Proof. The possible structures (i)–(vii) of ${\mathcal{Q}}_{\unicode[STIX]{x1D703}}$ are given by [Reference Payne and Thas21, 2.4.1]. We verify the claimed upper bounds for $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|$ . The bounds in cases (iii) and (iv) are immediate, because every line of ${\mathcal{Q}}$ is incident with exactly $s+1$ points, and every point of ${\mathcal{Q}}$ is incident with exactly $t+1$ lines. For case (ii), note [Reference Payne and Thas21, Section 2.7] that the maximum size of a set of pairwise noncollinear points in ${\mathcal{Q}}$ is $st+1$ . For (v), if ${\mathcal{Q}}_{\unicode[STIX]{x1D703}}$ is a dual grid with parameters $t_{1}$ and $t_{2}$ , then $t_{1}\leqslant t$ and $t_{2}\leqslant t$ , and hence $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|\leqslant 2(t+1)$ .

Now suppose that ${\mathcal{Q}}_{\unicode[STIX]{x1D703}}$ is a grid with parameters $s_{1}$ and $s_{2}$ , noting that $s_{1}\leqslant s$ and $s_{2}\leqslant s$ , and assuming (without loss of generality) that $s_{1}\geqslant s_{2}$ . If $s_{1}=s_{2}=s$ then $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|=(s+1)^{2}$ , and ${\mathcal{Q}}_{\unicode[STIX]{x1D703}}$ is a generalized quadrangle of order $(s,1)$ , so [Reference Payne and Thas21, 2.2.2(i)] implies that $s\leqslant t$ . The case $s_{2}=s-1$ cannot occur, because if $\unicode[STIX]{x1D703}$ fixes $s$ points incident with a line then it must also fix the final point; and if $s_{2}\leqslant s-2$ then $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|\leqslant (s+1)(s-1)<s^{2}$ . Finally, suppose that ${\mathcal{Q}}$ is a thick finite generalized quadrangle, and let $(s^{\prime },t^{\prime })$ denote its order. Then $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|=(s^{\prime }+1)(s^{\prime }t^{\prime }+1)$ by Lemma 2.1(i). If $t^{\prime }=t$ then $s^{\prime }<s$ because $\unicode[STIX]{x1D703}\neq 1$ , so [Reference Payne and Thas21, 2.2.1] implies that $s^{\prime }t=s^{\prime }t^{\prime }\leqslant s$ , and hence $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|\leqslant (s/t+1)(s+1)\leqslant (t^{2}/t+1)(s+1)=(s+1)(t+1)$ , where for the second inequality we use Lemma 2.1(ii). If $t^{\prime }<t$ then the dual statement of [Reference Payne and Thas21, 2.2.1] yields $s^{\prime }t^{\prime }\leqslant t$ , so $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|=(s^{\prime }+1)(s^{\prime }t^{\prime }+1)\leqslant (s+1)(t+1)$ .

The final assertion is deduced by comparing the upper bounds on $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|$ established in each case. We observe that $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|\leqslant (s+1)(t+1)$ except possibly in the second case of (v), where our bound is $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|<s^{2}$ . However, if $s\leqslant t+2$ then in this case we have $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|<s^{2}<(s+1)(t+1)$ .◻

Proof of Theorem 1.3.

Let $(s,t)$ be the order of ${\mathcal{Q}}$ , and let ${\mathcal{Q}}_{\unicode[STIX]{x1D703}}=({\mathcal{P}}_{\unicode[STIX]{x1D703}},{\mathcal{L}}_{\unicode[STIX]{x1D703}})$ be the substructure of ${\mathcal{Q}}$ fixed by $\unicode[STIX]{x1D703}$ . We must show that either $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|<|{\mathcal{P}}|^{4/5}$ , or $(s,t)=(2,4)$ and $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|=15$ for ${\mathcal{Q}}=\mathsf{Q}^{-}(5,2)$ .

First suppose that $s\neq 2$ . By Lemma 2.2, we have either $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|<s^{2}$ or  $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|\leqslant (s+1)(t+1)$ . If $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|<s^{2}$ then $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|<|{\mathcal{P}}|^{4/5}$ since $|{\mathcal{P}}|=(s+1)(st+1)>s^{2}t\geqslant s^{5/2}$ by Lemma 2.1. If $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|\leqslant (s+1)(t+1)$ then it suffices to show that the function $f(s,t)=((s+1)(st+1))^{4/5}-(s+1)(t+1)$ is positive for all $s\geqslant 3$ , for all $s^{1/2}\leqslant t\leqslant s^{2}$ . This is readily checked when $s\in \{3,4\}$ , so assume that $s\geqslant 5$ . Regarding $s$ and $t$ as real variables, we have

$$\begin{eqnarray}\frac{\unicode[STIX]{x2202}f}{\unicode[STIX]{x2202}t}(s,t)=\frac{(s+1)(4s-h(s,t))}{h(s,t)},\quad \text{where }h(s,t)=5((s+1)(st+1))^{1/5}.\end{eqnarray}$$

As $s,t$ are positive, this derivative is positive if and only if $4s-h(s,t)>0$ . Since $s\geqslant 5$ and $2\leqslant t\leqslant s^{2}$ , we have $h(s,t)\leqslant 5({\textstyle \frac{6}{5}}s)^{1/5}({\textstyle \frac{11}{10}}st)^{1/5}\leqslant 5({\textstyle \frac{33}{25}})^{1/5}s^{4/5}$ . Hence, $4s-h(s,t)\geqslant s^{4/5}(4s^{1/5}-5({\textstyle \frac{33}{25}})^{1/5})$ . The right-hand side of this inequality is positive if $s>({\textstyle \frac{5}{4}})^{5}({\textstyle \frac{33}{25}})={\textstyle \frac{4125}{1024}}\approx 4.028$ , and so certainly $(\unicode[STIX]{x2202}f/\unicode[STIX]{x2202}t)(s,t)>0$ when $s\geqslant 5$ and $s^{1/2}\leqslant t\leqslant s^{2}$ . Since $f(s,t)\geqslant f(s,s^{1/2})$ and $f(s,s^{1/2})>0$ for $s\geqslant 5$ , it follows that $f(s,t)>0$ for all $s\geqslant 5$ , for all $s^{1/2}\leqslant t\leqslant s^{2}$ .

Now suppose that $s=2$ . Then $t\in \{2,4\}$ by Lemma 2.1. There exist unique generalized quadrangles of orders $(2,2)$ and $(2,4)$ , namely the symplectic space $\mathsf{W}(3,2)$ and the elliptic quadric $\mathsf{Q}^{-}(5,2)$ , respectively [Reference Payne and Thas21, 5.2.3 and 5.3.2]. The full collineation groups of these generalized quadrangles are $\text{P}\unicode[STIX]{x1D6E4}\text{Sp}_{4}(2)$ and $\text{P}\unicode[STIX]{x1D6E4}\text{U}_{4}(2)$ , respectively. One may use the package FinInG [Reference Bamberg, Betten, Cara, De Beule, Lavrauw and Neunhöffer3] in the computer algebra system GAP [15] to check that every nonidentity collineation of $\mathsf{W}(3,2)$ fixes at most $7$ points. Since $\mathsf{W}(3,2)$ has a total of $15$ points and $15^{4/5}\approx 8.73>7$ , the claimed inequality $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|<|{\mathcal{P}}|^{4/5}$ holds for every nonidentity collineation $\unicode[STIX]{x1D703}$ in this case. On the other hand, there exist $36$ nonidentity collineations of $\mathsf{Q}^{-}(5,2)$ that fix $15$ points, but the total number of points of $\mathsf{Q}^{-}(5,2)$ is $27$ and $27^{4/5}\approx 13.97<15$ . We also remark that the substructure fixed by such a collineation is, in fact, a generalized quadrangle of order $(2,2)$ . Every other nonidentity collineation of $\mathsf{Q}^{-}(5,2)$ fixes at most $9$ points.◻

Proof of Theorem 1.2.

Since the action of $H_{1}$ on $\unicode[STIX]{x1D6FA}_{1}$ is not semiregular, there exists $h_{1}\in H_{1}\setminus \{1\}$ fixing at least one point of $\unicode[STIX]{x1D6FA}_{1}$ . Let $f_{1}$ be the number of points of $\unicode[STIX]{x1D6FA}_{1}$ fixed by $h_{1}$ . Let $\unicode[STIX]{x1D703}=(h_{1},1,\ldots ,1)\in N$ , and let $f$ be the number of points of ${\mathcal{Q}}$ fixed by $\unicode[STIX]{x1D703}$ . If $r=1$ then Theorem 1.3 implies that $f_{1}=f<|{\mathcal{P}}|^{4/5}=|\unicode[STIX]{x1D6FA}_{1}|^{4/5}$ . If $r\geqslant 2$ then Theorem 1.3 gives $f_{1}\prod _{i=2}^{r}|\unicode[STIX]{x1D6FA}_{i}|=f<|{\mathcal{P}}|^{4/5}=(\prod _{i=1}^{r}|\unicode[STIX]{x1D6FA}_{i}|)^{4/5}$ , so $f_{1}(\prod _{i=2}^{r}|\unicode[STIX]{x1D6FA}_{i}|)^{1/5}<|\unicode[STIX]{x1D6FA}_{1}|^{4/5}$ . Since $(\prod _{i=2}^{r}|\unicode[STIX]{x1D6FA}_{i}|)^{1/5}\geqslant |\unicode[STIX]{x1D6FA}_{1}|^{(r-1)/5}$ , it follows that $f_{1}<|\unicode[STIX]{x1D6FA}_{1}|^{1-r/5}$ . In particular, $1-r/5>0$ because $f_{1}\geqslant 1$ , and so $r\leqslant 4$ .◻

Proof. If $n\geqslant 19$ and $x\in T$ is a $3$ -cycle, then we have

$$\begin{eqnarray}|C_{T}(x)|={\textstyle \frac{3}{2}}(n-3)!>({\textstyle \frac{1}{2}}n!)^{4/5}=|T|^{4/5}.\end{eqnarray}$$

The remaining assertions are readily verified using GAP [15]. ◻

Proof. This is readily verified by checking maximal centralizer orders in the ATLAS [Reference Conway, Curtis, Norton, Parker and Wilson10]. ◻

Proof. (i) For $T\cong \text{E}_{8}(q)$ , $\text{F}_{4}(q)$ or $^{3}\text{D}_{4}(q)$ , take $x\in T$ to be a unipotent element of type $\text{A}_{1}$ in the sense of [Reference Liebeck and Seitz18, Tables 22.2.1 and 22.2.4] and [Reference Spaltenstein23], respectively. Then $|C_{T}(x)|=q^{57}|\text{E}_{7}(q)|$ , $q^{15}|\text{C}_{3}(q)|$ or $q^{12}(q^{6}-1)$ , respectively, and it is readily checked that $|C_{T}(x)|>|T|^{3/5}$ in each case. Now suppose that $T\cong \text{E}_{7}(q)$ or $\text{E}_{6}^{\unicode[STIX]{x1D700}}(q)$ , and write $G:=\text{Inndiag}(T)$ . Take $x\in T$ to be a unipotent element of type $\text{A}_{1}$ in the sense of [Reference Liebeck and Seitz18, Tables 22.2.2 and 22.2.3], respectively. Then $x^{T}=x^{G}$ by [Reference Burness and Giudici9, Corollary 17.10], so we have $|C_{T}(x)|=|C_{G}(x)|/|G:T|=q^{33}|\text{D}_{6}(q)|/\text{gcd}(2,q-1)$ or $q^{21}|\text{A}_{5}^{\unicode[STIX]{x1D700}}(q)|/\text{gcd}(3,q-\unicode[STIX]{x1D700})$ , respectively, and again one can check that $|C_{T}(x)|>|T|^{3/5}$ in each case.

(ii) If $T\cong ^{2}\text{B}_{2}(q)$ then $|C_{T}(x)|\leqslant q^{2}<(q^{2}(q^{2}+1)(q-1))^{2/5}=|T|^{2/5}$ for all $x\in T\setminus \{1\}$  [Reference Suzuki24]. It remains to check that $|C_{T}(x)|>|T|^{2/5}$ for some $x\in T\setminus \{1\}$ when $T$ has type $^{2}\text{F}_{4}$ , $\text{G}_{2}$ or $^{2}\text{G}_{2}$ . In these respective cases, take $x$ to be a unipotent element of type $(\tilde{\text{A}}_{1})_{2}$ , $\text{A}_{1}$ or $(\tilde{\text{A}}_{1})_{3}$ in the sense of [Reference Liebeck and Seitz18, Tables 22.2.5–22.2.7], so that $|C_{T}(x)|=q^{10}|^{2}\text{B}_{2}(q)|$ , $q^{5}|\text{A}_{1}(q)|$ or $q^{3}$ .◻

Proof. Throughout the proof, we write $q=p^{f}$ with $p$ a prime and $f\geqslant 1$ . First suppose that $T$ has type $\text{A}_{1}$ . That is, $T\cong \text{PSL}_{2}(q)$ , with $q\geqslant 4$ . The smallest nontrivial conjugacy class of $T$ has size $q(q-1)$ , ${\textstyle \frac{1}{2}}(q^{2}-1)$ or ${\textstyle \frac{1}{2}}q(q-1)$ according to whether $p=2$ , $q\equiv 1\;(\text{mod}\;4)$ or $q\equiv 3\;(\text{mod}\;4)$ . Hence, every nontrivial conjugacy class of $T$ has size greater than $|T|^{3/5}$ if and only if $q\neq 7$ . Equivalently, $|C_{T}(x)|<|T|^{2/5}$ for all $x\in T\setminus \{1\}$ if and only if $q\neq 7$ .

Now suppose that $T$ has type $\text{A}_{n}^{\unicode[STIX]{x1D700}}$ with $n\geqslant 2$ . That is, $T\cong \text{PSL}_{n+1}^{\unicode[STIX]{x1D700}}(q)$ (where we write $\text{L}^{+}:=\text{L}$ and $\text{L}^{-}:=\text{U}$ ). Let $x\in G:=\text{PGL}_{n+1}^{\unicode[STIX]{x1D700}}(q)$ be an element of order $p$ with one Jordan block of size $2$ and $n-1$ Jordan blocks of size $1$ . That is, $a_{1}=n-1$ , $a_{2}=1$ and $a_{3}=\cdots =a_{p}=0$ in the notation of [Reference Burness and Giudici9, Section 3.2.3]. Then $x\in T$ , and $x^{T}=x^{G}$ by [Reference Burness and Giudici9, Propositions 3.2.7 and 3.3.10], so by [Reference Burness and Giudici9, Tables B.3 and B.4], $|C_{T}(x)|=|C_{G}(x)|/|G:T|=(1/d)|C_{G}(x)|=(1/d)q^{2n-1}|\text{GL}_{n-1}^{\unicode[STIX]{x1D700}}(q)|$ , where $d:=\gcd (n+1,q-\unicode[STIX]{x1D700})$ . Hence

(1) $$\begin{eqnarray}|C_{T}(x)|=\frac{1}{d}q^{n(n+1)/2}\mathop{\prod }_{i=1}^{n-1}(q^{i}-\unicode[STIX]{x1D700}^{i})\qquad \text{and}\qquad |T|=\frac{1}{d}q^{n(n+1)/2}\mathop{\prod }_{i=2}^{n+1}(q^{i}-\unicode[STIX]{x1D700}^{i}).\end{eqnarray}$$

For $n\in \{2,3\}$ we must show that $|C_{T}(x)|>|T|^{2/5}$ . If $n=2$ then $d\leqslant 3$ , so $|C_{T}(x)|\geqslant {\textstyle \frac{1}{3}}q^{3}(q-\unicode[STIX]{x1D700})$ while $|T|\leqslant q^{3}(q^{2}-\unicode[STIX]{x1D700}^{2})(q^{3}-\unicode[STIX]{x1D700}^{3})$ . This implies that $|C_{T}(x)|>|T|^{2/5}$ for all $q\geqslant 7$ , and one may check directly that this inequality also holds for $q<7$ . If $n=3$ then $d\leqslant 4$ , so $|C_{T}(x)|\geqslant {\textstyle \frac{1}{4}}q^{6}(q-\unicode[STIX]{x1D700})(q^{2}-\unicode[STIX]{x1D700}^{2})$ while $|T|\leqslant q^{6}(q^{2}-\unicode[STIX]{x1D700}^{2})(q^{3}-\unicode[STIX]{x1D700}^{3})(q^{4}-\unicode[STIX]{x1D700}^{4})$ . This implies that $|C_{T}(x)|>|T|^{2/5}$ for all $q\geqslant 3$ , and a direct calculation shows that this inequality also holds for $q=2$ . Now suppose that $4\leqslant n\leqslant 8$ . We must show that $|C_{T}(x)|>|T|^{3/5}$ . Since $q\geqslant 2$ , (1) gives

$$\begin{eqnarray}|C_{T}(x)|\geqslant \frac{1}{d}\frac{q^{n^{2}}}{2^{n-1}}\qquad \text{and}\qquad |T|\leqslant \frac{1}{d}\biggl(\frac{3}{2}\biggr)^{n}q^{n^{2}+2n},\end{eqnarray}$$

and so it suffices to show that $q^{2n^{2}-6n}>d^{2}2^{2n-5}3^{3n}$ . Indeed, since $d\leqslant n+1$ , it suffices to show that $q^{2n^{2}-6n}>(n+1)^{2}2^{2n-5}3^{3n}$ . This inequality holds for all $q\geqslant 2$ if $n\in \{7,8\}$ , for all $q\geqslant 3$ if $n=6$ , for all $q\geqslant 4$ if $n=5$ , and for all $q\geqslant 11$ if $n=4$ . In the remaining cases, where $(n,q)=(6,2)$ , $(5,2)$ , $(5,3)$ , or $(4,q)$ with $q<11$ , one may check directly that $|C_{T}(x)|>|T|^{3/5}$ . It remains to show that $|C_{T}(x)|>|T|^{4/5}$ for all $q\geqslant 2$ when $n\geqslant 9$ . If $q\geqslant 3$ then (1) gives

$$\begin{eqnarray}|C_{T}(x)|\geqslant \frac{1}{d}\biggl(\frac{2}{3}\biggr)^{n-1}q^{n^{2}}\qquad \text{and}\qquad |T|\leqslant \frac{1}{d}\biggl(\frac{4}{3}\biggr)^{n}q^{n^{2}+2n},\end{eqnarray}$$

so it suffices to show that $q^{n^{2}-8n}>d\cdot 2^{3n+5}3^{n-5}$ . Indeed, since $d\leqslant n+1$ , we can just show that $q^{n^{2}-8n}>(n+1)2^{3n+5}3^{n-5}$ . This inequality holds for all $q\geqslant 3$ if $n\geqslant 11$ ; if $n=10$ , it holds for all $q\geqslant 5$ , and if $n=9$ , it holds for all $q\geqslant 29$ . For $n=10$ with $2\leqslant q<5$ , and $n=9$ with $2\leqslant q<29$ , one may check directly that $|C_{T}(x)|>|T|^{4/5}$ . Finally, we must check that $|C_{T}(x)|>|T|^{4/5}$ when $q=2$ and $n\geqslant 9$ . Since $n\geqslant 9$ , and since ${\textstyle \frac{255}{256}}q^{i}\leqslant q^{i}-\unicode[STIX]{x1D700}\leqslant {\textstyle \frac{257}{256}}q^{i}$ for $i\geqslant 8$ , (1) gives

$$\begin{eqnarray}|C_{T}(x)|\geqslant \frac{1}{d}\biggl(\frac{255}{256}\biggr)^{n-8}q^{n^{2}-28}\mathop{\prod }_{i=1}^{7}(q^{i}-1)\end{eqnarray}$$

and

$$\begin{eqnarray}|T|\leqslant \frac{1}{d}\biggl(\frac{257}{256}\biggr)^{n-8}q^{n^{2}+2n-27}\mathop{\prod }_{i=2}^{7}(q^{i}-1).\end{eqnarray}$$

Noting also that $d\leqslant 3$ , we see that it suffices to show that

$$\begin{eqnarray}2^{n^{2}-8n-32}255^{5n-40}\mathop{\prod }_{i=2}^{7}(q^{i}-1)>3\cdot 256^{n-8}257^{4n-32}.\end{eqnarray}$$

This holds for all $n\geqslant 9$ , and so the proof of the $\text{A}_{n}^{\unicode[STIX]{x1D700}}$ case is complete.

Next, suppose that $T$ has type $\text{C}_{n}$ , where $n\geqslant 2$ . That is, $T\cong \text{PSp}_{2n}(q)$ . Write $G:=\text{PGSp}_{2n}(q)$ , noting that $|G:T|=\gcd (2,q-1)$ . If $p>2$ , take $x\in G$ of order $p$ with one Jordan block of size $2$ and $2(n-1)$ Jordan blocks of size $1$ . That is, $a_{1}=2(n-1)$ , $a_{2}=1$ and $a_{3}=\cdots =a_{p}=0$ in the notation of [Reference Burness and Giudici9, Section 3.4.3]. Then $x\in T$ , and by [Reference Burness and Giudici9, Proposition 3.4.12], $x^{G}$ splits into two $T$ -conjugacy classes and hence $|C_{T}(x)|=2|C_{G}(x)|/|G:T|={\textstyle \frac{1}{2}}q^{2n-1}|\text{Sp}_{2(n-1)}(q)|$ . If $p=2$ then $T=G$ and we take $x$ to be an involution of type $b_{1}$ as in [Reference Burness and Giudici9, Table 3.4.1], so that $|C_{T}(x)|=q^{2n-1}|\text{Sp}_{2(n-1)}(q)|$ . Hence, for every $p$ , we have

(2) $$\begin{eqnarray}|C_{T}(x)|=\frac{1}{d}q^{n^{2}}\mathop{\prod }_{i=1}^{n-1}(q^{2i}-1)\qquad \text{and}\qquad |T|=\frac{1}{d}q^{n^{2}}\mathop{\prod }_{i=1}^{n}(q^{2i}-1),\end{eqnarray}$$

where $d=\gcd (2,q-1)\leqslant 2$ . If $n=2$ then $|C_{T}(x)|\geqslant {\textstyle \frac{1}{2}}q^{4}(q^{2}-1)$ and $|T|\leqslant q^{4}(q^{2}-1)(q^{4}-1)$ , and it follows that $|C_{T}(x)|>|T|^{2/5}$ for all $q\geqslant 2$ . Similarly, for $n\in \{3,4\}$ one may check that $|C_{T}(x)|>|T|^{3/5}$ for all $q\geqslant 2$ . Now suppose that $n\geqslant 5$ . Since $q^{2i}\geqslant 4$ , we have $q^{2i}-1\geqslant {\textstyle \frac{3}{4}}q^{2i}$ for all $i\geqslant 1$ , and so $|C_{T}(x)|\geqslant {\textstyle \frac{1}{2}}({\textstyle \frac{3}{4}})^{n-1}q^{2n^{2}-n}$ , while $|T|<q^{2n^{2}+n}$ . Hence, to show that $|C_{T}(x)|>|T|^{4/5}$ , it suffices to show that $({\textstyle \frac{3}{4}})^{5n-5}q^{2n^{2}-9n}>2$ . This inequality holds for all $q\geqslant 2$ when $n\geqslant 6$ , and for all $q\geqslant 4$ when $n=5$ ; for $(n,q)=(5,2)$ and $(5,3)$ , one may check directly that $|C_{T}(x)|>|T|^{4/5}$ .

Now suppose that $T$ has type $\text{B}_{n}$ , where $n\geqslant 2$ . That is, $T\cong \unicode[STIX]{x1D6FA}_{2n+1}(q)$ with $q$ odd. For $q\equiv 1$ or $3\;(\text{mod}\;4)$ , let $x\in G:=\text{PGO}_{2n+1}(q)$ be an involution of type $t_{n}$ or $t_{n}^{\prime }$ , respectively, in the sense of [Reference Burness and Giudici9, Sections 3.5.2.1 and 3.5.2.2]. Then $x\in T$ and $x^{T}=x^{G}$ , so $|C_{T}(x)|=|C_{G}(x)|/|G:T|={\textstyle \frac{1}{2}}|C_{G}(x)|=|\text{SO}_{2n}^{\pm }(q)|$ by [Reference Burness and Giudici9, Table B.8]. Now,

(3) $$\begin{eqnarray}|\text{SO}_{2n}^{\pm }(q)|=q^{n^{2}-n}(q^{n}\mp 1)\mathop{\prod }_{i=1}^{n-1}(q^{2i}-1)>\frac{1}{2}q^{n^{2}}\mathop{\prod }_{i=1}^{n-1}(q^{2i}-1),\end{eqnarray}$$

and the right-hand side above is the value of $|C_{T}(x)|$ that we obtained in the $\text{C}_{n}$ case. Since $|\unicode[STIX]{x1D6FA}_{2n+1}(q)|=|\text{PSp}_{2n}(q)|$ , we therefore reach the same conclusions as for type $\text{C}_{n}$ .

Now suppose that $T$ has type $\text{D}_{n}^{\unicode[STIX]{x1D700}}$ , namely $T\cong \text{P}\unicode[STIX]{x1D6FA}_{2n}^{\unicode[STIX]{x1D700}}(q)$ with $n\geqslant 4$ . Let $G:=\text{Inndiag}(\text{P}\unicode[STIX]{x1D6FA}_{2n}^{\unicode[STIX]{x1D700}}(q))$ , as defined on [Reference Burness and Giudici9, page 56]. Assume first that $p>2$ , noting that $|G:T|$ divides $4$ . Take $x\in G$ of order $p$ with one Jordan block of size $2(n-2)$ and two Jordan blocks of size $2$ . That is, $a_{1}=2(n-2)$ , $a_{2}=2$ and $a_{3}=\cdots =a_{p}=0$ in the notation of [Reference Burness and Giudici9, Section 3.5.3]. Then $x\in T$ , and [Reference Burness and Giudici9, Propositions 3.5.14(i) and (ii,b)] imply that $x^{T}=x^{G}$ . Therefore, [Reference Burness and Giudici9, Table B.12] gives $|C_{T}(x)|=|C_{G}(x)|/|G:T|\geqslant {\textstyle \frac{1}{4}}|C_{G}(x)|={\textstyle \frac{1}{8}}q^{4n-7}|\text{O}_{2(n-2)}^{\unicode[STIX]{x1D700}_{1}}(q)||\text{Sp}_{2}(q)|$ , where the value of $\unicode[STIX]{x1D700}_{1}=\pm$ depends on $n$ and $q$ as described there. Multiplying the inequality in (3) by $2$ to get a lower bound for $|\text{O}_{2(n-2)}^{\unicode[STIX]{x1D700}_{1}}(q)|$ , it follows that

(4) $$\begin{eqnarray}|C_{T}(x)|>\frac{1}{8}q^{n^{2}-2}(q^{2}-1)\mathop{\prod }_{i=1}^{n-3}(q^{2i}-1),\end{eqnarray}$$

while

(5) $$\begin{eqnarray}|T|=\frac{1}{d}q^{n(n-1)}(q^{n}-\unicode[STIX]{x1D700})\mathop{\prod }_{i=1}^{n-1}(q^{2i}-1),\end{eqnarray}$$

where $d=\gcd (4,q^{n}-\unicode[STIX]{x1D700})$ . Since $q^{2i}\geqslant 9$ for all $i\geqslant 1$ , and in particular $q^{n}\geqslant 3^{4}=81$ , we have

(6) $$\begin{eqnarray}|C_{T}(x)|>{\textstyle \frac{1}{8}}({\textstyle \frac{8}{9}})^{n-2}q^{2n^{2}-5n+6}\qquad \text{and}\qquad |T|<{\textstyle \frac{82}{81}}q^{2n^{2}-n}.\end{eqnarray}$$

For $4\leqslant n\leqslant 8$ we need $|C_{T}(x)|>|T|^{3/5}$ , so by (6) it suffices to show that $({\textstyle \frac{8}{9}})^{5n-10}q^{4n^{2}-22n+30}>8^{5}({\textstyle \frac{82}{81}})^{3}$ , which holds unless $(n,q)=(4,3)$ or $(4,5)$ . For $(n,q)=(4,5)$ , (4)–(5) imply that $|C_{T}(x)|>|T|^{3/5}$ ; for $(n,q)=(4,3)$ , a GAP [15] calculation shows that there exists an elements $x\in T\setminus \{1\}$ for which this inequality holds. For $n\geqslant 9$ we claim that $|C_{T}(x)|>|T|^{4/5}$ , and we now have $q^{n}\geqslant 3^{9}=19683$ , so we can replace the ${\textstyle \frac{82}{81}}$ in (6) by ${\textstyle \frac{19684}{19683}}$ to see that it suffices to show that $({\textstyle \frac{8}{9}})^{5n-10}q^{2n^{2}-21n+30}>8^{5}({\textstyle \frac{19684}{19683}})^{4}$ . If $n\geqslant 10$ then this inequality holds for all $q\geqslant 3$ , and if $n=9$ then it holds for $q\geqslant 127$ . For $n=9$ and $q<127$ , using the equality in (3) we obtain

$$\begin{eqnarray}|C_{T}(x)|\geqslant \frac{1}{4}q^{n^{2}-n}(q^{n-2}-1)(q^{2}-1)\mathop{\prod }_{i=1}^{n-3}(q^{2i}-1),\end{eqnarray}$$

which implies that $|C_{T}(x)|>|T|^{4/5}$ except when $q=3$ and $\unicode[STIX]{x1D700}=+$ . However, in this case we have $|G:T|=2$ (compare [Reference Burness and Giudici9, Figure 2.5.1 and Lemma 2.2.9], noting that the discriminant of a hyperbolic quadratic form on $\mathbb{F}_{q}^{2n}$ with $(n,q)=(9,3)$ is $\boxtimes$ , in the notation used there, because $n(q-1)/4=9$ is odd), so the ${\textstyle \frac{1}{4}}$ in the above estimate for $|C_{T}(x)|$ may be replaced by ${\textstyle \frac{1}{2}}$ , and we again obtain $|C_{T}(x)|>|T|^{4/5}$ .

Finally, suppose that $T\cong \text{P}\unicode[STIX]{x1D6FA}_{2n}^{\unicode[STIX]{x1D700}}(q)$ with $q$ even, noting that $T=G$ in this case. Take $x\in G$ to be an involution of type $a_{2}$ as in [Reference Burness and Giudici9, Table 3.5.1]. Then $|C_{T}(x)|=q^{4n-7}|\unicode[STIX]{x1D6FA}_{2(n-2)}^{\unicode[STIX]{x1D700}}(q)||\text{Sp}_{2}(q)|$ and $\gcd (4,q^{n}-\unicode[STIX]{x1D700})=1$ , so instead of (4)–(5) we have

(7) $$\begin{eqnarray}|C_{T}(x)|>\frac{1}{4}q^{n^{2}-2}(q^{2}-1)\mathop{\prod }_{i=1}^{n-3}(q^{2i}-1)\end{eqnarray}$$

and

(8) $$\begin{eqnarray}|T|=q^{n(n-1)}(q^{n}-\unicode[STIX]{x1D700})\mathop{\prod }_{i=1}^{n-1}(q^{2i}-1).\end{eqnarray}$$

(In the bound for $|C_{T}(x)|$ we drop a factor of ${\textstyle \frac{1}{4}}$ because $|G:T|=1$ for $q$ even, but pick up a factor of ${\textstyle \frac{1}{2}}$ because $\unicode[STIX]{x1D6FA}_{2(n-2)}^{\unicode[STIX]{x1D700}}(q)$ has index $2$ in $\text{SO}_{2(n-2)}^{\unicode[STIX]{x1D700}}(q)$ .) For $4\leqslant n\leqslant 8$ we need $|C_{T}(x)|>|T|^{3/5}$ . As $q^{i}\geqslant 2$ for all $i\geqslant 1$ , and in particular $q^{n}\geqslant 16$ , it suffices to show that $({\textstyle \frac{3}{4}})^{5n-10}q^{4n^{2}-22n+30}>4^{5}({\textstyle \frac{17}{16}})^{3}$ . This inequality holds unless $(n,q)=(4,2)$ or $(4,4)$ , in which cases a direct calculation shows that $|C_{T}(x)|>|T|^{3/5}$ . For $n\geqslant 9$ we must show that $|C_{T}(x)|>|T|^{4/5}$ . We now have $q^{n}\geqslant 512$ , and so it suffices to show that $({\textstyle \frac{3}{4}})^{5n-10}q^{2n^{2}-21n+30}>4^{5}({\textstyle \frac{513}{512}})^{4}$ . This inequality holds unless $(n,q)=(10,2)$ , or $n=9$ and $q\leqslant 2^{8}$ . One may use (7)–(8) to check that $|C_{T}(x)|>|T|^{4/5}$ in each of these cases except $(n,q)=(9,2)$ , in which case the desired inequality may be verified by a direct calculation.◻

Hypothesis 4.1. Let $T$ be a nonabelian finite simple group, let $k\geqslant 2$ , and write $H=T^{k}$ . Let $\unicode[STIX]{x1D6FA}=\{(y_{1},\ldots ,y_{k-1},1)\mid y_{1},\ldots ,y_{k-1}\in T\}\leqslant H$ , and let $H$ act on $\unicode[STIX]{x1D6FA}$ by

(9) $$\begin{eqnarray}(y_{1},\ldots ,y_{k-1},1)^{(x_{1},\ldots ,x_{k})}=(x_{k}^{-1}y_{1}x_{1},\ldots ,x_{k}^{-1}y_{k-1}x_{k-1},1).\end{eqnarray}$$

Suppose that $N=H^{r}$ is a collineation group of a thick finite generalized quadrangle ${\mathcal{Q}}=({\mathcal{P}},{\mathcal{L}},\text{I})$ of order $(s,t)$ , such that ${\mathcal{P}}=\unicode[STIX]{x1D6FA}^{r}$ and $N$ has the product action on ${\mathcal{P}}$ .

Proof. Note first that $|{\mathcal{P}}|=|\unicode[STIX]{x1D6FA}|^{r}=|T|^{(k-1)r}$ . In particular, the excluded case $(s,t)=(2,4)$ in Corollary 1.5 does not arise, because $|{\mathcal{P}}|\geqslant |T|\geqslant |\text{Alt}_{5}|=60>(2+1)(2\cdot 4+1)$ . If we take $x:=x_{1}=\cdots =x_{k}\neq 1$ in (9), then $(y_{1},\ldots ,y_{k-1},1)\in \unicode[STIX]{x1D6FA}$ is fixed if and only if $y_{1},\ldots ,y_{k-1}\in C_{T}(x)$ . That is, $(x,\ldots ,x)\in H$ fixes precisely $|C_{T}(x)|^{k-1}$ elements of $\unicode[STIX]{x1D6FA}$ (and, in particular, the action of $H$ on $\unicode[STIX]{x1D6FA}$ is not semiregular). Corollary 1.5 therefore implies that $r\leqslant 4$ and $|C_{T}(x)|^{k-1}<|\unicode[STIX]{x1D6FA}|^{1-r/5}=|T|^{(k-1)(1-r/5)}$ , namely $|C_{T}(x)|<|T|^{1-r/5}$ , for all $x\in T\setminus \{1\}$ . If $r=4$ then we have a contradiction because every nonabelian finite simple group $T$ contains a nonidentity element $x$ with $|C_{T}(x)|>|T|^{1/5}$ . For example, it is well known that $T$ contains an involution $x$ with $|C_{T}(x)|>|T|^{1/3}$ (indeed, every involution has this property [Reference Liebeck and Shalev19, Proposition 2.4]). Therefore, $r\leqslant 3$ . In particular, Proposition 1.6 tells us that $T$ must be one of the groups appearing in Table 2. To prove the final assertion, suppose toward a contradiction that $r=3$ and $(s,t)=(|\unicode[STIX]{x1D6FA}|-1,|\unicode[STIX]{x1D6FA}|+1)$ . Take any $x\in T$ with $|C_{T}(x)|>|T|^{1/3}$ . Then $((x,\ldots ,x),(1,\ldots ,1),(1,\ldots ,1))\in H^{r}=H^{3}=N$ fixes $|C_{T}(x)|^{k-1}|T|^{2(k-1)}>|T|^{7(k-1)/3}=|{\mathcal{P}}|^{7/9}$ points of ${\mathcal{Q}}$ , contradicting Proposition 2.6.◻

Hypothesis 5.1. Let $T$ be a nonabelian finite simple group and let $N=T^{r}\times T^{r}$ act on $T^{r}$ by

$$\begin{eqnarray}y^{(u_{1},u_{2})}=u_{2}^{-1}yu_{1}.\end{eqnarray}$$

Let $M=\{(u,1)\mid u\in T^{r}\}\leqslant N$ , so that $M$ may be identified with $T^{r}$ acting regularly on itself by right multiplication. Suppose that $N$ is a collineation group of a thick finite generalized quadrangle ${\mathcal{Q}}=({\mathcal{P}},{\mathcal{L}},\text{I})$ of order $(s,t)$ with ${\mathcal{P}}=T^{r}$ . Let ${\mathcal{P}}_{1}\subset {\mathcal{P}}$ denote the set of points collinear with but not equal to the identity element $1\in T^{r}={\mathcal{P}}$ , and let ${\mathcal{L}}_{1}\subset {\mathcal{L}}$ denote the set of lines incident with $1$ . Given a line $\ell \in {\mathcal{L}}$ , let $\bar{\ell }\subset {\mathcal{P}}$ denote the set of points incident with $\ell$ .

Proof. Let us first establish some notation. Given $u\in T^{r}={\mathcal{P}}$ , write

$$\begin{eqnarray}\displaystyle & \displaystyle \operatorname{fix}_{{\mathcal{P}}}(u)=\{P\in {\mathcal{P}}\mid P^{(u,1)}=P\}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \operatorname{coll}_{{\mathcal{P}}}(u)=\{P\in {\mathcal{P}}\mid P^{(u,1)}\text{ is collinear with but not equal to }P\}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \operatorname{fix}_{{\mathcal{L}}}(u)=\{\ell \in {\mathcal{L}}\mid \ell ^{(u,1)}=\ell \}, & \displaystyle \nonumber\\ \displaystyle & \displaystyle \operatorname{conc}_{{\mathcal{L}}}(u)=\{\ell \in {\mathcal{L}}\mid \ell ^{(u,1)}\text{ is concurrent with but not equal to }\ell \}. & \displaystyle \nonumber\end{eqnarray}$$

Since the subgroup $M=\{(u,1)\mid u\in T^{r}\}$ of $N$ acts regularly on ${\mathcal{P}}$ , $\operatorname{fix}_{{\mathcal{P}}}(u)$ is empty. Moreover, $P\in \operatorname{coll}_{{\mathcal{P}}}(u)$ if and only if the points $P^{(P^{-1},1)}=1$ and $(Pu)^{(P^{-1},1)}=PuP^{-1}$ are collinear, which is if and only if $PuP^{-1}\in u^{T^{r}}\cap {\mathcal{P}}_{1}$ . Since for $g,h\in T^{r}$ we have $gug^{-1}=huh^{-1}$ if and only if $g^{-1}h\in C_{T^{r}}(u)$ , it follows that

$$\begin{eqnarray}|\text{coll}_{{\mathcal{P}}}(u)|=|u^{T^{r}}\cap {\mathcal{P}}_{1}||C_{T^{r}}(u)|,\end{eqnarray}$$

as in the proof of [Reference Yoshiara27, Lemma 3]. Then (again, as in that proof) [Reference Payne and Thas21, 1.9.2] implies that

(10) $$\begin{eqnarray}|\text{coll}_{{\mathcal{P}}}(u)|=(s+1)|\text{fix}_{{\mathcal{L}}}(u)|+|\text{conc}_{{\mathcal{L}}}(u)|=|u^{T^{r}}\cap {\mathcal{P}}_{1}||C_{T^{r}}(u)|,\end{eqnarray}$$

for every $u\in T^{r}$ .

Now, since $x\in {\mathcal{P}}_{1}$ , we have $u^{-1}xu=x^{(u,u)}\in {\mathcal{P}}_{1}$ for every collineation of the form $(u,u)\in N$ , because such collineations (are precisely those that) fix the point $1$ . That is, every $T^{r}$ -conjugate of $x$ is in ${\mathcal{P}}_{1}$ . In other words, $x^{T^{r}}\cap {\mathcal{P}}_{1}=x^{T^{r}}$ , and so (10) implies that

(11) $$\begin{eqnarray}\displaystyle |\text{coll}_{{\mathcal{P}}}(x)| & = & \displaystyle (s+1)|\text{fix}_{{\mathcal{L}}}(x)|+|\text{conc}_{{\mathcal{L}}}(x)|\nonumber\\ \displaystyle & = & \displaystyle |x^{T^{r}}||C_{T^{r}}(x)|=|T^{r}|=|{\mathcal{P}}|=(s+1)(st+1).\end{eqnarray}$$

In particular, we have $\operatorname{coll}_{{\mathcal{P}}}(x)={\mathcal{P}}$ ; that is, every point of ${\mathcal{Q}}$ is mapped to a collinear point under the collineation $(x,1)\in M$ . We now claim that $\operatorname{conc}_{{\mathcal{L}}}(x)$ is empty. If not, then some line $\ell$ is concurrent with its image under the collineation $(x,1)$ . Let $P$ denote the unique point incident with both $\ell$ and $\ell ^{(x,1)}$ . Then $Px^{-1}$ is incident with $\ell$ , being the image of $P$ under the collineation $(x,1)^{-1}=(x^{-1},1)$ , and $Px^{-1}\neq P$ because $x\neq 1$ and $M$ acts regularly on ${\mathcal{P}}$ . Since ${\mathcal{Q}}$ is thick, there exists a third point $P_{3}$ incident with $\ell$ , distinct from $P$ and $Px^{-1}$ . Since $\operatorname{coll}_{{\mathcal{P}}}(x)={\mathcal{P}}$ , the points $P_{3}^{(x,1)}=P_{3}x$ and $P_{3}$ are collinear. Moreover, $P_{3}x$ is collinear with $P$ , because both of these points are incident with $\ell ^{(x,1)}$ . Therefore, $P_{3}x$ is collinear with two distinct points that are incident with $\ell$ , namely $P_{3}$ and $P$ , and so $P_{3}x$ is itself incident with $\ell$ because ${\mathcal{Q}}$ contains no triangles. This, however, means that $P_{3}x$ is incident with both $\ell$ and $\ell ^{(x,1)}$ , which forces $P_{3}x=P$ and hence $P_{3}=Px^{-1}$ , a contradiction. Therefore, $|\text{conc}_{{\mathcal{L}}}(x)|=0$ as claimed, and so (11) implies that

(12) $$\begin{eqnarray}|\text{fix}_{{\mathcal{L}}}(x)|=st+1.\end{eqnarray}$$

Next, we show that $(x^{i})^{T^{r}}\subseteq {\mathcal{P}}_{1}$ for all $i\in \{1,\ldots ,|x|-1\}$ . For each such $i$ , we certainly have $\operatorname{fix}_{{\mathcal{L}}}(x)\subseteq \operatorname{fix}_{{\mathcal{L}}}(x^{i})$ , because if the collineation $(x,1)$ fixes a line then so too does $(x,1)^{i}=(x^{i},1)$ . In particular, $|\text{fix}_{{\mathcal{L}}}(x^{i})|\geqslant |\text{fix}_{{\mathcal{L}}}(x)|=st+1$ , by (12). On the other hand, no two lines fixed by $(x^{i},1)$ can be concurrent, because if they were, then the unique point incident with both of them would be fixed by $(x^{i},1)$ , a contradiction since $M$ acts regularly on ${\mathcal{P}}$ . Hence, the total number of points that are incident with some line in $\operatorname{fix}_{{\mathcal{L}}}(x^{i})$ is $(s+1)|\text{fix}_{{\mathcal{L}}}(x^{i})|$ . As this number cannot exceed $|{\mathcal{P}}|=(s+1)(st+1)$ , we must also have $|\text{fix}_{{\mathcal{L}}}(x^{i})|\leqslant st+1$ . Therefore, $|\text{fix}_{{\mathcal{L}}}(x^{i})|=st+1$ . Now (10) implies, on the one hand, that

$$\begin{eqnarray}|\text{coll}_{{\mathcal{P}}}(x^{i})|=(s+1)|\text{fix}_{{\mathcal{L}}}(x^{i})|+|\text{conc}_{{\mathcal{L}}}(x^{i})|=|{\mathcal{P}}|+|\text{conc}_{{\mathcal{L}}}(x^{i})|.\end{eqnarray}$$

Since $|\text{coll}_{{\mathcal{P}}}(x^{i})|\leqslant |{\mathcal{P}}|$ , this implies that $|\text{conc}_{{\mathcal{L}}}(x^{i})|=0$ , and then in turn that $|{\mathcal{P}}|=|\text{coll}_{{\mathcal{P}}}(x^{i})|$ . Appealing again to (10), we now deduce that $|(x^{i})^{T^{r}}\cap {\mathcal{P}}_{1}||C_{T^{r}}(x^{i})|=|{\mathcal{P}}|=|T^{r}|$ , which implies that $(x^{i})^{T^{r}}\subseteq {\mathcal{P}}_{1}$ as required. The first assertion is therefore proved.

Finally, we must show that the collineation $(x,1)$ fixes the unique line $\ell _{x}\in {\mathcal{L}}_{1}$ incident with $x$ . If $|x|=2$ , then $(x,1)$ fixes $\ell _{x}$ because it fixes setwise the subset $\{1,x\}$ of points incident with $\ell _{x}$ . That is, it maps $1$ to $1^{(x,1)}=1x=x$ and $x$ to $x^{(x,1)}=x^{2}=1$ . Now suppose that $|x|>2$ . Then the point $1^{(x,1)^{2}}=x^{2}\neq 1$ is collinear with $x$ because $x$ is collinear with $1$ . On the other hand, $(x^{2})^{T_{r}}\subseteq {\mathcal{P}}_{1}$ by the first assertion, so in particular $x^{2}$ is collinear with $1$ . Therefore, $x^{2}$ is collinear with two distinct points incident with $\ell _{x}$ (namely $1$ and $x$ ), and so is itself incident with $\ell _{x}$ because ${\mathcal{Q}}$ contains no triangles. Hence, $(x,1)$ fixes $\ell _{x}$ because it maps a pair of points incident with $\ell _{x}$ , namely $1$ and $x$ , to another pair of points incident with $\ell _{x}$ , namely $x$ and $x^{2}$ .◻

Proof. (i) This follows immediately from Lemma 5.2.

(ii) If $u\in \bar{\ell }$ then the collineation $(u,1)\in M$ fixes $\ell$ by Lemma 5.2. Conversely, if $(u,1)$ fixes $\ell$ then, because $1\in {\mathcal{P}}$ is incident with $\ell$ , so too is $1^{(u,1)}=u$ ; that is, $u\in \bar{\ell }$ .

(iii) If $\ell \in {\mathcal{L}}_{1}$ is not incident with any involution, then $\bar{\ell }$ , which is a subgroup of  $T^{r}$ by (ii), must have odd order. That is, $s+1=|\bar{\ell }|$ must be odd. However, $(s+1)(st+1)=|T|^{r}$ is even by the Feit–Thompson theorem [Reference Feit and Thompson13], so $s$ must be odd and hence $s+1$ must be even, a contradiction.

(iv) If $\ell \in {\mathcal{L}}_{1}$ is incident with representatives of every conjugacy class of involutions in ${\mathcal{P}}_{1}$ , then $\ell$ can be mapped to any other line in ${\mathcal{L}}_{1}$ by some element of the stabilizer $N_{1}=\{(u,u)\mid u\in T^{r}\}$ in $N$ of the point $1\in {\mathcal{P}}=T^{r}$ . Since $N$ acts transitively on ${\mathcal{P}}$ , this means that $N$ acts transitively on the flags of ${\mathcal{Q}}$ , and hence on ${\mathcal{L}}$ . If $r=1$ , this contradicts [Reference Bamberg, Popiel and Praeger8, Theorem 1.1], because $N$ acts primitively on ${\mathcal{P}}$ in this case. Hence, $r\geqslant 2$ .

(v) Suppose toward a contradiction that $T^{r}$ contains at most two conjugacy classes of involutions. Then $r=1$ , because if $r\geqslant 2$ then any involution $x\in T$ gives rise to the three pairwise nonconjugate involutions $(x,1,\ldots ,1)$ , $(1,x,1,\ldots ,1)$ and $(x,x,1,\ldots ,1)$ in $T^{r}$ . Hence, by (iii), $T$ must have exactly two conjugacy classes of involutions, say $x^{T}$ and $y^{T}$ , and both must be contained in ${\mathcal{P}}_{1}$ . Without loss of generality, $x$ and $y$ commute, because at least one of them centralizes a Sylow $2$ -subgroup of $T$ . Therefore, $xy$ is an involution, and so must be collinear with $1\in {\mathcal{P}}$ . Since $1$ is collinear with $x$ , the images of $1$ and $x$ under the collineation $(y,1)\in M$ are collinear. That is, $1^{(y,1)}=y$ is collinear with $x^{(y,1)}=xy$ . Similarly, $1$ and $y$ are collinear, and hence so too are $1^{(x,1)}=x$ and $y^{(x,1)}=yx=xy$ . Since the involution $xy$ is also collinear with $1$ and ${\mathcal{Q}}$ contains no triangles, the points $1$ , $x$ , $y$ and $xy$ must be incident with a common line. In particular, $x$ and $y$ are incident with a common line in ${\mathcal{L}}_{1}$ . Since $r=1$ , this contradicts (iv).

(vi) Let $x$ , $y$ and $z$ denote representatives of the three $T^{r}$ -conjugacy classes of involutions. If ${\mathcal{P}}_{1}$ contains exactly one of these classes, then $N$ acts flag-transitively by (iv), and it follows from [Reference Bamberg, Popiel and Praeger8, Theorem 1.1] that $r\geqslant 2$ . Now suppose that ${\mathcal{P}}_{1}$ contains all three of $x^{T}$ , $y^{T}$ and $z^{T}$ . Without loss of generality, $x$ centralizes a Sylow $2$ -subgroup of  $T^{r}$ and both $y$ and $z$ commute with $x$ , so $xy=yx$ and $xz=zx$ are involutions. Arguing as in the proof of (iii), we deduce that $1$ , $x$ , $y$ and $xy$ are incident with a common line $\ell \in {\mathcal{L}}_{1}$ . Replacing $y$ by $z$ in this argument, we see that $z$ is also incident with $\ell$ , so (iv) again implies that $N$ acts flag-transitively, and it follows that $r\geqslant 2$ .

(vii) If $\gcd (s,t)>1$ then [Reference Yoshiara27, Lemma 6(1)] implies that every nontrivial $T^{r}$ -conjugacy class intersects ${\mathcal{P}}_{1}$ . However, assertion (i) then implies that ${\mathcal{P}}_{1}={\mathcal{P}}\setminus \{1\}$ , which is impossible. Therefore, $\gcd (s,t)=1$ . In particular, to show that $t\geqslant s+1$ it suffices to show that $t\geqslant s$ . The proof of this assertion is adapted from that of [Reference Bamberg, Popiel and Praeger8, Corollary 2.3]. Choose two distinct lines $\ell _{1},\ell _{2}\in {\mathcal{L}}_{1}$ , so that $\bar{\ell }_{1}$ and $\bar{\ell }_{2}$ are subgroups of $T^{r}$ by (ii). For brevity, we now abuse notation slightly and identify $\ell _{1}$ and $\ell _{2}$ with $\bar{\ell }_{1}$ and $\bar{\ell }_{2}$ , respectively, dropping the “bar” notation. Since $\ell _{1}$ is a subgroup of $T^{r}$ and right multiplication by any element of $T^{r}$ is a collineation of ${\mathcal{Q}}$ (identified with an element of $M$ ), we have in particular that every right coset $\ell _{1}g_{2}$ of $\ell _{1}$ with $g_{2}\in \ell _{2}$ corresponds precisely to the set of points incident with some line of ${\mathcal{Q}}$ . Similarly, left multiplications (identified with elements of $\{1\}\times T^{r}\leqslant N$ ) are collineations, so every left coset $g_{1}\ell _{2}$ of $\ell _{2}$ with $g_{1}\in \ell _{1}$ is a line of ${\mathcal{Q}}$ . Therefore, ${\mathcal{L}}^{\prime }=\{g_{1}\ell _{2}\mid g_{1}\in \ell _{1}\}\cup \{\ell _{1}g_{2}\mid g_{2}\in \ell _{2}\}$ is a subset of ${\mathcal{L}}$ . Consider also the subset ${\mathcal{P}}^{\prime }=\ell _{1}\ell _{2}$ of ${\mathcal{P}}=T^{r}$ , and let $\text{I}^{\prime }$ be the restriction of $\text{I}$ to $({\mathcal{P}}^{\prime }\times {\mathcal{L}}^{\prime })\cup ({\mathcal{L}}^{\prime }\times {\mathcal{P}}^{\prime })$ . If we can show that ${\mathcal{Q}}^{\prime }=({\mathcal{P}}^{\prime },{\mathcal{L}}^{\prime },\text{I}^{\prime })$ is a generalized quadrangle of ${\mathcal{Q}}$ of order $(s,1)$ , then [Reference Payne and Thas21, 2.2.2(i)] will imply that $t\geqslant s$ . Let us first check that ${\mathcal{Q}}^{\prime }$ satisfies the generalized quadrangle axiom. Take $\ell \in {\mathcal{L}}^{\prime }$ and $P\in {\mathcal{P}}^{\prime }$ not incident with $\ell$ . Since ${\mathcal{Q}}$ satisfies the generalized quadrangle axiom, there is a unique point $P_{0}\in {\mathcal{P}}$ incident with $\ell$ and collinear with $P$ . Since $\ell \subset {\mathcal{P}}^{\prime }$ , we have $P_{0}\in {\mathcal{P}}^{\prime }$ , and so ${\mathcal{Q}}^{\prime }$ also satisfies the generalized quadrangle axiom. It remains to check that ${\mathcal{Q}}^{\prime }$ has order $(s,1)$ . Every line in ${\mathcal{L}}^{\prime }$ is incident with $s+1$ points in ${\mathcal{P}}^{\prime }$ , being a coset of either $\ell _{1}$ or $\ell _{2}$ , so it remains to show that every point in ${\mathcal{P}}^{\prime }$ is incident with exactly two lines in ${\mathcal{L}}^{\prime }$ . Given $P=g_{1}g_{2}\in {\mathcal{P}}^{\prime }$ , where $g_{1}\in \ell _{1}$ , $g_{2}\in \ell _{2}$ , each line $\ell \in {\mathcal{L}}^{\prime }$ incident with $P$ is either of the form $h_{1}\ell _{2}$ for some $h_{1}\in \ell _{1}$ or $\ell _{1}h_{2}$ for some $h_{2}\in \ell _{2}$ , and since $P\in \ell$ , we must have $h_{1}=g_{1}$ or $h_{2}=g_{2}$ , respectively. Therefore, $P$ is incident with exactly two lines in ${\mathcal{L}}^{\prime }$ , namely $g_{1}\ell _{2}$ and $\ell _{1}g_{2}$ .◻

Proof. We have $|{\mathcal{P}}|=|T|^{r}$ , and $r\in \{1,2,3\}$ by Proposition 4.2. For each of the alternating and sporadic simple groups $T$ in Table 2, we check computationally for solutions of $|T|^{r}=(s+1)(st+1)$ satisfying $s\geqslant 2$ , $t\geqslant 2$ and properties (ii) and (iii) of Lemma 2.1 (see Remark 5.5). If $r=3$ then the only such solutions have the form $(s,t)=(|T|-1,|T|+1)=(|\unicode[STIX]{x1D6FA}|-1,|\unicode[STIX]{x1D6FA}|+1)$ , and this contradicts the final assertion of Proposition 4.2. If $r\in \{1,2\}$ then the possibilities for $T$ and $(s,t)$ are as in Table 3. By Lemma 5.3(i), ${\mathcal{P}}_{1}$ is a union of $T^{r}$ -conjugacy classes, and so we must be able to partition $|{\mathcal{P}}_{1}|=s(t+1)$ into a subset of the sizes of these classes (respecting multiplicities). When $r=1$ and $T\cong \text{Alt}_{7}$ or $\text{Alt}_{8}$ , this is impossible: the nontrivial conjugacy class sizes not exceeding $s(t+1)$ are $70$ , $105$ and $210$ in the first case, and $105$ , $112$ and $210$ in the second (with each occurring exactly once). Similarly, if $r=2$ and $T\cong \text{J}_{1}$ , one may check computationally that there is no partition of $s(t+1)$ , where $(s,t)=(419,175141)$ , into nontrivial $T^{2}$ -conjugacy class sizes. Hence, it remains to consider the cases where $r=2$ and $T\cong \text{Alt}_{6}$ or $\text{M}_{11}$ . Here we first determine computationally the possible partitions of $s(t+1)$ into nontrivial $T^{2}$ -conjugacy class sizes to obtain a list of possible partitions of ${\mathcal{P}}_{1}$ into $T^{2}$ -conjugacy classes. Now, because the point graph of ${\mathcal{Q}}$ is a strongly regular graph in which adjacent vertices have $\unicode[STIX]{x1D706}:=s-1$ common neighbors and nonadjacent vertices have $\unicode[STIX]{x1D707}:=t+1$ common neighbors, ${\mathcal{P}}_{1}$ must be a partial difference set of $T^{2}$ with these parameters. That is, each nonidentity element $y\in T^{2}$ must have exactly $\unicode[STIX]{x1D706}$ representations of the form $y=z_{i}z_{j}^{-1}$ for $z_{i},z_{j}\in {\mathcal{P}}_{1}$ if $y\in {\mathcal{P}}_{1}$ , and exactly $\unicode[STIX]{x1D707}$ such representations if $y\not \in {\mathcal{P}}_{1}$ . A computation verifies that this condition is violated for each of the partitions of ${\mathcal{P}}_{1}$ determined in the previous step.◻

Proof. By Propositions 4.2 and 5.4, $T$ is a Lie type group and $r\leqslant 3$ . We now show that $|C_{T}(x)|<|T|^{1-2r/9}$ for all $x\in T\setminus \{1\}$ and deduce from this that $r\neq 3$ . Suppose, toward a contradiction, that there exists $x\in T\setminus \{1\}$ with $|C_{T}(x)|\geqslant |T|^{1-2r/9}$ . Define $w=(x,1,\ldots ,1)\in T^{r}$ and let ${\mathcal{Q}}_{\unicode[STIX]{x1D703}}=({\mathcal{P}}_{\unicode[STIX]{x1D703}},{\mathcal{L}}_{\unicode[STIX]{x1D703}})$ be the substructure of ${\mathcal{Q}}$ fixed by $\unicode[STIX]{x1D703}=(w,w)\in N_{1}$ . Then ${\mathcal{P}}_{\unicode[STIX]{x1D703}}=C_{T}(x)\times T^{r-1}$ , and hence $|{\mathcal{P}}_{\unicode[STIX]{x1D703}}|\geqslant |T|^{(1-2r/9)+(r-1)}=|T|^{7r/9}=|{\mathcal{P}}|^{7/9}$ . Proposition 2.4 then says that either $s\geqslant t+3$ , or $(s,t)\in \{(2,4),(3,9)\}$ . The first of these conditions contradicts Lemma 5.3(vii); the second implies that $|T|^{r}=(s+1)(st+1)\in \{27,112\}$ , which is impossible because $|T|\geqslant 60$ . Hence, every $x\in T\setminus \{1\}$ must satisfy $|C_{T}(x)|<|T|^{1-2r/9}$ . For $r=3$ , this says that $|C_{T}(x)|<|T|^{1/3}$ for all $x\in T\setminus \{1\}$ , a contradiction because we can always find some $x$ with $|C_{T}(x)|>|T|^{1/3}$ (as noted in the proof of Proposition 4.2). Therefore, $r\leqslant 2$ .◻