Proof. As $(\alpha ,\dots ,\alpha )\pi $ fixes $\omega = D(\varphi _{t_1},\dots ,\varphi _{t_k})$ , there exists a unique $g\in T$ such that $t_i^{\alpha } = gt_{i^{\pi }}$ for all $i\in \{1,\dots ,k\}$ . Suppose $t_i = t_j$ for some $i\ne j$ (so i and j are in the same part of $\mathcal {P}_{\mathbf {x}}$ ). Then $t_{i^{\pi }} = g^{-1}t_i^{\alpha } = g^{-1}t_j^{\alpha } = t_{j^{\pi }}$ . This gives part (i).

For part (ii), it suffices to show that $g = 1$ . If $t_i = 1$ , then $t_{i^{\pi }} = g^{-1}$ , and we get $t_{j^{\pi }} = g^{-1}t_j^{\alpha } \ne g^{-1}$ if $t_j\ne 1$ . This implies that $|\mathcal {P}_{g^{-1}}| = |\mathcal {P}_1|$ , so $g = 1$ by our assumption.

Proof. We have $|\mathrm {Out}(T)|<|T|^{1/3}$ by Lemma 2.9 below. In particular, $|\mathrm {Out}(T)|<|T|/3$ , so there exists a prime s such that $|\mathrm {Out}(T)|<s<k$ (Bertrand’s postulate). Now, apply Lemma 2.5.

Proof. We use the same construction as in the proof of [Reference Fawcett25, Proposition 3.3]. By [Reference Seress53, Theorem 1], there exists a partition $\mathcal {P}=\{\Pi _1,\Pi _2,\Pi _3\}$ of $[k]$ such that each $\Pi _i$ is nonempty, $|\Pi _1|$ , $|\Pi _2|$ and $|\Pi _3|$ are distinct, and

(2) $$ \begin{align} \bigcap_{m = 1}^3 P_{\{\Pi_m\}} = 1. \end{align} $$

Let $x_1,x_2\in T$ be nontrivial elements of distinct orders. By the main theorem of [Reference Guralnick and Kantor33], there exist $y_1,y_2\in T$ such that $\langle x_i,y_i\rangle = T$ . Let $\Delta _i = \{D,D(\varphi _{t_{i,1}},\dots ,\varphi _{t_{i,k}})\}$ for $i\in \{1,2\}$ , where $t_{i,j} = 1$ if $j\in \Pi _1$ , $t_{i,j} = x_i$ if $j\in \Pi _2$ , and $t_{i,j} = y_i$ if $j\in \Pi _3$ . As explained in the proof of [Reference Fawcett25, Proposition 3.3], both $\Delta _1$ and $\Delta _2$ are bases for G.

Suppose $\Delta _1^{(\alpha ,\dots ,\alpha )\pi } = \Delta _2$ . Then there exists $g\in T$ such that $t_{1,j}^{\alpha } = gt_{2,j^{\pi }}$ for all $j\in [k]$ . If $t_{1,j} = t_{1,j'}$ for some $j'\in [k]$ , then $t_{2,j} = t_{2,j'}$ and

$$ \begin{align*} t_{2,j^{\pi}} = g^{-1}t_{1,j}^{\alpha} = g^{-1}t_{1,j'}^{\alpha} = t_{2,(j')^{\pi}}. \end{align*} $$

Hence, $\pi \in P_{\{\mathcal {P}\}}$ , and so $\pi \in P_{\{\Pi _m\}}$ for each $m\in \{1,2,3\}$ as $|\Pi _1|$ , $|\Pi _2|$ and $|\Pi _3|$ are distinct. This implies that $\pi = 1$ by (2), and so $g = 1$ . However, it follows that $x_1^{\alpha } = x_2$ , which is incompatible with $|x_1|\ne |x_2|$ . We conclude that $\Delta _1$ and $\Delta _2$ are in distinct $G_D$ -orbits, and thus $r(G)\geqslant 2$ .

Proof. This is [Reference Burness10, Proposition 3.9(i)] when T is a classical group, and the bounds for exceptional groups are clear.

Proof. If T is an exceptional group, then we take x and y to be $t_1$ and $t_2$ in [Reference Kantor, Lubotzky and Shalev38, Table 2], respectively, noting that $t_1$ is a generator of the maximal torus $T_1$ in that table. It is evident that $|t_1|\ne |t_2|$ in each case, and $\{t_1,t_2\}$ invariably generates T by [Reference Kantor, Lubotzky and Shalev38] (see [Reference Kantor, Lubotzky and Shalev38, p. 312]). Moreover, we observe that $\langle t_1\rangle $ and $\langle t_2\rangle $ are both maximal tori, which implies that each $t_i$ is regular semisimple.

To complete the proof, we may assume T is a classical group. Here, we will work with the corresponding quasisimple group $Q\in \{\mathrm {SL}_n^{\varepsilon }(q),\mathrm {Sp}_n(q),\Omega _n^{\varepsilon }(q)\}$ , noting that if Q is invariably generated by $\{t_1,t_2\}$ , with $t_1$ and $t_2$ regular semisimple, then T is invariably generated by $\{x,y\}$ , where x and y are the images of $t_1$ and $t_2$ in T, respectively (so x and y are also regular semisimple). Moreover, $|x| = |t_1|/a$ and $|y| = |t_2|/b$ for some integers $a,b$ dividing $|Q|/|T|$ , so $|x|\ne |y|$ if

(3) $$ \begin{align} |t_1| \mbox{ is indivisible by } |t_2||Q|/|T| \mbox{ and } |t_2| \mbox{ is indivisible by } |t_1||Q|/|T|. \end{align} $$

First, assume $Q\notin \{\mathrm {SL}_2(q),\Omega _8^+(q)\}$ . Here, we use the same $t_1$ and $t_2$ as presented in [Reference Kantor, Lubotzky and Shalev38, Table 1]. In each case, it is clear that $t_1$ and $t_2$ are semisimple elements satisfying (3), and $\{t_1,t_2\}$ invariably generates Q by [Reference Kantor, Lubotzky and Shalev38, Lemma 5.3]. Thus, it suffices to show that $t_1$ and $t_2$ are regular in every case, which is a straightforward exercise (for instance, we can work with the criterion for regularity in terms of the eigenvalues on $\overline {V}$ discussed as above). For example, consider the element $t_2\in Q = \Omega _{4m}^+(q)$ . Here, $t_2$ lifts to an element $\widehat {t_2}\in \mathrm {GL}(V)$ of the form

$$ \begin{align*} \widehat{t_2} = \begin{pmatrix} A&\\ &B \end{pmatrix}^{\delta} \end{align*} $$

with respect to a standard basis (see [Reference Kleidman and Liebeck39, Proposition 2.5.3]), where $\delta \in \{1,2\}$ , $A\in \mathrm {SO}_{4m-4}^-(q)$ has order $q^{2m-2}+1$ and $B\in \mathrm {SO}_4^-(q)$ has order $q^{2}+1$ . We only deal with the case where $\delta = 1$ since a similar argument holds for $\delta = 2$ . Then the eigenvalues of A over the algebraic closure K of $\mathbb {F}_q$ are

$$ \begin{align*} \lambda,\lambda^q,\dots,\lambda^{q^{4m-3}} \end{align*} $$

for some $\lambda \in K$ of order $q^{2m-2}+1$ . Similarly, the set of eigenvalues of B over K is $\{\mu ,\mu ^q,\mu ^{q^2},\mu ^{q^3}\}$ for some $\mu \in K$ of order $q^2+1$ . If $\mu = \lambda ^{q^i}$ for some $i\in \{0,\dots ,4m-3\}$ , then $\lambda ^{q^i(q^2+1)} = 1$ and so $q^{2m-2}+1$ divides $q^i(q^2+1)$ , which implies that $q^{2m-2}+1$ divides $q^2+1$ since $(q^{2m-2}+1,q^i) = 1$ . However, since $m\geqslant 3$ , this is impossible. It follows that the eigenvalues of $\widehat {t_2}$ over K are distinct, and so $t_2$ is a regular semisimple element.

Finally, let us handle the two excluded cases above. If $Q = \mathrm {SL}_2(q)$ with $q\notin \{4,5,7,9\}$ , then we take the same $t_1$ and $t_2$ as indicated in the proof of [Reference Kantor, Lubotzky and Shalev38, Lemma 5.3]. The group $\mathrm {L}_2(4)$ is invariably generated by an element of order $3$ and an element of order $5$ , and if $q = 9$ , then we take x and y to be of order $4$ and $5$ , respectively. If $Q = \Omega _8^+(q)$ with $q\notin \{2,3\}$ , then we take $t_1$ as in [Reference Kantor, Lubotzky and Shalev38, Table 1], and $t_2$ an element of order $(q^3-1)/(2,q-1)$ as described in the proof of [Reference Kantor, Lubotzky and Shalev38, Lemma 5.4], where it is denoted $t_3$ .

Proof of Theorem 2.12.

First, observe that we only need to consider prime order elements in $\mathrm {Aut}(T)$ , since $C_T(x)\leqslant C_T(x^m)$ for any integer m and $x\in \mathrm {Aut}(T)$ .

Assume $T = A_n$ is an alternating group. If $n = 5$ or $6$ , then the result can be checked using Magma. Now, assume $n\geqslant 7$ , so $\mathrm {Aut}(T) = S_n$ . It is easy to see that $|C_T(x)|$ is maximal when x is a transposition, in which case $C_{S_n}(x) \cong S_2\times S_{n-2}$ and thus $|C_T(x)| = (n-2)!$ . Hence, $h(T) = (n-2)!$ . If T is a sporadic group, then $|C_T(x)|$ can be read off from the character table of T, which can be accessed computationally via the GAP Character Table Library [Reference Breuer6].

For the remainder, we may assume T is a simple group of Lie type over $\mathbb {F}_q$ , where $q = p^f$ with p a prime. Assume $x\in \mathrm {Aut}(T)$ is of prime order r. If $x\in \mathrm {Inndiag}(T)$ , then x is semisimple if $p\ne r$ , otherwise x is unipotent. And if $x\notin \mathrm {Inndiag}(T)$ , then x is a field, graph or graph-field automorphism. Here, if x is a graph or graph-field automorphism, then $r\in \{2,3\}$ .

Assume T is an exceptional group. Here, we assume $T\ne {{}^2}G_2(3)'\cong \mathrm {L}_2(8)$ and $T\ne G_2(2)'\cong \mathrm {U}_3(3)$ as noted in (4). By [Reference Burness and Thomas19, Proposition 2.11], $|C_T(x)|$ is maximal when $x\in T$ is a long root element. Now, assume $x\in T$ is a long root element. If T is not ${{}^3}D_4(q)$ or ${{}^2}B_2(q)$ , then $|C_T(x)|$ can be read off from the tables in [Reference Liebeck and Seitz43, Chapter 22], noting that $x^{\mathrm {Inndiag}(T)} = x^T$ by [Reference Liebeck and Seitz43, Corollary 17.10]. If $T = {{}^3}D_4(q)$ or ${{}^2}B_2(q)$ , then we can find $|C_T(x)|$ in [Reference Spaltenstein55, p. 677] and [Reference Suzuki57], respectively.

For the remainder of the proof, we assume T is a classical group defined over $\mathbb {F}_q$ . Let V be the natural module of T, and write $\overline {V} = V\otimes K$ , where K is the algebraic closure of $\mathbb {F}_q$ . For $x\in \mathrm {PGL}(V)$ , let $\widehat {x}$ be a preimage of x in $\mathrm {GL}(V)$ . Following [Reference Burness10, Definition 3.16], we define

$$ \begin{align*} \nu(x) = \min\{\dim [\overline{V},\lambda \widehat{x}]:\lambda\in K^*\}, \end{align*} $$

where $[\overline {V},\lambda \widehat {x}] = \{v-\lambda \widehat {x}v:v\in \overline {V}\}$ . That is, $\nu (x)$ is the codimension of the largest eigenspace of $\widehat {x}$ on $\overline {V}$ , noting that $\nu (x)$ is independent of the choice of the preimage $\widehat {x}$ . Upper and lower bounds on $|x^T|$ in terms of n, q and $\nu (x)$ are given in [Reference Burness10, Section 3]. Similarly, if x is a field, graph or graph-field automorphism, then lower bounds for $|x^T|$ can be read off from [Reference Burness10, Table 3.11]. In addition, $|C_{\mathrm {Inndiag}(T)}(x)|$ , and a description of the splitting of $x^{\mathrm {Inndiag}(T)}$ into distinct T-classes, can be found in [Reference Burness and Giudici13, Chapter 3]. In particular, note that if $x\in \mathrm {Inndiag}(T)$ is a semisimple element of prime order, then $x^{\mathrm {Inndiag}(T)} = x^T$ (see [Reference Gorenstein, Lyons and Solomon31, Theorem 4.2.2(j)], also recorded as [Reference Burness and Giudici13, Theorem 3.1.12]).

We start with the case where $T = \mathrm {L}_2(q)$ . Let H be the normaliser in $\mathrm {PGL}_2(q)$ of a nonsplit maximal torus of T, so $H\cong D_{2(q+1)}$ . If q is odd, then we let x be the central involution in H, and if q is even, let $x\in H$ be an element of odd prime order. Then $|C_T(x)| = q+1$ , so $h(T)\geqslant q+1$ . Let $y\in \mathrm {Aut}(T)$ be an element of prime order. Note that if y is unipotent, then $|C_T(y)| = q$ , and $|C_T(y)|$ divides $q+1$ or $q-1$ if y is semisimple. Thus, we only need to consider field automorphisms, noting that $|C_{\mathrm {PGL}_2(q)}(y)| = |\mathrm {PGL}_2(q^{1/r})|$ if y is a field automorphism of prime order r. It follows that $|C_{\mathrm {PGL}_2(q)}(y)|> q+1$ only if $r = 2$ (so f is even). Indeed,

$$ \begin{align*} |C_T(y)| = |C_{\mathrm{PGL}_2(q)}(y)| = |\mathrm{PGL}_2(q^{1/2})|>q+1 \end{align*} $$

if y is an involutory field automorphism, and so we conclude that $h(T) = |\mathrm {PGL}_2(q^{1/2})|$ if f is even, and $h(T) = q+1$ if f is odd.

To complete the proof for linear and unitary groups, we assume $T = \mathrm {L}_n^{\varepsilon }(q)$ with $n\geqslant 3$ . Let $x\in T$ be a unipotent element with Jordan form $[J_2,J_1^{n-2}]$ on the natural module, noting that x is a long root element. Then $|C_{\mathrm {PGL}_n^{\varepsilon }(q)}(x)|$ can be read off from [Reference Burness and Giudici13, Tables B.3 and B.4], and we have $x^{\mathrm {PGL}_n^{\varepsilon }(q)} = x^T$ by [Reference Burness and Giudici13, Propositions 3.2.7 and 3.3.10]. More specifically,

$$ \begin{align*} |C_T(x)| = (n,q-\varepsilon)^{-1} q^{2n-3}|\mathrm{GL}_{n-2}^{\varepsilon}(q)| \end{align*} $$

and

$$ \begin{align*} |x^T| = |x^{\mathrm{PGL}_n^{\varepsilon}(q)}|= \frac{|\mathrm{PGL}_n^{\varepsilon}(q)|}{q^{2n-3}|\mathrm{GL}_{n-2}^{\varepsilon}(q)|} < \frac{2q^{2n-1}}{q-1}. \end{align*} $$

The cases where $n\in \{3,4\}$ require special attention, which will be treated separately.

Assume $T = \mathrm {L}_3^{\varepsilon }(q)$ , so $|C_T(x)| = (3,q-\varepsilon )^{-1}q^3(q-\varepsilon )$ , and let y be an element in $\mathrm {Aut}(T)$ of prime order that is not of Jordan form $[J_2,J_1]$ . If y is unipotent or semisimple and $\nu (y) = 2$ , then either y has Jordan form $[J_3]$ or $|y|$ is odd, so by [Reference Burness10, Propositions 3.22 and 3.36],

$$ \begin{align*} |y^T|> \frac{1}{2(3,q-\varepsilon)}\left(\frac{q}{q+1}\right)q^6 > (q^2-1)(q^2+\varepsilon q+1) = |x^T|. \end{align*} $$

If $\nu (y) = 1$ and y is semisimple, then a preimage $\widehat {y}$ of y in $\mathrm {GL}(V)$ is $[\omega I_1, I_{2}]$ , so $|C_T(y)| = (3,q-\varepsilon )^{-1}|\mathrm {GL}_{2}^{\varepsilon }(q)|$ . It is easy to see that $|C_T(y)| < |C_T(x)|$ . If y is a graph automorphism, then $|C_{\mathrm {PGL}_3^{\varepsilon }(q)}(y)| = |\mathrm {SL}_2(q)|$ , so $|C_T(y)|<|C_T(x)|$ evidently. If y is a field automorphism of odd prime order r, then by [Reference Burness and Giudici13, Propositions 3.2.9 and 3.3.12],

$$ \begin{align*} |C_{\mathrm{PGL}_3^{\varepsilon}(q)}(y)| =|\mathrm{PGL}_3^{\varepsilon}(q^{1/r})| \leqslant q(q^{2/3}-1)(q-\varepsilon), \end{align*} $$

so $|C_T(y)|\leqslant |C_{\mathrm {PGL}_3^{\varepsilon }(q)}(y)| < |C_T(x)|$ . Thus, we only need to consider involutory field or graph-field automorphisms, so we can assume $\varepsilon = +$ and f is even. Let $y_1$ be an involutory field automorphism. Then by [Reference Burness and Giudici13, Proposition 3.2.9],

$$ \begin{align*} |C_T(y_1)| = \frac{(3,q^{1/2}+1)}{(3,q-1)}|\mathrm{PGL}_3(q^{1/2})|. \end{align*} $$

Similarly, if $y_2$ is a graph-field automorphism, then

$$ \begin{align*} |C_T(y_2)| = \frac{(3,q^{1/2}-1)}{(3,q-1)}|\mathrm{PGU}_3(q^{1/2})| \end{align*} $$

by [Reference Burness and Giudici13, Proposition 3.2.15]. Note that

$$ \begin{align*} |\mathrm{PGL}_3(q^{1/2})| < q^3(q-1) < |\mathrm{PGU}_3(q^{1/2})| < 3|\mathrm{PGL}_3(q^{1/2})|. \end{align*} $$

Therefore, $h(T) = |C_T(x)|$ if f is odd or $\varepsilon = -$ , $h(T) = |C_T(y_1)|$ if $\varepsilon = +$ , f is even and $3\mid q^{1/2}+1$ , otherwise $h(T) = |C_T(y_2)|$ .

Next, assume $T = \mathrm {L}_4^{\varepsilon }(q)$ and let z be a graph automorphism of type $\gamma _1$ (see [Reference Burness and Giudici13, Sections 3.2.5 and 3.3.5]), so by [Reference Burness and Giudici13, Propositions 3.2.14 and 3.3.17], we have

$$ \begin{align*} |C_T(z)| = \frac{(2,q-\varepsilon)}{(4,q-\varepsilon)}|\mathrm{PGSp}_4(q)|> \frac{1}{(4,q-\varepsilon)}q^6(q^2-1)(q-\varepsilon) = |C_T(x)| \end{align*} $$

and we claim that $h(T) = |C_T(z)|$ . Note that

$$ \begin{align*} |z^T| = \frac{q^2(q^3-\varepsilon)}{(2,q-\varepsilon)}. \end{align*} $$

By [Reference Burness10, Propositions 3.22, 3.36, 3.37 and 3.48], we have

$$ \begin{align*} |y^T|> \frac{1}{2}\left(\frac{q}{q+1}\right)q^6 \end{align*} $$

for any unipotent, semisimple, field or graph-field element $y\in \mathrm {Aut}(T)$ of prime order. Hence, $|y^T|> |z^T|$ if $q\geqslant 4$ , and for $q\in \{2,3\}$ we can check that $|y^T|> |z^T|$ using Magma. Similarly, if y is a graph automorphism, then $|y^T|\geqslant |z^T|$ by inspecting [Reference Burness and Giudici13, Tables B.3 and B.4].

Finally, assume $T = \mathrm {L}_n^{\varepsilon }(q)$ and $n\geqslant 5$ . Then by applying the bounds in [Reference Burness10, Table 3.11] we see that

$$ \begin{align*} |y^T|> \frac{1}{2}\left(\frac{q}{q+1}\right)^{\frac{1}{2}(1-\varepsilon)}q^{\frac{1}{2}(n^2-n-4)} > \frac{2q^{2n-1}}{q-1} > |x^T| \end{align*} $$

if y is a field, graph or graph-field automorphism, unless $(n,q) = (5,2)$ or $(6,2)$ , in which cases one can check that $|y^T|>|x^T|$ with the aid of Magma. If y is a unipotent or semisimple element with $\nu (y)\geqslant 2$ , then

$$ \begin{align*} |y^T|> \frac{1}{2}\left(\frac{q}{q+1}\right)q^{4n-8} > \frac{2q^{2n-1}}{q-1} > |x^T| \end{align*} $$

by [Reference Burness10, Proposition 3.36]. Thus, we only need to consider the cases where $\nu (y) = 1$ and y is not $\mathrm {Aut}(T)$ -conjugate to x. In this setting, y is semisimple, and a preimage $\widehat {y}$ of y in $\mathrm {GL}(V)$ is $[\omega I_1,I_{n-1}]$ , where $\omega $ is a nontrivial r-th root of unity in $\mathbb {F}_q$ if $\varepsilon = +$ , or $\mathbb {F}_{q^2}$ if $\varepsilon = -$ , for some prime r. It follows that

$$ \begin{align*} |C_T(y)| = (n,q-\varepsilon)^{-1}|\mathrm{GL}_{n-1}^{\varepsilon}(q)|. \end{align*} $$

Note that $|C_T(y)|> |C_T(x)|$ if and only if $\varepsilon = -$ and n is even. This implies that

$$ \begin{align*} h(T) = (n,q-\varepsilon)^{-1}|\mathrm{GL}_{n-1}^{\varepsilon}(q)| \end{align*} $$

if $\varepsilon = -$ and n is even, otherwise $h(T) = |C_T(x)|$ .

This concludes the proof of Theorem 2.12 for linear and unitary groups. We can use a very similar approach to handle the symplectic and orthogonal groups and we omit the details. But let us remark that if $T = \mathrm {PSp}_n(q)$ is a symplectic group, then $|C_T(x)|$ is maximal when x is a long root element, unless $n = 4$ and q is odd, where an involution of type $t_1$ gives the maximal centraliser. If $T = \mathrm {P\Omega }_n^{\varepsilon }(q)$ , where n is odd or q is even, then $|C_T(x)|$ is maximal when x is an involution of type $t_1^{\prime }$ or $b_1$ , respectively. Finally, if $T = \mathrm {P\Omega }_n^{\varepsilon }(q)$ with n even and q odd, then a graph automorphism of type $\gamma _1$ has the maximal centraliser. All the relevant information about these elements can be found in [Reference Burness and Giudici13, Chapter 3].

Proof. First, assume (i) holds. If $t_i = t_j$ for some $i\ne j$ , then $(i,j)\in W$ stabilises D and $D(\varphi _{t_1},\dots ,\varphi _{t_{k}})$ , which is incompatible with (i). Thus, $t_1,\dots ,t_k$ are distinct. Suppose $g\alpha \in \mathrm {Hol}(T,\{t_1,\dots ,t_k\})$ . Then for any i we have

(5) $$ \begin{align} t_j = t_i^{g\alpha} = (g^{-1}t_i)^{\alpha} = (g^{-1})^{\alpha} t_i^{\alpha} \end{align} $$

for some j. That is, $g\alpha $ induces a permutation $\pi \in S_k$ by $(g^{-1})^{\alpha } t_i^{\alpha } = t_{i^{\pi }}$ . Now, it is easy to see that $(\alpha ,\dots ,\alpha )\pi $ fixes $D(\varphi _{t_1},\dots ,\varphi _{t_k})$ . Hence, $\alpha = 1$ and $\pi = 1$ , which implies that $g = 1$ by (5), noting that $i = j$ since $\pi = 1$ .

Conversely, suppose (ii) holds and $(\alpha ,\dots ,\alpha )\pi $ fixes D and $D(\varphi _{t_1},\dots ,\varphi _{t_k})$ . Then there exists $g\in T$ such that $t_{i^{\pi }}= g^{-1}t_i^{\alpha }$ for all i. It follows that $g^{\alpha ^{-1}}\alpha \in \mathrm {Hol}(T,\{t_1,\dots ,t_k\})$ , which implies that $g=1$ and $\alpha = 1$ . As $t_1,\dots ,t_k$ are distinct, this gives $\pi = 1$ and so (i) holds.

Proof. This follows directly from Lemma 2.15, noting that $\mathrm {Hol}(T,S) = \mathrm {Hol}(T,T\setminus S)$ .

Proof. We have $g^{-1}S_1 = S_2^{\alpha ^{-1}}$ , so $1\in g^{-1}S_1$ and thus $g\in S_1$ .

Proof. Suppose $g\alpha \in \mathrm {Hol}(T,S)$ , where $g\in T$ and $\alpha \in \mathrm {Aut}(T)$ . By Lemma 2.17, we have $g\in S$ . If $g = t_1 = 1$ , then $\alpha \in \mathrm {Aut}(T,\{t_2,\dots ,t_k\})$ and the condition (i) forces $\alpha = 1$ . If $g = t_i$ for some $2\leqslant i\leqslant k$ , then $t_i^{-1}S = S^{\alpha ^{-1}}$ , which implies that $\{|t_i^{-1}t_1|,\dots ,|t_i^{-1}t_k|\}=\{1,|t_2|,\dots ,|t_k|\}$ , which is incompatible with the condition (ii).

Proof. In view of Lemma 2.18, it suffices to show that the conditions (i) and (ii) imply that $\mathrm {Aut}(T,\{t_2,\dots ,t_k\}) = 1$ . Suppose $\alpha \in \mathrm {Aut}(T,\{t_2,\dots ,t_k\})$ . Then $\alpha \in C_{\mathrm {Aut}(T)}(t_i)$ for each i, as $t_2,\dots ,t_k$ have distinct orders. It follows that $\alpha $ centralises $\langle t_2,\dots ,t_k\rangle = M$ and so $\alpha \in C_{\mathrm {Aut}(T)}(M)$ . Since $\mathrm {Out}(T) = 1$ , this implies that $\alpha \in C_T(M)\leqslant N_T(M) = M$ since M is maximal, so $\alpha \in Z(M) = 1$ . This completes the proof.

Proof. This follows immediately from Lemma 2.17.

Proof. This follows by noting that any subset fixed by $\sigma $ is a union of some cycles comprising $\sigma $ .

Proof. Let $\sigma \in \mathrm {Hol}(T)$ be such that it fixes at least one element in T. We may assume $\sigma $ fixes $1\in T$ by the transitivity of $\mathrm {Hol}(T)$ . Thus, $\sigma \in \mathrm {Aut}(T)$ and hence $C_T(\sigma )$ is the set of fixed points of $\sigma $ , which completes the proof.

Proof. Put $\ell = 1$ and $m = n/t$ in Lemma 3.5.

Proof. First, if $T = A_5$ , then we construct the permutation group $\mathrm {Hol}(T)$ on T using the function Holomorph in Magma. Then we find two random k-subsets of T lying in distinct regular $\mathrm {Hol}(T)$ -orbits by random search.

Hence, we may assume $|T|\geqslant 168$ and thus $4\log |T| < |T|/4$ . First, assume $|T|/4\leqslant k\leqslant |T|/2$ . By Corollary 3.4, we have

$$ \begin{align*} |\mathrm{fix}(\sigma,\mathscr{P}_k)| \leqslant \sum_{u=0}^{\lfloor k/r\rfloor}{|T|/r\choose u}{h(T)\choose \lfloor h(T)/2\rfloor}\leqslant 2^{|T|/r}{h(T)\choose \lfloor h(T)/2\rfloor}\leqslant 2^{|T|/2}{h(T)\choose \lfloor h(T)/2\rfloor} \end{align*} $$

for every element $\sigma \in \mathrm {Hol}(T)$ of prime order. Hence, (8) holds if

(10) $$ \begin{align} {|T|\choose k}>|\mathrm{Hol}(T)|2^{|T|/2+1}{h(T)\choose \lfloor h(T)/2\rfloor}, \end{align} $$

and it suffices to consider $k = |T|/4$ . Now, we apply (9), which gives

$$ \begin{align*} {|T|\choose |T|/4}>\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{8}}\frac{4}{\sqrt{3|T|}}\left(\frac{4}{3^{3/4}}\right)^{|T|} \end{align*} $$

and

$$ \begin{align*} {h(T)\choose \lfloor h(T)/2\rfloor} < \frac{1}{\sqrt{2\pi}}\cdot \sqrt{\frac{4}{ h(T)}}\cdot 2^{ h(T)} \leqslant \frac{1}{\sqrt{2\pi}}\cdot \sqrt{\frac{40}{|T|}}\cdot 2^{|T|/10} \end{align*} $$

as $h(T)\leqslant |T|/10$ by Corollary 2.14. Combining the inequalities above, we see that (10) holds for $k = |T|/4$ if

$$ \begin{align*} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{8}}\frac{4}{\sqrt{3|T|}}\left(\frac{4}{3^{3/4}}\right)^{|T|}>|\mathrm{Hol}(T)|\cdot 2^{|T|/2+1}\cdot \frac{1}{\sqrt{2\pi}}\cdot \sqrt{\frac{40}{|T|}}\cdot 2^{|T|/10}. \end{align*} $$

Finally, since $|\mathrm {Out}(T)| < |T|^{1/3}$ by Lemma 2.9, it suffices to show that

(11) $$ \begin{align} t_0^{|T|}>\sqrt{30}e^{\frac{1}{8}}|T|^{\frac{7}{3}}, \end{align} $$

where

$$ \begin{align*} t_0 = 4\cdot 3^{-\frac{3}{4}}\cdot 2^{-\frac{1}{2}-\frac{1}{10}} = 1.1577.... \end{align*} $$

and it is easy to check that the inequality in (11) holds for all $|T|\geqslant 168$ .

Now, assume $4\log |T| < k < |T|/4$ and let $\sigma \in \mathrm {Hol}(T)$ be of prime order r. Observe that $ru\leqslant k< |T|/4$ for all $u\in \{0,\dots ,\lfloor k/r\rfloor \}$ , so

$$ \begin{align*} \begin{aligned} \sum_{u=0}^{\lfloor k/r\rfloor}{|T|/r\choose u}{ h(T)\choose k-ru}&<\sum_{u=0}^{\lfloor k/r\rfloor}{|T|/2\choose u}{ h(T)\choose k-ru}\\ &<\sum_{u=0}^{\lfloor k/r\rfloor}{|T|/2\choose ru}{ h(T)\choose k-ru} \\&< {|T|/2+ h(T)\choose k}, \end{aligned} \end{align*} $$

noting that the third inequality follows from the Vandermonde’s identity. Thus, (8) holds if

(12) $$ \begin{align} {|T|\choose k}>2|\mathrm{Hol}(T)|{|T|/2+ h(T)\choose k}. \end{align} $$

It is easy to see that (12) is equivalent to

$$ \begin{align*} \frac{|T|!}{(|T|-k)!}>2|\mathrm{Hol}(T)|\frac{(|T|/2+ h(T))!}{(|T|/2+ h(T)-k)!}. \end{align*} $$

Now,

$$ \begin{align*} \frac{|T|-m}{|T|/2+ h(T)-m}\geqslant \frac{|T|}{|T|/2+ h(T)} =: t \end{align*} $$

for every $m\in \{0,\dots ,k-1\}$ and thus (12) holds if $t^k>2|\mathrm {Hol}(T)|$ . By Corollary 2.14, we have $|T|/ h(T)\geqslant 10$ , and hence $t\geqslant 5/3$ . Therefore, (12) holds if $(5/3)^k> |T|^{8/3}$ (by applying Lemma 2.9), which implies the desired result.

Proof. If $8\log |T| < h(T)$ , then $k < h(T)/2$ and (8) follows via (13) and Corollary 3.4. By inspecting Table 1, we see that $8\log |T|\geqslant h(T)$ only if T is isomorphic to one of the following groups:

(14) $$ \begin{align} \mathrm{M}_{11},\ \mathrm{J}_1,\ {{}^2}B_2(8),\ \mathrm{L}_3(3),\ \mathrm{L}_2(q)\ (q\leqslant 167). \end{align} $$

Assume T is one of the groups in (14), and suppose $\sigma \in \mathrm {Hol}(T)$ has prime order r. We claim that

(15) $$ \begin{align} |\mathrm{fix}(\sigma,\mathscr{P}_k)| < \sum_{u=0}^{\lfloor k/2\rfloor}{|T|/2\choose u}{h(T)\choose k-2u}. \end{align} $$

To see this, first assume $\sigma $ is fixed-point-free on T. Here, $|\mathrm {fix}(\sigma ,\mathscr {P}_k)| = 0$ if $r\nmid k$ , and

$$ \begin{align*} |\mathrm{fix}(\sigma,\mathscr{P}_k)| = {|T|/r\choose k/r} \end{align*} $$

otherwise. In particular, the inequality in (15) holds. Now, assume $\sigma $ has a fixed point on T. Since $\sigma $ is conjugate to an element fixing the identity element in T, we may assume $\sigma \in \mathrm {Aut}(T)$ . Then with the aid of Magma and Corollary 3.4, it is easy to check that (15) holds when T is one of the groups in (14).

We conclude that the proof is complete by combining (13) and (15) with Lemma 3.1.

Proof. First, observe that (13) holds if

(17) $$ \begin{align} {|T|\choose k}>2|\mathrm{Hol}(T)|\lfloor k/2\rfloor{|T|/2\choose u}{ h(T)\choose k-2u} \end{align} $$

for every $u\in \{0,\dots ,\lfloor k/2\rfloor \}$ . Now,

$$ \begin{align*} \left(\frac{k}{k-2u}\right)^{k-2u} < e^{2u} \end{align*} $$

for all such u. Therefore, (17) follows by combining (16) and the well-known bounds on binomial coefficients

$$ \begin{align*} \frac{n^m}{m^m}<{n\choose m}<\frac{(en)^m}{m^m} \end{align*} $$

for any integers $n\geqslant m\geqslant 0$ , where we define $m^m = 1$ if $m = 0$ .

Proof. We first prove that (13) holds if $k = k_0$ . In view of Lemma 3.9, it suffices to verify the inequality in (16) for all $u\in \{0,\dots ,\lfloor k/2\rfloor \}$ and we will do this by induction. First assume $u = \lfloor k/2\rfloor $ and note that (18) is equivalent to (16) if k is even. For k odd, we have $u = (k-1)/2$ and the inequality in (16) is as follows:

(20) $$ \begin{align} \left(\frac{|T|(k-1)}{k^2e^3}\right)^{k}|T|> \frac{k-1}{k^2e}\cdot 4|\mathrm{Hol}(T)|^2\left(\frac{k-1}{2}\right)^2h(T)^2. \end{align} $$

In view of (19), we see that (20) holds if

$$ \begin{align*} \left(\frac{|T|}{ke^3}\right)^{k}\left(\frac{k-1}{k}\right)^{k-1}e> k^2|\mathrm{Hol}(T)|^2, \end{align*} $$

which is implied by (18) since $(\frac {k-1}{k})^{k-1}>e^{-1}$ . Therefore, (16) holds for $u = \lfloor k/2\rfloor $ and we have established the base case for the induction. Now, suppose (16) holds for $u=u_0$ , where $1\leqslant u_0\leqslant \lfloor k/2\rfloor $ . It suffices to show that (16) holds for $u = u_0-1$ . Here, the desired inequality holds if

$$ \begin{align*} 2^{-1}|T|\cdot\frac{(u_0-1)^{u_0-1}}{u_0^{u_0}}> k^{-2}e^{-1}\cdot h(T)^2, \end{align*} $$

but this is implied by (19), noting that $(\frac {u_0-1}{u_0})^{u_0-1}> e^{-1}$ and $2u_0\leqslant k$ . In conclusion, if $k = k_0$ , then (16) holds for all $u\in \{0,\dots ,\lfloor k/2\rfloor \}$ and thus (13) holds by Lemma 3.9.

Finally, we need to show that (13) holds when $k_0<k$ . By (19), we have $h(T)^2 < k_0|T| < k|T|$ , and by arguing as above, it suffices to show that

(21) $$ \begin{align} |T|^k> |\mathrm{Hol}(T)|^2k^{2+k}e^{3k}. \end{align} $$

Since $|T|> 4080$ and $5\leqslant k\leqslant 4\log |T|$ , we get

$$ \begin{align*} |T|> 2e^4(4\log|T| + 1) \geqslant 2e^4(k+1) > \left(\frac{k+1}{k}\right)^{k+2}e^3(k+1). \end{align*} $$

Therefore, (21) holds for all $k_0 \leqslant k\leqslant 4\log |T|$ by induction on k, and the proof is complete.

Proof. This is similar to the proof of Lemma 3.10, working with Lemma 3.9 to establish the inequality in (13). First, assume $k = k_0$ and note that (22) is equivalent to (16) with $u = 0$ . We now use induction to show that (16) holds for all $u\in \{0,\dots ,\lfloor k/2\rfloor \}$ . To do this, suppose (16) holds for $u = u_0$ , where $0\leqslant u_0\leqslant \lfloor k/2\rfloor -1$ . Then (23) implies that

$$ \begin{align*} 2|T|^{-1}\cdot \frac{(u_0+1)^{u_0+1}}{u_0^{u_0}}>k^2e\cdot h(T)^{-2}, \end{align*} $$

and thus (16) holds for $u = u_0+1$ and the result follows.

Finally, let us assume $k_0<k$ . It suffices to show that

$$ \begin{align*} |T|^{k}> 2|\mathrm{Hol}(T)|\lfloor k/2\rfloor e^{k}h(T)^k \end{align*} $$

for all $k_0\leqslant k\leqslant 4\log |T|$ . This is clear by induction on k, since we have

$$ \begin{align*} |T|> 2eh(T) \end{align*} $$

for every T by Corollary 2.14.

Proof. This is established in the proof of Theorem 1.5 in [Reference Fawcett25].

Proof. Let $f_p(\mathrm {Aut}(T))$ be the number of conjugacy classes of elements of prime order in $\mathrm {Aut}(T)$ . It follows from the proof of [Reference Fawcett25, Lemma 4.2] that

$$ \begin{align*} r_2(G)\leqslant |\mathrm{Out}(T)|f_p(\mathrm{Aut}(T))\left(\frac{ h(T)}{|T|}\right)^{k-2}. \end{align*} $$

Thus, it suffices to show that

(25) $$ \begin{align} |\mathrm{Out}(T)|f_p(\mathrm{Aut}(T))<\left(\frac{|T|}{ h(T)}\right)^2. \end{align} $$

First, assume $T = A_n$ is an alternating group. Then as discussed in the proof of [Reference Fawcett25, Lemma 4.2], we have $f_p(\mathrm {Aut}(T)) < \frac {n^2}{2}$ . This implies (25) since $h(T) = (n-2)!$ by Theorem 2.12.

Next, assume T is a sporadic group. Then $f_p(\mathrm {Aut}(T))$ can be read off from the character table of $\mathrm {Aut}(T)$ and it is easy to check that (25) holds in every case.

Finally, assume T is a simple group of Lie type over $\mathbb {F}_q$ . Let $f(T)$ be the number of conjugacy classes in T. As noted in [Reference Gallagher28], we have $f_p(\mathrm {Aut}(T))\leqslant |\mathrm {Out}(T)|f(T)$ . Thus, it suffices to show that

(26) $$ \begin{align} |\mathrm{Out}(T)|^2f(T) < \left(\frac{|T|}{ h(T)}\right)^2. \end{align} $$

We divide the proof into several cases.

Case 1. $T\ne \mathrm {L}_n^{\varepsilon }(q)$ .

In this setting, [Reference Garzoni and Gill29, Theorem 1.2] implies that $f(T)<|T|/h(T)$ , so in view of (26), it suffices to show that

(27) $$ \begin{align} h(T)|\mathrm{Out}(T)|^2<|T|. \end{align} $$

First, we assume $T\ne \mathrm {P\Omega }_{8}^+(q)$ . Here, $|\mathrm {Out}(T)|\leqslant 8\log q$ and by inspecting Table 1, one can see that $|T|/h(T)\geqslant q^3/2$ . It is straightforward to check that if $q\geqslant 13$ , then $128(\log q)^2 < q^3$ , which implies that (27) holds for $q\geqslant 13$ . Then there are only finitely many exceptional groups of Lie type to consider, and in each case we can use the precise value of $h(T)$ in Table 1 to verify (27). Hence, we may assume $q\leqslant 11$ and T is a classical group. By our assumption, $T = \mathrm {PSp}_n(q)$ , $\Omega _n(q)$ , $\mathrm {P\Omega }_n^-(q)$ , or $\mathrm {P\Omega }_n^+(q)$ with $n\geqslant 10$ in the latter case. In each case, we have $|T|/h(T)> q^{n-2}$ by inspecting Table 1, so if $n\geqslant 8$ we have

$$ \begin{align*} |\mathrm{Out}(T)|^2\leqslant 64(\log q)^2 \leqslant q^6 \leqslant q^{n-2} < |T|/h(T) \end{align*} $$

and thus (27) holds. There are finitely many groups remaining and we can check that (27) holds in each case.

Now, assume $T = \mathrm {P\Omega }_{8}^+(q)$ . Here, $|T|/h(T)> q^6$ and $|\mathrm {Out}(T)|\leqslant 24f\leqslant 24\log q$ . This shows that (27) holds for $q\geqslant 4$ since we have $24^2(\log q)^2<q^6$ . If $q = 2$ , then $|\mathrm {Out}(T)|^2 = 36 < 120 = |T|/h(T)$ , while if $q = 3$ , then $|\mathrm {Out}(T)|^2 = 576 < 1080 = |T|/h(T)$ .

Case 2. $T = \mathrm {U}_n(q)$ , $n\geqslant 3$ .

In this case, [Reference Garzoni and Gill29, Theorem 1.2] implies that $f(T)<\frac {1}{2}|T|/h(T)$ , except when $(n,q) = (3,3)$ or $(4,3)$ . In the latter two cases, it is easy to check (26). In other cases, we have $|T|/h(T)>q^n$ by inspecting Table 1, so (26) holds if

(28) $$ \begin{align} |\mathrm{Out}(T)|^2<2q^n. \end{align} $$

Notice that $|\mathrm {Out}(T)| \leqslant 2(q+1)\log q < q^2$ for $q\geqslant 7$ , and for $q \in \{3,5\}$ we still have $|\mathrm {Out}(T)|\leqslant 2(q+1)<q^2$ . This implies that if $q\notin \{2,4\}$ and $n\geqslant 4$ , then we have

$$ \begin{align*} |\mathrm{Out}(T)|^2<q^4\leqslant q^n<2q^n \end{align*} $$

and so (28) is satisfied. If $q = 2$ , then $|\mathrm {Out}(T)| \leqslant 6$ , so (28) holds if $n\geqslant 5$ , and if $q = 4$ , then $|\mathrm {Out}(T)|\leqslant 20$ , and thus (28) holds for $n\geqslant 4$ . It is straightforward to check (26) when $T = \mathrm {U}_4(2)$ , where we have $f(T) = 20$ .

Finally, assume $n = 3$ , so $|\mathrm {Out}(T)|\leqslant 6\log q$ . Here, (28) is satisfied for all $q>4$ since $(6\log q)^2<2q^3$ . By our assumption, the only remaining cases are $T = \mathrm {U}_3(3)$ with $f(T) = 14$ and $T=\mathrm {U}_3(4)$ with $f(T) = 22$ , so the inequality in (26) holds.

Case 3. $T = \mathrm {L}_n(q)$ .

Here, we assume $(n,q)\ne (2,4), (2,5),(2,9),(3,2),(4,2)$ as noted in (4). If $n = 2$ and $q\in \{7,11\}$ , then an easy computation using Magma shows that (25) holds, and the result follows.

In each of the remaining cases, we have $|T|/h(T)> q^{n-1}$ by inspecting Table 1. Moreover, [Reference Fulman and Guralnick27, Corollary 1.2] implies that $f_p(\mathrm {Aut}(T))<100|T|/h(T)$ , so (25) holds if

(29) $$ \begin{align} 100|\mathrm{Out}(T)|<q^{n-1}. \end{align} $$

Since $|\mathrm {Out}(T)|\leqslant 2(q-1)\log q<q^2$ for all q, (29) holds if $n\geqslant 10$ . Moreover, if $n\geqslant 4$ , then (29) holds if $q> 100$ , while for $q<100$ it is easy to check that (29) still holds in each case, unless $q=2$ and $n\leqslant 8$ , or $n\in \{5,6\}$ and $q\leqslant 4$ , or $n = 4$ and $q\leqslant 9$ . But in each of these cases, it is straightforward to check that (25) is satisfied, so to complete the proof we may assume $n\in \{2,3\}$

Suppose $n = 3$ , so $|\mathrm {Out}(T)|\leqslant 6\log q$ , and (29) holds if $600\log q <q^2$ . The latter holds if $q>59$ . In fact, by working with the precise value of $|\mathrm {Out}(T)|$ we see that (29) holds if $q>25$ . Finally, if $q\leqslant 25$ , then we can check (25) using Magma.

To complete the proof, we may assume $T = \mathrm {L}_2(q)$ , so $|\mathrm {Out}(T)| \leqslant 2\log q$ and $|T|/h(T)\geqslant (q+1)q^{1/2}/2$ . Thus, (25) holds if

$$ \begin{align*} 800\log q <(q+1)^2 \end{align*} $$

since we have $f_p(\mathrm {Aut}(T))<100q$ by [Reference Fulman and Guralnick27, Corollary 1.2]. In this way, we deduce that (25) holds if $q\geqslant 71$ . And for $q < 71$ , we can check that (25) holds with the aid of Magma.

Proof. First, let

$$ \begin{align*} \begin{aligned} R_4(G) &= \{(\alpha,\dots,\alpha)\pi\in R_3(G):\pi = (1,2)\},\\ R_4(T) &= \{\alpha\in\mathrm{Aut}(T):(\alpha,\dots,\alpha)\pi\in R_4(G)\} \end{aligned} \end{align*} $$

as in the proof of [Reference Fawcett25, Theorem 1.5]. Set $P = S_k$ and

$$ \begin{align*} r_4(G):=|(1,2)^P|\sum_{\alpha\in R_4(T)}\frac{|\alpha^{\mathrm{Aut}(T)}|}{|T|}\left(\frac{|C_{\mathrm{Inn}(T)}(\alpha)|}{|T|}\right)^{k-3}. \end{align*} $$

Then we have

(30) $$ \begin{align} \begin{aligned} r_4(G) &= {k\choose 2}\left(\frac{1}{|T|}+\sum_{\alpha\in R_4(T)\setminus\{1\}}\frac{|\alpha^{\mathrm{Aut}(T)}|}{|T|}\left(\frac{|C_{\mathrm{Inn}(T)}(\alpha)|}{|T|}\right)^{k-3}\right)\\ &\leqslant {k\choose 2}\left(\frac{1}{|T|}+|\mathrm{Out}(T)|\left(\frac{ h(T)}{|T|}\right)^{k-3}\right). \end{aligned} \end{align} $$

As noted in the proof of [Reference Fawcett25, Theorem 1.5], we have

(31) $$ \begin{align} r_3(G) \leqslant r_4(G)+\sum_{\pi\in R\setminus\{(1,2)\}}\frac{|\pi^P|}{|T|^{k-r_{\pi}-\frac{5}{3}}}, \end{align} $$

where R is a set of representatives for the conjugacy classes of elements of prime order in P and $r_{\pi }$ is the number of $\langle \pi \rangle $ -orbits in $[k]$ . Without loss of generality, we may assume $(1,2)\in R$ .

Let $x,y\in R$ be the representatives the P-classes $(1,2,3)^P$ and $(1,2)(3,4)^P$ , respectively. Note that $r_x = r_y = k-2$ and $r_z\leqslant k-3$ for all $z\in R\setminus \{(1,2),x,y\}$ . Then

$$ \begin{align*} \begin{aligned} \sum_{\pi\in R\setminus\{(1,2)\}}\frac{|\pi^P|}{|T|^{k-r_{\pi}-\frac{5}{3}}} &= \sum_{\pi\in R\setminus\{(1,2),x,y\}}\frac{|\pi^P|}{|T|^{k-r_{\pi}-\frac{5}{3}}} + |T|^{-\frac{1}{3}}\left(2{k\choose 3}+\frac{1}{2}{k\choose 2}{k-2\choose 2}\right)\\ &<\frac{k!}{|T|^{\frac{4}{3}}}+|T|^{-\frac{1}{3}}\left(2{k\choose 3}+\frac{1}{2}{k\choose 2}{k-2\choose 2}\right) \end{aligned} \end{align*} $$

and so the lemma follows by combining (30) and (31).

Proof. By (24) and (34), we have

$$ \begin{align*} \frac{1}{2}> Q_1(G) + Q_2(G) > 1-\frac{r(G)|G|}{|T|^{2k-2}}= 1-\frac{r(G)|\mathrm{Out}(T)|\cdot k!}{|T|^{k-2}}. \end{align*} $$

It suffices to prove that

$$ \begin{align*} 2|\mathrm{Out}(T)|\cdot k!\leqslant |T|^{k-2}, \end{align*} $$

which is clear since $k\leqslant 4\log |T|$ .

Proof. If $T\notin \{\mathrm {Ly},\mathrm {Th},\mathrm {J}_4,\mathbb {B},\mathbb {M}\}$ , then we can construct T as a permutation group in Magma using the function AutomorphismGroupSimpleGroup. Then the result follows by random search (see Remark 2.21(i)). If $T\in \{\mathrm {Ly},\mathrm {Th},\mathrm {J}_4,\mathbb {B},\mathbb {M}\}$ , then $|\mathrm {Out}(T)| = 1$ . Let M be a maximal subgroup of T with

(35) $$ \begin{align} (T,M) \in \{ (\mathrm{Ly},G_2(5)),\ (\mathrm{Th},\mathrm{AGL}_2(5)),\ (\mathrm{J}_4,\mathrm{M}_{22}.2),\ (\mathbb{B},\mathrm{Fi}_{23}),\ (\mathbb{M},\mathrm{L}_2(71))\}. \end{align} $$

In view of Corollary 2.19, the result follows by random search as in Remark 2.21(ii).

Proof. Suppose T is a Lie type group defined over $\mathbb {F}_q$ , where $q = p^f$ for some prime p. We will work with a quasisimple group Q with $Q/Z(Q) = T$ . Let m be the number of regular semisimple conjugacy classes in Q. Then T has at least $m|T|/|Q|$ regular semisimple T-classes, and thus T has at least eight regular semisimple $\mathrm {Aut}(T)$ -classes if

(36) $$ \begin{align} m|T|\geqslant 8|\mathrm{Out}(T)||Q|. \end{align} $$

First, assume Q is a simply connected quasisimple exceptional group. Then m has been computed by Lübeck [Reference Lübeck46], and one can see that (36) holds for every $T\notin \mathcal {C}$ by inspecting [Reference Lübeck46].

Next, assume $Q\in \{\mathrm {SL}_n^{\varepsilon }(q),\mathrm {Sp}_n(q)\}$ , so m is given in [Reference Fulman and Guralnick26]. The result now follows by inspecting [Reference Fulman and Guralnick26]. For example, if $Q = \mathrm {SL}_2(q)$ , then $|Q|/|T| = (2,q-1)$ , $|\mathrm {Out}(T)| = (2,q-1)f$ and

$$ \begin{align*} m = q-3+(2,q) \end{align*} $$

by [Reference Fulman and Guralnick26, Theorem 2.4]. Thus, (36) is valid if

$$ \begin{align*} q-3+(2,q)\geqslant 8(2,q-1)^2f, \end{align*} $$

which holds for all $q> 81$ . For the cases where $q\leqslant 81$ and $T\notin \mathcal {C}$ , one can check using Magma that there are at least eight regular semisimple $\mathrm {Aut}(T)$ -classes. We use an entirely similar argument to treat all the other cases and we omit the details.

To complete the proof, we assume $Q = \Omega _n^{\varepsilon }(q)$ , so Q has index $2$ in $\mathrm {SO}_n^{\varepsilon }(q)$ . First, assume q is even. Here, $Q = T$ and every semisimple element in $\mathrm {SO}_n^{\varepsilon }(q)$ has odd order, and so lies in Q. This implies that m is at least the number of regular semisimple $\mathrm {SO}_n^{\varepsilon }(q)$ -classes in $\mathrm {SO}_n^{\varepsilon }(q)$ , which is computed in [Reference Fulman and Guralnick26, Theorem 5.12], and the result follows by arguing as above.

Finally, assume $Q = \Omega _n^{\varepsilon }(q)$ and q is odd. Write $d=\lceil n/2\rceil -1$ . Let $A\in \mathrm {GL}_d(q)$ be of order $q^d-1$ and let

$$ \begin{align*} x = \begin{pmatrix} A&&\\ &(A^{-1})^T&\\ &&I_{n-2d} \end{pmatrix} \end{align*} $$

with respect to a standard basis (see [Reference Kleidman and Liebeck39, Proposition 2.5.3]). Then $x\in \mathrm {SO}_n^{\varepsilon }(q)$ , so $y:=x^2\in \Omega _n^{\varepsilon }(q)$ , noting that

$$ \begin{align*} y = \begin{pmatrix} B&&\\ &(B^{-1})^T&\\ &&I_{n-2d} \end{pmatrix}, \end{align*} $$

where $B = A^2$ . Let $\mu $ be an eigenvalue of B of order $(q^d-1)/2$ in the algebraic closure K of $\mathbb {F}_q$ . Then it is easy to show that $\mu \ne \mu ^{\pm q^t}$ for any $1\leqslant t\leqslant d-1$ , and the set of eigenvalues of y is

$$ \begin{align*} \{\mu,\mu^q,\dots,\mu^{q^{d-1}},\mu^{-1},\mu^{-q},\dots,\mu^{-q^{d-1}},1\}, \end{align*} $$

where $1$ has multiplicity $n-2d\in \{1,2\}$ and any other eigenvalue has multiplicity $1$ . It follows that $y^i$ is regular semisimple if $(i,(q^d-1)/2) = 1$ . This gives at least

$$ \begin{align*} \frac{\phi\left((q^d-1)/2\right)}{2d} \end{align*} $$

regular semisimple $\mathrm {GO}_n^{\varepsilon }(q)$ -classes in Q, where $\phi $ is the Euler’s totient function (note that two elements are not conjugate in $\mathrm {GL}_n(q)$ if they have distinct sets of eigenvalues). By arguing as above, T has at least eight regular semisimple $\mathrm {Aut}(T)$ -classes if

(37) $$ \begin{align} \phi\left((q^d-1)/2\right)\geqslant 32d\cdot |\mathrm{Aut}(T):\mathrm{PGO}_n^{\varepsilon}(q)|, \end{align} $$

noting that $|\mathrm {Aut}(T):\mathrm {PGO}_n^{\varepsilon }(q)|\leqslant f\leqslant \log q$ if $d\ne 3$ , while $|\mathrm {Aut}(T):\mathrm {PGO}_n^{\varepsilon }(q)|\leqslant 3f\leqslant 3\log q$ if $d = 3$ . It is easy to check that (37) holds unless

$$ \begin{align*} (d,q)\in\{(6,3),(5,3),(4,3),(4,5),(4,7),(3,3),(3,5),(3,7)\}. \end{align*} $$

For these remaining cases, one can use Magma to obtain m and so (36) holds unless $Q\in \{\Omega _{10}^-(3),\Omega _8^+(5),\Omega _8^{\varepsilon }(3),\Omega _7(3)\}$ , where we can directly check that there are at least eight regular semisimple $\mathrm {Aut}(T)$ -classes in T with the aid of Magma.

Proof. Let x and y be as described in Theorem 2.11. Let $z_1$ and $z_2$ be semisimple elements in T lying in distinct $\mathrm {Aut}(T)$ -classes and

$$ \begin{align*} z_1,z_2\notin x^{\mathrm{Aut}(T)}\cup (x^{-1})^{\mathrm{Aut}(T)}\cup y^{\mathrm{Aut}(T)}\cup (y^{-1})^{\mathrm{Aut}(T)}. \end{align*} $$

Note that the existence of $z_1$ and $z_2$ follows from Lemma 4.2. Then by applying [Reference Gow32, Theorem 2], which asserts that the product of any two regular semisimple T-classes contains all semisimple elements in T, there exist $g_i$ and $h_i$ in T such that $z_i = x^{g_i}y^{h_i}$ , and without loss of generality we may assume $g_i = 1$ , so $z_i = xy^{h_i}$ . It is easy to see that $\mathrm {Hol}(T,\{1,x^{-1},y^{h_i}\}) = 1$ , and so $b(G) = 2$ . By Lemma 2.16, it suffices to show that $S_1 = \{1,x^{-1},y^{h_1}\}$ and $S_2 = \{1,x^{-1},y^{h_2}\}$ are in distinct $\mathrm {Hol}(T)$ -orbits. Suppose $S_1^{g\alpha } = S_2$ for some $g\alpha \in \mathrm {Hol}(T)$ , and note that $g\in S_1$ by Lemma 2.17. If $g = 1$ , then $(x^{-1})^{\alpha } = x^{-1}$ and $(y^{h_1})^{\alpha } = y^{h_2}$ . However, this implies that

$$ \begin{align*} z_1^{\alpha} = (xy^{h_1})^{\alpha} = xy^{h_2} = z_2, \end{align*} $$

which is incompatible with our assumption $z_1^{\mathrm {Aut}(T)}\ne z_2^{\mathrm {Aut}(T)}$ . If $g = x^{-1}$ , then $(y^{h_1})^g = xy^{h_1} = z_1$ , which is not $\mathrm {Aut}(T)$ -conjugate to any element in $S_2$ , a contradiction. Finally, if $g = y^{h_1}$ , then $(x^{-1})^g = y^{-h_1}x^{-1} = z_1^{-1}$ . With the same reason, this is impossible. Therefore, there is no $g\alpha \in \mathrm {Hol}(T)$ such that $S_1^{g\alpha } = S_2$ , which completes the proof.

Proof. Let x and y be as in Theorem 2.11. By [Reference Gow32, Theorem 2], every semisimple element in T lies in $x^Ty^T$ , so we may assume that

(38) $$ \begin{align} x^{-1}y\notin x^{\mathrm{Aut}(T)}\cup (x^{-1})^{\mathrm{Aut}(T)}\cup y^{\mathrm{Aut}(T)}\cup (y^{-1})^{\mathrm{Aut}(T)}. \end{align} $$

Additionally, using Lemma 4.2, let $z_0$ be a regular semisimple element such that

(39) $$ \begin{align} z_0\notin x^{\mathrm{Aut}(T)}\cup (x^{-1})^{\mathrm{Aut}(T)}\cup y^{\mathrm{Aut}(T)}\cup (y^{-1})^{\mathrm{Aut}(T)}\cup (x^{-1}y)^{\mathrm{Aut}(T)}\cup (y^{-1}x)^{\mathrm{Aut}(T)}. \end{align} $$

Again, [Reference Gow32, Theorem 2] implies that $x^Tz_0^T$ contains all semisimple elements in T. Thus, by Lemma 4.2, there exists $z\in z_0^T$ such that

(40) $$ \begin{align} z^{-1}x\notin x^{\mathrm{Aut}(T)}\cup (x^{-1})^{\mathrm{Aut}(T)}\cup y^{\mathrm{Aut}(T)}\cup (y^{-1})^{\mathrm{Aut}(T)}\cup (x^{-1}y)^{\mathrm{Aut}(T)}\cup (y^{-1}x)^{\mathrm{Aut}(T)}. \end{align} $$

Set $S_1 = \{1,x,y,z\}$ and suppose $g\alpha \in \mathrm {Hol}(T,S_1)$ . If $g = 1$ , then $\alpha \in \mathrm {Aut}(T,S_1) = 1$ as $\langle x,y\rangle = T$ and $x,y,z$ are in distinct $\mathrm {Aut}(T)$ -classes. If $g = x$ , then $x^{-1}y\in x^{-1}S_1 = S_1^{\alpha ^{-1}}$ , which is incompatible with either (38) or (39). The case where $g = y$ can be eliminated using the same argument. If $g = z$ , then $z^{-1}S_1 = S_1^{\alpha ^{-1}}$ , and by using (39) and (40), both $z^{-1}$ and $z^{-1}x$ are $\mathrm {Aut}(T)$ -conjugate to z, which yields $z^{-1} = z^{\alpha } = z^{-1}x$ , a contradiction. Thus, we have $b(G) = 2$ .

Similarly, Lemma 4.2 implies that there exists a regular semisimple element $w\in T$ such that $w\ne z$ ,

$$ \begin{align*} w\notin x^{\mathrm{Aut}(T)}\cup (x^{-1})^{\mathrm{Aut}(T)}\cup y^{\mathrm{Aut}(T)}\cup (y^{-1})^{\mathrm{Aut}(T)}\cup (x^{-1}y)^{\mathrm{Aut}(T)}\cup (y^{-1}x)^{\mathrm{Aut}(T)} \end{align*} $$

and

$$ \begin{align*} w^{-1}x\notin x^{\mathrm{Aut}(T)}\cup (x^{-1})^{\mathrm{Aut}(T)}\cup y^{\mathrm{Aut}(T)}\cup (y^{-1})^{\mathrm{Aut}(T)}\cup (x^{-1}y)^{\mathrm{Aut}(T)}\cup (y^{-1}x)^{\mathrm{Aut}(T)}. \end{align*} $$

Set $S_2 = \{1,x,y,w\}$ . By arguing as above, we have $\mathrm {Hol}(T,S_2) = 1$ and it suffices to show that $S_1$ and $S_2$ are in distinct $\mathrm {Hol}(T)$ -orbits. Suppose $S_1^{g\alpha } = S_2$ , and note that $g\in S_1$ by Lemma 2.17. If $g=1$ , then $x^{\alpha } = x$ and $y^{\alpha } = y$ , which implies that $\alpha = 1$ . However, this is incompatible with $z\ne w$ . If $g = x$ , then