Proof. Let G be a primitive subgroup of $S_n$ that is in $\mathfrak {A}_r$ . Since G is soluble, M is an elementary abelian r-subgroup. By the O’Nan–Scott theorem [Reference Scott, Cooperstein and Mason10], $|M| = n = |G|$ , so $G = M \cong C_{r}$ and $n=r$ . Conversely, any transitive subgroup G of $S_{r}$ is primitive [Reference Wielandt15, Theorem 8.3]. Since n is prime, any subgroup of order n in $S_n$ will be generated by an n-cycle. Further, any two n-cycles are conjugate in $S_n$ . Thus, the primitive subgroups of $S_n$ that are also in $\mathfrak {A}_r$ form a single conjugacy class.

Proof of Theorem 1.3.

Let G be a subgroup of $S_{\Omega }$ , where $|\Omega |=n$ , and let ${G \in \mathfrak {A}_q\mathfrak {A}_r}$ . Then $G = Q \rtimes R$ , where Q is an elementary abelian Sylow q-subgroup, R is an elementary abelian Sylow r-subgroup and $|G| = q^\beta r^\gamma $ , with $\beta $ , $\gamma \in \mathbb {N}$ . Let M be a minimal normal subgroup of G. Then M is an elementary abelian u-group. Clearly, $|M|=u^{k}$ for some $k> 1$ and for some prime $u \in \{q,r\}$ .

Now $F(G)$ , the Fitting subgroup of G, is an abelian normal subgroup of G and so, by the O’Nan–Scott theorem, $n=|M|=|F(G)|$ . However, $M \leq F(G)$ , therefore, ${M=F(G)}$ and $n=u^k$ . If $\beta \geq 1$ , then $Q \leq F(G)$ and we have $n= q^{\beta } = u^k$ and $M=F(G) = Q$ . Let $H=G_{\alpha }$ be the stabiliser of an $\alpha \in \Omega $ . By [Reference Blackburn, Neumann and Venkataraman1, Proposition 6.13], G is a semidirect product of M by H and H acts faithfully by conjugation on M. By Maschke’s theorem, M is completely reducible. However, M is a minimal normal subgroup of G, so M is a nontrivial irreducible $\mathbb {F}_{q}H$ -module and H is an abelian group acting faithfully on M. By [Reference Venkataraman14, Corollary 4.1], $H\cong C_{r}$ and $ \beta = \dim M = \text {order } q \bmod r$ and the result follows. If $\gamma = 0$ or $\beta = 0$ , then $|G|$ is a power of u, where $u \in \{q, r\}$ . Thus, G is a primitive subgroup that is also in $\mathfrak {A}_u$ and the result follows by Lemma 2.1.

Proof of Theorem 1.4.

Let G be a primitive subgroup of $S_{\Omega }$ that is in $\mathfrak {A}_q\mathfrak {A}_r$ , where $|\Omega | = n$ , and let $|G|= q^{\beta }r^{\gamma }$ . Let M be a minimal normal subgroup of G. As seen in the proof of Theorem 1.3, $M = F(G)$ and $n=|M|$ is either a power of q or r. If $\gamma = 0$ or $\beta = 0$ , then $|G|$ is a power of u, where $u \in \{q, r\}$ . Thus, G is a primitive subgroup that is also in $\mathfrak {A}_u$ and the result follows by Lemma 2.1.

We know the structure of G when $\beta \geq 1$ from the proof of Theorem 1.3. Hence, H can be regarded as a soluble r-subgroup of $\mathrm{GL}(\beta ,q)$ and it is not difficult to show that the conjugacy class of G in $S_{n}$ is determined by the conjugacy class of H in $\mathrm{GL}(\beta ,q)$ . Let S be a Singer subgroup of $\mathrm{GL}(\beta ,q)$ , so that $|S| = q^{\beta }-1$ . Now, $|H| =r$ and r divides $|S|$ . Further, $\gcd (|\mathrm{GL}(\beta ,q)|/|S|, r)=1$ as $\beta $ is the least positive integer such that ${r \mid q^{\beta }-1}$ . From [Reference Hestenes3, Theorem 2.11], $H^{x} \leq S$ for some $x \in \mathrm{GL}(\beta ,q)$ . Since all Singer subgroups are conjugate in $\mathrm{GL}(\beta ,q)$ , the result follows.

Proof. Let G be a nontrivial irreducible subgroup of $\mathrm{GL}(\alpha , s)$ that is also in $\mathfrak {A}_r$ . Then G is an elementary abelian r-group of order $r^{\gamma }$ , say, where $\gamma \in \mathbb {N}$ . Since G is a faithful abelian irreducible subgroup of $\mathrm{GL}(\alpha , s)$ whose order is coprime to s, it follows that G is cyclic [Reference Venkataraman14, Lemma 4.2]. Thus, $|G|=r$ and $\alpha = d$ , where $d= \text {order } s \bmod r$ . From [Reference Short11, Theorem 2.3.3], the irreducible cyclic subgroups of order r in $\mathrm{GL}(\alpha , s)$ lie in a single conjugacy class.

Proof. Let G be maximal amongst subgroups of $\mathrm{GL}(\alpha , s)$ that are also in $\mathfrak {A}_r$ . Since $\text {char}(\mathbb {F}_{p}) = t \nmid |G|$ , by Maschke’s theorem, we can find groups $G_i$ such that $G \leq G_{1} \times G_{2} \times \cdots \times G_{k} = \hat {G} \leq \mathrm{GL}(\alpha , s)$ , where for each i, the group $G_i$ is a (maximal) irreducible subgroup of $\mathrm{GL}(\alpha _i, s)$ that is also in $\mathfrak {A}_r$ . Further, $\alpha = \alpha _1 + \cdots + \alpha _k$ . Clearly, $G_i \cong C_r$ and $\alpha _i = d = \text {order } s \bmod r$ for each i. Thus, we must have $\alpha = dk$ and by the maximality of G, we have $G = \hat {G}$ . Further, the conjugacy classes of $G_i$ in $\mathrm{GL}(\alpha _i, s)$ determine the conjugacy class of G in $\mathrm{GL}(\alpha , s)$ .

So if d does not divide $\alpha $ , then $\mathrm{GL}(\alpha , s)$ cannot have an elementary abelian r-subgroup. If $d \mid \alpha $ , then any G that is maximal amongst subgroups of $\mathrm{GL}(\alpha , s)$ that are also in $\mathfrak {A}_r$ must have order $r^{k}$ , where $k = \alpha /d$ . By Lemma 3.1, all such groups form a single conjugacy class.

Proof. Let $V = (\mathbb {F}_{s})^{\alpha }$ . Let G be a primitive subgroup of $\mathrm{GL}(\alpha , s)$ that is in $\mathfrak {A}_q\mathfrak {A}_r$ and let $|G| = q^{\beta } r^{\gamma }$ , where $\beta $ and $\gamma $ are natural numbers. If $\beta = 0$ or $\gamma = 0$ , then the result follows from Lemma 3.1. Assume that $\beta $ and $\gamma $ are at least 1. Let $F = F(G)$ be the Fitting subgroup of G. Since $G \in \mathfrak {A}_q\mathfrak {A}_r$ , it follows that F is abelian and $|F|=q^{\beta }r^{\gamma _1} =m$ , where $\gamma _1 \leq \gamma $ . By Clifford’s theorem, since G is primitive, $V = X_{1} \oplus X_{2} \oplus \cdots \oplus X_{a}$ as an F-module, where the $X_{i}$ are conjugates of X, an irreducible $\mathbb {F}_{s}F$ -submodule of V. Note that F acts faithfully on X.

Let E be the subalgebra generated by F in $\text {End}(V)$ . The $X_{i}$ are conjugates of X, so E acts faithfully and irreducibly on X and E is commutative. By [Reference Blackburn, Neumann and Venkataraman1, Proposition 8.2 and Theorem 8.3], E is a field. Thus, $E \cong \mathbb {F}_{s^{c}}$ as an $\mathbb {F}_{s}F$ -module, where $c = \dim (X)$ and $\alpha = ac$ . Note that F is an abelian group of order m acting faithfully and irreducibly on X. Consequently, F is cyclic and c is the least positive integer such that $m \mid s^{c} -1 $ . Clearly, $m=q$ or $m=qr$ and so $\beta =1$ . It is not difficult to show that G acts on E by conjugation. Hence, there exists a homomorphism from G to $\text {Gal}_{\mathbb {F}_{s}}(E)$ . Let N be the kernel of this map. Then $N = C_{G}(E) \leq C_{G}(F) \leq F$ . However, $F \leq N$ . Hence, $F=N$ . So ${G}/{F} \leq \text {Gal}_{\mathbb {F}_{s}}(E) \cong C_{c}$ and $|G| \leq cm$ .

Let G be an irreducible imprimitive subgroup of $\mathrm{GL}(\alpha , s)$ that is also in $\mathfrak {A}_q\mathfrak {A}_r$ . Then $G \leq G_{1}\, \mathrm{wr} \,G_{2} \leq \mathrm{GL}(\alpha , s)$ , where $G_{1}$ is a primitive subgroup of $\mathrm{GL}(\alpha _1, s)$ that is in $\mathfrak {A}_q\mathfrak {A}_r$ , and the group $G_{2}$ can be regarded as a transitive subgroup of $S_{k}$ that is in $\mathfrak {A}_q\mathfrak {A}_r$ . Further, $\alpha = \alpha _{1} k$ . By the previous part, $|G_1| \leq c'm'$ , where $c' = \text {order}\ s \bmod m' $ and $m' = |F(G_1)|$ is either r or q or $qr$ . Also $c' \mid \alpha _1$ . By [Reference Venkataraman13, Proposition 2.1], $|G_2| \leq {(6^{1/2})}^{k-1}$ . Using $c' \leq 2^{c'-1} \leq {(6^{1/2})}^{c' - 1}$ , we see that $|G| \leq {(6^{1/2})}^{\alpha - 1} \, {(m')}^{k}$ . Since $m' \mid p^{c'} -1$ , we have ${(m')}^k \leq d^{\alpha }$ , where $d = \min \{ qr, s \}$ .

Since t does not divide q or r, by Maschke’s theorem, any subgroup G of $\mathrm{GL}(\alpha , s)$ that is in $\mathfrak {A}_q\mathfrak {A}_r$ will be completely reducible. Thus, $G \leq G_{1} \times \cdots \times G_{k} \leq \mathrm{GL}(\alpha , s)$ , where the $G_{i}$ are irreducible subgroups of $\mathrm{GL}(\alpha _i, s)$ that are in $\mathfrak {A}_q\mathfrak {A}_r$ and $\alpha = \alpha _{1} + \cdots + \alpha _{k}$ . Hence, $|G| \leq {(6^{1/2})}^{\alpha - 1} \, {d}^{\alpha }$ , where $d = \min \{ qr, s \}$ .

Proof. Let G be a subgroup of $\mathrm{GL}(\alpha , s)$ that is maximal amongst irreducible subgroups of $\mathrm{GL}(\alpha , s)$ that are in $\mathfrak {A}_q\mathfrak {A}_r$ . Let $|G| = q^{\beta } r^{\gamma }$ , where $\beta $ and $\gamma $ are natural numbers. If $\beta = 0$ or $\gamma = 0$ , then the result follows from Lemma 3.1. Assume that $\beta $ and $\gamma $ are at least 1. Let $V = (\mathbb {F}_{s})^{\alpha }$ and $F = F(G)$ , the Fitting subgroup of G. Then $F=Q \times R_1$ , where Q is the unique Sylow q-subgroup of G and $R_1 \leq R$ , where R is a Sylow r-subgroup of G. So F is abelian and $|F|=q^{\beta }r^{\gamma _1} =m$ , where $\gamma _1 \leq \gamma $ .

From Clifford’s theorem, regarding V as an $\mathbb {F}_{s}F$ -module, $V = Y_{1} \oplus Y_{2} \oplus \cdots \oplus Y_{l}$ , where $Y_{i} = k X_{i}$ for all i, and $X_{1}, \ldots , X_{l}$ are irreducible $\mathbb {F}_{s}F$ -submodules of V. Further, for each $i,j$ , there exists $g_{ij} \in G$ such that $g_{ij}X_{i} = X_{j}$ and, for $i=1, \ldots , l$ , the $X_{i}$ form a maximal set of pairwise nonisomorphic conjugates. Also, the action of G on the $Y_{i}$ is transitive. It is not difficult to check that $ C_{F}(Y_{i}) = C_{F}(X_{i}) = K_i$ , say. Thus, $F/K_i$ acts faithfully on $Y_{i}$ and when its action is restricted to $X_{i}$ , it acts faithfully and irreducibly on $X_{i}$ . Since $X_{i}$ is a nontrivial irreducible faithful $\mathbb {F}_{s}F/K_i$ -module, and t is coprime to q and r, it follows that $F/K_i$ is cyclic and $\dim _{\mathbb {F}_{s}}(X_i) = d_i$ , where $d_i$ is the least positive integer such that $m_i$ divides $s^{d_i} -1$ , and where $m_i$ is the order of $F/K_i$ . Since the $X_i$ are conjugate, $\dim _{\mathbb {F}_{s}}(X_i) = d_i = d$ for all i.

Let $E_{i}$ be the subalgebra generated by $F/K_i$ in $\text {End}_{{\mathbb{F}}_{s}}(Y_{i})$ . Note that $E_{i}$ is commutative as $F/K_i$ is abelian. Further, $X_{i}$ is a faithful irreducible $E_{i}$ -module. So $E_{i}$ is simple and becomes a field such that $E_{i} \cong \mathbb{F}_{s{^d}}$ . We also observe that $\alpha = k l d$ .

Let $k, l, d$ be fixed such that $\alpha = kld$ . Next we find the number of choices for F up to conjugacy in $\mathrm{GL}(V)$ . Clearly,

$$ \begin{align*} F &\leq F/K_1 \times F/K_2 \times \cdots \times F/K_l \leq {E_1}^{*} \times {E_2}^{*} \times \cdots \times {E_l}^{*} \\ &\leq \mathrm{GL}(Y_{1}) \times \mathrm{GL}(Y_{2}) \times \cdots \times \mathrm{GL}(Y_{l}) \leq \mathrm{GL}(V), \end{align*} $$

where ${E_i}^*$ denotes the multiplicative group of the field $E_i$ . Let $E = {E_1}^{*} \times {E_2}^{*} \times \cdots \times {E_l}^{*}$ . Then $|E|= (s^{d} - 1)^{l}$ . Regarding V as an $\mathbb{F}_{s}E$ -module, $V = kX_{1} \oplus kX_{2} \oplus \cdots \oplus kX_{l}$ , where $E_{i}^{*}$ acts faithfully and irreducibly on $X_{i}$ and $\dim _{E_{i}}(X_{i}) = 1$ for all i. Further, for $i \neq j$ , $E_{i}^{*}$ acts trivially on $X_{j}$ . It is not difficult to show that there is only one conjugacy class of subgroups of type E in $\mathrm{GL}(V)$ .

So once $k,l$ and d are chosen such that $\alpha = k l d$ , up to conjugacy, there is only one choice for E. Since E is a direct product of l isomorphic cyclic groups, any subgroup of E can be generated by l elements. In particular, F can be generated by l elements. So the number of choices for F as a subgroup of E is at most $|E|^{l}= (s^{d} - 1)^{{l}^{2}}$ .

Since, G acts transitively on $\{Y_{1},\ldots , Y_{l}\}$ , there exists a homomorphism $\phi $ from G into $S_{l}$ . Let ${N} = \ker (\phi ) = \{g \in G \mid gY_{i} = Y_{i}\text { for all } i \}$ . Clearly, $F \leq {N}$ and $G/{N}$ is a transitive subgroup of $S_{l}$ that is in $\mathfrak {A}_r$ . If $g \in {N}$ , then $gE_{i}g^{-1}=E_{i}$ . Thus, there exists a homomorphism $\psi _{i}: {N} \rightarrow \text {Gal}_{\mathbb {F}_{s}}(E_{i})$ . This induces a homomorphism $\psi $ from N to $\text {Gal}_{\mathbb {F}_{s}}(E_{1}) \times \text {Gal}_{\mathbb {F}_{s}}(E_{2}) \times \cdots \times \text {Gal}_{\mathbb {F}_{s}}(E_{l})$ such that $\ker (\psi ) = \bigcap _{i=1}^{l} N_{i} = F$ , where $N_{i} = \ker (\psi _i) = C_{N}(E_{i})$ . So ${N}/F$ is isomorphic to a subgroup of $\text {Gal}_{\mathbb {F}_{s}}(E_{1}) \times \text {Gal}_{\mathbb {F}_{s}}(E_{2}) \times \cdots \times \text {Gal}_{\mathbb {F}_{s}}(E_{l})$ . Since $\text {Gal}_{\mathbb {F}_{s}}(E_{i}) \cong C_{d}$ for every i, it follows that ${N}/F$ can be generated by l elements.

Let $T=\mathrm{GL}(\alpha , s)$ . Let $\hat {N}=\{x \in N_{T}{(F)} \mid xY_{i}=Y_{i} \ \text {for all}\ i \}$ . Then $F \leq N \leq \hat {N} \leq N_{T}{(F)}$ . We will find the number of choices for N as a subgroup of $\hat {N}$ , given that F has been chosen. The group $\hat {N}$ acts by conjugation on $E_{i}$ and fixes the elements of $\mathbb {F}_s$ . So we have a homomorphism $\rho _{i}: \hat {N} \rightarrow \text {Gal}_{\mathbb {F}_{s}}(E_{i})$ with kernel $C_{\hat {N}}{(E_{i})}$ . Define $C = \bigcap _{i=1}^{l} C_{\hat {N}}{(E_{i})}$ . Note that $N \cap C = F$ . Also, $\hat {N}/C$ is isomorphic to a subgroup of $\text {Gal}_{\mathbb {F}_{s}}(E_{1}) \times \text {Gal}_{\mathbb {F}_{s}}(E_{2}) \times \cdots \times \text {Gal}_{\mathbb {F}_{s}}(E_{l})$ , where each $\text {Gal}_{\mathbb {F}_{s}}(E_{i}) $ is isomorphic to $C_{d}$ . So $|\hat {N}/C| \leq d^{l}$ . Clearly, C centralises $E_{i}$ for each i. Therefore, there exists a homomorphism from C into $\mathrm{GL}_{E_{i}}{(Y_{i})}$ for each i. Hence, C is isomorphic to a subgroup of $\mathrm{GL}_{E_{1}}(Y_{1}) \times \mathrm{GL}_{E_{2}}(Y_{2}) \times \cdots \times \mathrm{GL}_{E_{l}}(Y_{l})$ . As $\dim _{\mathbb {E}_{i}}(Y_i) = k$ and $E_{i} \cong \mathbb{F}_{s^{d}}$ for all i, it follows that $|C| \leq s^{dk^{2}l}$ . Hence, $|\hat {N}| \leq d^{l}s^{dk^{2}l}$ .

Now $NC/C \cong N/(N \cap C) = N/F$ . So $NC/C$ can be generated by l elements since $N/F$ can be generated by l elements. However, $|\hat {N}/C| \leq d^{l}$ , therefore, there are at most $d^{l^{2}}$ choices for $NC/C$ as a subgroup of $\hat {N}/C$ . Once we make a choice for $NC/C$ as a subgroup of $\hat {N}/C$ , we choose a set of l generators for $NC/C$ . As $N \cap C = F$ , we see that N is determined as a subgroup of $\hat {N}$ by F and l other elements that map to the chosen generating set of $NC/C$ . We have $|C|$ choices for an element of $\hat {N}$ that maps to any fixed element of $\hat {N}/C$ . Thus, there are at most $|C|^{l}$ choices for N as a subgroup of $\hat {N}$ once $NC/C$ has been chosen. So we have at most $d^{l^{2}}(s^{{dk^{2}l}})^{l} = d^{l^{2}}s^{dk^{2}l^{2}}$ choices for N as a subgroup of $\hat {N}$ , once F is fixed.

Next we find the number of choices for G given that F and N are fixed as subgroups of T and $\hat {N} \leq T$ , respectively. Let $\hat {Y}=\{y \in N_{T}{(F)} \mid y \text { permutes the } Y_i \}$ . Then $F \leq G \leq \hat {Y} \leq N_{T}{(F)} \leq \mathrm{GL}(V)$ . Also there exists a homomorphism from $\hat {Y}$ to $S_{l}$ with kernel ${\{y \in \hat {Y} \mid yY_{i} = Y_{i} \text { for all } i\} = \hat {N}}$ . Thus, $\hat {Y}/\hat {N}$ may be regarded as a subgroup of $S_{l}$ . However, $G \cap \hat {N} = N$ . Thus, $G/N = G/(G \cap \hat {N}) \cong G\hat {N}/\hat {N}$ . So ${G/N \cong G\hat {N}/\hat {N} \leq \hat {Y}/\hat {N} \leq S_{l}}$ . Note that $G/N$ is a transitive subgroup of $S_{l}$ that is in $ {\mathfrak {A}_{\mathrm{r}}}$ . By [Reference Lucchini5, Theorem 1], there exists a constant b such that $S_{l}$ has at most $ 2^{bl^{2} / {\sqrt {\log l}}}$ transitive subgroups for $l>1$ . Hence, the number of choices for $G\hat {N}/\hat {N}$ as a subgroup of $\hat {Y}/\hat {N}$ is at most $ 2^{bl^{2} / {\sqrt {\log l}}}$ .

By [Reference Lucchini, Menegazzo and Morigi6, Theorem 2], there exists a constant c such that any transitive permutation group of finite degree greater than $1$ can be generated by $\lfloor cl/\sqrt {\log l}\rfloor $ generators. Thus, $G\hat {N}/\hat {N}$ can be generated by $\lfloor cl/\sqrt {\log l}\rfloor $ generators for $l> 1$ . Once a choice for $G\hat {N}/\hat {N}$ is made as a subgroup of $\hat {Y}/\hat {N}$ and $\lfloor cl/\sqrt {\log l}\rfloor $ generators are chosen for $G\hat {N}/\hat {N}$ in $\hat {Y}/\hat {N}$ , then G is determined as a subgroup of $\hat {Y}$ by $\hat {N}$ and the elements of $\hat {Y}$ that map to the $\lfloor cl/\sqrt {\log l}\rfloor $ generators chosen for $G\hat {N}/\hat {N}$ . So we have at most $|\hat {N}|^{\lfloor cl/\sqrt {\log l}\rfloor }$ choices for G as a subgroup of $\hat {Y}$ once a choice of $G\hat {N}/\hat {N}$ in $\hat {Y}/\hat {N}$ is fixed. Hence, there are

$$ \begin{align*}2^{bl^{2} / {\sqrt{\log l}}}\, (d^{l}s^{dk^{2}l})^{\lfloor cl/\sqrt{\log l}\rfloor} \leq 2^{bl^{2} / {\sqrt{\log l}}} d^{cl^{2}/ \sqrt{\log l}} s^{{cdk^{2}l^{2}}/\sqrt{\log(l)}}\end{align*} $$

choices for G as a subgroup of $\hat {Y}$ assuming that choices for F and N have been made. Putting together all these estimates, the number of conjugacy classes of subgroups that are maximal amongst irreducible subgroups of $\mathrm{GL}(\alpha , s)$ that are in $\mathfrak {A}_q\mathfrak {A}_r$ is at most

$$ \begin{align*} \sum_{(k,l,d)}^{} {(s^{d} - 1)^{{l}^{2}} d^{l^{2}} s^{dk^{2}l^{2}} {2^{bl^{2} / {\sqrt{\log l}}}} {d^{cl^{2}/ \sqrt{\log l}}} {s^{{cdk^{2}l^{2}}/\sqrt{\log l}}}}, \end{align*} $$

where $(k,l,d)$ ranges over ordered triples of natural numbers which satisfy $\alpha = k l d$ and $l>1$ . We simplify the above expression as follows. Writing $\alpha = k l d$ ,

$$ \begin{align*} {(s^{d} - 1)^{{l}^{2}}d^{l^{2}}s^{dk^{2}l^{2}}{s^{{cdk^{2}l^{2}}/\sqrt{\log l}}}} \leq s^{(3 +c){\alpha}^2}. \end{align*} $$

Since $x/\sqrt {\log x}$ is increasing for $x> e^{1/2}$ , we have $l/\sqrt {\log l} \leq \alpha /\sqrt {\log \alpha }$ for $l \geq 2$ . Thus, $2^{bl^{2} / {\sqrt {\log l}}} d^{cl^{2}/ \sqrt {\log l}} \leq 2^{(b + c) {\alpha }^2/\sqrt {\log \alpha }}$ .

There are at most $2^{({5}/{6})\log \alpha +\log 6}$ choices for $(k,l,d)$ . Thus, there exist constants b and c such that the number of conjugacy classes of subgroups that are maximal amongst irreducible subgroups of $\mathrm{GL}(\alpha , s)$ that are in $\mathfrak {A}_q\mathfrak {A}_r$ is at most

$$ \begin{align*} 2^{(b+c) ({\alpha}^2/\sqrt{\log \alpha}) + (5/6) \log \alpha + \log6} \, s^{(3+c){\alpha}^2}\end{align*} $$

provided $\alpha> 1$ .

Proof of Theorem 1.5.

Let G be maximal amongst subgroups of $\mathrm{GL}(\alpha , s)$ that are also in $\mathfrak {A}_q\mathfrak {A}_r$ . As the characteristic of $\mathbb {F}_{s} = t$ and $t \nmid |G|$ , by Maschke’s theorem, $G \leq \hat {G_{1}}\times \cdots \times \hat {G_{k}} \leq \mathrm{GL}(\alpha , s)$ , where the $\hat {G_{i}}$ are maximal among irreducible subgroups of $\mathrm{GL}(\alpha _{i}, p)$ that are also in $\mathfrak {A}_q\mathfrak {A}_r$ , and where $\alpha = \alpha _{1} + \cdots + \alpha _{k}$ . By the maximality of G, we have $G = \hat {G_{1}}\times \cdots \times \hat {G_{k}}$ .

The conjugacy classes of $\hat {G_{i}} \in \mathrm{GL}(\alpha _{i}, s)$ determine the conjugacy class of ${G \in \mathrm{GL}(\alpha , s)}$ . So by Proposition 4.2, the number of conjugacy classes of subgroups that are maximal amongst the subgroups of $\mathrm{GL}(\alpha , s)$ that are also in $\mathfrak {A}_q\mathfrak {A}_r$ is at most

$$ \begin{align*} \sum_{(\alpha)} \prod_{i=1}^{k} 2^{(b+c) ({\alpha_i}^2/\sqrt{\log \alpha_i}) + (5/6) \log \alpha_i + \log6} s^{(3+c){\alpha_i}^2},\end{align*} $$

where the sum is over all unordered partitions $\alpha _{1}, \ldots , \alpha _{k}$ of $\alpha $ . We assume that if $\alpha _i =1$ for some i, then the part of the expression corresponding to it in the product is 1. Since $x/\sqrt {\log x}$ is increasing for $x> e^{1/2}$ and $\alpha = \alpha _{1} + \cdots + \alpha _{k}$ ,

$$ \begin{align*} \prod_{i=1}^{k} 2^{(b+c) ({\alpha_i}^2/\sqrt{\log \alpha_i}) + (5/6) \log \alpha_i + \log6} \leq 2^{(b+c) ({\alpha}^2/\sqrt{\log \alpha}) + (5/6) \alpha \log \alpha + \alpha \log6}. \end{align*} $$

It is not difficult to show that the number of unordered partitions of $\alpha $ is at most $2^{\alpha -1}$ . So the number of conjugacy classes of subgroups that are maximal amongst the subgroups of $\mathrm{GL}(\alpha , s)$ that are also in $\mathfrak {A}_q\mathfrak {A}_r$ is at most

$$ \begin{align*} 2^{(b+c) ({\alpha}^2/\sqrt{\log \alpha}) + (5/6) \alpha \log \alpha + \alpha (1+ \log 6))} s^{(3+c){\alpha}^2} \end{align*} $$

provided $\alpha> 1$ .

Proof of Theorem 1.2.

Let G be a group of order $n = p^{\alpha }q^{\beta }r^{\gamma }$ in $\mathfrak {A}_p\mathfrak {A}_q\mathfrak {A}_r$ . Then $G= P \rtimes H$ , where P is the unique Sylow p-subgroup of G and $H \in \mathfrak {A}_q\mathfrak {A}_r$ . So we can write $H =Q \rtimes R$ , where $|Q|=q^{\beta }$ and $|R|=r^{\gamma }$ . Let $G_{1}={G}/{O_{p'}{(G)}}$ , $G_{2}={G}/{O_{q'}{(G)}}$ and $G_{3}={G}/{O_{r'}{(G)}}$ . Clearly, each $G_{i}$ is a soluble A-group and $G \leq G_{1} \times G_{2} \times G_{3}$ as a subdirect product. Further, ${O_{p'}{(G_{1})}}=1={O_{q'}{(G_{2})}}={O_{r'}{(G_{3})}}$ .

Since $G_{1}={G}/{O_{p'}{(G)}}$ , we see that $G_{1} \in \mathfrak {A}_p\mathfrak {A}_q\mathfrak {A}_r$ and if $P_1$ is the Sylow p-subgroup of $G_1$ , then $P_{1} \cong P$ . Thus, $|G_1|=p^{\alpha }q^{\beta _{1}}r^{\gamma _{1}}$ and we can write $G_{1}= P_{1} \rtimes H_{1}$ , where ${H_{1} \in \mathfrak {A}_q\mathfrak {A}_r}$ . So $H_{1}= Q_{1} \rtimes R_{1}$ , where $Q_{1}\in \mathfrak {A}_q$ and $|Q_1|=q^{\beta _{1}}$ , $R_{1}\in \mathfrak {A}_r$ and $|R_1|=r^{\gamma _{1}}$ . Further, $H_{1}$ acts faithfully on $P_{1}$ . Hence, we can regard $H_1 \leq \mathrm{Aut}(P_{1}) \cong \mathrm{GL}(\alpha , p)$ . Let $M_{1}$ be a subgroup that is maximal amongst $p'$ -A-subgroups of $\mathrm{GL}(\alpha , p)$ that are also in $\mathfrak {A}_q\mathfrak {A}_r$ and such that $H_{1} \leq M_{1}$ . Let $\hat {G_{1}}=P_{1}M_{1}$ . The number of conjugacy classes of the $M_{1}$ in $\mathrm{GL}(\alpha , p)$ is at most $p^{5\alpha ^{2}} 6^{\alpha (\alpha -1)/4} 2^{\alpha - 1 + (23/6) \alpha \log \alpha + \alpha \log 6}$ by Remark 4.3.

Since $G_{2}={G}/{O_{q'}{(G)}}$ , we see that $G_{2} \in \mathfrak {A}_q\mathfrak {A}_r$ and if $Q_2$ is the Sylow q-subgroup of  $G_2$ , then $Q_{2} \cong Q$ . Thus, $|G_2|=q^{\beta }r^{\gamma _{2}}$ and we can write $G_{2}=Q_{2} \rtimes H_{2}$ , where $H_{2} \in \mathfrak {A}_r$ . So $|H_2|=r^{\gamma _{2}}$ . Also, $H_2 \leq \mathrm{Aut}(Q_{2}) \cong \mathrm{GL}(\beta , q)$ . Let $M_{2}$ be a subgroup that is maximal amongst $q'$ -A-subgroups of $\mathrm{GL}(\beta , q)$ that are also in $\mathfrak {A}_r$ and such that $H_{2} \leq M_{2}$ . Let $\hat {G_{2}} = Q_{2}M_{2}$ . The number of conjugacy classes of $M_{2}$ in $\mathrm{GL}(\beta , q)$ is at most $1$ by Proposition 3.2.

Since $G_{3}={G}/{O_{r'}{(G)}}$ , we see that $G_{3} \in \mathfrak {A}_r\mathfrak {A}_q$ and if $R_3$ is the Sylow r-subgroup of $G_3$ , then $R_{3} \cong R$ . Thus, $|G_3|=q^{\beta _3}r^{\gamma }$ and we can write $G_{3}=R_{3} \rtimes H_{3}$ , where $H_{3} \in \mathfrak {A}_r$ . So $|H_3|=q^{\beta _{3}}$ . Also, $H_3 \leq \mathrm{Aut}(R_{3}) \cong \mathrm{GL}(\gamma , r)$ . Let $M_{3}$ be a subgroup that is maximal amongst $r'$ -A-subgroups of $\mathrm{GL}(\gamma , r)$ that are also in $\mathfrak {A}_q$ and such that $H_{3} \leq M_{3}$ . Let $\hat {G_{3}} = R_{3}M_{3}$ . The number of conjugacy classes of the $M_{3}$ in $\mathrm{GL}(\gamma , r)$ is at most $1$ by Proposition 3.2.

Let $\hat {G} = \hat {G_{1}} \times \hat {G_{2}} \times \hat {G_{3}}$ . Then $G \leq \hat {G}$ . The choices for $P_{1}, Q_{2}$ and $R_{3}$ are unique, up to isomorphism. We enumerate the possibilities for $\hat {G}$ up to isomorphism and then find the number of subgroups of $\hat {G}$ of order n up to isomorphism. For the former, we count the number of $\hat {G_{i}}$ up to isomorphism which depends on the conjugacy class of the $M_{i}$ . Hence, the number of choices for $\hat {G}$ up to isomorphism is $\prod _{i=1}^{3} \{\text {number of choices for } \hat {G_{i}} \text { up to isomorphism}\}$ . Now we estimate the choices for G as a subgroup of $\hat {G}$ using a method of ‘Sylow systems’ introduced by Pyber in [Reference Pyber9] .

Let $\hat {G}$ be fixed. We count the number of choices for G as a subgroup of $\hat {G}$ . Let ${\cal S} = \{S_{1}, S_{2}, S_{3}\}$ be a Sylow system for G, where $S_{1}$ is the Sylow p-subgroup of G, $S_{2}$ is a Sylow q-subgroup of G and $S_{3}$ is a Sylow r-subgroup of G such that $S_{i}S_{j} = S_{j}S_{i}$ for all $i, j=1,2,3$ . Then $G = S_{1} S_{2} S_{3}$ . By [Reference Blackburn, Neumann and Venkataraman1, Theorem 6.2, page 49], there exists ${{\cal B} = \{B_{1}, B_{2}, B_{3}\}}$ , a Sylow system for $\hat {G}$ such that $S_{i} \leq B_{i}$ , where $B_{1}$ is the Sylow p-subgroup of $\hat {G}$ , $B_{2}$ is a Sylow q-subgroup of $\hat {G}$ and $B_{3}$ is a Sylow r-subgroup of $\hat {G}$ . Note that $|B_1|=p^{\alpha }$ . Further, any two Sylow systems for $\hat {G}$ are conjugate. Hence, the number of choices for G as a subgroup of $\hat {G}$ and up to conjugacy is at most

$$ \begin{align*}\mid \!\{S_1, S_2, S_3 \mid S_{i} \leq B_{i}, |S_1|=p^{\alpha}, |S_2|=q^{\beta}, |S_3|=r^{\gamma}\! \} \mid \,\leq \, |B_1|^{\alpha} |B_2|^{\beta} |B_3|^{\gamma}. \end{align*} $$

We observe that $B_2= T_{21} \times T_{22} \times T_{23}$ , where $T_{2i}$ are Sylow q-subgroups of $\hat {G_{i}}$ for $i=1,2,3$ . From [Reference Venkataraman13, Proposition 3.1], $|T_{21}| \leq |M_1| \leq (6^{1/2})^{{\alpha }-1} p^{\alpha }$ and $|T_{23}| = |M_3| \leq (6^{1/2})^{{\gamma }-1} r^{\gamma }$ . Further, $|T_{22}|=|Q_2|=q^{\beta }$ . Hence, $|B_2| \leq (6^{1/2})^{{\alpha + \gamma }-2} p^{\alpha } q^{\beta } r^{\gamma } \leq (6^{1/2})^{{\alpha + \gamma }} n$ and so $|B_2|^{\beta } \leq (6^{1/2})^{({\alpha + \gamma })\beta } n^{\beta }$ . Similarly, we can show that $|B_3| \leq (6^{1/2})^{{\alpha + \beta }-2} p^{\alpha } q^{\beta } r^{\gamma }$ . So $|B_3|^{\gamma } \leq (6^{1/2})^{({\alpha + \beta })\gamma } n^{\gamma }$ . Putting all the estimates together, the number of choices for G as a subgroup of $\hat {G}$ up to conjugacy is at most $|B_1|^{\alpha } |B_2|^{\beta } |B_3|^{\gamma }$ , which is at most

$$ \begin{align*}p^{{\alpha}^2}\, (6^{1/2})^{({\alpha + \gamma})\beta} n^{\beta} (6^{1/2})^{({\alpha + \beta})\gamma} n^{\gamma} \leq p^{{\alpha}^2}\, (6^{1/2})^{{({\alpha + \gamma})\beta} + {({\alpha + \beta})\gamma}}n^{\beta + \gamma}.\end{align*} $$

Therefore, the number of groups of order $p^{\alpha }q^{\beta }r^{\gamma }$ in $\mathfrak {A}_p\mathfrak {A}_q\mathfrak {A}_r$ up to isomorphism is at most

$$ \begin{align*} & p^{5\alpha^{2}} \, 6^{\alpha(\alpha-1)/4} \,2^{\alpha - 1 + (23/6) \alpha \log\alpha + \alpha \log6}\, p^{{\alpha}^2}\, (6^{1/2})^{{({\alpha + \gamma})\beta} + {({\alpha + \beta})\gamma}}\, n^{\beta + \gamma} \\ &\quad= p^{6\alpha^{2}} \,2^{\alpha - 1 + (23/6) \alpha \log\alpha + \alpha \log6} (6^{1/2})^{{({\alpha + \gamma})\beta} + {({\alpha + \beta})\gamma} +{\alpha(\alpha-1)/2}}\, n^{\beta + \gamma}.\\[-36pt] \end{align*} $$