A polynomial bound for the number of maximal systems of imprimitivity of a finite transitive permutation group

TL;DR

This paper establishes an upper bound on the number of maximal systems of imprimitivity in finite transitive permutation groups, showing it grows at most proportionally to n^{3/2}, with tighter bounds for soluble groups.

Contribution

It provides a polynomial bound on the number of maximal systems of imprimitivity, generalizing previous results and improving understanding of subgroup structures in finite groups.

Findings

Maximal systems of imprimitivity are bounded by a constant times n^{3/2}.

For soluble groups, the bound is linear in the index, at most |G:H|-1.

The results unify bounds for all finite groups and soluble groups.

Abstract

We show that, there exists a constant $a$ such that, for every subgroup $H$ of a finite group $G$, the number of maximal subgroups of $G$ containing $H$ is bounded above by $a|G:H|^{3/2}$. In particular, a transitive permutation group of degree $n$ has at most $an^{3/2}$ maximal systems of imprimitivity. When $G$ is soluble, generalizing a classic result of Tim Wall, we prove a much stroger bound, that is, the number of maximal subgroups of $G$ containing $H$ is at most $|G:H|-1$.

Equations

$$\max (H,G):=|\{M\mid M\text{ maximal subgroup of }G\text{ with }H\le M\}|,$$

$${\mathrm{core}}_G(M):=\bigcap _{g\in G}{M}^g$$

$$L_k:=\{(l_1,\dots ,l_k)\in L^k\mid l_1\equiv \cdots \equiv l_k(\mathrm{mod}A)\}.$$

$$g_1{a}_1\cdot g_2{a}_2=g_1{g}_2{a}_1^{g_2}{a}_2,$$

$$(a^g{)}^{\phi }=(a^{\phi }{)}^g,$$

$$G=UX\text{ and }Y=U\cap X.$$

$$L_A:=\{\begin{array}{ll}A⋊(G/C_G(A))& \text{ if A is abelian},\\ G/C_G(A)& \text{ otherwise}.\end{array}$$

$$G/N\cong L_A\text{ and }\mathrm{soc}(G/N){\sim }_{G}A.$$

$$R_G(A):=\bigcap _{N\in {\mathcal{N}}_A}N$$

$$a^{\prime }:=\sum _{u=1}^{\infty }\frac{1}{u^{3/2}}.$$

$$a:=\frac{11c{a}^{\prime }}{1-1/2^{3/2}}.$$

$$\tilde{H}:=\bigcap _{\begin{array}{c}H\le M<G\\ M\text{ max. in }G\end{array}}M.$$

$$H\text{ is an intersection of maximal subgroups of }G.$$

$${\mathrm{core}}_G(H)=1.$$

$$H=X_1\cap \cdots \cap X_{\rho }\cap Y_1\cap \cdots \cap Y_{\sigma },$$

$$X:=X_1\cap \cdots \cap X_{\rho }\text{ and }Y:=Y_1\cap \cdots \cap Y_{\sigma }.$$

$$R\le X.$$

$$G/R\cong L_k,$$

$$\frac{T}{R}=\frac{X}{R}\cap \frac{I}{R}.$$

$$H\cap D=(X\cap Y)\cap D=X\cap (Y\cap D)=X\cap D=X\cap (I\cap D)=(X\cap I)\cap D=T\cap D.$$

$$|G:HD|=\frac{|G:H|}{|HD:H|}=\frac{|G:H|}{|D:H\cap D|}=\frac{|G:H|}{|D:T\cap D|}.$$

$$\sigma =\max (HD/D,G/D)\le a|G/D:HD/D{|}^{3/2}=a|G:HD{|}^{3/2}=a{\left(\frac{|G:H|}{|D:D\cap T|}\right)}^{3/2}\le \frac{a}{2^{3/2}}|G:H{|}^{3/2};$$

$$\sigma =\max (HD/D,G/D)\le |G/D:HD/D|-1=|G:HD|-1=\frac{|G:H|}{|D:D\cap T|}-1\le \frac{|G:H|}{2}-1.$$

$$\rho \le \max (HR,G)\le a|G:HR{|}^{3/2}=a{\left(\frac{|G:H|}{|HR:H|}\right)}^{3/2}\le \frac{a}{2^{3/2}}|G:H{|}^{3/2};$$

$$\rho \le \max (HR,G)\le |G:HR|-1=\frac{|G:H|}{|HR:H|}-1\le \frac{|G:H|}{2}-1.$$

$$\max (H,G)=\sigma +\rho \le \frac{2a}{2^{3/2}}\cdot |G:H{|}^{3/2}<a|G:H{|}^{3/2};$$

$$\max (H,G)=\sigma +\rho \le \frac{|G:H|}{2}-1+\frac{|G:H|}{2}-1<|G:H|-1.$$

$$\mathcal{C}:=\{{\mathrm{core}}_G(X_i)\mid i\in \{1,\dots ,\rho \}\}$$

$${\mathcal{M}}_C:=\{X_i\mid i\in \{1,\dots ,\rho \},C={\mathrm{core}}_G(X_i)\}.$$

$$|\mathcal{C}|\le k^2.$$

Full text

1. Introduction

Tim Wall in 1961 [12] has conjectured that the number of maximal subgroups of a finite group $G$ is less than the group order $|G|$. Wall himself proved the conjecture under the additional hypothesis that $G$ is soluble. The first remarkable progress towards a good understanding of Wall’s conjecture is due to Liebeck, Pyber and Shalev [11]; they proved that all, but (possibly) finitely many, simple groups satisfy Wall’s conjecture. Actually, Liebeck, Pyber and Shalev prove [11, Theorem $1.3$] a polynomial version of Wall’s conjecture: there exists an absolute constant $c$ such that, every finite group $G$ has at most $c|G|^{3/2}$ maximal subgroups. Based on the conjecture of Guralnick on the dimension of certain first cohomology groups [6] and on some computer computations of Frank Lübeck, Wall’s conjecture was disproved in 2012 by the participants of an AIM workshop, see [7].

The question of Wall can be generalised in the context of finite permutation groups and this was done by Peter Cameron, see [3]. (See [3] also for the motivation for this question.)

Question 1.1** (Cameron [3]).**

Is the number of maximal blocks of imprimitivity through a point for a transitive group $G$ of degree $n$ bounded above by a polynomial of degree n? Find the best bound!

To see that this question extends naturally the question of Wall we fix some notation. Given a finite group $G$ and a subgroup $H$ of $G$, we denote by

[TABLE]

the number of maximal subgroups of $G$ containing $H$. Now, if $\Omega$ is the domain of a transitive permutation group $G$ and $\omega\in\Omega$, then there exists a one-to-one correspondence between the maximal systems of imprimitivity of $G$ and the maximal subgroups of $G$ containing the point stabiliser $G_{\omega}$ and hence Question 1.1 asks for a polynomial upper bound for $\max(G_{\omega},G)$ as a function of $n=|G:G_{\omega}|$. When $n=|G|$, that is, $G$ acts regularly on itself, the question of Cameron reduces to the question of Wall and [11, Theorem $1.3$] yields a positive solution in this case, with exponent $3/2$.

The main result of this paper is a positive solution to Question 1.1.

Theorem 1.2**.**

There exists a constant $a$ such that, for every finite group $G$ and for every subgroup $H$ of $G$, we have $\max(H,G)\leq a|G:H|^{3/2}$. In particular, a transitive permutation group of degree $n$ has at most $an^{3/2}$ maximal systems of imprimitivity.

In the case of soluble groups we actually obtain a much tighter bound, which extends the result of Wall [12, (8.6), page $58$] for soluble groups on his own conjecture.

Theorem 1.3**.**

If $G$ is a finite soluble group and $H$ is a proper subgroup of $G$, then $\max(H,G)\leq|G:H|-1$. In particular, a soluble transitive permutation group of degree $n\geq 2$ has at most $n-1$ maximal systems of imprimitivity.

2. Preliminaries

We start by reviewing some basic results on $G$-groups, on monolithic primitive groups and on crowns tailored to our proof of Theorem 1.2. For the first part we follow [5], for the second part we follow [8] and for the third part we follow [1, Chapter 1] and [5]. This section will also help for setting some notation. All groups in this paper are finite.

2.1. Monolithic primitive groups and crown-based power

Recall that an abstract group $L$ is said to be primitive if it has a maximal subgroup with trivial core. Incidentally, given a group $G$ and a subgroup $M$ be denote by

[TABLE]

the core of $M$ in $G$. The socle $\mathrm{soc}(L)$ of a primitive group $L$ is either a minimal normal subgroup, or the direct product of two non-abelian minimal normal subgroups. A primitive group $L$ is said to be monolithic if the first case occurs, that is, $\mathrm{soc}(L)$ is a minimal normal subgroup of $L$ and hence (necessarily) $L$ has a unique minimal normal subgroup.

Let $L$ be a monolithic primitive group and let $A:=\mathrm{soc}(L)$. For each positive integer $k$, let $L^{k}$ be the $k$-fold direct product of $L$. The crown-based power of $L$ of size $k$ is the subgroup $L_{k}$ of $L^{k}$ defined by

[TABLE]

Equivalently, if we denote by $\mathrm{diag}(L^{k})$ the diagonal subgroup of $L^{k}$, then $L_{k}=A^{k}\mathrm{diag}(L^{k})$.

For the proof of the next lemma we need some basic terminology, which we borrow from [9, Section 4.3 and 4.4]. Let $\kappa$ be a positive integer and let $A$ be a direct product $S_{1}\times\cdots\times S_{\kappa}$, where the $S_{i}$s are pair-wise isomorphic non-abelian simple groups. We denote by $\pi_{i}:A\to S_{i}$ the natural projection onto $S_{i}$. A subgroup $X$ of $A$ is said to be a strip, if $X\neq 1$ and, for each $i\in\{1,\ldots,\kappa\}$, either $X\cap\mathrm{Ker}(\pi_{i})=1$ or $\pi_{i}(X)=1$. The support of the strip $X$ is the set $\{i\in\{1,\ldots,\kappa\}\mid\pi_{i}(X)\neq 1\}$. The strip $X$ is said to be full if $\pi_{i}(X)=S_{i}$, for all $i$ in the support of $X$. Two strips $X$ and $Y$ are disjoint if their supports are disjoint. A subgroup $X$ of $A$ is said to be a subdirect subgroup if, for each $i\in\{1,\ldots,\kappa\}$, $\pi_{i}(X)=S_{i}$.

Scott’s lemma (see for instance [9, Theorem 4.16]) shows (among other things) that if $X$ is a subdirect subgroup of $A$, then $X$ is a direct product of pairwise disjoint full strips of $A$.

Lemma 2.1**.**

Let $L_{k^{\prime}}$ be a crown-based power of $L$ of size $k^{\prime}$ having non-abelian socle $N^{k^{\prime}}$ and let $H^{\prime}$ be a core-free subgroup of $L_{k^{\prime}}$ contained in $N^{k^{\prime}}$. Then $|N^{k^{\prime}}:H^{\prime}|\geq 5^{k^{\prime}}$.

Proof.

We argue by induction on $k^{\prime}$. If $k^{\prime}=1$, then the result is clear because $N^{k^{\prime}}=N$ has no proper subgroups having index less then $5$. Suppose that $k^{\prime}\geq 2$ and write $N:=N_{1}\times\cdots\times N_{k^{\prime}}$, where $N_{1},\ldots,N_{k^{\prime}}$ are the minimal normal subgroups of $L_{k^{\prime}}$ contained in $N^{k^{\prime}}$. For each $i\in\{1,\ldots,k^{\prime}\}$, we denote by $\pi_{i}:N^{k^{\prime}}\to N_{i}$ the natural projection onto $N_{i}$.

Suppose that there exists $i\in\{1,\ldots,k^{\prime}\}$ with $\pi_{i}(H^{\prime})<N_{i}$. Then, $N_{i}H^{\prime}/N_{i}$ is a core-free subgroup of $L_{k^{\prime}}/N_{i}\cong L_{k^{\prime}-1}$ and is contained in $N^{k^{\prime}}/N_{i}$. Therefore, by induction, $|N^{k^{\prime}}:H^{\prime}N_{i}|=|N^{k^{\prime}}/N_{i}:H^{\prime}N_{i}/N_{i}|\geq 5^{k^{\prime}-1}$. Furthermore, $|H^{\prime}N_{i}:H^{\prime}|=|N_{i}:H^{\prime}\cap N_{i}|\geq 5$ because $N_{i}$ has no proper subgroups having index less then $5$. Therefore, $|N^{k^{\prime}}:H^{\prime}|\geq 5^{k^{\prime}}$.

Suppose that, for every $i\in\{1,\ldots,k^{\prime}\}$, $\pi_{i}(H^{\prime})=N_{i}$. Since $N$ is non-abelian, we may write $N_{i}=S_{i,1}\times\cdots\times S_{i,\ell}$, for some pair-wise isomorphic non-abelian simple groups $S_{i,j}$ of cardinality $s$. For each $i\in\{1,\ldots,k^{\prime}\}$ and $j\in\{1,\ldots,\ell\}$, we denote by $\pi_{i,j}:N^{k^{\prime}}\to S_{i,j}$ the natural projection onto $S_{i,j}$. As $\pi_{i}(H^{\prime})=N_{i}$, we deduce $\pi_{i,j}(H^{\prime})=S_{i,j}$, for every $i\in\{1,\ldots,k^{\prime}\}$ and $j\in\{1,\ldots,\ell\}$. In particular, $H^{\prime}$ is a subdirect subgroup of $S_{1,1}\times\cdots\times S_{k^{\prime},\ell}$ and hence (by Scott’s lemma) $H^{\prime}$ is a direct product of pair-wise disjoint full strips. Since no $N_{i}$ is contained in $H^{\prime}$, there exist two distinct indices $i_{1},i_{2}\in\{1,\ldots,k^{\prime}\}$ and $j_{1},j_{2}\in\{1,\ldots,\ell\}$ such that $(i_{1},j_{1})$ and $(i_{2},j_{2})$ are involved in the same full strip of $H^{\prime}$. If we now consider the projection $\pi_{i_{1},i_{2}}:N^{k^{\prime}}\to N_{i_{1}}\times N_{i_{2}}$, we obtain $|N_{i_{1}}\times N_{i_{2}}:\pi_{i_{1},i_{2}}(H^{\prime})|\geq s\geq 60\geq 5^{2}$. The inductive hypothesis applied to $\mathrm{Ker}(\pi_{i_{1},i_{2}})\cap H^{\prime}$ yields $|\mathrm{Ker}(\pi_{i_{1},i_{2}}):\mathrm{Ker}(\pi_{i_{1},i_{2}})\cap H^{\prime}|\geq 5^{k^{\prime}-2}$ and hence $|N^{k^{\prime}}:H^{\prime}|\geq 5^{k^{\prime}}$. ∎

In the proof of Theorem 1.2 and 1.3, we use without mention the following basic fact.

Lemma 2.2**.**

Let $M$ be a normal subgroup of a crown-based power $L_{k}$ with socle $N^{k}$. Then either $M\leq N^{k}$ or $N^{k}\leq M$.

Proof.

For each $i\in\{1,\ldots,k\}$, we write $N_{i}:=\{(n_{1},\ldots,n_{k})\in N^{k}\mid n_{j}=1,\,\forall j\in\{1,\ldots,k\}\setminus\{i\}\}$. In particular, $N=N_{1}\times\cdots\times N_{k}$.

Let $M$ be a normal subgroup of the crown based power $L_{k}$ with socle $N^{k}$ and with $M\nleq N^{k}$. Let $m\in M\setminus N^{k}$. For each $i\in\{1,\ldots,k\}$, since $M$ does not centralize $N_{i}$, we deduce $1\neq[M,N_{i}]\leq M\cap N_{i}$. As $N_{i}$ is one of the minimal normal subgroups of $L_{k}$, we must have $N_{i}\leq M$. Therefore, $N^{k}=N_{1}\times\cdots\times N_{k}\leq M$. ∎

2.2. Basic facts on $G$-groups

Given a group $G$, a $G$-group $A$ is a group $A$ together with a group homomorphism $\theta:G\to\mathrm{Aut}(A)$. (For simplicity, we write $a^{g}$ for the image of $a\in A$ under the automorphism $\theta(g)$.) Given a $G$-group $A$, we have the corresponding semi-direct product $A\rtimes_{\theta}G$ (or simply $A\rtimes G$ when $\theta$ is clear from the context), where the multiplication is given by

[TABLE]

for every $a_{1},a_{2}\in A$ and for every $g_{1},g_{2}\in G$. A $G$-group $A$ is said to be irreducible if $G$ leaves invariant no non-identity proper normal subgroup of $A$.

Two $G$-groups $A$ and $B$ are said to be $G$-isomorphic (and we write $A\cong_{G}B$), if there exists an isomorphism $\varphi:A\to B$ such that

[TABLE]

for every $a\in A$ and for every $g\in G$. Similarly, we say that $A$ and $B$ are $G$-equivalent (and we write $A\sim_{G}B$), if there exist two isomorphisms $\varphi:A\to B$ and $\Phi:A\rtimes G\rightarrow B\rtimes G$ such that the following diagram commutes.

{1}$${A}$${A\rtimes G}$${G}$${1}$${1}$${B}$${B\rtimes G}$${G}$${1}$$\scriptstyle{\varphi}$$\scriptstyle{\Phi}

Being “$G$-equivalent” is an equivalence relation among $G$-groups coarser than the “$G$-isomorphic” equivalence relation, that is, two $G$-isomorphic $G$-groups are necessarily $G$-equivalent. The converse is not necessarily true: for instance, if $A$ and $B$ are two isomorphic non-abelian simple groups and $G:=A\times B$ acts on $A$ and on $B$ by conjugation, then $A\ncong_{G}B$ and $A\sim_{G}B$. However, when $A$ and $B$ are abelian, the converse is true, that is, if $A$ and $B$ are abelian, then $A\sim_{G}B$ if and only if $A\cong_{G}B$, see [8, page $178$].

Let $G$ be a group and let $A:=X/Y$ be a chief factor of $G$, where $X$ and $Y$ are normal subgroups of $G$. Clearly, the action by conjugation of $G$ endows $A$ of the structure of $G$-group and, in fact, $A$ is an irreducible $G$-group. On the set of chief factors, the $G$-equivalence relation is easily described. Indeed, it is proved in [8, Proposition $1.4$] that two chief factors $A$ and $B$ of $G$ are $G$-equivalent if and only if either

• •

$A$ and $B$ are $G$-isomorphic, or

• •

there exists a maximal subgroup $M$ of $G$ such that $G/\mathrm{core}_{G}(M)$ has two minimal normal subgroups $N_{1}$ and $N_{2}$ $G$-isomorphic to $A$ and $B$ respectively.

(The example in the previous paragraph witnesses that the second possibility does arise.) From this, it follows that, for every monolithic primitive group $L$ and for every $k\in\mathbb{N}$, the minimal normal subgroups of the crown-based power $L_{k}$ are all $L_{k}$-equivalent.

2.3. Crowns of a finite group

Let $X$ and $Y$ be normal subgroups of $G$ with $A=X/Y$ a chief factor of $G$. A complement $U$ to $A$ in $G$ is a subgroup $U$ of $G$ such that

[TABLE]

We say that $A=X/Y$ is a Frattini chief factor if $X/Y$ is contained in the Frattini subgroup of $G/Y$; this is equivalent to saying that $A$ is abelian and there is no complement to $A$ in $G$. The number $\delta_{G}(A)$ of non-Frattini chief factors $G$-equivalent to $A$ in any chief series of $G$ does not depend on the series and hence $\delta_{G}(A)$ is a well-defined integer depending only on the chief factor $A$.

We denote by $L_{A}$ the monolithic primitive group associated to $A$, that is,

[TABLE]

If $A$ is a non-Frattini chief factor of $G$, then $L_{A}$ is a homomorphic image of $G$. More precisely, there exists a normal subgroup $N$ of $G$ such that

[TABLE]

Consider now the collection $\mathcal{N}_{A}$ of all normal subgroups $N$ of $G$ with $G/N\cong L_{A}$ and $\operatorname{soc}(G/N)\sim_{G}A$: the intersection

[TABLE]

has the property that $G/R_{G}(A)$ is isomorphic to the crown-based power $(L_{A})_{\delta_{G}(A)}$, that is, $G/R_{G}(A)\cong(L_{A})_{\delta_{G}(A)}$.

The socle $I_{G}(A)/R_{G}(A)$ of $G/R_{G}(A)$ is called the $A$-crown of $G$ and it is a direct product of $\delta_{G}(A)$ minimal normal subgroups all $G$-equivalent to $A$.

We conclude this preliminary section with two technical lemmas and one of the main results from [11].

Lemma 2.3**.**

[1, Lemma 1.3.6]** Let $G$ be a finite group with trivial Frattini subgroup. There exists a chief factor $A$ of $G$ and a non-identity normal subgroup $D$ of $G$ with $I_{G}(A)=R_{G}(A)\times D.$

Lemma 2.4**.**

[5, Proposition 11]** Let $G$ be a finite group with trivial Frattini subgroup, let $I_{G}(A),R_{G}(A)$ and $D$ be as in the statement of Lemma 2.3 and let $K$ be a subgroup of $G$. If $G=KD=KR_{G}(A),$ then $G=K.$

Theorem 2.5**.**

[11*, Theorem $1.4$]*There exists a constant $c$ such that every finite group has at most $cn^{3/2}$ core-free maximal subgroups of index $n$.

Theorem 2.5 is an improvement of [10, Corollary 2]. We warn the reader that the statement of Theorem 2.5 is slightly different from that of Theorem $1.4$ in [11]: to get Theorem 2.5 one should take into account Theorem $1.4$ in [11] and the remark following its statement.

3. Proof of Theorems 1.2 and 1.3

In this section we prove Theorems 1.2 and 1.3. Our proofs are inspired from some ideas developed in [4]. Moreover, our proofs have some similarities and hence we start by deducing some general facts holding for both.

We start by defining the universal constant $a$. Observe that the series $\sum_{u=1}^{\infty}u^{-3/2}$ converges. We write

[TABLE]

Let $c$ be the universal constant arising from Theorem 2.5. We define

[TABLE]

Recall that $\max(H,G)$ is the number of maximal subgroups of $G$ containing $H$. For the proof of Theorems 1.2 and 1.3 we argue by induction on $|G:H|+|G|$. The case $|G:H|=1$ for the proof of Theorem 1.2 is clear because $\max(H,G)=0$. Similarly, the case that $H$ is maximal in $G$ for the proof of Theorem 1.3 is clear because $\max(H,G)=1$. In particular, for the proof of Theorem 1.2, we suppose $|G:H|>1$ and, for the proof of Theorem 1.3, we suppose that $H$ is not maximal in $G$.

Consider

[TABLE]

Observe that $\max(H,G)=\max(\tilde{H},G)$. In particular, when $H<\tilde{H}$, we have $|G:\tilde{H}|<|G:H|$ and hence, by induction, we have $\max(H,G)=\max(\tilde{H},G)\leq a|G:\tilde{H}|^{3/2}<a|G:H|^{3/2}$; moreover, when $G$ is soluble, we have $\max(H,G)=\max(\tilde{H},G)\leq|G:\tilde{H}|-1<|G:H|-1$. Therefore, we may suppose $H=\tilde{H}$, that is,

[TABLE]

Suppose that $H$ contains a non-identity normal subgroup $N$ of $G$. Since $\max(H,G)=\max(H/N,G/N)$ and $|G/N|<|G|$, by induction, we have $\max(H,G)=\max(H/N,G/N)\leq a|G/N:H/N|^{3/2}=a|G:H|^{3/2}$; moreover, when $G$ is soluble, we have $\max(H,G)=\max(H/N,G/N)\leq|G/N:G/N|-1=|G:H|-1$. Therefore, we may suppose

[TABLE]

Let $F$ be the Frattini subgroup of $G$. From (3.1), we have $F\leq H$ and hence, from (3.2), $F=1$. In particular, we may now apply Lemma 2.3 to the group $G$.

Choose $I$, $R$ and $D$ as in Lemma 2.3. From (3.1), we may write

[TABLE]

where $X_{1},\dots,X_{\rho}$ are the maximal subgroups of $G$ not containing $D$ and $Y_{1},\dots,Y_{\sigma}$ are the maximal subgroups of $G$ containing $D.$ We define

[TABLE]

Thus $H=X\cap Y$.

For every $i\in\{1,\ldots,\rho\}$, since $D\nleq X_{i}$, we have $G=DX_{i}$ and hence Lemma 2.4 (applied with $K:=X_{i}$) yields $R\leq X_{i}$. In particular,

[TABLE]

Since $R=R_{G}(A)$ for some chief factor $A$ of $G$, Section 2.3 yields

[TABLE]

for some monolithic primitive group $L$ and for some positive integer $k$. We let $N$ denote the minimal normal subgroup (a.k.a. the socle) of $L$. From the definition of $I$ and $R$, we have $I/R=\operatorname{soc}(G/R)\cong\operatorname{soc}(L_{k})=N^{k}$. Finally, let $T:=X\cap I$. In particular,

[TABLE]

We have

[TABLE]

It follows

[TABLE]

If $D\leq T$, then $D\leq X$ and hence $D\leq X\cap Y=H$ because $D\leq Y$; however this is a contradiction because $D\neq 1$ and hence, from (3.2), $D\not\leq H.$ Therefore $D\nleq T$ and $|D:T\cap D|>1$.

Applying our inductive hypothesis, we obtain

[TABLE]

moreover, when $G$ is soluble and $HD$ is a proper subgroup of $G$, we obtain

[TABLE]

(Observe that, when $G$ is soluble and $G=HD$, we have $\sigma=0$ and hence the inequality $\sigma\leq|G:H|/2-1$ is valid also in this degenerate case.)

From (3.3), we deduce $\rho\leq\max(HR,G)$. If $R\nleq H$, then $|G:HR|<|G:H|$ and hence, applying our inductive hypothesis, we obtain

[TABLE]

moreover, when $G$ is soluble and $HR$ is a proper subgroup of $G$, we obtain

[TABLE]

(As above, when $G$ is soluble and $G=HR$, we have $\rho=0$ and hence the inequality $\rho\leq|G:H|/2-1$ is valid also in this degenerate case.)

Now, from (3.4) and (3.6), we have

[TABLE]

similarly, when $G$ is soluble, from (3.5) and (3.7), we have

[TABLE]

In particular, for the rest of the proof, we may assume that $R\leq H$. Now, (3.2) yields $R=1$ and hence $G\cong L_{k}$ and $D=I$. Therefore, we may identify $G$ with $L_{k}$ and $D$ with $N^{k}$.

Set

[TABLE]

and, for every $C\in\mathcal{C}$, set

[TABLE]

For the rest of our argument for proving Theorems 1.2 and 1.3, we prefer to keep the proofs separate.

Proof of Theorem 1.2.

Case 1: Suppose that $N$ is non-abelian.

Since $N$ is non-abelian, the group $G=L_{k}$ has exactly $k$ minimal normal subgroups. We denote by $N_{1},\ldots,N_{k}$ the minimal normal subgroups of $G$. In particular, $I=N^{k}=N_{1}\times N_{2}\times\cdots\times N_{k}$.

We claim that, for every $i\in\{1,\ldots,\rho\}$, there exist $x,y\in\{1,\ldots,k\}$ such that $N_{\ell}\leq X_{i}$, for every $\ell\in\{1,\ldots,k\}\setminus\{x,y\}$, that is, $X_{i}$ contains all but possibly at most two minimal normal subgroups of $G$.

We argue by induction on $k$. The statement is clearly true when $k\leq 2$. Suppose then $k\geq 3$ and let $C:=\mathrm{core}_{G}(X_{i})$. If $C=1$, then $X_{i}$ is a maximal core-free subgroup of $G$ and hence the action of $G$ on the right cosets of $X_{i}$ gives rise to a faithful primitive permutation representation. Since a primitive permutation group has at most two minimal normal subgroups [2, Theorem $4.4$] and since $G$ has exactly $k$ minimal normal subgroups, we deduce that $k\leq 2$, which is a contradiction. Therefore $C\neq 1$.

Since $N_{1},\ldots,N_{k}$ are the minimal normal subgroups of $L_{k}$, we deduce that there exists $\ell\in\{1,\ldots,k\}$ with $N_{\ell}\leq C$. Now, the proof of the claim follows applying the inductive hypothesis to $G/N_{\ell}\cong L_{k-1}$ and to its maximal subgroup $X_{i}/N_{\ell}$.

The previous claim shows that, for every $C\in\mathcal{C}$, $C$ contains all but possibly at most two minimal normal subgroups of $N^{k}=I$. Therefore,

[TABLE]

Let $C\in\mathcal{C}$ and let $M\in\mathcal{M}_{C}$. The reader might find useful to see Figure 1, where we have drawn a fragment of the subgroup lattice of $G$ relevant to our argument.

Let $k^{\prime}$ be the number of minimal normal subgroups of $G$ contained in $M$. In particular, $I\cap M\cong N^{k^{\prime}}$. Observe that $I\cap H$ is contained in $I\cap M$ and is core-free in $G$. Applying Lemma 2.1 (with $H^{\prime}$ replaced by $I\cap H$ in a crowned-based group isomorphic to $L_{k^{\prime}}$), we get $|I\cap M:I\cap H|\geq 5^{k^{\prime}}$. As $k^{\prime}\geq k-2$, we deduce $t\geq 5^{k-2}$.

Now, $M/C$ is a core-free maximal subgroup of $G/C$. From Theorem 2.5, when $C=\mathrm{core}_{G}(M)$ and $z=|G:C|$ are fixed, we have at most $cz^{3/2}$ choices for $M$. As $t\geq 5^{k-2}$, we have $z\leq|G:H|/5^{k-2}$. Thus

[TABLE]

Therefore,

[TABLE]

Finally, it is easy to verify that, for every $k$, $k^{2}/5^{3(k-2)/2}\leq 11$. Summing up,

[TABLE]

From (3.4), (3.8) and from the definition of $a$, we have

[TABLE]

Case 2: Suppose that $N$ is abelian.

As $N$ is abelian, the action of $L$ by conjugation on $N$ endows $N$ of the structure of an $L$-module. Since $L$ is primitive, $N$ is irreducible. Set $q:=|\mathrm{End}_{L}(N)|$. Now, $N$ is a vector space over the finite field $\mathbb{F}_{q}$ with $q$ elements, and hence $|N|=q^{k^{\prime}}$, for some positive integer $k^{\prime}$.

Let $C\in\mathcal{C}$ and let $M\in\mathcal{M}_{C}$. From Lemma 2.2, $C\leq I$. Now, the action of $G/C$ on the right cosets of $M/C$ is a primitive permutation group with point stabilizer $M/C$. Observe that in this primitive action, $I/C$ is the socle of $G/C$. In particular, $G/C$ acts irreducibly as a linear group on $I/C$ and hence $C$ is a maximal $L$-submodule of $I$. Since $I$ is the direct sum of $k$ pairwise isomorphic irreducible $L$-modules, we deduce that we have at most $(q^{k}-1)/(q-1)$ choices for $C$. Moreover, $|G:M|=|G/C:M/C|=|N|=q^{k^{\prime}}$. From Theorem 2.5, when $C$ is fixed, we have at most $c|G:M|^{3/2}=c(q^{k^{\prime}})^{3/2}$ choices for $M\in\mathcal{M}_{C}$. This yields

[TABLE]

As we have observed above, $M\cap I=C$ is an $L$-submodule of $G$. Since an intersection of $L$-submodules is an $L$-submodule, we deduce that

[TABLE]

is an $L$-submodule of $I$ and hence $H\cap I\unlhd G$. Since $H$ is core-free in $I$, we deduce $H\cap I=1$ and hence $|I|=|N|^{k}=q^{kk^{\prime}}$ divides $|G:H|$. In particular, $|G:H|\leq q^{kk^{\prime}}$. Therefore, from (3.9), we obtain

[TABLE]

When $k\neq 1$ or when $(k,k^{\prime})\neq(2,1)$, we have $\frac{k+3k^{\prime}/2}{kk^{\prime}}\leq 3/2$. When $k=1$, by refining (3.9), we obtain the sharper bound $\rho\leq cq^{3k^{\prime}/2}\leq c|G:H|^{3/2}$. When $(k,k^{\prime})=(2,1)$, we may refine again (3.9): $\rho\leq c(q+1)q^{3/2}\leq c\cdot 2q\cdot q^{3/2}=2cq^{5/2}\leq 2c|G:H|^{5/4}\leq 2c|G:H|^{3/2}$. Summing up, in all cases we have

[TABLE]

From (3.4) and (3.10), we have

[TABLE]

∎

The rest of the proof of Theorem 1.3 follows the same idea as in the “Case 2” above, but taking in account that the whole group $G$ is soluble.

Proof of Theorem 1.3.

Since $G=L_{k}$ and $I=N^{k}$, we may write $G=I\rtimes K$, where $K$ is a complement of $N$ in $L$. As in the proof of Theorem 1.2 for the case that $N$ is abelian, we have that the action of $L$ by conjugation on $N$ endows $N$ of the structure of an $L$-module. Since $L$ is primitive, $N$ is irreducible. Set $q:=|\mathrm{End}_{L}(N)|$. Now, $N$ is a vector space over the finite field $\mathbb{F}_{q}$ with $q$ elements, and hence $|N|=q^{k^{\prime}}$, for some positive integer $k^{\prime}$.

Let $C\in\mathcal{C}$ and let $M\in\mathcal{M}_{C}$. As we have observed above (for the proof of “Case 2”), $M\cap I=C$ is a maximal $L$-submodule of $G$, $H\cap I=1$ and $|I|=|N|^{k}=q^{kk^{\prime}}$ divides $|G:H|$. In particular, $|G:H|=\ell q^{kk^{\prime}}$, for some positive integer $\ell$.

Since $G$ is soluble and since $M$ is a maximal subgroup of $G$ supplementing $I$, we have $M=C\rtimes K^{x}$, for some maximal $L$-submodule $C$ of $I$ and some $x\in I$. Arguing as in the proof of Theorem 1.2 for the case that $N$ is abelian, we deduce that we have at most $(q^{k}-1)/(q-1)$ choices for $C$. Moreover, we have at most $|I/C|=|G:M|=|N|=q^{k^{\prime}}$ choices for $x$. This yields

[TABLE]

Now, (3.5) gives $\sigma\leq|G:H|/|D:D\cap T|-1$: recall that $D=I=N^{k}$ and $D\cap T=D\cap H=I\cap H=1$. Thus $\sigma\leq|G:H|/|D|-1=|G:H|/q^{kk^{\prime}}-1=\ell-1$. Therefore,

[TABLE]

When $\ell\geq 2$, a computation shows that the right hand side of (3.12) is less than or equal to $\ell q^{kk^{\prime}}-1=|G:H|-1$. In particular, we may suppose that $\ell=1$. In this case, $|G:H|=q^{kk^{\prime}}=|I|$ and hence $G=IH=I\rtimes H$. Moreover, $\sigma=0$. Since $H$ is not a maximal subgroup of $G$ (recall the base case for our inductive argument), $k\geq 2$.

Assume also $k^{\prime}=1$. Since $|\mathrm{End}_{L}(N)|=q=|N|$, we deduce that $L/N$ is isomorphic to a subgroup of the multiplicative group of the field $\mathbb{F}_{q}$ and hence $|L:N|$ is relatively prime to $q$. Therefore $|G:I|$ is relatively prime to $q$ and hence so is $|H|$. Therefore, replacing $H$ by a suitable $G$-conjugate, we may suppose that $K=H$. Using this information, we may now refine our earlier argument bounding $\rho$. Let $C\in\mathcal{C}$ and let $M\in\mathcal{M}_{C}$. Since $G=I\rtimes H$ is soluble, $M$ is a maximal subgroup of $G$ supplementing $I$ and $H\leq M$, we have $M=C\rtimes H$, for some maximal $L$-submodule $C$ of $I$. We deduce that we have at most $(q^{k}-1)/(q-1)$ choices for $C$ and hence we have at most $(q^{k}-1)/(q-1)$ choices for $M$. This yields

[TABLE]

and the result is proved in this case.

Assume $k^{\prime}\geq 2$. A computation (using $\ell=1$ and $k,k^{\prime}\geq 2$) shows that the right hand side of (3.12) is less than or equal to $q^{kk^{\prime}}-1=|G:H|-1$. ∎