Proof. Assume that $N=S^u,$ with $u\in \mathbb N$ . We may identify G with a subgroup of the wreath product $\operatorname {\mathrm {Aut}} S\wr \operatorname {\mathrm {Sym}}(u).$ In this identification, any element of G can be written in the form $(\alpha _1,\ldots ,\alpha _u)\rho ,$ with $\alpha _1,\ldots ,\alpha _u \in \operatorname {\mathrm {Aut}} S$ and $\rho \in \operatorname {\mathrm {Sym}}(u)$ . Moreover, we identify S with the inner automorphism group $\operatorname {\mathrm {Inn}}(S)$ and, therefore, we can write $N=\{(s_1,\ldots ,s_u)\mid s_1,\ldots ,s_u \in S\}.$

Now, consider $g_1,g_2\in G$ given by

$$ \begin{align*}g_1=(a_1,\ldots,a_u)\sigma,\quad g_2=(b_1,\ldots,b_u)\tau.\end{align*} $$

Write

$$ \begin{align*}\sigma=\sigma_1\cdots \sigma_c, \quad \tau=\tau_1\cdots \tau_d,\end{align*} $$

as products of disjoint cyclic permutations, and assume that 1 belongs to the supports of $\sigma _1$ and $\tau _1.$ Suppose

$$ \begin{align*}\sigma_1=(1,i_2,\ldots,i_e), \quad \tau_1=(1,j_2,\ldots,j_f).\end{align*} $$

Let

$$ \begin{align*}X=\{(\alpha_1,\ldots,\alpha_u)\rho \in G \mid \rho(1)=1\}.\end{align*} $$

The map $\phi \colon X\to \operatorname {\mathrm {Aut}} S$ sending $(\alpha _1,\ldots ,\alpha _u)\rho $ to $\alpha _1$ is a group homomorphism. Now, let $n_1=(s_1,\ldots ,s_u), n_2=(t_1,\ldots ,t_u) \in N$ . Then, $(n_1g_1)^e, (n_2g_2)^f \in X$ and

$$ \begin{align*}\phi((n_1g_1)^e)=s_1a_1s_{i_2}a_{i_2}\cdots s_{i_e}a_{i_e},\quad \phi((n_2g_2)^f)=t_1b_1t_{j_2}b_{j_2}\cdots t_{j_f}b_{j_f}. \end{align*} $$

Let

$$ \begin{align*}\gamma_1=a_{1}s_{i_2}a_{i_2}\cdots s_{i_e}a_{i_e},\quad \gamma_2=b_{1}t_{j_2}b_{j_2}\cdots t_{j_f}b_{j_f}.\end{align*} $$

If $(s_1,t_1)\notin \mathcal N_{\gamma _1,\gamma _2}(\operatorname {\mathrm {Aut}} S,S),$ then

$$ \begin{align*}\phi(\langle (n_1g_1)^{e}, (n_2g_2)^f\rangle)= \langle s_1\gamma_1,t_1\gamma_2\rangle\end{align*} $$

is not nilpotent. In particular, $\langle n_1g_1, n_2g_2\rangle $ is not nilpotent. Thus, given $s_i, t_j$ for $i\neq 1$ and $j\neq 1$ , there are at least

$$ \begin{align*}|S|^2(1-\nu_{\gamma_1,\gamma_2}(\operatorname{\mathrm{Aut}} S, S))\geq |S|^2(1-\tilde\nu(S))\end{align*} $$

choices for $s_1$ and $t_1$ such that $\langle n_1g_1, n_2g_2\rangle $ is not nilpotent. It follows that there are at least $|S|^{2u}(1-\tilde \nu (S))$ pairs $(n_1,n_2)\in N^2$ such that $\langle n_1g_1,n_2g_2\rangle $ is not nilpotent. This is equivalent to saying that $\nu _{g_1,g_2}(G,N)\leq \tilde \nu (S)$ for every $(g_1,g_2) \in G^2.$ Hence

$$ \begin{align*} \nu(G)|G|^2 & =\sum_{(Ng_1,Ng_2)\in (G/N)^2}\nu_{g_1,g_2}(G,N)\\&\leq \sum_{(Ng_1,Ng_2)\in (G/N)^2}|N|^2\tilde\nu(S)\\&\leq |G|^2\tilde\nu(S) \end{align*} $$

and, consequently, $\nu (G)\leq \tilde \nu (S).$

Proof. Given $a_1,a_2 \in \operatorname {\mathrm {Aut}} S$ , let $\mathcal P_{a_1,a_2}(S)=\{(s_1,s_2)\in S^2\mid S\leq \langle s_1a_1,s_2a_2\rangle \}$ . We define

$$ \begin{align*}\pi_{a_1,a_2}(S):=\frac{|\mathcal P_{a_1,a_2}(S)|}{|S|^2}, \quad \tilde \pi(S)=\min_{a_1,a_2\in \operatorname{\mathrm{Aut}} S}\pi_{a_1,a_2}(S).\end{align*} $$

Clearly, if $(s_1,s_2) \in \mathcal P_{a_1,a_2}(S),$ then $\langle s_1a_1, s_2a_2\rangle $ is not nilpotent and, therefore, $(s_1,s_2) \notin \mathcal N_{a_1,a_2}(\operatorname {\mathrm {Aut}} S,S).$ Thus,

$$ \begin{align*}\pi_{a_1,a_2}(S)\leq 1-\nu_{a_1,a_2}(\operatorname{\mathrm{Aut}}(S),S)\end{align*} $$

for every $a_1, a_2\in \operatorname {\mathrm {Aut}} S$ and, therefore,

$$ \begin{align*}\tilde \nu(S)\leq 1-\tilde\pi(S).\end{align*} $$

Results of Dixon [Reference Dixon1], Kantor and Lubotzky [Reference Kantor and Lubotzky5], and Liebeck and Shalev [Reference Liebeck and Shalev6] establish that $\tilde \pi (S) \to 1$ as $|S| \to \infty .$ Explicit lower bounds for $\tilde \pi (S)$ are given in [Reference Menezes, Quick and Roney-Dougal7], where, in particular, all nonabelian simple groups S with $\tilde \pi (S)\leq 9/10$ are listed. Since $1-9/10=1/10>1/12$ , this is not enough for our purposes, but more detailed information can be found in [Reference Menezes8]. In particular, combining [Reference Menezes8, Table 5.1, Theorem 6.01 and Lemma 7.1.1], it turns out that if S is not an alternating group, then either $\tilde {\pi }(S)> {11}/{12}$ (and, consequently, $\tilde {\nu }(S) < {1}/{12})$ or S belongs to the family $\mathcal S$ of the following groups:

$$ \begin{align*}&\text{PSL}(2,7), \text{PSL}(2,8), \text{PSL}(2,11), \text{PSL}(2,13), \text{PSL}(3,3),\\& \text{PSL}(3,4), \text{PSp}(6,2), \text{PSU}(4,2), \mathrm{M}_{11},\text{M}_{12}.\end{align*} $$

The GAP Library of Tables of Marks [9] contains the tables of marks of all almost simple groups whose socle belongs to $\mathcal S$ and, using them, the exact value of $\tilde {\nu }(S)$ , reported in Table 1, can be easily obtained (see Section 3 for more details).

Table 1

$\tilde \nu (S)$ for $S\in \mathcal S$ .

Table 1 Long description

The table has two columns. The left column is labeled bold S and lists group names: P S L open parenthesis 2 comma 7 close parenthesis, P S L open parenthesis 2 comma 8 close parenthesis, P S L open parenthesis 2 comma 11 close parenthesis, P S L open parenthesis 2 comma 13 close parenthesis, P S L open parenthesis 3 comma 3 close parenthesis, P S L open parenthesis 3 comma 4 close parenthesis, P S U open parenthesis 4 comma 2 close parenthesis, P S p open parenthesis 6 comma 2 close parenthesis, M sub 11, and M sub 12. The right column is labeled nu tilde of bold S and lists the corresponding values: 3 all over 56 is approximately 0.0536, 1 all over 56 is approximately 0.0179, 2 all over 165 is approximately 0.0121, 3 all over 364 is approximately 0.0082, 1 all over 234 is approximately 0.0043, 5 all over 4032 is approximately 0.0032, 67 all over 23760 is approximately 0.0026, 1 all over 4536 is approximately 0.0002, 1 all over 440 is approximately 0.0023, and 7 all over 11880 is approximately 0.0006. Each value is aligned with its group name in the same row.

Now, assume that $S=\mathrm {Alt}(n)$ . Using GAP and, in particular, the library of tables of marks, it can be checked that

$$ \begin{align*} \tilde\nu(\mathrm{Alt}(5))=\dfrac{1}{12}, \quad \tilde\nu(\mathrm{Alt}(6))=\dfrac{1}{36}, \quad \tilde\nu(\mathrm{Alt}(7))=\dfrac{1}{210},\quad \tilde\nu(\mathrm{Alt}(8))=\dfrac{19}{9720}, \quad \tilde\nu(\mathrm{Alt}(9))=\dfrac{1}{2160}. \end{align*} $$

Now, suppose $n\geq 10$ and, for $1\leq i\leq n,$ let

$$ \begin{align*}X_i= \{s\in \operatorname{\mathrm{Sym}}(n)\mid s(i)=i\}\cong \operatorname{\mathrm{Sym}}(n-1),\quad Y_i=X_i\cap \mathrm{Alt}(n)\cong \mathrm{Alt}(n-1).\end{align*} $$

Consider $a_1,a_2\in \operatorname {\mathrm {Sym}}(n)$ and let $Y_{a_1,a_2}=\langle a_1,a_2,\mathrm {Alt}(n)\rangle .$ Moreover, let

$$ \begin{align*}\Omega_0=\{(s_1,s_2)\in (\mathrm{Alt}(n))^2\mid \langle s_1a_1,s_2a_2\rangle=Y_{a_1,a_2}\}\end{align*} $$

and, for $1\leq i\leq n$ , let

$$ \begin{align*}\Omega_i=\{(s_1,s_2)\in (\mathrm{Alt}(n))^2\mid \langle s_1a_1,s_2a_2\rangle = Y_{a_1,a_2}\cap X_i\}.\end{align*} $$

Notice that $\Omega _0, \Omega _1,\ldots ,\Omega _n$ are disjoint subsets of $(\mathrm {Alt}(n))^2$ and

$$ \begin{align*}\Omega_0\cup \Omega_1\cup \cdots \cup \Omega_n\subseteq (\mathrm{Alt}(n))^2\setminus \mathcal N_{a_1,a_2}(\operatorname{\mathrm{Sym}}(n),\mathrm{Alt}(n)).\end{align*} $$

It follows that

$$ \begin{align*}\nu_{a_1,a_2}(\operatorname{\mathrm{Sym}}(n),\mathrm{Alt}(n))\leq 1-\sum_{0\leq i\leq n}\frac{|\Omega_i|}{|\mathrm{Alt}(n)|^2}. \end{align*} $$

For $1\leq i\leq n,$ there exist $s_1, s_2\in \mathrm {Alt}(n)$ such that $\langle s_1a_1, s_2a_2\rangle \leq X_i$ and

$$ \begin{align*}|\Omega_i|=|Y_i|^2\pi_{s_1a_1,s_2a_2}(X_i,Y_i)\geq |\mathrm{Alt}(n-1)|^2\tilde\pi(\mathrm{Alt}(n-1)).\end{align*} $$

Moreover,

$$ \begin{align*}|\Omega_0|\geq |\mathrm{Alt}(n)|^2\tilde\pi(\mathrm{Alt}(n)).\end{align*} $$

Hence,

$$ \begin{align*}\begin{aligned}\nu_{a_1,a_2}(\operatorname{\mathrm{Sym}}(n),\mathrm{Alt}(n))&\leq 1-\tilde\pi(\mathrm{Alt}(n))-n\bigg(\frac{|\mathrm{Alt}(n-1)|^2\tilde\pi(\mathrm{Alt}(n-1))}{|\mathrm{Alt}(n)|^2}\bigg)\\&=1-\tilde \pi(\mathrm{Alt}(n))-\frac{\tilde\pi(\mathrm{Alt}(n-1))}{n}.\end{aligned}\end{align*} $$

It follows from [Reference Menezes, Quick and Roney-Dougal7] that if $m\geq 9$ , then $\tilde \pi (\mathrm {Alt}(m))\geq \tilde \pi (\mathrm {Alt}(9))= {15403}/{18144}$ and, therefore, for $n\geq 10,$

$$ \begin{align*} \tilde\nu(\mathrm{Alt}(n))&\leq 1-\tilde \pi(\mathrm{Alt}(n))-\frac{\tilde\pi(\mathrm{Alt}(n-1))}{n}\\&\leq 1-\frac{15403}{18144}\bigg(1+\frac{1}{10}\bigg)=\frac{12007}{181440}\approx 0.0622. \\[-42pt]\end{align*} $$

Proof of Theorem 1.1

Assume that G is a nonsolvable finite group. Then, G admits a nonabelian chief factor $X/Y$ . In particular, $G/C_G(X/Y)$ has a unique minimal normal subgroup and this minimal normal subgroup is isomorphic to $X/Y.$ It can be easily proved that $\nu (G)\leq \nu (G/N)$ for every normal subgroup N of G (see, for example, [Reference Jafarian Amiri, Madadiand and Rostami4, Lemma 2.1]). This implies that $\nu (G)\leq \nu (G/C_G(X/Y)).$ Moreover, by Proposition 2.3, $\nu (G/C_G(X/Y))\leq \tilde \nu (S),$ where S is a composition factor of $X/Y.$ Since, by Proposition 2.4, $\tilde \nu (S)\leq 1/12$ , we conclude

$$ \begin{align*} \nu(G)\leq \nu(G/C_G(X/Y))\leq \tilde \nu(S)\leq \frac{1}{12}.\\[-42pt] \end{align*} $$

Proof. Denote by $\mathcal H$ the set of 2-generated subgroups H of G that are nilpotent and such that $G=HN$ . Assume that $(g_1,g_2)\in \Omega $ and, for every $H\in \mathcal H$ , let $\Delta (g_1,g_2,H)$ be the set of $(n_1,n_2)$ such that $H=\langle n_1g_1,n_2g_2\rangle .$ Then, $\mathcal N_{g_1,g_2}(G,N)$ is the disjoint union of the subsets $\Delta (g_1,g_2,H)$ , with H ranging over $\mathcal H.$

Fix $H\in \mathcal H$ . Since $HN=G,$ there exist $n_1,n_2\in N$ such that $g_1=n_1h_1$ and $g_2=n_2h_2$ , and therefore, $(n_1^{-1},n_2^{-1})\in \Delta (g_1,g_2,H)$ . Moreover,

$$ \begin{align*}\Delta(g_1,g_2,H)=\{(m_1,m_2)\in (N\cap H)^2\mid \langle (m_1n_1)(n_1^{-1}g_1),(m_2n_2)(n_2^{-1}g_2)\rangle=H\}.\end{align*} $$

It follows from the main result in [Reference Gaschütz2] that $|\Delta (g_1,g_2,H)|/|N\cap H|^2$ does not depend on the choice of $(g_1,g_2)$ and coincides with the conditional probability that two randomly chosen elements of H generate H given that they generate H modulo $H\cap N.$ Consequently, the cardinality of the sets $\Delta (g_1,g_2,H)$ , and hence also that of their disjoint union, does not depend on the choice of $(g_1, g_2)$ .