Proof. First observe that

$$\begin{align*}a(n,1) = 1-\delta(A_n),\;\; b(n,1) = 2(1-\delta(S_n)) - a(n,1) \end{align*}$$

with respect to the natural actions of $A_n$ and $S_n$ on $[n]$ . By [Reference Boston, Dabrowski, Foguel, Gies, Leavitt and Ose9, Corollary 2.6], we have

$$\begin{align*}\delta(A_n) = \delta(S_n) - \frac{(-1)^n(n-1)}{n!},\;\; \delta(S_n) = \frac{[n!/e]}{n!}, \end{align*}$$

and it is easy to check that $c(n,1) \leqslant 2/3$ for all $n \geqslant 5$ .

Proof. First we compute $c(3,2) = 1, \ c(4,2) = 1/2, \ c(5,2) = 3/5, \ c(6,2) = 2/3$ and

$$\begin{align*}c(7,2) = \frac{38}{63}, \;\; c(8,2) = \frac{7}{12}, \;\; c(9,2) = \frac{3691}{6480}. \end{align*}$$

In addition, we check that $c(n,2) \leqslant 0.63$ for $n=10,11,12$ . Now assume $n \geqslant 13$ and suppose we have $c(m,2) \leqslant 0.63$ for all $7 \leqslant m \leqslant n-1$ . Then (2.1) gives

$$\begin{align*}c(n,2) \leqslant \frac{1}{n}\left(2+\sum_{m=2}^9c(m,2) + \sum_{j=3}^{n-10}c(n-j,2)\right) \leqslant \frac{1}{n}\left(\frac{341233}{45360}+\frac{63}{100}(n-12)\right)< 0.63, \end{align*}$$

and the result follows.

Proof. First assume $k=3$ , so $n \geqslant 7$ , and we compute

$$\begin{align*}c(4,3)=\frac{3}{4}, \; c(5,3)=\frac{3}{5},\; c(6,3)=\frac{3}{8},\; c(7,3) = \frac{18}{35},\; c(8,3) = \frac{25}{48}. \end{align*}$$

Suppose $n \geqslant 9$ and $c(m,3) \leqslant 0.63$ for all $5 \leqslant m \leqslant n-1$ . Then by applying the upper bound in (2.1), we deduce that

$$\begin{align*}c(n,3) \leqslant \frac{1}{n}\left(3+1+\frac{3}{4} + \sum_{j=4}^{n-5}c(n-j,3)\right) \leqslant \frac{1}{n}\left(\frac{19}{4}+\frac{63}{100}(n-8)\right)< 0.63. \end{align*}$$

The case $k=4$ is very similar.

Finally, let us assume $k \geqslant 5$ . By Lemmas 2.3 and 2.4, we have $c(k+1,k) = c(k+1,1) \leqslant 2/3$ and $c(k+2,k) = c(k+2,2) \leqslant 0.63$ . Let us assume $c(m,k) \leqslant 0.63$ for all $k+2 \leqslant m \leqslant n-1$ . Then (2.1) gives

$$\begin{align*}c(n,k) \leqslant \frac{1}{n}\left(k+1+\frac{2}{3}+\frac{63}{100}(n-2k-2)\right) = \frac{63}{100}-\frac{39k-61}{150n}<0.63, \end{align*}$$

and the result follows.

Proof of Proposition 2.2.

The cases $n \in \{5,6\}$ can be checked directly, so let us assume $n \geqslant 7$ . Then Lemmas 2.4 and 2.5 imply that $f(n,k) \leqslant 0.63$ for all $2 \leqslant k < n/2$ , while we have

$$\begin{align*}f(n,1) = 1-\frac{[n!/e]}{n!}+\frac{(-1)^n(n-1)}{n!}> 0.63. \end{align*}$$

The result follows.

Proof. Part (ii) follows immediately from the information in Table 1, so we just need to consider part (i). As before, we may assume G is primitive, which means that H is a maximal subgroup of G.

To begin with, let us assume G is not the Monster group $\mathbb {M}$ . Then in each case, the character tables of G and H are available in the GAP Character Table Library [Reference Breuer12]. In addition, the corresponding fusion map from H-classes to G-classes is also available, with the single exception of the case where $G = \mathbb {B}$ is the Baby Monster and $H = (2^2 \times F_4(2)){:}2$ . Putting the latter case to one side for now, we can use the fusion map to determine the G-classes comprising $\Delta (G)$ , which in turn allows us to compute $\delta (G)$ precisely. We can then read off the minimum over all maximal subgroups, which gives the value of $\alpha (G)$ recorded in Table 1.

Now assume $G = \mathbb {B}$ and $H = (2^2 \times F_4(2)){:}2$ . Here we use the function PossibleClassFusions to produce a list of $64$ candidate fusion maps, and one checks that each candidate map produces the same permutation character $\chi = 1^G_H$ . So once again, we can calculate $\delta (G)$ precisely, and the result follows.

Finally, let us assume $G = \mathbb {M}$ is the Monster. By the main theorem of [Reference Dietrich, Lee and Popiel28], there are $46$ conjugacy classes of maximal subgroups of G. For $31$ of these classes, we can access the character tables of G and a representative H via the GAP function NamesOfFusionSources, as well as the corresponding fusion maps. So in each of these cases, we can compute $\delta (G)$ precisely, just as we did above. In particular, if $\mathcal {M}_1$ denotes this specific collection of maximal subgroups, then

$$\begin{align*}\min\{\delta(G,H) \,:\, H \in \mathcal{M}_1\} = \frac{26707770823339783801504}{49722462258718251877875} = \gamma, \end{align*}$$

with equality if and only if $H = 2^{1+24}.\mathrm {Co}_1$ . So in order to complete the proof, we need to show that $\delta (G,H)> \gamma $ for all of the remaining maximal subgroups H. To do this, let $\omega (H)$ be the spectrum of H, which is simply the set of element orders in H, and consider the crude lower bound

(2.2) $$ \begin{align} \delta(G) \geqslant \frac{|\{ x \in G \,:\, |x| \not\in \omega(H)\}|}{|G|}. \end{align} $$

First assume H is one of the following maximal subgroups:

$$\begin{align*}\mathrm{L}_2(13){:}2,\, \mathrm{L}_2(19){:}2,\, \mathrm{L}_2(29){:}2, \, \mathrm{U}_3(4){:}4,\, (\mathrm{ L}_2(11) \times \mathrm{L}_2(11)).4, \, 11^2{:}(5 \times 2A_5), \end{align*}$$

$$\begin{align*}7^2{:}\mathrm{SL}_2(7), \, 3^8.\mathrm{P\Omega}_{8}^{-}(3).2, \, 3^{3+2+6+6}.(\mathrm{L}_3(3) \times \mathrm{ SD}_{16}),\, 3^{2+5+10}.(\mathrm{M}_{11} \times 2S_4), \end{align*}$$

$$\begin{align*}(3^2{:}2 \times \mathrm{P\Omega}_8^{+}(3)).S_4, \, 2^{3+6+12+18}.(\mathrm{L}_3(2) \times 3S_6), \, 2^{2+11+22}.(\mathrm{M}_{24} \times S_3). \end{align*}$$

In each of these cases, a permutation representation of H is given in the Web Atlas [Reference Wilson84], and with the aid of Magma [Reference Bosma, Cannon and Playoust8], it is easy to determine $\omega (H)$ . We can then verify the desired bound via (2.2). For example, if $H = 2^{3+6+12+18}.(\mathrm {L}_3(2) \times 3S_6)$ , then the Web Atlas provides a representation of H on $1032192$ points, and we can use Magma to compute

$$\begin{align*}\omega(H) = \left\{ \begin{array}{c} 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 16, 20,21, 24, \\ 28, 30, 32, 35, 40, 42, 48, 56, 60, 70, 84, 105 \end{array} \right\}. \end{align*}$$

And then by inspecting the character table of G, the bound in (2.2) yields

$$\begin{align*}\delta(G) \geqslant \frac{74617454008173302577265307}{105784031083359216398221125}> \gamma, \end{align*}$$

and the result follows.

Finally, let us assume $H = 2^{5+10+20}.(S_3 \times \mathrm {L}_5(2))$ or $2^{10+16}.\Omega _{10}^{+}(2)$ . In these two cases, the Web Atlas does not provide a permutation representation of H, and so we need to modify the argument. In the first case, one can check that the trivial bound

(2.3) $$ \begin{align} \delta(G) \geqslant \frac{|\{x \in G \,:\, |H| \mbox{ is indivisible by } |x|\}|}{|G|} \end{align} $$

is sufficient, so we may assume $H = 2^{10+16}.\Omega _{10}^{+}(2)$ . Here the same bound yields $\delta (G)> 1/2$ , but we need to work a little bit harder to force $\delta (G)> \gamma $ .

To do this, we first observe that

$$\begin{align*}\{n \in \omega(G) \,:\, n \nmid |H|\} = \left\{ \begin{array}{c} 11, 13, 19, 22, 23, 26, 29, 33, 38, 39, 41, 44, 46, 47,52, \\ 55, 57, 59, 66, 69, 71, 78, 87, 88, 92, 94, 95, 104, 110 \end{array} \right\} = A. \end{align*}$$

In addition, with the aid of Magma, we compute

$$\begin{align*}\omega(\Omega_{10}^{+}(2)) = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 17, 18, 20, 21, 24, 30, 31, 42, 45,51, 60 \}. \end{align*}$$

Since $H/N = \Omega _{10}^{+}(2)$ for some normal $2$ -subgroup N, it follows that every element in G of order $93, \ 105$ and $119$ is a derangement. So if we now set $B = A \cup \{93,105,119\}$ , then

$$\begin{align*}\delta(G) \geqslant \frac{|\{ x\in G \,:\, |x| \in B\}|}{|G|}> \gamma, \end{align*}$$

and the proof is complete.

Proof of Theorem C(i).

Let $G \leqslant \mathrm {Sym}(\Omega )$ be a finite primitive permutation group with socle $N = T^k$ , where T is simple. The possibilities for G and N are described by the O’Nan-Scott theorem, and we will refer to the five types briefly described above. Our goal is to prove that $\delta (N) \geqslant \varepsilon $ , where $\varepsilon>0$ is the constant in Theorem 2.1.

Of course, if G is an affine group or a twisted wreath product, then N is regular, and thus

$$\begin{align*}\delta(N)=\frac{\Delta(N)}{|N|}=1 - \frac{1}{|N|} \geqslant \frac{1}{2}. \end{align*}$$

And if G is almost simple, then N is a simple transitive permutation group on $\Omega $ , and thus Theorem 2.1 implies that $\delta (N) \geqslant \varepsilon $ .

Next assume G is a diagonal type group, so $k \geqslant 2$ . Let

$$\begin{align*}D = \{(t, \ldots , t) \in N \,:\, t \in T \} \end{align*}$$

be the diagonal subgroup of N and note that we may identify $\Omega $ with the set $N/D$ of right cosets of D in N. Given $x = (x_1 ,\ldots , x_k) \in N$ and $\omega = D(t_1,\ldots ,t_k) \in \Omega $ , we have $\omega ^x = \omega $ if and only if $(t_1x_1t_1^{-1}, \ldots , t_kx_kt_k^{-1}) \in D$ . Therefore, x is a derangement if and only if at least two of the components of x are not T-conjugate, and thus

$$\begin{align*}\delta(N) = \frac{|N| - \left(\sum_{i=1}^m |t_i^T|^k\right)}{|N|} = 1 - \sum_{i=1}^m|C_T(t_i)|^{-k} \geqslant 1 - \sum_{i=1}^m|C_T(t_i)|^{-2}, \end{align*}$$

where $\{t_1, \ldots , t_m\}$ is a complete set of representatives of the conjugacy classes in T. Since $|C_T(t_i)| \geqslant 3$ for all i (recall that a finite group with a self-centralising involution is soluble; see [Reference Blackburn, Britnell and Wildon7, Proposition 4.2], for example), we conclude that

$$\begin{align*}\delta(N) \geqslant 1 - \frac{1}{3}\sum_{i=1}^m |C_T(t_i)|^{-1} = \frac{2}{3}. \end{align*}$$

Finally, let us assume G is a product type group. Here $G \leqslant L \wr S_b$ , where $L \leqslant \mathrm { Sym}(\Gamma )$ is a primitive group with socle $S = T^a, \ b \geqslant 2$ and L is either almost simple or diagonal type. Then $N = S^b = T^{ab}$ and G acts on $\Omega = \Gamma ^b$ with the product action. In particular, if $x = (x_1, \ldots , x_b) \in N$ , where each $x_i$ is contained in S, then

$$\begin{align*}(\gamma_1, \ldots, \gamma_b)^{x} = (\gamma_1^{x_1}, \ldots, \gamma_b^{x_b}) \end{align*}$$

for all $\omega = (\gamma _1, \ldots , \gamma _b) \in \Omega $ . It follows that x is a derangement on $\Omega $ if and only if at least one $x_i$ is a derangement on $\Gamma $ , whence

(2.4) $$ \begin{align} \delta(N)> \frac{|\Delta(S,\Gamma)||S|^{b-1}}{|S|^b} = \delta(S,\Gamma). \end{align} $$

If L is almost simple, then Theorem 2.1 implies that $\delta (N) \geqslant \varepsilon $ . And if L is a diagonal type group, then our previous argument gives $\delta (N) \geqslant 2/3$ .

Proof. This is a special case of a well-known formula of Frobenius, which follows from the familiar orthogonality relations satisfied by the character table of G. See [Reference James and Liebeck54, Theorem 30.4], for example.

Proof. Let $\mathcal {C}(G)$ be the set of conjugacy classes C in G with $G = C^2$ . Using the GAP Character Table Library [Reference Breuer12] and Lemma 3.2, it is easy to determine all of the conjugacy classes in $\mathcal {C}(G)$ . For example, if $G = \mathbb {M}$ is the Monster, then $146$ of the $194$ conjugacy classes in G are contained in $\mathcal {C}(G)$ .

First assume $G \ne \mathbb {B},\mathbb {M}$ . In each of these cases, we can use the GAP function Maxes to access the character table of every maximal subgroup H of G, together with the fusion map from H-classes to G-classes. This allows us to determine all the conjugacy classes of derangements for the action of G on $G/H$ , and it is a routine exercise to check whether or not one of these classes is contained in $\mathcal {C}(G)$ . We find that there is always at least one such class, unless $(G,H)$ is one of the following:

(3.1) $$ \begin{align} \begin{array}{rl} \mathrm{M}_{11}{:} & H = S_5 \\ \mathrm{M}_{22}{:} & H = 2^4{:}A_6 \text{ or } 2^4{:}S_5 \\ \mathrm{M}_{23}{:} & H = \mathrm{M}_{22}, \ \mathrm{L}_3(4).2, \ 2^4{:}A_7, \ \mathrm{M}_{11} \text{ or } 2^4{:}(3 \times A_5).2 \\ \mathrm{M}_{24}{:} &H =\mathrm{M}_{12}.2 \end{array} \end{align} $$

In each of these cases, we first check that we can always find two classes of derangements C and D with $G = CD$ , which is easy to verify using GAP and Lemma 3.2. In addition, we check that $G = \{1\} \cup C^2 = \{1\} \cup D^2$ .

So to complete the proof for $G \ne \mathbb {B}, \mathbb {M}$ , we just need to inspect the special cases $(G,H)$ in (3.1) in order to determine whether or not H has a proper subgroup K that meets every class in $\mathcal {C}(G)$ . To do this, we use Magma to construct $G = \mathrm {M}_n$ and H in the natural permutation representation on n points, and we inspect the maximal subgroups of H. In this way, we deduce that if $G = \mathrm {M}_{11}, \ \mathrm {M}_{22}$ or $\mathrm {M}_{24}$ , then every proper subgroup of H fails to intersect at least one of the classes in $\mathcal {C}(G)$ . Now assume $G = \mathrm {M}_{23}$ . Here G has four maximal subgroups

$$\begin{align*}H \in \{ \mathrm{M}_{22}, \, \mathrm{L}_3(4).2, \, 2^4{:}A_7, 2^4{:}(3 \times A_5).2\} \end{align*}$$

with a maximal subgroup $K<H$ meeting every class in $\mathcal {C}(G)$ . These second maximal subgroups of G are recorded in the second row for $G = \mathrm {M}_{23}$ in Table 2, up to conjugacy in G. In turn, one checks that if $J<K$ is maximal, then $J \cap C$ is empty for at least one class $C \in \mathcal {C}(G)$ , so no further examples arise.

Next assume $G = \mathbb {B}$ is the Baby Monster. As above, we can use the GAP function Maxes to access the character table of each maximal subgroup H of G. And as explained in the proof of Proposition 2.7, we can also compute the corresponding permutation character $1^G_H$ , which allows us to check that there is always at least one class of derangements in $\mathcal {C}(G)$ .

Finally, suppose $G = \mathbb {M}$ is the Monster. Recall that G has $46$ conjugacy classes of maximal subgroups (see [Reference Dietrich, Lee and Popiel28]). Now G has a unique class C of elements of order $41$ , and our earlier computation shows that $C \in \mathcal {C}(G)$ . By inspecting [Reference Dietrich, Lee and Popiel28, Table 1], we deduce that the only maximal subgroups intersecting C are $3^8.\mathrm {P\Omega }_{8}^{-}(3).2$ and $41{:}40$ . In addition, G has a unique class D of elements of order $19$ , and we note that $D \in \mathcal {C}(G)$ , so the result follows since neither $3^8.\mathrm {P\Omega }_{8}^{-}(3).2$ nor $41{:}40$ contains elements in D.

Proof. Part (i) is the main theorem of [Reference Bertram6], while part (ii) is a recent result of Larsen and Tiep [Reference Larsen and Tiep59, Theorem 5(i)]. Finally, part (iii) is due to Brenner [Reference Brenner11].

Proof of Proposition 3.5.

The groups with $5 \leqslant n \leqslant 18$ can be checked using Magma [Reference Bosma, Cannon and Playoust8]. To do this, we first construct the character table of G, and we use the Frobenius formula (see Lemma 3.2) to determine the set $\mathcal {C}(G)$ of conjugacy classes C with $G = C^2$ . Next we construct a representative H of each conjugacy class of maximal subgroups of G, and it is straightforward to check that every class in $\mathcal {C}(G)$ meets H if and only if $(G,H)$ is one of the cases recorded in part (ii)(a) of Proposition 3.5. And in each of these cases, it is routine to check that if $K<H$ is any maximal subgroup, then there exists a class $C \in \mathcal {C}(G)$ such that $C \cap K$ is empty. For the remainder, we will assume $n \geqslant 19$ .

Suppose H acts intransitively on $[n] = \{1, \ldots , n\}$ , so $H \leqslant (S_{k} \times S_{n-k}) \cap G$ for some positive integer $k \leqslant n/2$ . If n is even, then by applying parts (i) and (iii) of Proposition 3.7, we deduce that $G = D_{n-3}^2$ if $k \geqslant 4$ and $G = D^2$ if $k \leqslant 3$ , where $D = x^G$ and $x = (1, \ldots , 4)(5, \ldots , n)$ . If , then Proposition 3.7(ii) gives $G = D_n^2$ , so we may assume . If $k \geqslant 3$ , then every $(n-2)$ -cycle in G is a derangement, and we have $G = D_{n-2}^2$ by Proposition 3.7(i). And if $k=2$ , then Proposition 3.7(iii) implies that $G = D^2$ , where $D = x^G$ and $x = (1,\ldots ,4)(5, \ldots , n-1)$ . Finally, suppose $k=1$ . As above, let $C = x^G$ with $x = (1, \ldots , n)$ and set $D = (x^{-1})^G \ne C$ . Then [Reference Larsen and Tiep59, Theorem 1] implies that (3.2) holds.

Next assume H is imprimitive, so $H \leqslant (S_a \wr S_b) \cap G$ , where $n=ab$ and $a,b \geqslant 2$ . If n is odd, then $a \geqslant 3$ , and every $(n-2)$ -cycle is a derangement because it does not preserve a partition of $[n]$ into b blocks of size a. The result follows since $G = D_{n-2}^2$ by Proposition 3.7(i). Now assume n is even and set $\ell =5$ if $a=3$ , and $\ell =3$ otherwise. Then every $(n-\ell )$ -cycle is a derangement, and the desired conclusion holds since $G = D_{n-\ell }^2$ (note that if $a=3$ , then $n \geqslant 21$ , and thus $n-5 \geqslant 3n/4$ ).

Finally, let us assume H acts primitively on $[n]$ . Fix an odd integer $\ell $ such that $3n/4 \leqslant \ell \leqslant n-3$ . Then Proposition 3.7(i) gives $G = D_{\ell }^2$ , and the main theorem of [Reference Jones55] implies that every $\ell $ -cycle in G is a derangement (indeed, no proper primitive subgroup of G contains an $\ell $ -cycle). The result follows.

Proof. This is a straightforward computation. First we use Magma [Reference Bosma, Cannon and Playoust8] to construct the character table of G, and we then apply Lemma 3.2 to determine the set $\mathcal {C}'(G)$ of pairs of classes $(C,D)$ such that $G = C^2 \cup CD$ . We then construct a representative H of each conjugacy class of core-free maximal subgroups in $A_n$ and G, and in each case, it is easy to check that there is at least one pair $(C,D) \in \mathcal {C}'(G)$ such that $C \cap H$ and $D \cap H$ are both empty (including the special case $\Omega = \{1, \ldots , n\}$ ).

Proof. Suppose $x \in G$ has $n-t$ fixed points and c nontrivial orbits on $\{1, \ldots , n\}$ . Then $t+c \leqslant 3n/2$ , and the condition $n \geqslant 16$ allows us to appeal to Theorems 2.02 and 4.02 in [Reference Brenner11], which immediately give $L = C^2$ and $G \setminus L = CD$ in both (i) and (ii).

Proof. We may assume x is nontrivial. Since the set of derangements on $\{1, \ldots , n\}$ is a normal subset of G, it suffices to show that some conjugate of x has the desired property. By considering the disjoint cycle decomposition of x, it is easy to see that there exists a conjugate z of x such that $n^{z} \ne 1$ and $i^{z} \ne i+1$ for all $i \in \{1, \ldots , n-1\}$ . Then $z(n, \ldots , 1)$ has no fixed points on $\{1, \ldots , n\}$ , and the result follows.

Proof of Theorem 3.8.

In view of Lemma 3.10, we may assume $n \geqslant 16$ . Also recall that we may assume H is maximal in $L = A_n$ or $G = S_n$ . Set $[n] = \{1, \ldots , n\}$ .

First assume H is intransitive on $[n]$ , which means that we may assume $H = S_k \times S_{n-k}$ with $1 \leqslant k < n/2$ , and we identify $\Omega $ with the set of k-element subsets of $[n]$ . If $k=1$ , then Lemma 3.12 implies that $G = \Delta (G)C$ , where C is the class of n-cycles, and the result follows since every n-cycle is a derangement. And for $k \geqslant 2$ , we can define C and D as in part (i) of Proposition 3.11 with $\ell =n-1$ . Then the elements in C and D are derangements, and the proposition gives $G = C^2 \cup CD$ , as required.

Next suppose H is transitive and imprimitive on $[n]$ . Here we may assume $H = S_a \wr S_b$ for integers a and b with $n = ab$ and $a \geqslant 2$ . If $a \geqslant 3$ , then every $(n-2)$ -cycle and every $(n-1)$ -cycle is a derangement, and Proposition 3.11 yields $G = C^2 \cup CD$ , where C and D are defined as in part (i) of the proposition, with $\ell = n-2$ . And if $a=2$ , then the same conclusion holds if we define C and D as in part (ii) of Proposition 3.11.

Finally, let us assume H acts primitively on $[n]$ . By the main theorem of [Reference Jones55], every $(n-4)$ -cycle and every $(n-3)$ -cycle is a derangement. So the result follows by defining C and D as in part (i) of Proposition 3.11 with $\ell =n-4$ .

Proof. Parts (i) and (ii) follow immediately from [Reference Guralnick and Malle47, Theorem 7.1], noting that every semisimple conjugacy class in ${}^2B_2(q)$ and ${}^2G_2(q)$ is real (for example, this follows from Lemma 4.7 in Section 4.3 below). Now assume $G = \mathrm {L}_2(q)$ with q even. Here [Reference Guralnick and Malle47, Theorem 7.1] gives $G = C^2$ , so it just remains to show that $G = \{1\} \cup AB$ , where A and B are the conjugacy classes of z and $z^2$ , respectively, and we have $|z| = |z^2| = q+1$ .

Given a nontrivial element $g \in G$ , let $N(g)$ be the number of solutions to the equation $g = ab$ with $a \in A$ and $b \in B$ and recall Lemma 3.2, which states that

$$\begin{align*}N(g) = \frac{|A||B|}{|G|}\sum_{\chi \in \mathrm{Irr}(G)}\frac{\chi(z)\chi(z^2)\overline{\chi(g)}}{\chi(1)}, \end{align*}$$

where $\mathrm {Irr}(G)$ is the set of complex irreducible characters of G. By inspecting the character table of G (see [Reference Dornhoff30, Section 38], for example), it is a straightforward exercise to show that

$$\begin{align*}N(g) = \left\{ \begin{array}{ll} q & \mbox{if } |g| = 2 \\ q-1 & \mbox{if } |g| \mbox{ divides } q-1 \\ q+1 \mbox{ or } 1 & \mbox{if } |g| \mbox{ divides } q+1, \end{array}\right. \end{align*}$$

and we conclude that $|N(g)| \geqslant 1$ for all $1 \ne g \in G$ (this can also be checked computationally, using the GAP package Chevie [Reference Geck, Hiss, Lübeck, Malle and Pfeiffer43]). Therefore, $G = \{1\} \cup AB$ , as claimed.

Proof of Theorem 3.13.

As usual, we may assume H is a maximal subgroup. For now, let us assume $G \ne \mathrm {L}_2(q)$ if q is even (we will handle this special case at the end of the proof) and define the classes $C = x^G$ and $D = y^G$ as in Table 3. In view of Proposition 3.14, it suffices to show that either x or y is a derangement.

Suppose $G = {}^2B_2(q)$ . Here $N_G(\langle x \rangle )$ is the unique maximal overgroup of x, so x is a derangement unless $H = (q+\sqrt {2q}+1).4$ . And in the latter case, y is a derangement since $|H|$ is indivisible by $|y|$ , and the result follows. An entirely similar argument handles the groups $G = {}^2G_2(q)$ with $q \geqslant 27$ .

Next assume $G = \mathrm {U}_3(q)$ with $q \geqslant 3$ , and note that $1 \not \in C^2 \cup D^2$ since neither C nor D are real classes. Suppose $q \not \in \{3,5\}$ . Since $|x| = (q^2-q+1)/(3,q+1)$ , it follows that x is a Singer cycle, and the main theorem of [Reference Bereczky5] implies that x is a derangement unless H is a field extension subgroup of type $\mathrm {GU}_1(q^3)$ . But in the latter case, y is a derangement, and once again, the result follows. Finally, if $q=3$ , then x is a derangement unless $H = \mathrm {L}_2(7)$ , in which case y is a derangement. Similarly, if $q=5$ , then x is a derangement unless H is isomorphic to $A_7$ (there are three conjugacy classes of such subgroups), and in the latter situation, it is clear that y is a derangement.

Now suppose $G = \mathrm {L}_2(q)$ with $q \geqslant 7$ odd. If $q \geqslant 11$ , then [Reference Bereczky5] implies that x is a derangement unless $H = D_{q+1}$ is a field extension subgroup of type $\mathrm {GL}_1(q^2)$ , in which case y is a derangement. If $q=9$ , then x is a derangement unless $H = A_5$ (there are two classes of such subgroups). But in the latter case, H does not contain any elements of order $|y|=4$ , so y is a derangement. Now assume $q=7$ , so $|x| = 4, \ |y|=3$ and $C,D$ are the unique conjugacy classes in G containing elements of order $4,3$ , respectively. If $H = 7{:}3$ , then x is a derangement. However, if $H = S_4$ (there are two such classes), then x and y both have fixed points. Indeed, $\Delta (G) = A \cup B$ is the union of the two conjugacy classes of elements of order $7$ . It is easy to check that

$$\begin{align*}A^2 = B^2 = \{ z \in G \,:\, |z| \ne 1,4\},\;\; AB = \{ z \in G \,:\, |z| \ne 2\}, \end{align*}$$

and thus G is not equal to $\{1\} \cup A^2, \ \{1\} \cup B^2$ nor $\{1\} \cup AB$ , but we do get $G = A^2 \cup AB = B^2 \cup AB$ as required. Moreover, if we consider the action of G on $G/K$ , where K is a proper subgroup of H, then we find that either x or y is a derangement, so we have $G = E^2$ for some conjugacy class E of derangements.

To complete the proof, we may assume $G = \mathrm {L}_2(q)$ with $q \geqslant 4$ even. Set $C = x^G$ and $A = z^G, \ B = (z^2)^G$ , where $|x| = q-1$ and $|z|=q+1$ , so Proposition 3.14(iii) implies that $G = C^2 = \{1\} \cup AB$ . By [Reference Bereczky5], z and $z^2$ are derangements unless $H = D_{2(q+1)}$ is a field extension subgroup of type $\mathrm {GL}_1(q^2)$ , and in the latter case, we see that x is a derangement. The result follows.

Proof. This follows from [Reference Larsen, Shalev and Tiep58, Theorem 2.4(i)] if $n \geqslant 33$ . By slightly tweaking some of the estimates in the proof of this result, we will show that the same argument also works for $n = 30,31$ and $32$ .

In view of Lemma 3.2, we need

$$\begin{align*}\left| \sum_{\chi \in \mathrm{Irr}(G)} \frac{\chi(x)\chi(y)\overline{\chi(g)}}{\chi(1)}\right|> 0 \end{align*}$$

for each non-identity element $g \in G$ . As explained in the proof of [Reference Larsen, Shalev and Tiep58, Theorem 2.4(i)], it suffices to show that

$$\begin{align*}\sum_{i=2}^8 \frac{|\chi_i(g)|}{\chi_i(1)} < 1, \end{align*}$$

where the $\chi _i$ are the non-principal unipotent characters of G listed in [Reference Larsen, Shalev and Tiep58, (2.3)]. By repeating the argument in the proof of [Reference Larsen, Shalev and Tiep58, Theorem 2.4(i)], setting $q=2$ and applying [Reference Larsen, Shalev and Tiep58, Lemma 2.3] and [Reference Guralnick, Larsen and Tiep46, Theorem 1.6(i)], we get

$$\begin{align*}\sum_{i=2}^4 \frac{|\chi_i(g)|}{\chi_i(1)} < \frac{2^{n-1}+4}{2^n-2} + 0.1254 + \frac{1.76}{2^{(4n-15)/n}} < 0.781 \end{align*}$$

for all $n \geqslant 30$ . As in [Reference Larsen, Shalev and Tiep58, (2.8)], the combined contribution from the four remaining unipotent characters is at most

$$\begin{align*}4\cdot 2^{(21-n)/2} < 0.177, \end{align*}$$

so

$$\begin{align*}\sum_{i=2}^8 \frac{|\chi_i(g)|}{\chi_i(1)} < 0.781 + 0.177 < 1, \end{align*}$$

and the result follows.

Proof. Here $G = \mathrm {L}_3(q)$ with $q \geqslant 3$ , and we set $d = (3,q-1)$ . By [Reference Guralnick and Malle47, Theorem 7.3], we have $G = \{1\} \cup C^2 = \{1\} \cup D^2$ , where $C = x^G$ and $D = y^G$ with $|x| = (q^2+q+1)/d$ and $|y| = (q^2-1)/d$ . Note that x is a Singer cycle.

For $q \ne 4$ , the main theorem of [Reference Bereczky5] implies that x is a derangement unless H is a field extension subgroup of type $\mathrm {GL}_1(q^3)$ . But in the latter case, y is clearly a derangement, and the result follows. Finally, if $q=4$ , then $|x| = 7, \ |y| = 5$ , and one checks that no maximal subgroup of G has order divisible by $35$ , so in each case, either x or y is a derangement.

Proof. If $q \geqslant 3$ , then [Reference Lev61, Theorem 1] implies that $G = CC^{-1}$ , and the result follows since $C = C^{-1}$ by [Reference Newman73, Theorem 1] (recall that $A \in \mathrm {GL}_2(q)$ is irreducible). Similarly, if $q=2$ , then C is the unique conjugacy class of cyclic matrices in G all of whose eigenvalues are contained in $\mathbb {F}_2$ , and once again, the result follows from [Reference Lev61, Theorem 1] (also see [Reference Lev61, p.1243]).

Proof. Suppose $x \not \in \Delta (G)^2$ . Then $x\Delta (G)^{-1} \cap \Delta (G)$ is empty, and thus $|\Delta (G)| \leqslant |G|/2$ , which is a contradiction.

Proof of Theorem 3.16.

In view of Lemma 3.19, we may assume $n \geqslant 4$ . In addition, we may assume $H \ne P_1,P_{n-1}$ is a maximal subgroup of G. The groups $\mathrm {L}_6(2)$ and $\mathrm {L}_7(2)$ can be handled using Magma, and in both cases, we check that $G = C^2$ for some conjugacy class C of derangements. So for the remainder, we may assume $(n,q) \ne (6,2), (7,2)$ .

Write $q=p^f$ , where p is a prime. Recall that if $e \geqslant 2$ is an integer, then a prime divisor r of $q^e-1$ is a primitive prime divisor if $q^i-1$ is indivisible by r for all $i<e$ . By a theorem of Zsigmondy [Reference Zsigmondy85], such a divisor exists unless $(e,q) = (6,2)$ , or if $e=2$ and q is a Mersenne prime. Define the conjugacy classes $C = x^G$ and $D = y^G$ as in Proposition 3.20 and set

$$\begin{align*}m = \min\{|x|,|y|\}. \end{align*}$$

Since $(n,q) \ne (6,2), (7,2)$ , it follows that $|x|$ and $|y|$ are divisible by primitive prime divisors of $p^{fn}-1$ and $p^{f(n-1)}-1$ , respectively, so the maximal overgroups of x and y are described in [Reference Guralnick, Penttila, Praeger and Saxl49], and we can work through the possibilities arising in [Reference Guralnick, Penttila, Praeger and Saxl49, Examples 2.1-2.9]. In doing so, it will be convenient to adopt the standard notation $\mathcal {C}_1 \cup \cdots \cup \mathcal {C}_8 \cup \mathcal {S}$ for the nine collections of maximal subgroups of G arising in Aschbacher’s subgroup structure theorem [Reference Aschbacher2], appealing to [Reference Kleidman and Liebeck57] for a detailed description of the subgroups arising in each collection. We will often refer to the type of H, which gives a rough description of its structure (our usage is consistent with [Reference Kleidman and Liebeck57, p.58]). Let V be the natural module for G.

First assume $H \in \mathcal {C}_1$ , so $H = P_k$ is a maximal parabolic subgroup with $2 \leqslant k \leqslant n-2$ , and we may identify $\Omega $ with the set of k-dimensional subspaces of V. Since x acts irreducibly on V, we have $x \in \Delta (G)$ . Similarly, y fixes a decomposition $V = U \oplus W$ , where U is $1$ -dimensional and y acts irreducibly on W. This means that U and W are the only proper nonzero subspaces of V fixed by y, so y is also a derangement on $\Omega $ , and we conclude by applying Proposition 3.20.

Next suppose $H \in \mathcal {C}_2$ . By [Reference Guralnick, Penttila, Praeger and Saxl49], either x and y are both derangements, or H is of type $\mathrm {GL}_1(q) \wr S_n$ with $q \geqslant 5$ and $(n,q) \ne (4,5)$ (see [Reference Guralnick, Penttila, Praeger and Saxl49, Example 2.3], [Reference Kleidman and Liebeck57, Table 3.5.A] and [Reference Bray, Holt and Roney-Dougal10, Table 8.8]). Since x and y have at most $2$ composition factors on V, containment in (a conjugate of) H implies that $m \leqslant n^2(q-1)/4$ (see [Reference Breuer, Guralnick and Kantor13, Remark 5.1(i)]), but it is easy to check that $m> n^2(q-1)/4$ , so x and y are derangements as required.

Now assume $H \in \mathcal {C}_5$ is a subfield subgroup of type $\mathrm {GL}_n(q_0)$ , where $q=q_0^k$ with k a prime divisor of f. We have $H \leqslant \mathrm {PGL}_n(q_0)$ , and it is easy to see that $|\mathrm { PGL}_n(q_0)|$ is not divisible by a primitive prime divisor of $p^{f(n-\alpha )}-1$ for $\alpha =0$ or $1$ , whence x and y are derangements and the result follows.

Next assume $H \in \mathcal {C}_6$ . Then by inspecting [Reference Guralnick, Penttila, Praeger and Saxl49, Example 2.5] and [Reference Kleidman and Liebeck57, Proposition 4.6.6], we deduce that y is a derangement and that x has a fixed point only if $n = 2^k$ and for some $k \geqslant 2$ . Since $n/2$ is even, we note that $|x| = (q^{n/2}+1)/(2,q-1)$ is odd. But $H \leqslant 2^{2k}.\mathrm {Sp}_{2k}(2)$ and $|\mathrm { Sp}_{2k}(2)|$ is indivisible by $|x|$ , so x must also be a derangement, and this completes the argument in this case.

By the main theorem of [Reference Guralnick, Penttila, Praeger and Saxl49], both x and y are derangements if H is one of the tensor product subgroups in $\mathcal {C}_4 \cup \mathcal {C}_7$ , so in order to complete the proof, we may assume

$$\begin{align*}H \in \mathcal{C}_3 \cup \mathcal{C}_8 \cup \mathcal{S}. \end{align*}$$

Suppose $H \in \mathcal {C}_3$ , in which case H is a field extension subgroup of type $\mathrm {GL}_{n/k}(q^k)$ for some prime divisor k of n. First assume $q \geqslant 3$ and set $E = z^G$ , where $z = (A,J_{n-2})$ and $A \in \mathrm {SL}_2(q)$ has order $q+1$ . Then $G = E^2$ by Proposition 3.21, and we note that $z^{q+1}$ has Jordan form $(J_{n-2},J_1^2)$ . But there are no such elements in H, so $z^{q+1}$ , and hence z itself, is a derangement and the result follows. Similarly, if $q=2$ , then $G = E^2$ for the class E of regular unipotent elements (see Proposition 3.21), and once again, the result follows since $E \cap H$ is empty.

Next assume $H \in \mathcal {C}_8$ . There are several cases to work through:

1. (a) H is of type $\mathrm {U}_n(q_0)$ with $q = q_0^2$ ;

2. (b) H is of type $\mathrm {O}_n^{\varepsilon }(q)$ with q odd;

3. (c) H is of type $\mathrm {Sp}_n(q)$ with n even and $q \geqslant 2$ .

First consider case (a), where we have $H \leqslant \mathrm {PGU}_n(q_0)$ . We claim that x and y are derangements, where x and y are defined as in Proposition 3.20. To see this, first observe that every semisimple element in H has order at most $\alpha = q^{(n-1)/2} +(-1)^n$ by [Reference Guest, Morris, Praeger and Spiga44, Lemma 2.15]. If n is odd, then $m> \alpha $ , and so the claim holds. Now assume n is even, in which case $|y|>\alpha $ and thus y is a derangement. And since $|x|$ is divisible by a primitive prime divisor of $p^{fn}-1$ , but $|H|$ is not, we conclude that x is also a derangement and the claim follows.

Finally, let us consider cases (b) and (c). Suppose $q=2$ and $H = \mathrm {Sp}_n(2)$ , so $n \geqslant 8$ is even. If $n \geqslant 30$ , then Lemma 3.18 gives $G = \{1\} \cup CD$ , where C and D are both conjugacy classes of elements of order at least $2^{n-3}-1$ (see (3.3)). And since $|z| \leqslant 2^{n/2+1}$ for all $z \in H$ (see [Reference Guest, Morris, Praeger and Spiga44, Theorem 2.16]), we deduce that C and D comprise derangements and the result follows. For $8 \leqslant n \leqslant 28$ , we can use Magma to determine all the conjugacy classes in G and H, which allows us to show that

$$\begin{align*}\delta(G) \geqslant \frac{|\{ x \in G \,:\, |x| \not\in \omega(H)\}|}{|G|}> \frac{1}{2}, \end{align*}$$

where $\omega (H) = \{ |x| \,:\, x \in H\}$ is the spectrum of H. Therefore, $G = \Delta (G)^2$ by Lemma 3.22.

Now assume $q \geqslant 3$ and write $\mathbb {F}_q^{\times } = \langle \lambda \rangle $ . Consider the conjugacy class $E = z^G$ , where $z \in G$ is defined as follows:

$$\begin{align*}z = \left\{\begin{array}{ll} (A,J_1(\lambda^{-1}),J_{n-3}(1)) & \mbox{if } H \mbox{ is of type } \mathrm{O}_n(q) \mbox{ or } \mathrm{Sp}_n(q) \\ (B,J_{n-2}(1)) & \mbox{if } H \mbox{ is of type } \mathrm{O}_n^{\pm}(q). \end{array}\right. \end{align*}$$

Here $A,B \in \mathrm {GL}_2(q)$ are irreducible, with $\det (A) = \lambda $ and $\det (B)=1$ . By Proposition 3.21, we have $G = E^2$ , and we claim that z is a derangement.

To see this, first assume H is of type $\mathrm {O}_n(q)$ , so n is odd. Now $z^{q^2-1}$ has Jordan form $(J_{n-3},J_1^3)$ , but there are no such unipotent elements in H since all even size unipotent Jordan blocks must occur with an even multiplicity (see [Reference Liebeck and Seitz68, Theorem 3.1(ii)], for example). Therefore, z is a derangement, and the result follows. Similarly, if H is of type $\mathrm { Sp}_n(q)$ with $q \geqslant 3$ , then z is a derangement since odd size Jordan blocks have even multiplicity in the Jordan form of any unipotent element of H (see [Reference Liebeck and Seitz68, Theorem 3.1(ii)] and [Reference Liebeck and Seitz68, Lemma 6.2]). Finally, if n is even and H is of type $\mathrm { O}_{n}^{\varepsilon }(q)$ , then z is a derangement because $z^{q^2-1}$ has Jordan form $(J_{n-2},J_1^2)$ , which is not compatible with the form of any unipotent element in H since the even-size Jordan block $J_{n-2}$ has multiplicity $1$ .

To complete the proof, we may assume $H \in \mathcal {S}$ , so H is almost simple with socle $H_0$ . We claim that the elements x and y defined in Proposition 3.20 are derangements. Recall that $m = \min \{|x|,|y|\}$ .

For the groups with $4 \leqslant n \leqslant 12$ , we can read off the possibilities for H by inspecting the relevant tables in [Reference Bray, Holt and Roney-Dougal10, Chapter 8], and it is straightforward to show that H does not contain any elements of order $|x|$ or $|y|$ . For example, if $n=6$ , then $m = (q^3+1)/(2,q-1)$ , and by inspecting [Reference Bray, Holt and Roney-Dougal10, Table 8.25], we see that either

$$\begin{align*}H_0 \in \{ \mathrm{M}_{12}, A_7, \mathrm{L}_2(11), \mathrm{L}_3(4), \mathrm{U}_4(3) \}, \end{align*}$$

or $H_0 = \mathrm {L}_3(q)$ with q odd. In the latter case, [Reference Guest, Morris, Praeger and Spiga44, Theorem 2.16] gives $|z| \leqslant q^2+q+1 < m$ for all $z \in \mathrm {Aut}(H_0)$ , so x and y are derangements. Similarly, in the remaining cases, one can check that $|z| \leqslant 28$ for all $z \in \mathrm {Aut}(H_0)$ , which reduces the problem to $q \in \{2,3\}$ . Closer inspection of [Reference Bray, Holt and Roney-Dougal10, Table 8.25] shows that $(q,H) = (3,\mathrm {M}_{12})$ is the only possibility, and the result follows since $|z| \leqslant 11< 14 =m$ for all $z \in H$ .

So for the remainder, we may assume $n \geqslant 13$ . We now apply the main theorem of [Reference Guralnick, Penttila, Praeger and Saxl49], using the fact that $|x|$ and $|y|$ are divisible by primitive prime divisors of $q^n-1$ and $q^{n-1}-1$ , respectively. We consider [Reference Guralnick, Penttila, Praeger and Saxl49, Examples 2.6-2.9] in turn.

First observe that Ex. 2.6(b,c) and Ex. 2.8 do not arise since $n \geqslant 13$ . In Ex. 2.6(a), $H_0 = A_d$ is an alternating group and V is the fully deleted permutation module over $\mathbb {F}_p$ . However, this representation embeds $A_d$ in a symplectic or orthogonal group, so it does not arise. The relevant sporadic groups occurring in Ex. 2.7, as well as the cases in [Reference Guralnick, Penttila, Praeger and Saxl49, Table 7], can all be eliminated by considering element orders, so it just remains to handle the cases recorded in [Reference Guralnick, Penttila, Praeger and Saxl49, Table 8]. Here $H_0$ is a simple classical group over $\mathbb {F}_t$ , where $(t,q) = 1$ , and these possibilities can also be eliminated by considering element orders. For example, if $H_0 = \mathrm {PSp}_d(t)$ with t odd, then $n \geqslant (t^{d/2}-1)/2$ and [Reference Guest, Morris, Praeger and Spiga44, Theorem 2.16] gives $|z| \leqslant (t^{d/2}+1)/(t-1)$ for all $z \in H$ . But we have $m \geqslant 2^{(n-1)/2}+1$ for all q, so $|z| < m$ , and we conclude that x and y are derangements. Similarly, if $H_0 = \mathrm {L}_2(t)$ and $n = (t-1)/2$ , then $t \geqslant 27$ since $n \geqslant 13$ , and we have $|z| \leqslant t+1$ for all $z \in H$ . The result now follows since $m \geqslant 2^{(n-1)/2}+1> t+1$ .

Proof of Theorem F.

We consider the various possibilities for G in turn. If G is almost simple, then N is simple and transitive, so $N = \Delta (N)^2$ since we are assuming Conjecture 3.1 holds. If G is an affine group, or a twisted wreath product, then N is regular, and we deduce that $\delta (N) = 1-|N|^{-1}$ . So if $|N|>2$ , then $\delta (N)> 1/2$ , and thus $N = \Delta (N)^2$ by Lemma 3.22. And if $|N| = 2$ , then $G = N = S_2, \ \Omega = \{1,2\}, \ \Delta (N) = \{(1,2)\}$ and $\Delta (N)^2 = \{1\} \ne N$ , which explains why we assume $|\Omega | \geqslant 3$ in Theorem F.

If G is a diagonal type group, then our proof of Theorem C(i) at the end of Section 2 shows that $\delta (N) \geqslant 2/3$ , and thus $N = \Delta (N)^2$ by Lemma 3.22. Finally, let us assume $G \leqslant L \wr S_b$ is a product type group, where $b \geqslant 2$ and $L \leqslant \mathrm { Sym}(\Gamma )$ is a primitive almost simple or diagonal type group with socle $S = T^a$ . Then $N = S^b = T^{ab}$ and G acts on $\Omega = \Gamma ^b$ with the product action. Now

$$\begin{align*}\Delta(N) = \{ (x_1, \ldots, x_b) \in S^b \,:\, \mbox{at least one } x_i \mbox{ is a derangement on } \Gamma \}, \end{align*}$$

and thus $\Delta (N)$ contains $\Delta (S) \times S^{b-1}$ . The result now follows since we have already shown that $\Delta (S)^2 = S$ .

Proof. The first claim $G = \Delta (G)^2$ follows from Proposition 3.4. More precisely, either $G = C^2$ for some conjugacy class C of derangements, or $(G,H) = (\mathrm {M}_{23}, 2^4.(15{:}4))$ and

$$\begin{align*}G = \{1\} \cup C^2 = \{1\} \cup D^2 = CD, \end{align*}$$

where C and D are the two conjugacy classes in G of elements of order $23$ .

Now let us turn to the lower bound on $\delta (G)$ . It will be convenient to define

(4.1) $$ \begin{align} \alpha_{\mathrm{s}}(G) = \min\{ \delta(G,H) \,:\, H < G \mbox{ is soluble}\}. \end{align} $$

Since $\alpha _s(G) \geqslant \alpha (G)$ , we only need to consider the groups with $\alpha (G) \leqslant 21/55$ . By inspecting Table 1, it follows that $G \in \mathcal {A} \cup \mathcal {B}$ , where

$$ \begin{align*} \mathcal{A} & = \{ \mathrm{M}_{11}, \mathrm{M}_{12}, \mathrm{M}_{22}, \mathrm{M}_{23}, \mathrm{M}_{24}, \mathrm{ J}_3, \mathrm{McL}, \mathrm{HS}, \mathrm{Co}_3, \mathrm{O'N} \} \\ \mathcal{B} & = \{ \mathrm{Ly}, \mathrm{Th}\}. \end{align*} $$

For the groups in $\mathcal {A}$ , we can use Magma to construct G as a primitive permutation group of minimal degree, together with a set of representatives of the conjugacy classes of elements and maximal subgroups of G. For each maximal subgroup H, it is then straightforward to determine the set of derangements on $G/H$ , and we can read off $\delta (G,H)$ . In particular, if H is soluble, then we find that $\delta (G,H) \geqslant 21/55$ , with equality if and only if $G = \mathrm {M}_{11}$ and $H = \mathrm {U}_3(2).2$ or $2.S_4$ . And if H is insoluble and $\delta (G,H) \leqslant 21/55$ , then we construct a set of representatives of the conjugacy classes of maximal subgroups K of H, and in every case, one can check that $\delta (G,K)> 21/55$ .

Finally, let us assume $G = \mathrm {Ly}$ or $\mathrm {Th}$ , and let H be a maximal subgroup of G. First we use GAP (as in the proof of Proposition 2.7) to show that $\delta (G,H) \leqslant 21/55$ if and only if $(G,H) = (\mathrm {Ly}, 2.A_{11})$ or $(\mathrm {Th}, 2^{1+8}.A_9)$ .

Suppose $G = \mathrm {Ly}$ . Here we use the Magma function MatrixGroup to obtain G as a subgroup of $\mathrm {GL}_{111}(5)$ , and we construct $H = 2.A_{11}$ via the function MaximalSubgroups. We then construct a set of representatives of the conjugacy classes of maximal subgroups K of H, and in each case, we check that

$$\begin{align*}\delta(G,K) \geqslant \frac{|\{ x \in G \,:\, |x| \not\in \omega(K)\}|}{|G|}> \frac{21}{55}, \end{align*}$$

where $\omega (K) = \{ |x| \,:\, x \in K\}$ is the spectrum of K. The case $G = \mathrm {Th}$ is entirely similar, working with a $248$ -dimensional matrix representation of G over $\mathbb {F}_2$ .

Proof. Part (i) is a special case of Proposition 3.5. More precisely, either $G = C^2$ for some conjugacy class C of derangements, or

$$\begin{align*}(G,H) \in \{ (A_5,A_4), \, (A_5,S_3), \, (A_8, 2^4{:}(S_3\times S_3)) \} \end{align*}$$

and $G = \{1\} \cup CD$ , where C and D are the two classes of elements order $5$ (for $G = A_5$ ) or $7$ (for $G = A_8$ ).

Part (ii) can be checked using Magma and the approach described in Remark 4.4. In particular, we can compute $\alpha _{\mathrm {s}}(G)$ precisely for all $5 \leqslant n \leqslant 16$ (see (4.1) and Table 4).

For the remainder, we may assume $n \geqslant 17$ and $H \in \mathcal {S}^{+}$ . Then the solubility of H implies that H acts primitively on $\{1, \ldots , n\}$ , which in turn means that $n = p^d$ is a prime power and $H = \mathrm {AGL}_d(p) \cap G$ . Moreover, since $n \geqslant 17$ , it follows that $d=1$ and $H = \mathrm {AGL}_1(p) \cap G = p{:}(p-1)/2$ .

Suppose $x \in H$ has order r. Then either $r=p$ and x is a p-cycle, or r divides $(p-1)/2$ and x has cycle-type $(r^{(p-1)/r},1)$ as an element of G. (Note that H contains a Sylow p-subgroup of G, so every p-cycle in G has fixed points on $\Omega $ .) So if N denotes the number of elements in G with fixed points on $\Omega $ , then

$$\begin{align*}N = 1 + (p-1)! + \sum_{r \in \Lambda} \frac{p!}{((p-1)/r)!r^{(p-1)/r}}, \end{align*}$$

where $\Lambda $ is the set of divisors $r \geqslant 2$ of $(p-1)/2$ . For $17 \leqslant p<100$ , it is now entirely straightforward to check that the lower bound in part (iii) is satisfied (with equality if $p=17$ ), so we may assume $p>100$ .

For $r \in \Lambda $ , set

$$\begin{align*}f(r) = \frac{p!}{((p-1)/r)!r^{(p-1)/r}}. \end{align*}$$

We claim that $f(r)$ is maximal when $r = (p-1)/2$ . To see this, simply observe that $((p-1)/r)! \geqslant 2$ and $r^{(p-1)/r}$ is decreasing as a function of r, so we have $r^{(p-1)/r} \geqslant ((p-1)/2)^2$ , and the claim follows. Therefore,

$$\begin{align*}f(r) \leqslant 2\left(\frac{p!}{(p-1)^2}\right), \end{align*}$$

and using the crude bound $|\Lambda | \leqslant 2\sqrt {(p-1)/2}$ , we deduce that

$$\begin{align*}N \leqslant 1+(p-1)! + 2\sqrt{(p-1)/2} \cdot 2\left(\frac{p!}{(p-1)^2}\right). \end{align*}$$

In turn, this yields

$$\begin{align*}\delta(G) \geqslant \frac{2\left(\frac{p!}{2} - 1 - (p-1)! - 2\sqrt{(p-1)/2} \cdot \frac{2(p!)}{(p-1)^2}\right)}{p!} = 1-\frac{2}{p!} - \frac{2}{p} - \frac{8\sqrt{(p-1)/2}}{(p-1)^2}, \end{align*}$$

and we conclude that $\delta (G)> 0.97$ for all $p> 100$ .

Proof. The groups ${}^2G_2(3)'$ and ${}^2F_4(2)'$ can be checked directly. In each of the remaining cases, the result follows from [Reference Tiep and Zalesski80, Proposition 3.1] since the Weyl group of G contains a central involution.