leqs (C_{127})^2{:}(D_{16} \times C_7)$. Since there are exactly $4$ involutions in $D_{16} \setminus D_8$, we deduce that $\a_1 \leqs d^2/b$ with $d = 4.127^2$ and $b = |G_0: {}^2B_2(q)| = 34626060288$. One checks that the resulting bound on $\what{Q}(G)$ is good enough. To complete the proof, let us assume $q \geqs 2^6$ and $H$ is of type ${\rm O}_{2}^{-}(q^2)$, so \[ H_0 = {\rm O}_{2}^{-}(q^2).2 = C_{q^2+1}{:}C_4 \] and we will estimate the contribution to $\what{Q}(G)$ from the various elements of prime order (the details in this case were omitted in the proof of \cite[Lemma 6.9]{Burness2020base}). First let $x \in H$ be a unipotent involution. Then $x$ embeds in $G$ as an involution of type $c_2$ (in the notation of Aschbacher and Seitz \cite{AS}), whence \[ |x^G \cap H| = i_2(H_0) = q^2+1=a_1,\;\; |x^G| = (q^2-1)(q^4-1) = b_1. \] If $x$ is semisimple, then $|x^G| \geqs |{\rm Sp}_{4}(q):\GU_1(q^2)| = q^4(q^2-1)^2 = b_2$ and we note that there are at most $a_2=q^2+1$ such elements in $H$. Next suppose $x$ is a field automorphism of odd order. Then $|x^G|>q^{20/3}=b_3$ and $H$ contains fewer than $4(q^2+1)\log q = a_3$ such elements. Finally, suppose $x$ is an involutory field or graph automorphism (note that $G$ cannot contain elements of both types). If $\log q$ is even then every involution in $H$ is contained in $H_0$, so we may assume $\log q$ is odd and $x$ is a graph automorphism. Then $|x^G| = q^2(q+1)(q^2-1) = b_4$ and we note that $|x^G \cap H| \leqs |H_0| = 4(q^2+1) = a_4$. Therefore, by applying Lemma \ref{l:calc} we deduce that \[ \what{Q}(G) < \sum_{i=1}^3a_i^2/b_i + \a a_4^2/b_4, \] where $\a=1$ if $\log q$ is odd, otherwise $\a =0$, and we conclude that $\what{Q}(G)<1/4$. \end{proof} \subsection{Orthogonal groups}\label{ss:orth} In order to complete the proof of Theorem \ref{t:random}, we may assume $G_0 = {\rm P\O}_{n}^{\e}(q)$ with $n \geqs 7$. \begin{prop}\label{p:orth} The conclusion to Theorem \ref{t:random} holds if $G_0 = {\rm P\O}_{n}^{\e}(q)$ with $n \geqs 7$. \end{prop} \begin{proof} By inspecting Table \ref{tab:cases} we observe that either $n$ is even and $\e=+$, or $(n,q) = (7,3)$. First assume $n \in \{12,16\}$, $q=3$ and $H$ is of type ${\rm O}_4^+(q)\wr S_{n/4}$. For $n=16$, the upper bound in the proof of \cite[Proposition 6.3]{Burness2020base} gives $\what{Q}(G)<1/4$. On the other hand, if $n=12$ then we can construct $G$ and $H$ in {\sc Magma} (see \cite[Example 2.4]{Burness2020base}) and it is straightforward to check that $\what{Q}(G)<1/4$. The relevant cases with $G_0 = \O_7(3)$ or $\O_8^{+}(2)$ can also be handled using {\sc Magma} and the exceptions with $Q(G) \geqs 1/4$ are recorded in Table \ref{tab:random2}. To complete the proof, we may assume $G_0 = {\rm P\Omega}_{8}^{+}(q)$ with $q \geqs 3$. Suppose $q=3$ and $H$ is of type ${\rm O}_{4}^{+}(3) \wr S_2$, noting that $|G:G_0|<6$ since $b(G)=2$. Even though $|G:H|=14926275$ is large, we can still analyse this case in the usual way using {\sc Magma}, working with a set of $(H,H)$ double coset representatives to compute $r$ (and hence $Q(G)$) via \eqref{e:reg}. The results are presented in Table \ref{tab:random2}. Next assume $H$ is of type ${\rm O}_{2}^{\e'}(q) \wr S_4$. If $q \in \{3,4\}$ then $\e'=-$ and using {\sc Magma} one can check that either $Q(G)<1/4$, or $q = 3$, $G = \Aut(G_0)$, $r=823$ and \[ Q(G) = \frac{17810761}{44778825}. \] For example, if $q=3$ and $G = G_0.A_4$ then using \texttt{DoubleCosetCanonical} we can verify the bound $r \geqs 3075$, which forces $Q(G)<1/4$. We thank Eamonn O'Brien for his assistance with the precise calculation of $r$ when $G = \Aut(G_0)$. For $q \geqs 5$, we seek to apply the upper bound on $\what{Q}(G)$ presented in the proof of \cite[Lemma 6.7]{Burness2020base}. If $q \geqs 9$ then \[ \what{Q}(G) < 2q^{-1}+ q^{-2} + q^{-3}+q^{-7} <\frac{1}{4} \] and the result follows. One can check that the bounds in the proof of \cite[Lemma 6.7]{Burness2020base} are also sufficient when $q \in \{7,8\}$, so we may assume $q=5$. Here we have $H = N_G(K)$, where $K27$. By \cite[Lemma 4.7]{Burness2020base} we have $b(G,H) \leqs 3$, with equality if and only if ${\rm PGL}_{2}(q)0$ if $i>0$. It suffices to show that $x = \pm I_2$. Since $x$ fixes $\a$, the matrix of $A$ with respect to the basis $\{e_1,e_2\}$ is either diagonal or anti-diagonal. First assume $x$ fixes the two $1$-spaces comprising $\a$, so $A={\rm diag}(a,a^{-1})$ is diagonal. If $x$ also fixes the two spaces in $\beta$, then \begin{align*} (e_1+be_2)^x&=a\mu^i e_1+a^{-1}b^{p^j}e_2=\eta_1(e_1+be_2)\\ (e_1+ce_2)^x&=a\mu^i e_1+a^{-1}c^{p^j}e_2=\eta_2(e_1+ce_2) \end{align*} for some $\eta_1,\eta_2\in\mathbb{F}_q^{\times}$. Therefore \begin{equation}\label{e:c21} a^2\mu^i = b^{p^j-1} = c^{p^j-1} \end{equation} and thus~\ref{it:4.3.3} implies that $j=0$, so $i=0$ and $a^2=1$, which gives $x = \pm I_2$ as required. Similarly, if $x$ interchanges the spaces in $\beta$, then \begin{equation}\label{e:c22} a^2\mu^i = b^{p^j}c^{-1} = c^{p^j}b^{-1}. \end{equation} Here $b^{p^{2j}-1} = c^{p^{2j}-1}$, so~\ref{it:4.3.3} implies that $2j=0$ or $f$. Suppose $2j=0$, so $i=0$ and $a^2 = bc^{-1} = cb^{-1}$ and thus $bc^{-1}=\pm 1$. But $b \ne c$, so $bc^{-1}=-1$, which is incompatible with~\ref{it:4.3.2}. Now assume $2j=f$, so $q \equiv 1 \imod{4}$ and $-1$ is a square in $\mathbb{F}_q$. In addition, \eqref{e:c22} gives $(bc^{-1})^{p^{f/2}+1} =1$, so $bc^{-1}\in\langle \mu^{p^{f/2}-1}\rangle$ and thus $bc^{-1}$ is a square. Therefore, $-bc^{-1}$ is a square, which once again is incompatible with~\ref{it:4.3.2}. Now assume $A=\begin{pmatrix} 0&a\\ -a^{-1}&0 \end{pmatrix}$ is anti-diagonal. If $x$ fixes both spaces in $\beta$ then \begin{align*} (e_1+be_2)^x&=ab^{p^j}e_1-a^{-1}\mu^ie_2=\eta_1(e_1+be_2)\\(e_1+ce_2)^x&=ac^{p^j}e_1-a^{-1}\mu^ie_2=\eta_2(e_1+ce_2) \end{align*} for some $\eta_1,\eta_2\in\mathbb{F}_q^{\times}$. This gives \begin{equation}\label{e:c23} -a^2\mu^{-i}=b^{-p^j-1}=c^{-p^j-1}. \end{equation} Here $b^{p^{2j}-1}=c^{p^{2j}-1}$ and thus $2j=0$ or $f$ by~\ref{it:4.3.3}. If $2j=0$ then $i=0$ and $-a^2=b^{-2}=c^{-2}$, which implies that $bc^{-1}=\pm 1$. As noted above, this is incompatible with~\ref{it:4.3.2}. Now assume $2j=f$, so $q\equiv 1\pmod 4$ and $-1$ is a square in $\mathbb{F}_q$ once again. Then (\ref{e:c23}) gives $(bc^{-1})^{p^{f/2}+1}=1$ and as above we deduce that $bc^{-1}$ is a square. Hence, $-bc^{-1}$ is also a square, which contradicts~\ref{it:4.3.2}. Finally, suppose $A$ is anti-diagonal as above and assume $x$ interchanges the $1$-spaces in $\beta$. Here we get \begin{equation}\label{e:c24} -a^2\mu^{-i}=b^{-p^j}c^{-1}=c^{-p^j}b^{-1}, \end{equation} so $b^{p^j-1}=c^{p^j-1}$ and the condition in~\ref{it:4.3.3} implies that $j=0$ and $i=0$. Therefore $-bc^{-1} = (ab)^2$, which is incompatible with~\ref{it:4.3.2}. We conclude that if the scalars $b$ and $c$ satisfy the conditions in~\ref{it:4.3.1},~\ref{it:4.3.2} and~\ref{it:4.3.3}, then $\{\a,\b\}$ is a base. \end{proof} \begin{lemm} \label{l:c2_cond_PSigmaL} Let $G=\PSigmaL_2(q)$ with $q$ odd and set $\alpha$ and $\beta$ as in Lemma \ref{l:c2_cond}. Then $\{\alpha,\beta\}$ is a base for $G$ if and only if the scalars $b$ and $c$ satisfy the conditions \ref{it:4.3.1}--\ref{it:4.3.3} in Lemma \ref{l:c2_cond}. \end{lemm} \begin{proof} By Lemma \ref{l:c2_cond}, it suffices to show that if any of the conditions in~\ref{it:4.3.1},~\ref{it:4.3.2} or~\ref{it:4.3.3} fail to hold, then there exists an element $x \ne \pm I_2$ in ${\rm \Sigma L}_{2}(q) = \la \SL_2(q),\phi\ra$ that fixes $\a$ and $\b$. We proceed by inspecting the proof of Lemma \ref{l:c2_cond}, noting that $i=0$ in each of the equations \eqref{e:c21}--\eqref{e:c24}. As explained in the discussion preceding Lemma \ref{l:c2_cond}, if $bc=0$ then $\{\a,\b\}$ is not a base. Next assume $-bc^{-1}$ is a square in $\mathbb{F}_q$, say $d^2=-bc^{-1}$. Then setting $a=db^{-1}$ gives $-a^2 = b^{-1}c^{-1}$ and we get a solution to \eqref{e:c24} with $j=0$. Finally, suppose $b^{p^k-1} = c^{p^k-1}$ for some $027$. Recall that $b(G)=2$ if and only if $G$ does not contain $\PGL_2(q)$ as a proper subgroup. If $G=\PGL_2(q)$, then $\Sigma(G)$ is isomorphic to the Johnson graph $J(q+1,2)$ and we immediately deduce that \eqref{e:star} holds (as noted in \cite[Example 3.9]{SaxlGraph}). For the remainder, we may assume that $G\cap \PGL_2(q)=G_0$ and $q$ is odd. In view of Corollary \ref{c:c2_min}, we only need to consider the group $G = \PSigmaL_2(q)$. Fix $\alpha=\{\la e_1\ra, \la e_2\ra \}$ as before. By Theorem \ref{t:cd}, it suffices to show that if $\{\alpha,\beta\}$ is a base, then there exists $\gamma \in \O$ such that both $\{\alpha,\gamma\}$ and $\{\beta,\gamma\}$ are bases. By Lemma \ref{l:c2_cond_PSigmaL} we have $\beta=\{\la e_1+be_2\ra,\la e_1+ce_2\ra\}$, where $b,c \in \mathbb{F}_q$ are nonzero scalars such that $-bc^{-1}$ is a non-square and is not contained in any proper subfield of $\mathbb{F}_q$. Set $\gamma=\{\la e_1-be_2\ra,\la e_1-ce_2\ra \} \in \O$ and note that $\{\a,\gamma\}$ is a base by Lemma \ref{l:c2_cond}. By Corollary \ref{c:c2_cond}, $\{\beta,\gamma\}$ is a base if and only if there exists $d,e \in \mathbb{F}_q$ with $d,e \ne b-c$ such that \begin{equation}\label{e:uw} \{-b,-c\}=\left\{\frac{b(c-b)+dc}{c-b+d},\frac{b(c-b)+ec}{c-b+e}\right\} \end{equation} and $d,e$ satisfy the conditions in ~\ref{it:4.3.1},~\ref{it:4.3.2}, and~\ref{it:4.3.3} of Lemma \ref{l:c2_cond}. Set $d=\frac{2b(b-c)}{b+c}$ and $e=\frac{b^2-c^2}{2c}$. Then $d,e \ne b-c$, \eqref{e:uw} holds and $de \ne 0$. In addition, \[ -de^{-1}=-bc^{-1}\left(\frac{2c}{b+c}\right)^2=-4(bc^{-1}+cb^{-1}+2)^{-1} \] and we immediately deduce that $-de^{-1}$ is a non-square in $\mathbb{F}_q$. Finally, we claim that $de^{-1}$ is not contained in a proper subfield of $\mathbb{F}_q$. To do this, it suffices to show that $\eta=bc^{-1}+cb^{-1}$ is not contained in such a subfield. With this aim in mind, it will be useful to observe that \begin{align*} \eta^{p^k}-\eta & =(bc^{-1})^{p^k}+(bc^{-1})^{-p^k}-bc^{-1}-(bc^{-1})^{-1}\\ & = (bc^{-1})^{-p^k}((bc^{-1})^{p^k+1}-1)((bc^{-1})^{p^k-1}-1) \end{align*} for $1 \leqs k < f$, so $\eta$ is contained in the subfield $\mathbb{F}_{p^k}$ of $\mathbb{F}_q$ if and only if this expression is $0$. Now since $b$ and $c$ satisfy the condition in part~\ref{it:4.3.3} of Lemma \ref{l:c2_cond}, it follows that $(bc^{-1})^{p^k-1}-1 \ne 0$, whence $\eta \in \mathbb{F}_{p^k}$ if and only if $(bc^{-1})^{p^k+1}=1$. If the latter equality holds, then $bc^{-1}\in\mathbb{F}_{p^{2k}}$ and thus $2k=f$. In particular, this implies that both $-1$ and $bc^{-1}$ are squares, which contradicts ~\ref{it:4.3.2} in Lemma \ref{l:c2_cond}. This justifies the claim and we conclude that $d$ and $e$ satisfy the conditions in parts ~\ref{it:4.3.1},~\ref{it:4.3.2}, and~\ref{it:4.3.3} of Lemma \ref{l:c2_cond}. In particular, $\{\beta,\gamma\}$ is a base and the result follows. \end{proof} Next we turn to the problem of determining when $G$ has a unique regular suborbit on $\O$. We will need the following number-theoretic result, where $\phi$ and $\gamma= 0.57721...$ denote Euler's totient function and Euler's constant, respectively. \begin{lemm}\label{l:euler} For every integer $n \geqs 3$, \[ \phi(n) > \frac{n}{e^{\gamma}\log\log n+\frac{3}{\log\log n}}. \] \end{lemm} \begin{proof} See \cite[Theorem 15]{RS}. \end{proof} \begin{prop}\label{p:psl2_c2reg} Suppose $G_0 = {\rm L}_{2}(q)$ and $H$ is of type ${\rm GL}_{1}(q) \wr S_2$. Then $G$ has a unique regular suborbit if and only if $G = {\rm PGL}_{2}(q)$ and $q \geqs 4$, $q \ne 5$. \end{prop} \begin{proof} In view of Proposition \ref{p:comp}, we may assume $q>27$ and we recall that $G = {\rm PGL}_2(q)$ has a unique regular suborbit. For the remainder we may assume $q$ is odd and $G \cap {\rm PGL}_{2}(q) = G_0$; our aim is to show that $G$ has at least two regular suborbits. By Corollary \ref{c:c2_min}, we may assume that $G = \PSigmaL_2(q)$, in which case $G$ has $m/2f$ regular suborbits by Corollary \ref{c:c2_val}, where $m$ is the number of non-squares in $\mathbb{F}_q$ that are not contained in any proper subfield of $\mathbb{F}_q$. Any primitive element of $\mathbb{F}_q$ has this property and there are $\phi(q-1)$ such elements in $\mathbb{F}_q$. By applying the lower bound in Lemma \ref{l:euler} we deduce that $\phi(q-1) \geqs 4f$ for all $q>27$ and the result follows. \end{proof} Finally, we turn to the clique number of $\Sigma(G)$. If $G = {\rm PGL}_{2}(q)$ then $\omega(G) = q$ since $\Sigma(G)$ is isomorphic to the Johnson graph $J(q+1,2)$, whence $\omega(G) \geqs q$ if $G \leq \mathrm{PGL}_2(q)$ and the conclusion to Theorem~\ref{t:main11} holds. For $55$, but we have not been able to prove this. \begin{rema}\label{r:c2_clique} Let us say more about the difficulties that arise when trying to construct a clique of size $5$ when $G = \PSigmaL_2(q)$. By Proposition \ref{p:psl2_c2star} we see that $\omega(G)\geqs 3$. More precisely, $\{\alpha,\beta,\gamma\}$ is a clique of size $3$, where \[ \alpha = \{\la e_1\ra,\la e_2\ra\}, \; \beta = \{\la e_1+be_2\ra,\la e_1+ce_2\ra\}, \; \gamma = \{\la e_1-be_2\ra,\la e_1-ce_2\ra\} \] and $b,c \in \mathbb{F}_{q}^{\times}$ satisfy the conditions in parts~\ref{it:4.3.1},~\ref{it:4.3.2}, and~\ref{it:4.3.3} of Lemma \ref{l:c2_cond}. Similarly, if we choose different scalars $b',c' \in \mathbb{F}_q^{\times}$ then we can construct another clique $\{\a,\b',\gamma'\}$, where $\beta' = \{\la e_1+b'e_2\ra,\la e_1+c'e_2\ra\}$ and $\gamma' = \{\la e_1-b'e_2\ra,\la e_1-c'e_2\ra\}$. Then $\{\a,\b,\b',\gamma,\gamma'\}$ is a clique of size $5$ if $\{\beta, \beta'\}$ and $\{\beta, \gamma'\}$ are bases. So in view of Corollary \ref{c:c2_cond}, we need to find scalars $d,e \ne b-c$ satisfying conditions ~\ref{it:4.3.1}--\ref{it:4.3.3} in Lemma \ref{l:c2_cond} such that \[ \{b',c'\}=\left\{\frac{b(c-b)+dc}{c-b+d}, \frac{b(c-b)+ec}{c-b+e}\right\}, \] together with another pair of scalars $d',e' \ne b-c$ satisfying the same conditions with \[ \{-b',-c'\}=\left\{\frac{b(c-b)+d'c}{c-b+d'}, \frac{b(c-b)+e'c}{c-b+e'}\right\}. \] Here the main difficulty arises in verifying the required conditions in Lemma \ref{l:c2_cond}. For example, if we fix $b'$ and $c'$, then we need to show that $de^{-1}$ is not contained in a proper subfield of $\mathbb{F}_q$, which is not an easy condition to check. At the same time, we also need to verify the corresponding condition for $d'e'^{-1}$, which is an additional complication. \end{rema} \subsection{$\mcC_3$-actions}\label{ss:psl2_c3} In this section we assume $H$ is of type ${\rm GL}_{1}(q^2)$, so $H_0 = D_{2(q+1)/h}$ and $|\O| = \frac{1}{2}q(q-1)$, where $h=(2,q-1)$. By Proposition \ref{p:comp} we may assume $q>27$ and we note that \cite[Lemma 4.8]{Burness2020base} gives $b(G) \leqs 3$, with equality if and only if ${\rm PGL}_{2}(q) \leqs G$. Therefore, we may assume $q$ is odd and $G\cap \PGL_2(q)=G_0$, so either \begin{enumerate}[label = (\alph*)] \item $G=\langle G_0,\phi^j\rangle$ for some $j$ in the range $0\leqs j0$ if $i>0$. In order to prove that $\{\a,\omega_b\}$ is a base for $G$, it suffices to show that if $x$ fixes $\a$ and $\omega_b$, then $i=j = 0$ and $A = \pm I_2$. So let us assume $x$ fixes $\a$ and $\omega_b$, which means that $A$ is either diagonal or anti-diagonal with respect to the basis $\{u,v\}$ for the natural ${\rm SU}_{2}(q)$-module $U$. First assume $A={\rm diag}(a,a^{-1})$ is diagonal, so $a^{q+1}=1$. If $x$ fixes the two spaces in $\omega_b$, then \begin{align*} (u+bv)^x & = a\lambda^{i(q-1)}u+a^{-1}b^{p^j}v=\eta_1(u+bv)\\ (u-b^{-q}v)^x & = a\lambda^{i(q-1)}u-a^{-1}b^{-qp^j}v=\eta_2(u-b^{-q}v) \end{align*} for some $\eta_1,\eta_2\in\mathbb{F}_{q^2}^{\times}$, whence \begin{equation}\label{e:c31} a^2\lambda^{i(q-1)}=b^{p^j-1}=b^{q(1-p^j)}. \end{equation} Since $a^{q+1}=1$ we get $(b^{q+1})^{(p^j-1)}=1$, which implies that $(b^{q+1})^{\frac{1}{2}(p^{2j}-1)}= 1$ and thus $2j=0$ or $2f$ are the only possibilities. But we are assuming $j\ne f$, hence $j=0$ and thus $i=0$. Therefore, \eqref{e:c31} gives $a^2=1$ and we conclude that $x=\pm I_2$. Similarly, if $x$ interchanges the spaces in $\omega_b$, then \begin{align*} (u+bv)^x & = a\lambda^{i(q-1)}u+a^{-1}b^{p^j}v = \eta_1(u-b^{-q}v)\\ (u-b^{-q}v)^x & = a\lambda^{i(q-1)}u-a^{-1}b^{-qp^j}v=\eta_2(u+bv) \end{align*} for some $\eta_1,\eta_2\in\mathbb{F}_{q^2}^{\times}$ and we deduce that \begin{equation}\label{e:c32} -a^2\lambda^{i(q-1)}=b^{p^j+q}=b^{-qp^j-1}. \end{equation} In particular, since $a^{q+1}=1$, it follows that \[ (b^{q+1})^{\frac{1}{2}(p^{j+f}+1)}=(-1)^{\frac{1}{2}(q+1)}\lambda^{\frac{1}{2}i(q^2-1)}=(-\lambda^{i(q-1)})^{\frac{1}{2}(q+1)} \] and thus \[ (b^{q+1})^{\frac{1}{2}(p^{2j}-1)}=(b^{q+1})^{\frac{1}{2}(p^{2(j+f)}-1)}=(-\lambda^{i(q-1)})^{\frac{1}{2}(q+1)(p^{j+f}-1)}=1. \] Since \eqref{e:bcon} holds and $j \ne f$ we deduce that $j=0$ is the only possibility, implying $i=0$. Then \eqref{e:c32} gives $b^{q+1}=b^{-q-1}$ and thus $b^{q+1}=\pm 1$. By construction we have $b^{q+1} \ne -1$ (since $\omega_b \in \O$), while \eqref{e:bcon} implies that $b^{q+1} \ne 1$. Therefore, we have reached a contradiction and this case does not arise. Now let us assume $x$ interchanges the two spaces in $\a$, so \[ A=\begin{pmatrix} 0&a\\ -a^{-1}&0 \end{pmatrix} \] is anti-diagonal and $a^{q+1}=1$. If $x$ fixes the spaces in $\omega_b$, then \begin{align*} (u+bv)^x & = ab^{p^j}u-a^{-1}\lambda^{i(q-1)}v = \eta_1(u+bv)\\ (u-b^{-q}v)^x & = -ab^{-qp^j}u-a^{-1}\lambda^{i(q-1)}v = \eta_2(u-b^{-q}v) \end{align*} for some $\eta_1,\eta_2\in\mathbb{F}_{q^2}^{\times}$ and we get \begin{equation}\label{e:c33} -a^2\lambda^{-i(q-1)}=b^{-p^j-1}=b^{q+qp^j}. \end{equation} This implies that $(b^{q+1})^{\frac{1}{2}(p^{2j}-1)}=1$, which leads to a contradiction as above. Finally, suppose $x$ interchanges the two spaces in $\omega_b$. Here we get \begin{equation}\label{e:c34} a^2\lambda^{i(q-1)}=b^{q-p^j}=b^{qp^j-1} \end{equation} and thus $(b^{q+1})^{\frac{1}{2}(p^{f+j}-1)}=\lambda^{\frac{1}{2}i(q^2-1)}=\pm 1$ and so $(b^{q+1})^{\frac{1}{2}(p^{2(f+j)}-1)}=1$. It follows that $j=0$ and $i=0$, so $(b^{q+1})^{\frac{1}{2}(p^f-1)}=1$ and this is incompatible with \eqref{e:bcon}. \end{proof} \begin{lemm}\label{l:c3_cond_PSigmaL} Let $G=\PSigmaL_2(q)$ with $q$ odd. Then $\{\alpha,\omega_b\}$ is a base for $G$ if and only if \textup{(}\ref{e:bcon}\textup{)} holds for all $027$ (see Proposition \ref{p:comp}) and we recall that $b(G)=2$ if and only if $q$ is odd and $G\cap \PGL_2(q) = G_0$. In view of Corollary \ref{c:c3_min}, we may assume that $G = \PSigmaL_2(q)$. By Theorem \ref{t:cd}, it suffices to show that if $\{\a,\omega_b\}$ is a base for $G$, then there exists $c \in \mathbb{F}_{q^2}^{\times}$ with $c^{q+1}\ne -1$ such that both $\{\a,\omega_{c}\}$ and $\{\omega_b,\omega_{c}\}$ are also bases. Note that $b^{q+1} \ne -1$ and $b$ satisfies the condition in \eqref{e:bcon} for all $027$ is odd and $G=\PSigmaL_2(q)$. Let $r$ be the number of regular suborbits of $G$. If $q$ is a prime then $G = G_0$ and \cite[Lemma 7.9]{BurnessHarper} gives $r=(q-\ell)/4$, where $q \equiv \ell \imod{4}$ with $\ell \in \{1,3\}$. For the remainder, we may assume $q>p$. Let $\l$ be a primitive element of $\mathbb{F}_{q^2}$. Then by Lemma \ref{l:c3_cond}, we see that $\{\a,\omega_\l\}$ is a base for $G$. Therefore, the valency of $\Sigma(G)$ is at least $\phi(q^2-1)/2$, where $\phi$ is Euler's function and thus $r \geqs \phi(q^2-1)/2f(q+1)$ since $|H|=f(q+1)$. By applying the lower bound in Lemma \ref{l:euler} we deduce that \[ \frac{\phi(q^2-1)}{2f(q+1)} \geqs 2 \] for $q>27$ and the desired result follows. \end{proof} Finally, we consider the clique number $\omega(G)$ of $\Sigma(G)$. Our main result is the following, which establishes a strong form of Theorem \ref{t:main11} when $G = G_0$ is simple. \begin{prop} \label{p:psl2_c3cliq_G0} Suppose $G = \LL_2(q)$ and $H$ is of type $\GL_1(q^2)$, where $q$ is odd. Then $\omega(G)\geqs \frac{1}{2}(q-1)$. \end{prop} \begin{proof} Let $b$ be a non-square in $\mathbb{F}_{q^2}$ with $b^{q+1}\ne -1$. Then $bx$ is a non-square for all $x \in \mathbb{F}_q$ and thus $\{\alpha,\omega_{bx}\}$ is a base for $G$ if and only if $(bx)^{q+1}\ne -1$ (see Remark \ref{r:c3g0}). We claim that \[ C=\{\a\}\cup \{\omega_{bx} \suchthat x\in\mathbb{F}_q^\times, \litspace (bx)^{q+1}\ne -1\} \] is a clique in $\Sigma(G)$ with $|C| \geqs \frac{1}{2}(q-1)$. To see this, first note that if $x \in \mathbb{F}_q^\times$ and $(bx)^{q+1} = -1$, then $b^{q+1} = -x^{-2}$ and there are at most two possibilities for $x$. Let us also observe that if $y = -b^{-(q+1)}x^{-1}$ then $by = -(bx)^{-q}$ and thus $\omega_{bx} = \omega_{by}$. This shows that $|C| \geqs 1+\frac{1}{2}(q-3) = \frac{1}{2}(q-1)$. To complete the proof of the proposition, it suffices to show that $\{\omega_{bx},\omega_{by}\}$ is a base for all distinct points $\omega_{bx}, \omega_{by}$ in $C$. To this end, set $c = bx$ and note that $c$ is a non-square such that $c^{q+1}\ne -1$ and $by = cyx^{-1}\in c\mathbb{F}_q^\times$. In this way, the problem is reduced to showing that $\{\omega_b,\omega_{by}\}$ is a base for all $y\in\mathbb{F}_q^\times$ with $(by)^{q+1}\ne -1$ and $by \ne -b^{-q}$ (the latter condition forces $\omega_b \ne \omega_{by}$). By Corollary \ref{c:c3_cond}, it suffices to show that there exist scalars $a_1,d \in \mathbb{F}_{q^2}^{\times}$ such that \begin{equation}\label{e:bzz} by = \frac{ba_1^{-2}(b+b^{-q})+b^{-q}d}{a_1^{-2}(b+b^{-q})-d}, \end{equation} where $a_1^{q+1} = 1+b^{q+1}$ and $d$ satisfies the conditions in parts~\ref{it:4.16.2} and~\ref{it:4.16.3} of the corollary. In fact, in view of Remark \ref{r:c3g0}, we can replace the condition in~\ref{it:4.16.3} by the property that $d$ is a non-square. Let $a_1 \in \mathbb{F}_{q^2}^{\times}$ be any scalar such that $a_1^{q+1} = 1+b^{q+1}$, noting that $a_1$ exists since $b^{q+1} \in \mathbb{F}_q$. Since $by \ne -b^{-q}$, it follows that $b^{-(q+1)}+y \ne 0$ and we may define \[ d = \frac{ba_1^{-2}(1+b^{-(q+1)})(y-1)}{b^{-(q+1)}+y}. \] Since $b^{q+1} \ne -1$ we see that $d \ne a_1^{-2}(b+b^{-q})$ and by rearranging we deduce that \eqref{e:bzz} holds. In particular, since $by \ne 0$ we have $d \ne -b^{q+1}a_1^{-2}(b+b^{-q})$. If $d^{q+1}=-1$, then $(y-1)^2 = -b^{q+1}(b^{-(q+1)}+y)^2$, which forces $(b^{q+1}y^2+1)(b^{q+1}+1) = 0$. But this is false because $b^{q+1}y^2 = (by)^{q+1}\ne -1$ and $b^{q+1}\ne -1$, whence $d^{q+1}\ne -1$ and $d$ satisfies all of the conditions in part~\ref{it:4.16.2} of Corollary \ref{c:c3_cond}. Finally, let us observe that $a_1^{-2}$ is a square and $\frac{(1+b^{-(q+1)})(y-1)}{b^{-(q+1)}+y}\in\mathbb{F}_q$, which is also a square. Therefore, $d$ is a non-square since $b$ is a non-square. We conclude that $\{\omega_b,\omega_{by}\}$ is a base for $G$ and this completes the proof of the proposition. \end{proof} \begin{coro} \label{c:psl2_c3cliq_G0} If $G = G_0 = \LL_2(q)$ and $H$ is of type $\GL_1(q^2)$, then either $\omega(G)\geqs 5$, or $q = 5$ and $\omega(G) = 4$. \end{coro} \begin{proof} By Proposition \ref{p:comp} we may assume $q>27$. Now apply Proposition \ref{p:psl2_c3cliq_G0}. \end{proof} \begin{rema}\label{r:c3_clique} As for the $\mcC_2$-actions from the previous section, we have been unable to extend Corollary \ref{c:psl2_c3cliq_G0} to the base-two groups with $G \ne G_0$. Using a computational approach (see Section \ref{ss:clique}), it is straightforward to show that $\omega({\rm P\Sigma L}_{2}(q)) \geqs 5$ when $55$. In order to illustrate some of the difficulties that arise, let us assume $G = \PSigmaL_2(q)$. As explained in the proof of Proposition \ref{p:psl2_c3star}, if $b \in \mathbb{F}_{q^2}$ is a scalar such that $b^{q+1} \ne -1$ and \eqref{e:bcon} holds for all $011$. By combining Proposition \ref{p:bound} with Theorem \ref{t:random}, we may assume that either $G \in \mathcal{L}$ or a point stabiliser $H$ is insoluble. So in view of Propositions \ref{p:psl2_c2star} and \ref{p:psl2_c3star}, we may assume $H$ is insoluble. By inspecting \cite[Tables 8.1 and 8.2]{Low-Dimensional}, one of the following holds: \begin{enumerate}[label = (\alph*)] \item $H = S_5$ or $A_5$, $q \in \{p,p^2\}$ and $p \equiv \pm 1, \pm 3 \imod{10}$; \item $H$ is a subfield subgroup of type ${\rm GL}_{2}(q_0)$, where $q = q_0^k$, $k$ is a prime and $q_0 \geqs 4$. \end{enumerate} First consider case (a) and recall that we write $i_m(H)$ for the number of elements of order $m$ in $H$. In view of Proposition \ref{p:bound}, it suffices to show that $\what{Q}(G)<1/2$ (see Section \ref{ss:prob}). We refer the reader to \cite[Section 3.2]{BG_book} for detailed information on the conjugacy classes of elements of prime order in $G$. Let $x \in H$ be an element of prime order $m$ and note that $m \in \{2,3,5\}$. If $m=2$ then $|x^G| \geqs \frac{1}{2}q^{1/2}(q+1) = b_1$ (minimal if $x$ is an involutory field automorphism) and we note that $i_2(H) \leqs 25 = a_1$. Similarly, if $m \in \{3,5\}$ then $|x^G| \geqs q(q-1) = b_2$ and $i_3(H)+i_5(H) = 44 = a_2$. Therefore, Lemma \ref{l:calc} implies that \begin{equation}\label{e:qq} \what{Q}(G) \leqs a_1^2/b_1+a_2^2/b_2 \end{equation} and we deduce that $\what{Q}(G)<1/2$ if $q \geqs 197$. The remaining cases with $q<197$ can be verified using {\sc Magma}. Finally, let us consider the subfield subgroups in (b). First assume $k=2$. We claim that $b(G) \geqs 3$. If $p$ is odd then \[ |\O| = |G_0: {\rm PGL}_{2}(q_0)| = \frac{1}{2}q_0(q_0^2+1) \] and the claim follows since $b(G) \geqs \log_{|\O|}|G|>2$. Similarly, if $p=2$ and $G \ne G_0$ then $|\O| = q_0(q_0^2+1)$ and \[ b(G) \geqs \frac{\log |G|}{\log |\O|} \geqs \frac{\log 2q_0^2(q_0^4-1)}{\log q_0(q_0^2+1)} > 2. \] Finally, suppose $p=2$ and $G = G_0$. Here $\log_{|\O|}|G|<2$ and the subdegrees of $G$ are recorded in \cite[p.354]{FI}. By inspection, we see that $G$ does not have a regular suborbit and thus $b(G) \geqs 3$. Next assume $k \geqs 5$ and let $x \in H$ be an element of prime order. If $x$ is an involutory field automorphism of $G_0$, then $|x^G| \geqs \frac{1}{2}q^{1/2}(q+1) = b_1$ and we note that there are at most $a_1 = q_0^{1/2}(q_0+1)$ such elements in $H$. In each of the remaining cases, we have $|x^G| \geqs \frac{1}{2}q(q-1) = b_2$ and we observe that $|H| \leqs q_0(q_0^2-1)\log q = a_2$. By applying Lemma \ref{l:calc} we deduce that \eqref{e:qq} holds and this gives $\what{Q}(G)<1/2$ as required. To complete the proof, we may assume $k=3$. This case requires a more refined treatment. First let $x \in H \cap {\rm PGL}_2(q)$ be an element of prime order $m$. If $x$ is unipotent (so $m=p$) then $|x^G| \geqs \frac{1}{2}(q^2-1) = b_1$ and we note that there are exactly $a_1 = q_0-1$ such elements in $H$. Similarly, if $x$ is a semisimple involution then $|x^G| \geqs \frac{1}{2}q(q-1) = b_2$ and we have $i_2({\rm PGL}_2(q_0)) = q_0^2 = a_2$. Next suppose $m \ne p$ and $m \geqs 3$, so $m$ divides $q_0^2-1$ and there are $\frac{1}{2}(m-1)$ distinct $G_0$-classes of such elements in $G$ (and the same number of ${\rm L}_{2}(q_0)$-classes in $H \cap {\rm PGL}_{2}(q)$). If $\{x_1, \ldots, x_t\}$ is a set of representatives of the distinct $G$-classes of these elements, then there exist positive integers $k_i$ such that $\sum_i k_i = \frac{1}{2}(m-1)$ and \[ |x_i^G \cap H| = k_iq_0(q_0+\e),\;\; |x_i^G| = k_iq_0^3(q_0^3+\e), \] where $\e =1$ if $m$ divides $q_0-1$, otherwise $\e=-1$ (here $|x_i^{G_0}| = |x_i^{{\rm PGL}_2(q)}| = q_0^3(q_0^3+\e)$ and $k_i$ denotes the number of distinct $G_0$-classes that are fused under the action of field automorphisms in $G$). Therefore, the contribution to $\what{Q}(G)$ from elements of order $m$ is equal to \[ \sum_{i=1}^t\frac{(k_iq_0(q_0+\e))^2}{k_iq_0^3(q_0^3+\e)} = \frac{1}{2}(m-1)\cdot\frac{(q_0+\e)^2}{q_0(q_0^3+\e)}. \] If $m$ divides $q_0+1$ then $m-1 \leqs q_0$ and there are at most $\log(q_0+1)$ possibilities for $m$, so the total contribution to $\what{Q}(G)$ from these elements is at most \[ f_1(q_0) = \log(q_0+1) \cdot \frac{1}{2}q_0 \cdot \frac{(q_0-1)^2}{q_0(q_0^3-1)}. \] Similarly, the contribution from the elements with $m$ dividing $q_0-1$ is no more than \[ f_2(q_0) = \log(q_0-1) \cdot \frac{1}{2}(q_0-2) \cdot \frac{(q_0+1)^2}{q_0(q_0^3+1)}. \] Finally, let us assume $x \in G$ is a field automorphism of order $m$. As above, if $m=2$ then $|x^G| \geqs \frac{1}{2}q^{1/2}(q+1) = b_3$ and there are at most $a_3 = q_0^{1/2}(q_0+1)$ of these elements in $H$. Next suppose $m=3$. Here $|x^G| \geqs q_0^2(q_0^4+q_0^2+1) = b_4$ and we may assume $H = C_G(x)$, which implies that $H$ contains at most \[ 2\left(1+i_3({\rm L}_2(q_0))\right) \leqs 2\left(1+\frac{|{\rm GL}_2(q_0)|}{(q_0-1)^2}\right) = 2q_0(q_0+1)+2 = a_4 \] such elements. If $m=5$ then $|x^G|>q_0^{36/5} = b_5$ and there are at most $8q_0^{12/5} = a_5$ of these elements in $H$. Finally, if $m \geqs 7$ then $|x^G|>q_0^{54/7} = b_6$ and we observe that $|H| \leqs q_0(q_0^2-1)\log q = a_6$. Set $\a=1$ if $q_0 = q_1^2$ for some $q_1$, otherwise $\a=0$. Similarly, let $\b = 1$ if $q_0 = q_1^5$ and $\gamma = 1$ if $q_0 = q_1^m$ for some prime $m \geqs 7$ (otherwise $\b=0$ and $\gamma=0$, respectively). Then by bringing all of the above estimates together, we conclude that \[ \what{Q}(G) < f_1(q_0)+f_2(q_0) + \left(a_1^2/b_1+a_2^2/b_2+ a_4^2/b_4\right) + \a a_3^2/b_3+\b a_5^2/b_5 + \gamma a_6^2/b_6 \] and one checks that this upper bound is less than $1/2$ for all $q_0 \geqs 7$. Finally, the cases with $q_0 \in \{4,5\}$ can be checked using {\sc Magma}. \end{proof} \section{Proof of Theorem \ref{t:main1}}\label{s:main1} In this section we complete the proof of Theorem \ref{t:main1}. Let $G \leqs {\rm Sym}(\O)$ be a permutation group in $\mathcal{G}$ with socle $G_0$ and point stabiliser $H$. Recall that our goal is to show that the Saxl graph $\Sigma(G)$ has the following property: \[ \mbox{\emph{Any two vertices in $\Sigma(G)$ have a common neighbour}} \tag{$\star$} \] which immediately implies that $\Sigma(G)$ has diameter $2$. In view of Propositions \ref{p:bound}, \ref{p:psl2_c2star} and \ref{p:psl2_c3star}, we may assume $G \in \mathcal{G}\setminus \mathcal{L}$ and $Q(G) \geqs 1/2$, so the relevant groups can be determined by inspecting Tables \ref{tab:random} and \ref{tab:random2} (see Theorem \ref{t:random}). In every one of these cases, we can verify property \eqref{e:star} using {\sc Magma}. To do this, we first construct $G$ and $H$ using the functions \[ \mbox{\texttt{AutomorphismGroupSimpleGroup} and \texttt{MaximalSubgroups}.} \] Next we identify a set $R$ of $(H,H)$ double coset representatives and then for each $x \in R$ we seek an element $y \in G$ (by random search) such that $H \cap H^y = H^x \cap H^y=1$. Notice that \eqref{e:star} holds if and only if such an element $y$ exists for each $x \in R$. As demonstrated by the following example, it is easy to implement this approach in {\sc Magma}. \begin{exam} Suppose $G_0 = \O_{8}^{+}(2)$, $G = G_0.3$ and $H$ is of type ${\rm O}_{2}^{-}(2) \times \GU_3(2)$. Here $Q(G) = 2071/2800>1/2$ and so this is one of the cases we need to consider. We proceed as follows, noting that $G$ has a unique conjugacy class of maximal subgroups of order $11664$: \vspace{2mm} {\small \begin{verbatim} G:=AutomorphismGroupSimpleGroup("O+",8,2); S:=LowIndexSubgroups(G,2); G:=S[1]; M:=MaximalSubgroups(G:OrderEqual:=11664); H:=M[1]`subgroup; R,T:=DoubleCosetRepresentatives(G,H,H); z:=0; for x in R do if exists(y){y : y in G | #(H meet H^y) eq 1 and #(H^x meet H^y) eq 1} then z:=z+1; end if; end for; z eq #R; \end{verbatim}} \end{exam} \vspace{1mm} \noindent This returns \texttt{true} and we conclude that \eqref{e:star} holds. An entirely similar approach is effective for all of the relevant groups in Tables \ref{tab:random} and \ref{tab:random2}. \section{Proof of Theorems \ref{t:main11} and \ref{t:main12}}\label{s:main11} Let $G \leqs {\rm Sym}(\O)$ be a permutation group in $\mathcal{G}$ with socle $G_0$ and point stabiliser $H$. Let $\omega(G)$ and $\a(G)$ denote the clique and independence numbers of $\Sigma(G)$, respectively. \subsection{Clique number}\label{ss:clique} In view of Proposition \ref{p:bound} and Theorem \ref{t:random}, together with our work in Sections \ref{ss:psl2_c2} and \ref{ss:psl2_c3}, it remains to verify the bound $\omega(G) \geqs 5$ for the groups appearing in Tables \ref{tab:random} and \ref{tab:random2}. With the aid of {\sc Magma}, this is a straightforward exercise, working with a suitable permutation representation of $G$ and $H$. For example, the following function for checking the bound $\omega(G) \geqs 5$ can be used effectively in every case. This uses the fact that $\omega(G) \geqs 5$ if and only if there exist elements $x_2, \ldots, x_5$ in $G$ such that $H^{x_i} \cap H^{x_j} = 1$ for all distinct $i,j \in \{1, \ldots, 5\}$, where $x_1 = 1$. \vspace{2mm} {\small \begin{verbatim} clique:=function(G,H); exists(x2){y : y in G | #(H meet H^y) eq 1}; exists(x3){y : y in G | #(H meet H^y) eq 1 and #(H^x2 meet H^y) eq 1}; exists(x4){y : y in G | #(H meet H^y) eq 1 and #(H^x2 meet H^y) eq 1 and #(H^x3 meet H^y) eq 1}; exists(x5){y : y in G | #(H meet H^y) eq 1 and #(H^x2 meet H^y) eq 1 and #(H^x3 meet H^y) eq 1 and #(H^x4 meet H^y) eq 1}; return "omega(G) is at least 5"; end function; \end{verbatim}} \subsection{Independence number}\label{ss:indep} Now let us turn to the independence number $\a(G)$ of $\Sigma(G)$. Here we apply work of Magaard and Waldecker \cite{MagaardIndependence2,MagaardIndependence3} and we begin with the following trivial observation. \begin{lemm}\label{l:pt} Let $G \leqs {\rm Sym}(\O)$ be a permutation group and let $c$ be a positive integer. Then $\a(G) \leqs c$ only if every $(c+1)$-point stabiliser in $G$ is trivial. \end{lemm} \begin{prop}\label{p:ag2} Let $G \leqs {\rm Sym}(\O)$ be a group in $\mathcal{G}$ with socle $G_0$ and point stabiliser $H$. Then $\a(G) = 2$ if and only if $G = A_5$ and $\O$ is the set of $2$-element subsets of $\{1, \ldots, 5\}$. \end{prop} \begin{proof} First assume $G = G_0$ is simple. We apply \cite[Theorem 3.20]{MagaardIndependence2}, which gives a complete list of all the finite simple transitive groups with the property that every $3$-point stabiliser is trivial. The theorem implies that $G = {\rm L}_2(q)$, ${}^2B_2(q)$ or ${\rm L}_{3}(4)$, and by inspection (using \cite[Theorem 2]{Burness2020base}) we deduce that $G = {\rm L}_{2}(4)$ with $H = D_6$ is the only example with $G$ primitive and $b(G)=2$. Here $\Sigma(G)$ is the Johnson graph $J(5,2)$ and $\a(G)=2$ (note that $G$ is permutation isomorphic to $A_5$ acting on the set of $2$-element subsets of $\{1, \ldots, 5\}$). Now assume $G \ne G_0$. Then $G_0 \leqs {\rm Sym}(\O)$ is a transitive group and every $3$-point stabiliser is trivial, so as above the possibilities for $G_0$ and $\O$ are described in \cite[Theorem 3.20]{MagaardIndependence2} and by appealing to \cite[Theorem 1.3]{MagaardIndependence2} we see that $G_0 = {\rm L}_{2}(q)$ with $q=p^f$. More precisely, either $p$ is odd and $G = {\rm PGL}_{2}(q)$ or $G_0.2= \la G_0, \delta\phi^{f/2}\ra$ (in the notation of Section \ref{ss:psl2}), or $p=2$, $f$ is a prime and $G = \Aut(G_0)$. The cases with $q \in \{4,5\}$ can be checked using {\sc Magma} \cite{Magma} and we find that there are no groups $G\ne G_0$ with $\alpha(G) = 2$. Finally, let us assume $q\geqs 7$, in which case the relevant possibilities are labelled (a)--(d) in \cite[Theorem 3.20(1)]{MagaardIndependence2}. In (a) and (d), it is easy to check that $b(G_0)>2$, while the groups in (b) and (c) are imprimitive. Therefore, none of these cases arise and the proof is complete. \end{proof} \begin{prop}\label{p:ag3} Let $G \leqs {\rm Sym}(\O)$ be a group in $\mathcal{G}$ with socle $G_0$ and point stabiliser $H$. Then $\a(G) \ne 3$. \end{prop} \begin{proof} Seeking a contradiction, suppose $\a(G)=3$, in which case every $4$-point stabiliser is trivial. We may also assume that there exists a nontrivial $3$-point stabiliser. If $G=G_0$ then the possibilities for $G$ are described in \cite[Theorem 1.1(i)]{MagaardIndependence3}, but we find that there are no compatible examples with $G$ primitive, $H$ soluble and $b(G)=2$. The cases with $G \ne G_0$ are recorded in \cite[Theorem 1.2]{MagaardIndependence3} and once again we see that there are no valid examples. \end{proof} \vspace{2mm} This completes the proof of Theorems \ref{t:main11} and \ref{t:main12}. \section{Proof of Theorem \ref{t:main2}}\label{s:main2} In this final section we prove Theorem \ref{t:main2} on the groups $G \in \mathcal{G}$ with a unique regular suborbit. If $G \in \mathcal{L}$ then we refer the reader to Propositions \ref{p:psl2_c2reg} and \ref{p:psl2_c3reg}, so we may assume $G \in \mathcal{G} \setminus \mathcal{L}$. If $Q(G) \geqs 1/4$ then we can simply read off the relevant groups in Tables \ref{tab:random} and \ref{tab:random2}, so we may assume $Q(G)<1/4$. Then in view of \eqref{e:QG}, it follows that $G$ has a unique regular suborbit only if \[ |H|^2 > \frac{3}{4}|G|, \] where $H$ is a point stabiliser. The following result reveals that there are no such groups and this completes the proof of Theorem \ref{t:main2}. \begin{prop}\label{p:size} Let $G$ be a group in $\mathcal{G}\setminus \mathcal{L}$ with point stabiliser $H$. If $Q(G)<1/4$ then $|H|^2 \leqs \frac{3}{4}|G|$. \end{prop} \begin{proof} First assume $G_0 = A_m$ is an alternating group. If $m \leqs 12$ then it is easy to verify the desired bound with the aid of {\sc Magma}. On the other hand, if $m>12$ then by inspecting \cite[Table 14]{SolubleStabiliser} and \cite[Table 4]{Burness2020base} we deduce that $m$ is a prime, $H = {\rm AGL}_1(m)\cap G$ and we have \[ \frac{|H|^2}{|G|} \leqs \frac{m(m-1)}{(m-2)!}<\frac{3}{4} \] as required. Similarly, if $G_0$ is a sporadic group then the possibilities for $G$ and $H$ can be determined by combining the information in Table \ref{tab:random} and \cite[Table 4]{Burness2020base} with the tables of maximal subgroups in \cite{Wilson}. In each case, it is a straightforward exercise to show that $|H|^2 \leqs \frac{3}{4}|G|$. Next suppose $G_0$ is an exceptional group of Lie type. Then either $H=N_G(T)$ for some maximal torus $T$ of $G_0$ (see \cite[Table 5.2]{MaxExceptional}), or $(G,H)$ is recorded in Table \ref{tab:ex}. One can now verify the bound $|H|^2 \leqs \frac{3}{4}|G|$ by inspection. For example, if $G_0 ={}^2B_2(q)$ with $q = 2^f$ and $f \geqs 3$ odd, then $|G| \geqs q^2(q^2+1)(q-1)$ and by inspecting \cite[Table 5.2]{MaxExceptional} we deduce that $|H|\leqs 4(q+\sqrt{2q}+1)\log q$. A routine calculation shows that $|H|^2 \leqs \frac{3}{4}|G|$. Finally, let us assume $G_0$ is a classical group. Then $(G,H)$ is one of the cases recorded in Table \ref{tab:cases} and once again the result follows by inspection (recall that in each case, the precise structure of $H$ is given in \cite{KleidmanLiebeckClassicalGroups}). For instance, if $G_0 = {\rm L}_4^{\e}(q)$ and $H$ is of type $\GL_1^{\e}(q)\wr S_4$ with $q \geqs 3$, then $|H| \leqs 48(q+1)^3\log q$ and the result follows since $|G|>\frac{1}{8}q^{15}$. In fact, one can check that the only case in Table \ref{tab:cases} with $|H|^2>\frac{3}{4}|G|$ is where $G = {\rm PGSp}_6(3)$ and $H$ is of type ${\rm Sp}_2(3) \wr S_3$. But $b(G)=3$ in this case (see \cite[Table 7]{Burness2020base}), so this is not a group in $\mathcal{G}$. \end{proof} \nocite{*} \bibliographystyle{mersenne-plain} \bibliography{ALCO_Burness_633.bib} \end{document}