Proof. Assume that H does not lie in a maximal subgroup of type (I*), (II), (III) or (IV*). Let $q=p^f$ , where p is prime, and let $\mathbb {F}_q^n$ be the natural module for G. Since (I*), (II) and (III) cover the entire geometric class $\mathcal {C}$ of maximal subgroups of G, we know that H is not a subgroup of any maximal subgroup M contained in the geometric class $\mathcal {C}$ . Therefore, by the main theorem of [Reference Aschbacher1], we deduce that H is contained in $\mathcal {S}$ . Since H is not in (IV*), H must be a group of Lie type defined over $\mathbb {F}_{p^e}$ for some e. By [Reference Kleidman and Liebeck18, Proposition 5.4.6] (which is an application of Steinberg’s twisted tensor product theorem), f divides $de$ , for some $d \in \{1,2,3\}$ , and the embedding of H in G affords an irreducible $\mathbb {F}_{p^e}H$ -module V of dimension $n^{f/de}$ . Since H is not contained in any maximal twisted tensor product subgroup, by [Reference Seitz28, Corollary 6], we must have $f=de$ , so H is defined over $\mathbb {F}_{p^{f/d}}$ and $\dim {V} = n$ .

It remains to prove that $\mathrm {rank}(H) < \mathrm {rank}(G)$ . Referring to the bounds on the dimension of the minimal module in [Reference Kleidman and Liebeck18, Proposition 5.4.13], since H has an irreducible module of dimension n either $\mathrm {rank}(H) < \mathrm {rank}(G)$ or $(H,V)$ is one of a small number of possibilities all of which arise in the geometric class $\mathcal {C}$ or case (II). For instance, if $G = \mathrm {PSp}_n(q)$ with $n> 8$ , then n is strictly smaller than the dimension of the minimal module of any finite simple group of Lie type of $\mathrm {rank}(G) = n/2$ except $\mathrm {PSL}^\pm _{n/2+1}(q)$ or $\mathrm {P}\Omega ^\pm _n(q)$ , and, by [Reference Kleidman and Liebeck18, Proposition 5.4.11], the former groups have no irreducible modules of dimension n and latter groups embed in G as $\mathcal {C}_8$ subgroups if at all. However, H is not in $\mathcal {C}$ or (II), so $\mathrm {rank}(H) < \mathrm {rank}(G)$ .

Proof. First, assume that $\mathrm {soc}(M)$ is sporadic. Then $\ell (\mathrm {soc}(M))$ is given in [Reference Cameron, Solomon and Turull6, Tables III & IV] (the ‘probable values’ have since been verified), whence we deduce that $\ell (M) \leqslant 52$ .

Next, assume that $\mathrm {soc}(M) = A_d$ for some $d \geqslant 5$ . If G is exceptional, then $d \leqslant 18$ by [Reference Liebeck and Seitz22, Theorem 8], so $\ell (M) \leqslant \ell (S_d) + 1 \leqslant 28$ by equation (2.1), and if G is classical in dimension n, then [Reference Kleidman and Liebeck18, Proposition 5.3.7] implies that $d \leqslant 2n-1$ , so $\ell (M) \leqslant 3n-1 \leqslant 6r+2$ , again by equation (2.1).

Finally, assume that $\mathrm {soc}(M)$ is a group of Lie type of rank $r_0$ over $\mathbb {F}_{q_0}$ , where $q_0$ is a power of a prime $p_0$ . If G is classical in dimension n, then $p_0 \neq p$ and [Reference Kleidman and Liebeck18, Theorem 5.3.9] implies that $q_0^{r_0} \leqslant n^2$ , so using the facts that $|M| \leqslant q_0^{12r_0^2}$ and $n \leqslant 4r$ , we have

$$\begin{align*}\ell(M) \leqslant \log_2|M| \leqslant \log_2 (q_0^{12r_0^2}) \leqslant 12 (r_0 \log_2 q_0)^2 \leqslant 48 (\log_2 n)^2 \leqslant 48 (\log_2 4r)^2 \leqslant 192r. \end{align*}$$

If G is exceptional, then there are only finitely many possibilities for M. Consulting [Reference Liebeck and Seitz21, Tables 10.3 and 10.4] for the case $p_0 \neq p$ and [Reference Liebeck and Seitz22, Theorem 8] for the case $p_0 = p$ , we see that in all cases $|M| < 2^{200}$ , so $\ell (M) \leqslant 200 \leqslant 100r$ .

Proof. Write $\mathcal {M}(X)$ for the set of maximal subgroups of X and $\mathrm {Hom}(X,Y)$ for the set of homomorphisms from X to Y. Goursat’s lemma (see [Reference Scott31, (4.3.1)], for example) implies that

(2.2) $$ \begin{align} |\mathcal{M}(G)| = |\mathcal{M}(H)|+|\mathcal{M}(C_n)|+\sum_{\text{prime } p \mid n}(|\mathrm{Hom}(H,C_p)|-1). \end{align} $$

Note that $|\mathcal {M}(C_n)| = \omega (n)$ and $|\mathrm {Hom}(H,C_p)|=1$ unless $p \in \{2,3\}$ . It remains to check that $|\mathcal {M}(H)|+|\mathrm {Hom}(H,C_2)|+|\mathrm {Hom}(H,C_3)| \leqslant 11$ , which is easy to do for each $H \leqslant S_4$ .

Proof. For part (i), write $G = \langle a, b \mid a^m, b^n, a^ba^{-k} \rangle $ and $H = \langle b\rangle \cong C_n$ , and for part (ii) write $G = \langle a, b,c \mid a^m, b^n, c^2, a^ba^{-k}, a^ca, [b,c] \rangle $ and $H = \langle b, c\rangle \cong C_n \times C_2$ (in both cases, we assume that $\mathrm {gcd}(k,m)=1$ and $m \mid (k^n-1)$ ).

Let M be a maximal subgroup of G, and write $A = M \cap \langle a\rangle $ . The possibilities for M correspond to the maximal subgroups of $G/A$ . If $A = \langle a\rangle $ , then $G/A = H$ , so applying equation (2.2) as in the proof of Lemma 2.5, we see that the number of possibilities for M is at $\omega (n)$ and $\omega (n)+2$ in cases (i) and (ii), respectively. Now, assume that $A < \langle a\rangle $ . First, note that M projects into H, for otherwise $M < \langle a, M_0\rangle $ for a maximal subgroup $M_0$ of H, contradicting the maximality of M. We next claim that $|\langle a\rangle :A|$ is prime. For a contradiction, suppose otherwise. Then $A < \langle a^p\rangle $ for some prime divisor p of m. Now, $M < \langle M, a^p \rangle $ . If $\langle M, a^p \rangle = G$ , then $a \in \langle M, a^p\rangle = \langle a^p\rangle M$ which is impossible since $M \cap \langle a \rangle = A \leqslant \langle a^p\rangle $ , so $M < \langle M, a^p \rangle < G$ , which contradicts the maximality of M. Therefore, $|\langle a\rangle :A|$ is a prime divisor of m. To finish, we divide into the cases (i) and (ii).

For (i), if $|\langle a\rangle :A| = p$ , then $M = \langle A, a^ib\rangle $ , where $0 \leqslant i < p$ , so there are at most p possibilities for M. This means that if $m = p_1^{e_1} \dots p_k^{e_k}$ , where $p_1, \dots , p_k$ are the distinct prime divisors of m, there are at most $p_1+\dots +p_k+\omega (n) \leqslant m+\omega (n)$ maximal subgroups of G.

For (ii), if $|\langle a\rangle :A| = p$ , then $M = \langle A, a^ib, a^jc\rangle $ , where $0 \leqslant i,j < p$ . Now, $[a^ib,a^jc] \in \langle a\rangle \cap M = A$ , but $[a^ib,a^jc] = a^{(1-k)j-2ik}$ , so $Aa^{2ik} = Aa^{(1-k)j}$ , and thus there are at most two choices for i for each choice of j. Since there are at most p choices for j, there are at most $2p$ choices for M. As in the previous case, if $m = p_1^{e_1} \dots p_k^{e_k}$ , then there are at most $2p_1+\dots +2p_k+\omega (n)+2 \leqslant 2m+\omega (n)+2$ maximal subgroups of G.

Proof. Let t be the largest integer such that $n=s^t$ for some integer s. Then every maximal $\mathcal {C}_7$ subgroup of G has type $\mathrm {GL}^\varepsilon _m(q) \wr S_k$ , where k divides t and $n=m^k$ and by [Reference Kleidman and Liebeck18, Tables 3.5.A and 3.5.B], there are at most $n/m \ G$ -classes of subgroups of a given type. Therefore, the number of G-classes of $\mathcal {C}_7$ subgroups is at most $\sum _{k \mid t} s^{t-t/k} \leqslant \sum _{i=0}^{t-1} s^i = (s^t-1)/(s-1) < s^t = n$ .

Proof of Proposition 2.4.

First, consider part (i). For now, assume that $\mathrm {soc}(G) \neq \mathrm {PSL}^\pm _n(q)$ . Then $\mathrm {Out}(\mathrm {soc}(G)) = H \times C_{df}$ , where $H \leqslant S_4$ and $d \leqslant 3$ ; see [Reference Burness, Guralnick and Harper4, Table 2] for exceptional groups and [Reference Harper15, Section 5.2] for classical groups. Therefore, Lemma 2.5 implies that $\mathrm {Out}(\mathrm {soc}(G))$ has at most $\omega (f)+10$ maximal subgroups. It remains to assume that $\mathrm {soc}(G) = \mathrm {PSL}^\pm _n(q)$ . In this case, Lemma 2.6 implies that $\mathrm {Out}(\mathrm {PSL}_n(q)) = C_{\mathrm {gcd}(q-1,n)}{:}(C_2 \times C_f)$ has at most $2n+\omega (f)+2 = 2r+\omega (f)+4$ maximal subgroups and $\mathrm {Out}(\mathrm {PSU}_n(q)) = C_{\mathrm {gcd}(q+1,n)}{:}C_{2f}$ has at most $n+\omega (2f) \leqslant r+\omega (f)+2$ maximal subgroups. This proves part (i).

For parts (ii) and (iii), an $\mathrm {Inndiag}(\mathrm {soc}(G))$ -class yields at most $|\mathrm {Inndiag}(\mathrm {soc}(G)):\mathrm {soc}(G)|$ classes in G, and $|\mathrm {Inndiag}(\mathrm {soc}(G)):\mathrm {soc}(G)| \leqslant 4$ unless $\mathrm {soc}(G) = \mathrm {PSL}^\pm _n(q)$ , in which case we have $|\mathrm {Inndiag}(\mathrm {soc}(G)):\mathrm {soc}(G)| \leqslant n = r+1$ . Part (iii) now follows by the observation that there are at most $\omega (f)+2$ classes in $\mathrm {Inndiag}(\mathrm {soc}(G))$ . Part (ii) is easily verified by consulting [Reference Liebeck and Seitz19, Theorem 2] for exceptional groups and [Reference Kleidman and Liebeck18, Chapter 3] and [Reference Schaffer27] for classical groups (we use Lemma 2.7 in the one slightly more difficult case).

Proof. Let $\Gamma \subseteq \Omega $ be an independent set for G of size $H(G,\Omega )$ . Fix $\Delta \subseteq \Gamma $ such that $\Delta $ is independent for N and $N_{(\Delta )} = N_{(\Gamma )}$ , so, in particular, $|\Delta | \leqslant H(N,\Omega )$ . This is always possible by [Reference Gill, Lodá and Spiga12, Lemma 2.4], but the argument is short so we give it: If $\Gamma $ is independent for N, then let $\Delta = \Gamma $ ; otherwise, there exists a proper subset $\Gamma ' \subseteq \Gamma $ such that $N_{(\Gamma ')} = N_{(\Gamma )}$ , and we repeat the argument replacing $\Gamma $ with $\Gamma '$ .

Let $\varphi \colon G \to G/N$ be the quotient map, and let $p_1 < \dots < p_k$ be the prime divisors of $|G/N|$ , so, in particular, $k = \omega (|G/N|)$ . Write $|\varphi (G_{(\Gamma )})| = p_1^{e_1} \cdots p_k^{e_k}$ . Fix $1 \leqslant i \leqslant k$ . Suppose that for all $\alpha \in \Gamma \setminus \Delta $ , the $p_i$ -part of $|\varphi (G_{(\Delta \cup \{\alpha \})})|$ strictly exceeds $p_i^{e_i}$ . Then since $G_{(\Gamma )} = \bigcap _{\alpha \in \Gamma \setminus \Delta } G_{(\Delta \cup \{\alpha \})}$ and $\varphi (G)$ is cyclic, the $p_i$ -part of $|\varphi (G_{(\Gamma )})|$ strictly exceeds $p_i^{e_i}$ , which is a contradiction. Therefore, there exists $\alpha _i \in \Gamma \setminus \Delta $ such that the $p_i$ -part of $|\varphi (G_{(\Delta \cup \{\alpha _i\})})|$ is $p_i^{e_i}$ . Thus, $|\varphi (G_{(\Delta \cup \{\alpha _1, \dots , \alpha _k\})})| = p_1^{e_1} \cdots p_k^{e_k} = |\varphi (G_{(\Gamma )})|$ . However, $N_{(\Delta \cup \{\alpha _1, \dots , \alpha _k\})} = N_{(\Gamma )}$ , so we have $G_{(\Delta \cup \{\alpha _1, \dots , \alpha _k\})} = G_{(\Gamma )}$ . Since $\Gamma $ is independent for G, we deduce that $\Gamma = \Delta \cup \{\alpha _1, \dots , \alpha _k\}$ , which implies that $H(G,\Omega ) = |\Gamma | \leqslant |\Delta | + k \leqslant H(N,\Omega ) + \omega (f)$ , as sought.

Proof of Theorem 3.

Let G be an almost simple group of Lie type of rank r over $\mathbb {F}_{p^f}$ , where p is prime, acting primitively on $\Omega $ . Let $G_0 = \mathrm {soc}(G)$ , so $G_0 \leqslant G \leqslant \mathrm {Aut}(G_0)$ . Now, [Reference Gorenstein, Lyons and Solomon13, Theorem 2.5.12] implies that $\mathrm {Aut}(G_0)$ has a normal subgroup N such that $\mathrm {Aut}(G_0)/N = C_f$ and $|N/G_0| \leqslant 6r$ . Since $G/(G \cap N) \cong GN/N$ , by Lemma 3.1,

$$\begin{align*}H(G,\Omega) \leqslant H(G \cap N,\Omega) + \omega(|GN/N|) \leqslant H(G \cap N,\Omega) + \omega(f), \end{align*}$$

and, by [Reference Gill, Lodá and Spiga12, Lemma 2.8],

$$\begin{align*}H(G \cap N,\Omega) \leqslant H(G_0,\Omega) + \ell((G \cap N)/G_0) \leqslant H(G_0,\Omega) + \ell(N/G_0). \end{align*}$$

Now, $\ell (N/G_0) \leqslant \log _2{|N/G_0|} \leqslant \log _2(6r) \leqslant 3r^3$ , so $H(G,\Omega ) \leqslant H(G_0,\Omega ) + 3r^3 + \omega (f)$ . While the action of $G_0$ on $\Omega $ need not be primitive, as explained in the final paragraph of the proof of [Reference Gill and Liebeck10, Corollary 3], we still have $H(G_0,\Omega ) \leqslant 174r^8$ , so $H(G_0,\Omega ) \leqslant 177r^8 + \omega (f)$ .

Proof of Theorem 2.

By Theorem 3 and Proposition 2.3, there exist constants $A,B,C> 0$ such that for all almost simple groups of Lie type G of rank r over $\mathbb {F}_{p^f}$ , where p is prime, the following both hold

1. (i) if G acts faithfully and primitively on a set $\Omega $ , then $H(G,\Omega ) \leqslant Ar^B+\omega (f)$

2. (ii) if M is a maximal subgroup of G of type (IV*), then $\ell (M) \leqslant C r$ .

Define $\alpha = \max \{ 100A, C \}$ and $\beta =B+2$ (note that $(\alpha ,\beta ) = (17700,10)$ is a valid choice here since $(A,B) = (177,8)$ and $C=192$ are valid for Theorem 3 and Proposition 2.3).

Let G be an almost simple group of Lie type of rank r over $\mathbb {F}_{p^f}$ where p is prime. Then we claim that

(3.1) $$ \begin{align} m(G) \leqslant \alpha (r + \omega(f))^\beta. \end{align} $$

Let X be a minimal generating set for G. For each $x \in X$ , write $H_x = \langle X \setminus \{x\} \rangle $ and let $M_x$ be a maximal subgroup of G such that $H_x \leqslant M_x$ . For distinct $x,y \in X$ note that $M_x \neq M_y$ , for otherwise $\langle X \setminus \{x\} \rangle \leqslant M_x$ and $\langle X \setminus \{y\} \rangle \leqslant M_x$ , so $G = \langle X \rangle \leqslant M_x$ , which is impossible.

First, assume that for all $x \in X$ the maximal subgroup $M_x$ contains $\mathrm {soc}(G)$ or has type (I*), (II), (III) or (V). For a contradiction, suppose that $|X|> \alpha (r+\omega (f))^\beta $ . This means that

$$ \begin{gather*} |X|> \alpha(r+\omega(f))^\beta \geqslant 100A(r+\omega(f))^{B+2} \geqslant (Ar^B + \omega(f)) \cdot 100(r+\omega(f))^2 \\ \geqslant (Ar^B + \omega(f)) \cdot (100r + (r+1)(\omega(f)+2)) + (2r + \omega(f) + 10). \end{gather*} $$

By Proposition 2.4(i), $\mathrm {soc}(G) \leqslant M_x$ for at most $2r + \omega (f) + 10$ elements x of X. Therefore, $M_x$ is core-free for strictly greater than $(Ar^B + \omega (f)) \cdot (100r + (r+1)(\omega (f)+2))$ elements x of X. Now, Proposition 2.4(ii)–(iii) together with the pigeonhole principle implies that there exists a core-free maximal subgroup M of G and a subset $Y \subseteq X$ such that $|Y|> Ar^B + \omega (f)$ and for all $y \in Y$ there exists $g_y \in G$ such that $M_y = M^{g_y}$ .

We claim that $\bigcap _{y \in Y} M^{g_y} < \bigcap _{y \in Y \setminus \{y_0\}} M^{g_y}$ for all $y_0 \in Y$ . To see this, it suffices to fix $y_0 \in Y$ and show that $\bigcap _{y \in Y \setminus \{y_0\}} M^{g_y} \nleqslant M^{g_{y_0}}$ . For a contradiction, suppose otherwise. First, note that $\langle X \setminus \{y_0\} \rangle = H_{y_0} \leqslant M_{y_0} = M^{g_{y_0}}$ , Second note that for all $y \in Y \setminus \{y_0\}$ we have $y_0 \in \langle X \setminus (Y \setminus \{y_0\}) \rangle \leqslant \langle X \setminus \{y\} \rangle = H_y \leqslant M_y = M^{g_y}$ , so $y_0 \in \bigcap _{y \in Y \setminus \{y_0\}} M^{g_y}$ . Therefore, under the supposition that $\bigcap _{y \in Y \setminus \{y_0\}} M^{g_y} \leqslant M^{g_{y_0}}$ , we deduce that $G = \langle X \rangle \leqslant M^{g_{y_0}}$ , which is absurd. This establishes the claim.

This means that $\{ Mg_y \mid y \in Y \}$ is an independent set for the action of G on $G/M$ , so Theorem 3 implies that $|Y| \leqslant Ar^B + \omega (f)$ (see (ii) above), but this directly contradicts the fact that $|Y|> Ar^B + \omega (f)$ . Therefore, we deduce that $|X| \leqslant \alpha (r+\omega (f))^\beta $ .

Next, assume that there exists $x \in X$ such that $M_x$ has type (IV*). We clearly have the inequalities $|X| \leqslant m(H_x) + 1 \leqslant \ell (H_x) + 1 \leqslant \ell (M_x) + 1$ . Proposition 2.3 implies that $\ell (M_x) \leqslant Cr \leqslant \alpha r$ (see (i) above), so $|X| \leqslant \alpha r+1 \leqslant \alpha (r+\omega (f))^\beta $ .

We now pause to observe that we have proved equation (3.1) when $\mathrm {soc}(G) = \mathrm {PSL}_2(p^f)$ since in this case the (I) coincides with (I*) and (IV) coincides with (IV*).

Having established the result in the base case where the rank r is $1$ , we now complete the proof by induction. Suppose that $r = s> 1$ and that equation (3.1) holds for all groups with $r < s$ .

By Proposition 2.2, it remains to assume that G is classical and $H_x$ is an almost simple group of Lie type defined over $\mathbb {F}_{p^e} \subseteq \mathbb {F}_{p^f}$ such that $\mathrm {rank}(M_x) < r$ . Now, by induction,

$$\begin{align*}|X| \leqslant m(H_x) + 1 \leqslant \alpha (\mathrm{rank}(H_x) + \omega(e))^\beta + 1 \leqslant \alpha(r-1+\omega(f))^\beta + 1 \leqslant \alpha(r+\omega(f))^\beta. \end{align*}$$

Therefore, in all cases $|X| \leqslant \alpha (r+\omega (f))^\beta $ , as desired.

Proof. Let B be a Borel subgroup of G containing a maximal torus T, let $\Phi $ be the corresponding root system of G and let $\Delta = \{ \alpha _1, \dots , \alpha _r \}$ be a set of simple roots. We will now construct a minimal generating set for G. To refer to elements of G, we will use the standard Lie theoretic notation $x_\alpha (t)$ and $h_\alpha (t)$ ; see [Reference Gorenstein, Lyons and Solomon13, Theorem 1.12.1], for example.

For $1 \leqslant i \leqslant r$ , let $x_i$ and $y_i$ be the root elements $x_{\alpha _i}(1)$ and $x_{-\alpha _i}(1)$ , respectively. Write $f = e_1^{a_1} \cdots e_k^{a_k}$ , where $e_1, \dots , e_k$ are the distinct prime divisors of f (so $k = \omega (f)$ ). For $1 \leqslant i \leqslant k$ , let $f_i = e_i^{a_i}$ , let $\lambda _i$ be a primitive element of the subfield $\mathbb {F}_{p^{f_i}}$ and let $z_i = h_{\alpha _1}(\lambda _i)$ .

We claim that $X = \{ x_1, \dots , x_r, y_1, \dots , y_r, z_1, \dots , z_k \}$ is a minimal generating set for G (since $|X| = 2r+\omega (f)$ this establishes the result). The fact that X generates G follows from [Reference Gorenstein, Lyons and Solomon13, Theorem 1.12.1] and [Reference Malle and Testerman25, Corollary 24.2]. To see that X is minimal, note that for all $1 \leqslant i \leqslant k$ , the set $X \setminus \{z_i\}$ is contained in the subfield subgroup defined over the subfield $\mathbb {F}_{p^{f/e_i}}$ , and for all $1 \leqslant i \leqslant r$ , both of the sets $X \setminus \{x_i\}$ and $X \setminus \{y_i\}$ are contained in parabolic subgroups of type $P_i$ (corresponding to deleting node i from the Dynkin diagram of $\Phi $ ).

Proof. Let $\gamma = 52$ . If $\mathrm {soc}(G)$ is sporadic, then $m(G) \leqslant \ell (G) \leqslant 52$ (see [Reference Cameron, Solomon and Turull6, Tables III and IV]), and it is easy to check that the same bound holds when $\mathrm {soc}(G)$ is $A_6$ or ${}^2F_4(2)'$ .

We can now assume that $\mathrm {soc}(G) = A_n$ for $n \neq 6$ . In this case, G is $A_n$ or $S_n$ and Whiston proved that $m(G)$ is $n-2$ or $n-1$ , respectively [Reference Whiston29]. If $n \leqslant 53$ , then $m(G) \leqslant n-1 \leqslant 52$ . Otherwise, by [Reference Rosser and Schoenfeld26, Corollary 1] we know that $\pi (n)> n/\log {n}$ , where $\pi $ is the prime-counting function and $\log $ is the natural logarithm. Noting that $\log {n} < \sqrt {n}$ , these bounds give

$$\begin{align*}m(G) \leqslant n-1 < (n/\log{n})^2 < \pi(n)^2 = \omega(|\mathrm{soc}(G)|)^2.\\[-37pt] \end{align*}$$

Proof. It is easy to check that $|\mathrm {soc}(G)|$ is divisible by $p^{fd_1}-1$ , $p^{fd_2}-1$ , … $p^{fd_k} -1$ for some $d_1 < d_2 < \dots < d_k$ with $k \geqslant \max (1,\frac {1}{2}(r-1))$ . Moreover, if $e_1, \dots , e_l$ are the distinct prime divisors of f, then $|\mathrm {soc}(G)|$ is divisible by $p-1$ , $(p^{e_1}-1)/(p-1)$ , $(p^{e_1 e_2}-1)/(p-1)$ …, $(p^{e_1 e_2 \dots e_l}-1)/(p-1)$ . Note that $1 < e_1 < e_1e_2 < \dots < e_1e_2 \dots e_l < fd_2 < fd_3 \dots < fd_k$ . By Zsigmondy’s theorem [Reference Zsigmondy32], for all but at most one $i \in \{1, e_1, \dots , e_1e_2 \dots e_l, fd_2, fd_3, \dots , fd_k \}$ we may fix a primitive prime divisor of $p^i-1$ . Noting that $|\mathrm {soc}(G)|$ is also divisible by p, we deduce that $\omega (|\mathrm {soc}(G)|) \geqslant \max (1,\frac {1}{2}(r-1)) + \omega (f)$ .

Proof of Theorem 1.

Let $\alpha , \beta , \gamma> 0$ be constants satisfying Theorems 2 and 3.5. Let us define $a = \max \{\alpha \cdot 3^\beta , \gamma \}$ and $b = \max \{\beta ,2\}$ (note that $b=10$ is a valid choice here since $\beta =10$ is a valid choice in Theorem 2, and using $\alpha =10^5$ and $\gamma =52$ gives $a < 10^{10}$ ).

Let G be a finite group, let $S_0$ be a composition factor of G and let S be an almost simple group with socle $S_0$ . First assume that $S_0$ is alternating, sporadic or the Tits group. Then Theorem 3.5 implies that $m(S) \leqslant \gamma \cdot \omega (|S_0|)^2$ . Now, assume that $S_0$ is finite simple group of Lie type of rank r over $\mathbb {F}_{p^f}$ , where p is prime. Then Theorem 2 gives us the bound $m(S) \leqslant \alpha (r+\omega (f))^\beta $ . By Lemma 3.6, $\omega (|S_0|) \geqslant \max (1,\frac {1}{2}(r-1))+\omega (f) \geqslant \frac {1}{3}(r+\omega (f))$ , so $m(S) \leqslant \alpha \cdot 3^\beta \cdot \omega (|S_0|)^\beta $ . Therefore, in both cases, $m(S) \leqslant a \cdot \omega (|S_0|)^b$ . By Theorem 3.4, this establishes that $m(G) \leqslant a \cdot \delta (G)^b$ .