Permutation Groups and Set-Orbits on the Power Set

Abstract

A permutation group G acting on a set $\Omega$ induces a permutation group on the power set ${\mathscr {P}}(\Omega )$ (the set of all subsets of $\Omega$). Let G be a finite permutation group of degree n and s(G) denote the number of orbits of G on ${\mathscr {P}}(\Omega )$. It is an interesting problem to determine the lower bound $\inf \left( \frac{\log _2 s(G)}{n}\right)$ over all groups G that do not contain any alternating group $\mathrm{A}_\ell$ (where $\ell >t$ for some fixed $t\geqslant 4)$ as a composition factor. The second author obtained the answer for the case $t=4$ in Yang (J Algebra Appl 19:2150005, 2020). In this paper, we continue this investigation and study the cases when $t\geqslant 5$, and give the explicit lower bounds $\inf \left( \frac{\log _2 s(G)}{n}\right)$ for each positive integer $5\leqslant t\leqslant 166$.

1 Introduction

A permutation group G acting on a set $\Omega$ induces a permutation group on the power set ${\mathscr {P}}(\Omega )$. It is an important subject in representation theory to study this particular action. For example, Gluck [4] showed that if the group G is solvable and the action is primitive, then G always has a regular orbit on the power set ${\mathscr {P}}(\Omega )$ except for a few exceptional cases. Later, Seress generalized this to arbitrary groups in [8]. We define the orbits of this action to be set-orbits, and let s(G) denote the number of set-orbits of G. Since sets of different cardinalities belong to different orbits, it is clear that $s(G)\geqslant |\Omega | + 1$. Groups with the property $s(G)= |\Omega | + 1$ (i.e., set transitive groups) have been classified by Beaumont and Peterson in [2]. They showed that apart from a few exceptional cases of degree at most 9, a set-transitive group of degree n always contains the alternating group $\mathrm{A}_n$.

Let G be a permutation group of degree n. In [1], Babai and Pyber showed that if G has no large alternating composition factors then s(G) is exponential in n. More precisely they proved the following result [1, Theorem 1]. Let G be a permutation group of degree n. If G does not contain any $\mathrm{A}_\ell$ (where $\ell >t$ for some fixed $t\geqslant 4)$ as a composition factor, then $\frac{\log _2 s(G)}{n} \geqslant \frac{c}{t}$ for some positive constant c (unspecified). It appears that this result has many applications. In [5], Keller applied this result to find a lower bound for the number of conjugacy classes of a solvable group. In [7], Nguyen used this result to study the multiplicities of conjugacy class sizes of finite groups.

In [1], Babai and Pyber also raised the following question: what is $\inf \left( \frac{\log _2 s(G)}{n}\right)$ over all solvable groups G? This question was answered in a recent paper of the second author in [9]. Clearly, a more interesting question is to answer the following: what is $\inf \left( \frac{\log _2 s(G)}{n}\right)$ over all groups G that do not contain any $\mathrm{A}_\ell$ (where $\ell >t$ for some fixed $t\geqslant 4)$ as a composition factor? The second author studied a special case of this question in [10], and showed that when $t=4$, the answer is

$$\begin{aligned} \inf \left( \frac{\log _2 s(G)}{n}\right) = \lim \limits _{k \mapsto \infty } \frac{\log _2 s(\mathrm{M}_{24} \wr \mathrm{M}_{12} \wr \overbrace{S_4 \wr \cdots \wr S_4}^\text {k terms})}{24\cdot 12\cdot 4^k}. \end{aligned}$$

In this paper, we continue this investigation and study the cases when $t\geqslant 5$, and give the explicit lower bounds $\inf \left( \frac{\log _2 s(G)}{n}\right)$ for each positive integer $5 \leqslant t \leqslant 166$. In fact, we show that

$$\begin{aligned}\begin{array}{ll} \inf \left( \frac{\log _2 s(G)}{n}\right) = \lim \limits _{k \mapsto +\infty } \frac{\log _2 s(\mathrm{M}_{24}\wr \overbrace{\mathrm{S}_t \wr \cdots \wr \mathrm{S}_t}^{\mathrm {k terms}})}{24 \cdot t^k} if t\in [5, 16], \quad \mathrm{and}\\ \inf \left( \frac{\log _2 s(G)}{n}\right) = \lim \limits _{k \mapsto +\infty } \frac{\log _2 s(\overbrace{\mathrm{S}_t \wr \cdots \wr \mathrm{S}_t }^{\mathrm {k terms}})}{t^k} if t\in [17, 166]. \end{array} \end{aligned}$$

We also ask some related questions at the end of the paper. Our main results are the following.

Theorem 1.1

Let t be an integer with $5\leqslant t\leqslant 16$. Then

$$\begin{aligned} \inf \left( \frac{\log _2 s(G)}{n}\right) = \lim \limits _{k \mapsto +\infty } \frac{\log _2 s(\mathrm{M}_{24} \wr \overbrace{\mathrm{S}_t \wr \cdots \wr \mathrm{S}_t}^\text {k terms})}{24 \cdot t^k}, \end{aligned}$$

where the infimum is taken over all permutation groups G not containing any composition factor $\mathrm{A}_\ell$ with $\ell >t$, and n denotes the degree of G.

Theorem 1.2

Let t be an integer with $17\leqslant t\leqslant 166$. Then

$$\begin{aligned} \inf \left( \frac{\log _2 s(G)}{n}\right) =\lim \limits _{k \mapsto +\infty } \frac{\log _2 s(\overbrace{\mathrm{S}_t\wr \cdots \wr \mathrm{S}_t}^\text {k terms})}{t^k}, \end{aligned}$$

where the infimum is taken over all permutation groups G not containing any composition factor $\mathrm{A}_\ell$ with $\ell >t$, and n denotes the degree of G.

2 Preliminary Results

Some of the following definitions and lemmas appear in [10], we include those here for the convenience of the reader.

Let T be a finite group and S a permutation group. We denote by $T \wr S$ the wreath product of T with S. Let G be a permutation group of degree n. We use s(G) to denote the number of set-orbits of G and we denote $rs(G)=\frac{\log _2 s(G)}{n}$.

We recall some basic facts about the decompositions of transitive groups. Let G be a transitive permutation group acting on a set $\Omega$, where $|\Omega |=n$. A system of imprimitivity is a partition of $\Omega$, invariant under G. A primitive group has no non-trivial system of imprimitivity. Let $\{\Omega _1,\dots ,\Omega _m \}$ denote a system of imprimitivity of G with maximal block-size b $(1 \leqslant b <n; b = 1$ if and only if G is primitive; $bm=n$). Let N be the intersection of the stabilizers of the blocks. Then G/N is a primitive group of degree m acting upon the set of blocks $\Omega _i$. If $G_i$ denotes the permutation group of degree b induced on $\Omega _i$ by the set-wise stabilizer of $\Omega _i$ in G, then the groups $G_i$ are permutationally equivalent transitive groups and $N\leqslant G_1\times \cdots \times G_m\leqslant \mathrm{Sym}(\Omega )$.

Let G be a transitive group of degree n and assume that G is not primitive. Let us consider a system of imprimitivity of G that consists of $m \geqslant 2$ blocks of size b, where b is maximal. Thus G can be embedded in $K \wr P_1$, written $G \lesssim K \wr P_1$, where K is a permutation group of degree n/m and $P_1$ is the primitive quotient group of G that acts upon the m blocks. We may keep doing this, and after re-index for convenience, and we can get that $G\lesssim H\wr P_1\wr \cdots \wr P_k$, where H is a permutation group and the $P_i$ are all primitive groups. If this happens, we say that G is induced from the permutation group H.

In what follows, we need to make some preparations for the proof of Theorems 1.1 and 1.2, and we begin with two important lemmas.

Lemma 2.1

[1, Prop. 1] If $H \leqslant G \leqslant \mathrm{Sym}(\Omega )$, then $s(G) \leqslant s(H) \leqslant s(G) \cdot |G:H|$.

Lemma 2.2

[1, Prop. 2] Assume that G is intransitive on $\Omega$ and has orbits $\Omega _1,\dots ,\Omega _m$. Let $G_i$ be the restriction of G to $\Omega _i$. Then

$$\begin{aligned} s(G)\geqslant s(G_1)\times \cdots \times s(G_m). \end{aligned}$$

In view of Lemma  2.2 and the fact that $\frac{a+b}{c+d}\geqslant \min \{\frac{a}{c},\frac{b}{d} \}$ for positive integers a, b, c and d, it suffices to consider transitive permutation groups in order to find $\inf \left( \frac{\log _2 s(G)}{n}\right)$. To see this, assume that the action of G on $\Omega$ is not transitive, then we may assume G has two blocks $\Omega _1$ and $\Omega _2$, where $\Omega$ is a disjoint union of $\Omega _1$ and $\Omega _2$. We assume $|\Omega _1|=m_1$ and $|\Omega _2|=m_2$. Let $G_i$ be the restriction of G to $\Omega _i$ for $i=1,2$. Then $s(G) \ge s(G_1) \cdot s(G_2)$. We see that

$$\begin{aligned} rs(G)=\frac{\log _2 s(G)}{m_1 + m_2} \ge \frac{\log _2 s(G_1) + \log _2 s(G_2)}{m_1+m_2}\geqslant \min \{rs(G_1),rs(G_2)\}. \end{aligned}$$

Lemma 2.3 is basic and also very useful to find $\inf \left( \frac{\log _2 s(G)}{n}\right)$. This lemma and its proof are almost identical to [1, Lemma 1].

Lemma 2.3

Let G be a transitive permutation group acting on a set $\Omega$ where $|\Omega |=n$. Let $\{\Omega _1, \dots , \Omega _m \}$ denote a system of imprimitivity of G with maximal block-size b, where $1\leqslant b <n$ and $bm=n$. In particular, $b=1$ if and only if G is primitive. Let N denote the normal subgroup of G stabilizing each of the blocks $\Omega _i$. Let $G_i=\mathrm{Stab}_G(\Omega _i)$ and $s=s(G_1)$. Then

• (i) $s(G)\geqslant s^m/|G/N|$.

• (ii) $s(G)\geqslant {{s+m-1}\atopwithdelims ()s-1}$. Moreover, the equality holds if $G/N \cong \mathrm{S}_m$.

Proof

By Lemmas 2.1 and 2.2, it is easy to verify that part (i) follows.

Let A be a subset of $\Omega$ and let $\alpha _j~(0\leqslant j\leqslant s)$ denote the number of intersections of A with $\Omega _i$ that lie in the j-th orbit of $G_i$ on the powerset of $\Omega _i$. Let B be another subset of $\Omega$ with the number $\beta _j$ defined similarly. If A and B are in the same orbit of G, then $\alpha _j=\beta _j$ for $(0 \leqslant j \leqslant s)$. Therefore, s(G) is at least the number of partitions of m into s nonnegative integers (where the order of the summands is taken into consideration). It is well-known that this number is

$$\begin{aligned} {{s+m-1} \atopwithdelims ()s-1} \end{aligned}$$

which proves the first part of (ii). Since $\mathrm{S}_m$ is set-transitive on all the subsets of the same size, we know the equation will hold.$\square$

We need the following estimates of the order of the primitive permutation groups.

Lemma 2.4

Let G be a primitive permutation group of degree n where G does not contain $\mathrm{A}_n$. Then

• (i) $|G| < 50 \cdot n^{\sqrt{n}}$.

• (ii) $|G| < 3^n$. Moreover, if $n> 24$, then $|G| < 2^n$.

• (iii) $|G|\leqslant 2^{0.76 n}$ when $n\geqslant 25$ and $n\not =32$.

Proof

Part (i) is from [6, Corollary 1.1 (ii)], and part (ii) is from [6, Corollary 1.2]. Part (iii) follows from part (i) for $n\geqslant 89$, and we may check the remaining results using Gap [3] for $25\leqslant n\leqslant 88$. $\square$

Table 1 The lower bounds and corresponding primitive groups

Lemma 2.5

Let G be a primitive permutation group of degree n where G does not contain any $\mathrm{A}_\ell$ ($\ell > 4$) as a composition factor. If $n\leqslant 24$ or $n=32$, then the lower bound of s(G) and corresponding primitive permutation group can be determined.

Proof

The results can be easily checked by Gap [3]. For convenience, we list the results in Table 1. We also need to mention the following cases which will be used in Sects.  3 and  4.

If $n=23$, the group with the largest order is $G \cong \mathrm{M}_{23}$, and $s(G)=72$. However, the group with the second largest order has order 506 and $s(G) \geqslant 16770$.

If $n=24$, the group with the second largest order has order 12144 and $s(G)\geqslant 1674$. The group with the largest order is $G\cong \mathrm{M}_{24}$ and $s(G)=49$.

If $n=32$, the group with the second largest order has order 29760 and $s(G) \geqslant 144321$. The group with the largest order is $G\cong$ PrimitiveGroup(32,3) $\cong \mathrm{ASL}(5,2)$, and $s(G)\geqslant 361$. We remark here that 361 is a lower bound obtained by Gap [3] using random search, but we do not get the best possible lower bound since the current one works for our purpose. $\square$

Remark 2.6. Using Gap [3], we can easily obtain the maximum order of primitive groups of degree n which do not contain the simple group $\mathrm{A}_{n}$ as a composition factor. For convenience, we list some results in Table 2.

Table 2 The maximal order of primitive groups not containing $\mathrm{A}_n$

3 Proof of Theorem 1.1

In this section, we shall give the explicit lower bounds $\inf \left( \frac{\log _2 s(G)}{n}\right)$ where G does not contain $\mathrm{A}_\ell$ with $\ell >t$ for $t\in [5, 16])$ as a composition factor. In what follows, we only provide the detailed proof for the case $t=5$. The proof for the remaining cases is analogous up to replacing a few numbers appropriately.

Proposition 3.1

We have the following equality

$$\begin{aligned} \inf \left( \frac{\log _2 s(G)}{n}\right) = \lim _{k \mapsto +\infty } \frac{\log _2 s(\mathrm{M}_{24} \wr \overbrace{\mathrm{S}_5\wr \cdots \wr \mathrm{S}_5}^\text {k terms})}{24 \cdot 5^k}, \end{aligned}$$

where the infimum is taken over all permutation groups G not containing any composition factor $\mathrm{A}_\ell$ with $\ell > 5$, and n denotes the degree of G.

To complete the proof of Proposition 3.1, we need a number of preparatory lemmas. Set

$$\begin{aligned} a_k= \frac{\log _2 s(\mathrm{M}_{24}\wr \mathrm{S}_5\wr \cdots \wr \mathrm{S}_5)}{24 \cdot 5^k}\,with\, k\geqslant 0\, \hbox {and}\, M={\lim \limits _{k \rightarrow +\infty }a_{k}}. \end{aligned}$$

For convenience of notation, we first define a sequence $\{s_k\}_{k\geqslant 0}$, where

$$\begin{aligned} s_{0}=s(\mathrm{M}_{24})=49 \,\hbox {and}\, s_{k+1}={{s_{k}+4}\atopwithdelims ()5} for \,k\geqslant 0. \end{aligned}$$

It is clear that the sequence $\{s_k\}_{k\geqslant 0}$ is strictly increasing. By the definition, $a_k=\frac{\log _2{s_k}}{24 \cdot 5^k}$. It is easy to see that $a_k>0$. Now, we consider the sequence $\{a_k\}_{k\geqslant 0}$. Since

$$\begin{aligned} a_{k+1}=\frac{\log _2{{{s_{k}+4} \atopwithdelims ()5}}}{24 \cdot 5^{k+1}}=\frac{\log _2 \left( \frac{(s_{k}+4)(s_{k}+3)(s_{k}+2)(s_{k}+1) s_{k}}{120}\right) }{24 \cdot 5^{k+1}} < \frac{\log _2 (s_k)^5}{24\cdot 5^{k+1}}=a_k, \end{aligned}$$

the sequence $\{a_k\}_{k \geqslant 0}$ is strictly decreasing, and so the $\lim \limits _{k \mapsto \infty } a_k$ exists.

Applying Lemma 2.3, we calculate that

$$\begin{aligned} a_{0}\approx 0.233946243505,~ a_{1}\approx 0.178770507941, ~ a_{2}\approx 0.167259031994. \end{aligned}$$

Lemma 3.2

Let G be a primitive permutation group of degree n. If G does not contain any $\mathrm{A}_\ell$ ($\ell > 5$) as a composition factor, then $rs(G)\geqslant a_{2}$.

Proof

If $n\geqslant 25$ and $n\ne 32$, then by Lemma 2.4 (ii), we have $s(G)\geqslant 2^{n}/|G|\geqslant 2^{n}/2^{0.76n}=2^{0.24n}$, and so $rs(G)=\frac{\log _{2}s(G)}{n}\geqslant 0.24\geqslant a_{2}$.

If $n\leqslant 24$, then $s(G)\geqslant n+1$. Thus $rs(G)\geqslant 0.193494007907\geqslant a_{2}$.

If $n=32$, then by Table 1, we have $s(G)\geqslant 361$. In this case, one has

$$\begin{aligned} rs(G)\geqslant 0.265495469577\geqslant a_{2}.\end{aligned}$$

This completes the proof. $\square$

Lemma 3.3

Let G be a transitive permutation group of degree n induced from a permutation group H of degree m, where G does not contain any $\mathrm{A}_\ell$ ($\ell > 5$) as a composition factor. Let $\alpha _1=120^{\frac{1}{4}}$. If $\frac{\log _{2}s(H)}{m}-\frac{\log _{2}\alpha _1}{m}\geqslant \beta$, then $rs(G)\geqslant \beta$.

Proof

We may assume that $G\lesssim H\wr P_{1}\wr \cdots \wr P_z$, where the $P_{i}$ are primitive permutation groups and $\deg (P_{i})=k_{i}$ for $1\leqslant i\leqslant z$. Using Gap [3] and Lemma 2.4, it is straightforward to verify that $|P_{1}|\leqslant \alpha _1^{k_{1}-1}$. Since $\frac{\log _{2}s(H)}{m}-\frac{\log _{2}\alpha _1}{m}\geqslant \beta$, we have $s(H)\geqslant \alpha _1\cdot 2^{m\beta }$. On the other hand, Lemma 2.3 implies

$$\begin{aligned} s(H\wr P_{1})\geqslant s(H)^{k_{1}}/\alpha _1^{k_{1}-1}\geqslant \alpha _1\cdot 2^{mk_{1}\beta }. \end{aligned}$$

Consequently,

$$\begin{aligned} \frac{\log _{2}s(H\wr P_{1})}{mk_{1}}-\frac{\log _{2}\alpha _1}{mk_{1}}\geqslant \frac{\log _{2}\alpha _1\cdot 2^{mk_{1}\beta }}{mk_{1}}-\frac{\log _{2}\alpha _1}{mk_{1}}=\beta . \end{aligned}$$

Now, let $K=H\wr P_{1}$. Then by Lemma 2.3, one has

$$\begin{aligned} s(K\wr P_{2})\geqslant s(K)^{k_{2}}/\alpha _1^{k_{2}-1}\geqslant \alpha _1\cdot 2^{m(k_{1}k_{2})\beta }. \end{aligned}$$

This implies that

$$\begin{aligned} \frac{\log _{2}s(H\wr P_{1}\wr P_{2})}{mk_{1}k_{2}}- \frac{\log _{2}\alpha _1}{mk_{1}k_{2}}\geqslant \frac{\log _{2}\alpha _1\cdot 2^{m(k_{1}k_{2})\beta }}{mk_{1}k_{2}}-\frac{\log _{2}\alpha _1}{mk_{1}k_{2}}=\beta . \end{aligned}$$

In a similar fashion, one can prove $\frac{\log _{2}s(H\wr P _1\wr \cdots \wr P_z)}{mk_{1}\cdots k_z}-\frac{\log _{2}\alpha _1}{mk_{1}\cdots k_z}\geqslant \beta$, where $n=mk_{1}\cdots k_z$. Since $G\lesssim H\wr P_{1}\wr \cdots \wr P_z$, we derive from Lemma 2.2 that $s(G)\geqslant s(H\wr P_{1}\wr \cdots \wr P_z)$, and thereby $rs(G)\geqslant \frac{\log _{2}s(H\wr P_{1}\wr \cdots \wr P_z)}{n}\geqslant \beta$. This completes the proof. $\square$

Lemma 3.4

Let G be a transitive permutation group of degree n, which is induced from a primitive permutation group H of degree m. If G does not contain any $\mathrm{A}_\ell$ ($\ell > 5$) as a composition factor, and $H\ncong \mathrm{M}_{24}$, then $rs(G)\geqslant a_{2}$.

Proof

In view of Proposition 3.3, it suffices to prove that:

$$\begin{aligned} \frac{\log _{2}s(H)}{m}-\frac{\log _{2}\alpha _1}{m}\geqslant a_{2}. \end{aligned}$$

(1)

Suppose that $m\geqslant 25$ and $m\ne 32$. By Lemma 2.4, we have $|H|\leqslant 2^{0.76m}$. Since $s(H)\geqslant 2^{m}/|H|$, it follows that

$$\begin{aligned} \frac{\log _{2}s(H)}{m}-\frac{\log _{2}\alpha _1}{m}\geqslant \frac{\log _{2}(2^{m}/2^{0.76m})}{m}-\frac{\log _{2}120}{4\cdot m}\geqslant 0.170931094044\geqslant a_{2}. \end{aligned}$$

In this case, inequality (1) follows.

Now, we denote by $f(H, m, \alpha _1)$ the number $\frac{\log _{2}s(H)}{m}-\frac{\log _{2}\alpha _1}{m}$. If $12\leqslant m\leqslant 24$ or $m=32$, then by Table 1, we can check all the minimum values of $f(H, m, \alpha _1)$ using Table 3.

Table 3 $f(H, m, \alpha _1)$ ($12\leqslant m \leqslant 24$ or $m=32$)

If $4\leqslant m\leqslant 11$, then $s(H)\geqslant m+1$, and thereby

$$\begin{aligned} \frac{\log _{2}s(H)}{m}-\frac{\log _{2}\alpha _1}{m} \geqslant \frac{\log _{2}(m+1)}{m}-\frac{\log _{2}120}{4m} \geqslant 0.168930895619\geqslant a_{2}. \end{aligned}$$

If $m=3$ or 2, then by Lemma 3.2, we may assume that

$$\begin{aligned} G\lesssim H\wr P_{1}\wr P_{2}\cdots \wr P_z, where \deg (P_{1})=m_{1}. \end{aligned}$$

Let $K=H\wr P_{1}$. By Proposition 3.3, it is sufficient to check that inequality (2) holds.

$$\begin{aligned} \frac{\log _{2}s(K)}{mm_{1}} -\frac{\log _{2}\alpha _1}{mm_{1}}\geqslant a_{2}. \end{aligned}$$

(2)

Denote by $f(K, m, m_{1}, \alpha _1)$ the number $\frac{\log _{2}s(K)}{mm_{1}}-\frac{\log _{2}\alpha _1}{mm_{1}}$. Then we need to check the inequality $f(K,m,m_{1},\alpha _1)\geqslant a_{2}$ holds. In what follows, we distinguish two cases.

Case 1  $m=3$. In this case, we know that $s(H)\geqslant 4$. By Lemma 2.3, one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 4^{m_{1}}/|P_{1}|$. Thus

$$\begin{aligned} f(K, 3, m_{1}, \alpha _1)\geqslant \frac{m_{1}\log _{2}4 -\log _{2}|P_{1}|}{3m_{1}}-\frac{\log _{2}120}{12m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K, 3, m_{1},\alpha _1)\geqslant \frac{m_{1}\log _{2}4 -\log _{2}2^{0.76m_{1}}}{3m_{1}} -\frac{\log _{2}120}{12m_{1}}\geqslant 0.390310364681\geqslant a_{2}. \end{aligned}$$

If $m_{1}\leqslant 16$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}3+m_{1}\\ 3\end{array}}\right)$, and thereby

$$\begin{aligned} f(K,3,m_1, \alpha _1)\geqslant \frac{\log _{2} \left( {\begin{array}{c}3+m_{1}\\ 3\end{array}}\right) }{3m_{1}}-\frac{\log _{2}120}{12m_{1}} \geqslant 0.170700629302\geqslant a_{2}. \end{aligned}$$

If $17\leqslant m_{1}\leqslant 24$ or $m_{1}=32$, then by Table 2, we can calculate the minimum values of $f(K,3,m_1,\alpha _1)$ as follows (Table 4).

Table 4 $f(K, 3, m_{1}, \alpha _1)$ ($17\leqslant m_{1}\leqslant 24$ or $m_{1}=32$)

Case 2  $m=2$. In this case, we know that $s(H)\geqslant 3$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 3^{m_{1}}/|P_{1}|$. Consequently,

$$\begin{aligned} f(K,2,m_{1},\alpha _1)\geqslant \frac{m_{1}\log _{2}3-\log _{2}|P_{1}|}{2m_{1}} -\frac{\log _{2}120}{8m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,2,m_1,\alpha _1)\geqslant \frac{m_{1}\log _{2}3 -\log _{2}2^{0.76m_{1}}}{2m_{1}} -\frac{\log _{2}120}{8m_{1}}\geqslant 0.277946797383\geqslant a_{2}. \end{aligned}$$

If $m_{1}\leqslant 17$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}2+m_{1}\\ 2\end{array}}\right)$. Consequently

$$\begin{aligned} f(K,2,m_{1},\alpha _1)\geqslant \frac{\log _{2}\left( {\begin{array}{c}2+m_{1}\\ 2\end{array}}\right) }{8m_{1}} -\frac{\log _{2}120}{8m_{1}}\geqslant 0.167386172529\geqslant a_{2}. \end{aligned}$$

If $18\leqslant m_{1}\leqslant 24$ or $m_{1}=32$, then by Table 2, we can calculate the minimum values of $f(K,2,m_{1}, \alpha _1)$ as follows (Table 5).

Table 5 $f(K, 2, m_{1}, \alpha _1)$ ($18\leqslant m_{1}\leqslant 24$ or $m_{1}=32$)

Lemma 3.5

Let $G\lesssim H\wr P_{1}\wr \cdots \wr P_z$ be a transitive permutation group of degree n which does not contain any $\mathrm{A}_\ell$ $(\ell >5)$ as a composition factor, where $H\cong \mathrm{M}_{24}$, and all the $P_{i}$ are primitive groups. If $\deg (P_{1})\ne 5$, then $rs(G)\geqslant a_{2}$.

Proof

By Table 1, we have $s(H)=49$. Let $K=H\wr P_1$, where $\deg (P_{1})\ne 5$. By Proposition 3.3, it suffices to prove that

$$\begin{aligned} \frac{\log _{2}s(K)}{24m_{1}} -\frac{\log _{2}\alpha _1}{24m_{1}}\geqslant a_{2}. \end{aligned}$$

Denote by $f(K,m_{1},\alpha _1)$ the number $\frac{\log _{2}s(K)}{24m_{1}} -\frac{\log _{2}\alpha _1}{24m_{1}}$. By Lemma 2.3 (i), we have that $s(K)\geqslant 49^{m_{1}}/|P_{1}|$, and thereby

$$\begin{aligned} f(K,m_{1},\alpha _1)\geqslant \frac{m_{1}\log _{2}49 -\log _{2}|P_{1}|}{24m_{1}}-\frac{\log _{2}120}{96m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,m_{1},\alpha _1)\geqslant \frac{m_{1}\log _{2}49 -\log _{2}2^{0.76m_{1}}}{24m_{1}}-\frac{\log _{2}120}{96m_{1}}\geqslant 0.199401705757\geqslant a_{2}. \end{aligned}$$

If $m_{1}=2$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}50\\ 48\end{array}}\right)$, and so

$$\begin{aligned} f(K,2,\alpha _1)\geqslant \frac{\log _{2}\left( {\begin{array}{c}50\\ 48\end{array}}\right) }{48} -\frac{\log _{2}120}{192}\geqslant 0.177746737187\geqslant a_{2}. \end{aligned}$$

If $3\leqslant m_{1}\leqslant 24$ or $m_{1}=32$, then we apply Table 2 to calculate the minimum values of $f(K,m_{1},\alpha _1)$ as follows (Table 6).

Table 6 $f(K, m_{1}, \alpha _1)$ ($3\leqslant m_{1}\leqslant 24$ or $m_{1}=32$)

This completes the proof of this proposition. $\square$

Lemma 3.6

Let $H\cong \mathrm{M}_{24}\wr \mathrm{S}_5\wr \cdots \wr \mathrm{S}_5$ be a transitive permutation group of degree $24\cdot 5^k$ with $k\geqslant 0$. Let $P_{1},\dots ,P_\ell$ be primitive permutation groups. If $\deg (P_{1})\ne 5$, then $rs(H\wr \mathrm{S}_{5}\wr \mathrm{S}_{5})\leqslant rs(H\wr P_{1}\wr \cdots \wr P_\ell )$.

Proof

Let $s(H)=A$ and $n=\deg (H)=24\cdot 5^{k}$. Then $A\geqslant s(\mathrm{M}_{24})=49$. Let $B=s(H\wr \mathrm{S}_{5})$. By Lemma 2.3 (ii) we have $B=\left( {\begin{array}{c}A+4\\ 5\end{array}}\right) =\frac{(A+4)(A+3)(A+2)(A+1)A}{120}$, and thus $B+4\leqslant \frac{(A+4)^{5}}{120}$. Again using Lemma 2.3 (ii), one has

$$\begin{aligned} s(H\wr \mathrm{S}_{5}\wr \mathrm{S}_{5})=\left( {\begin{array}{c}B+4\\ 5\end{array}}\right) =\frac{(B+4)(B+3)(B+2)(B+1)B}{120}, \end{aligned}$$

and so $s(H\wr \mathrm{S}_{5}\wr \mathrm{S}_{5})\leqslant \frac{[\frac{(A+4)^{5}}{120}]^{5}}{120}$. Thus, we have the following inequality

$$\begin{aligned} rs(H\wr \mathrm{S}_{5}\wr \mathrm{S}_{5})=\frac{\log _{2}s(H\wr \mathrm{S}_{5}\wr \mathrm{S}_{5})}{n\cdot 5^{2}}\leqslant \frac{\log _{2} \frac{[\frac{(A+4)^{5}}{120}]^{5}}{120}}{n\cdot 5^{2}}\\ =\frac{\log _{2}(A+4)}{n}-\frac{\log _{2}120}{5n} -\frac{\log _{2}120}{25n}. \end{aligned}$$

On the other hand, let $\deg (P_{1})=m_{1}$. By Lemma 2.3 (i), we have $s(H\wr P_{1})\geqslant s(H)^{m_{1}}/|P_{1}|=A^{m_{1}}/|P_{1}|$, and hence

$$\begin{aligned} \frac{\log _{2}s(H\wr P_{1})}{nm_{1}}-\frac{\log _{2}\alpha _1}{nm_{1}}\geqslant \frac{\log _{2}(A)}{n}-\frac{\log _{2}|P_{1}|}{nm_{1}} -\frac{\log _{2}\alpha _1}{nm_{1}}. \end{aligned}$$

Let $G=H\wr P_{1}\wr \cdots \wr P_\ell$. Then it is enough to verify the following inequality:

$$\begin{aligned} \frac{\log _{2}(A+4)}{n}-\frac{\log _{2}120}{5n} -\frac{\log _{2}120}{25n}\leqslant \frac{\log _{2}(A)}{n}-\frac{\log _{2}|P_{1}|}{nm_{1}} -\frac{\log _{2}\alpha _1}{nm_{1}}. \end{aligned}$$

(3)

Consequently, (3) holds if and only if

$$\begin{aligned} \frac{\log _{2}|P_{1}|}{m_{1}}+\frac{\log _{2}\alpha _1}{m_{1}}\leqslant \frac{\log _{2}120}{5}+\frac{\log _{2}120}{25} -\log _{2}(\frac{A+4}{A}). \end{aligned}$$

Since $A\geqslant 49$, we have $\log _{2}(\frac{A+7}{A})\leqslant 0.113210611045$. Thus, we only need to check the following inequality:

$$\begin{aligned}\frac{\log _{2}|P_{1}|}{m_{1}} +\frac{\log _{2}\alpha _1}{m_{1}}\leqslant 1.544443132498.\end{aligned}$$

Denote by $f(P_{1},m_{1},\alpha _1)$ the number $\frac{\log _{2}|P_{1}|}{m_{1}}+\frac{\log _{2}\alpha _1}{m_{1}}$. If $m_{1}>24$, then by Lemma 2.4 (ii), we have $|P_{1}|<2^{m_{1}}$, and so

$$\begin{aligned} f(P_{1}, m_{1}, \alpha _1)<\frac{\log _{2}2^{m_{1}}}{m_{1}}+\frac{\log _{2}\alpha _1}{m_{1}}< 1+\frac{\log _{2}120}{24\times 4}\approx 1.071946777038. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Table 2, we can check each minimum value of function $f(P_{1},m_{1},\alpha _1)$ using Table 7.

Table 7 $f(P_{1}, m_{1}, \alpha _1)$ ($2\leqslant m_{1}\leqslant 24$)

This completes the proof of this lemma. $\square$

We are now ready to complete the proof of proposition 3.1.

Proof of Proposition 3.1: Note that $a_{k}={\frac{\log _{2}s(\mathrm{M}_{24}\wr \mathrm{S}_{5}\wr \cdots \wr \mathrm{S}_{_{5}})}{24\cdot 5^{k}}}$ and $M={\lim \limits _{k \rightarrow +\infty }a_{k}}$. Since the sequence $\{a_k\}_{k \geqslant 0}$ is strictly decreasing, this implies that $M<a_{2}$. If G is primitive, then by Lemma 3.2, we have $rs(G)\geqslant a_{2}$. Now, if G is not primitive, then G is induced from $\mathrm{M}_{24}$ by Lemma 3.4. Combining Lemma 3.5 and Lemma 3.6, one has

$$\begin{aligned} \inf \Big (\frac{\log _{2}s(G)}{n}\Big )={\lim _{k \rightarrow +\infty }\frac{\log _{2}s(\mathrm{M}_{24}\wr \overbrace{\mathrm{S}_{5}\wr \cdots \wr \mathrm{S}_{_{5}}}^{\text{ k } \ terms})}{24\cdot 5^{k}}}. \end{aligned}$$

This completes the proof of this proposition. $\square$

4 Proof of Theorem 1.2

In this section, we shall give the explicit lower bounds $\inf \left( \frac{\log _2 s(G)}{n}\right)$ where G does not contain any alternating group $\mathrm{A}_\ell$ with $\ell >t$ for $t\in [17,166]$ as a composition factor. We give the detailed proof for the case where $t=17$. Essentially the identical proof works in the remaining cases up to replacing a few numbers appropriately.

Proposition 4.1

We have

$$\begin{aligned} \inf \left( \frac{\log _2 s(G)}{n}\right) = \lim _{k \mapsto +\infty } \frac{\log _2 s(\overbrace{\mathrm{S}_{17}\wr \mathrm{S}_{17} \wr \cdots \wr \mathrm{S}_{17}}^ \text {k terms})}{17^k}. \end{aligned}$$

where the infimum is taken over all permutation groups G not containing any composition factor $\mathrm{A}_\ell$ with $\ell > 17$, and n denotes the degree of G.

For the remaining of section, we prove Proposition 4.1 by a series of lemmas. Let

$$\begin{aligned} b_k= \frac{\log _2 s(\mathrm{S}_{17}\wr \mathrm{S}_{17} \wr \cdots \wr \mathrm{S}_{17})}{17^k} and M={\lim \limits _{k \rightarrow +\infty }b_{k}}, where k\geqslant 1. \end{aligned}$$

We first define a sequence $\{s_k\}_{k \geqslant 1}$ where $s_{1}=s(\mathrm{S}_{17})=18$ and $s_{k+1}={{s_{k}+16} \atopwithdelims ()17}$ for $k\geqslant 1$. Clearly the sequence $\{s_k\}_{k\geqslant 1}$ is strictly increasing. By the definition, we have $b_k=\frac{\log _2{s_k}}{17^k}$. It is easy to see $b_k>0$. Noting that

$$\begin{aligned} b_{k+1}=\frac{\log _2{{{s_{k}+16} \atopwithdelims ()17}}}{17^{k+1}}=\frac{\log _2 (\frac{(s_{k}+16)(s_{k}+15)\cdots (s_{k}+2)(s_{k}+1) s_{k}}{17!})}{17^{k+1}} < \frac{\log _2 (s_k)^{17}}{{17}^{k+1}}=b_k, \end{aligned}$$

the sequence $\{b_k\}_{k\geqslant 1}$ is strictly decreasing, and hence M exists.

Applying Lemma 2.3, we calculate that

$$\begin{aligned} b_{1}\approx 0.245289705967 \hbox {~and~} b_{2}\approx 0.107681363290.\end{aligned}$$

Lemma 4.2

Let G be a primitive permutation group of degree n. If G does not contain any $\mathrm{A}_\ell ~(\ell >17)$ as a composition factor, then $rs(G)\geqslant b_{2}$.

Proof

If $n\geqslant 25$ and $n\ne 32$, then by Theorem 2.4 (ii), we have $s(G)\geqslant 2^{n}/|G|\geqslant 2^{0.24n}$, and so $rs(G)=\frac{\log _{2}s(G)}{n}\geqslant 0.24\geqslant b_{2}$.

If $n\leqslant 24$, then $s(G)\geqslant n+1$, and thus

$$\begin{aligned}rs(G)=\frac{\log _{2}s(G)}{n}\geqslant \frac{\log _{2}(n+1)}{n}\geqslant 0.193494007907\geqslant b_{2}.\end{aligned}$$

If $n=32$, then by Table 1, we have $s(G)\geqslant 361$. In this case, one has

$$\begin{aligned} rs(G)=\frac{\log _{2}s(G)}{32} \geqslant 0.265495469577\geqslant b_{2}. \end{aligned}$$

This completes the proof. $\square$

Arguing as in Proposition 3.3, we have the following conclusion.

Lemma 4.3

Let G be a transitive permutation group of degree n induced from a permutation group H of degree m, where G does not contain any $\mathrm{A}_\ell ~ (\ell >17)$ as a composition factor. Let $\alpha _2=(17!)^{\frac{1}{16}}$. If $\frac{\log _{2}s(H)}{m}- \frac{\log _{2}\alpha _2}{m}\geqslant \beta$, then $rs(G)\geqslant \beta$.

Lemma 4.4

Let G be a transitive permutation group of degree n, which be induced from a primitive permutation group H of degree m. If G does not contain any $\mathrm{A}_\ell ~(\ell >17)$ as a composition factor, and $H\ncong \mathrm{S}_{17}$, then $rs(G)\geqslant b_{2}$.

Proof

In view of Lemma 4.3, it suffices to prove the following inequality:

$$\begin{aligned} \frac{\log _{2}s(H)}{m}-\frac{\log _{2}\alpha _2}{m}\geqslant b_{2}. \end{aligned}$$

(4)

Suppose that $m\geqslant 25$ and $m\ne 32$. By Lemma 2.4 (ii), we have $|H|\leqslant 2^{0.76m}$. Since $s(H)\geqslant 2^{m}/|H|$, one has

$$\begin{aligned} \frac{\log _{2}s(H)}{m}-\frac{\log _{2}\alpha _2}{m}\geqslant \frac{\log _{2}(2^{m}/2^{0.76m})}{m}-\frac{\log _{2}17!}{16m} \geqslant 0.119155991722>b_{2}. \end{aligned}$$

In this case, inequality (4) follows.

Now, we use $f(H,m,\alpha _2)$ to denote $\frac{\log _{2}s(H)}{m}-\frac{\log _{2}\alpha _2}{m}$. If $13\leqslant m\leqslant 24$ or $m=32$, then by Table 2, we can verify $f(H,m,\alpha _2)\geqslant 0.108067068215$, and so (4) holds.

If $2\leqslant m\leqslant 12$, then by Lemma 4.2, we may assume that $G\lesssim H\wr P_{1}\wr P_{2}\wr \cdots \wr P_k$, where $\deg (P_{1})=m_{1}$. Let $K=H\wr P_{1}$. By Lemma 3.3, it is enough to check the following inequality:

$$\begin{aligned} \frac{\log _{2}s(K)}{mm_{1}}-\frac{\log _{2}\alpha _2}{mm_{1}}\geqslant b_{2}. \end{aligned}$$

(5)

Denote by $f(K,m,m_{1},\alpha _2)$ the number $\frac{\log _{2}s(K)}{mm_{1}}-\frac{\log _{2}\alpha _2}{mm_{1}}$. Then we only need to check the following inequality:

$$\begin{aligned} f(K,m,m_{1},\alpha _2)=\frac{\log _{2}s(K)}{mm_{1}} -\frac{\log _{2}17!}{16mm_{1}}\geqslant b_{2}. \end{aligned}$$

In what follows, we separate the argument into eleven cases.

Case 1  $m=9$. In this case, we know that $s(H)\geqslant 10$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 10^{m_{1}}/|P_{1}|$, and thereby

$$\begin{aligned}f(K,9,m_{1},\alpha _2)\geqslant \frac{m_{1}\log _{2}10 -\log _{2}|P_{1}|}{9m_{1}}-\frac{\log _{2}17!}{144m_{1}}.\end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K, 9, m_{1}, \alpha _2)\geqslant 0.27123156517\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 21$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}9+m_{1}\\ 9\end{array}}\right)$, and thereby

$$\begin{aligned} f(K,9,m_{1},\alpha _2)\geqslant 0.109783771278\geqslant b_{2}. \end{aligned}$$

If $22\leqslant m_{1}\leqslant 24$ or $m_{1}=32$, then by Table 2, we can check all the values of $f(K,9,m_{1}, \alpha _2)\geqslant 0.226101898040$, and so (5) follows.

Case 2  $m=8$. In this case, we know that $s(H)\geqslant 10$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 10^{m_{1}}/|P_{1}|$, and so

$$\begin{aligned} f(K,8,m_{1},\alpha _2)\geqslant \frac{m_{1}\log _{2}10 -\log _{2}|P_{1}|}{8m_{1}}-\frac{\log _{2}17!}{128m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,8,m_{1},\alpha _2)\geqslant 0.305135510826\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}9+m_{1}\\ 9\end{array}}\right)$, and so

$$\begin{aligned} f(K,8,m_{1},\alpha _2)\geqslant 0.115519620447\geqslant b_{2}. \end{aligned}$$

Case 3  $m=7$. In this case, we know $s(H)\geqslant 10$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 10^{m_{1}}/|P_{1}|$, and thus

$$\begin{aligned} f(K, 7, m_{1}, \alpha _2)\geqslant \frac{m_{1}\log _{2}10 -\log _{2}|P_{1}|}{7m_{1}}-\frac{\log _{2}17!}{112m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,7,m_{1},\alpha _2)\geqslant 0.348726298087\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}9+m_{1}\\ 9\end{array}}\right)$, and thereby

$$\begin{aligned} f(K,7,m_{1},\alpha _2)\geqslant 0.132022423368\geqslant b_{2}. \end{aligned}$$

Case 4  $m=12$. For this case, we know that $s(H)\geqslant 14$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 14^{m_{1}}/|P_{1}|$, and consequently

$$\begin{aligned} f(K,12,m_{1}, \alpha _2)\geqslant \frac{m_{1}\log _{2}14-\log _{2}|P_{1}|}{12m_{1}} -\frac{\log _{2}17!}{192m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K, 12, m_{1}, \alpha _2)\geqslant 0.243875909482\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 20$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}13+m_{1}\\ 13\end{array}}\right)$, and so

$$\begin{aligned} f(K,12,m_{1},\alpha _2)\geqslant 0.108638661215\geqslant b_{2}. \end{aligned}$$

If $21\leqslant m_{1}\leqslant 24$ or $m_{1}=32$, then by Table 2, we can calculate

$$\begin{aligned} f(K,9,m_{1},\alpha _2)=0.210028659127\geqslant b_{2}. \end{aligned}$$

Case 5  $m=11$. In this case, we have $s(H)\geqslant 14$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 14^{m_{1}}/|P_{1}|$, and thus

$$\begin{aligned} f(K,11,m_{1},\alpha _2)\geqslant \frac{m_{1}\log _{2}14 -\log _{2}|P_{1}|}{11m_{1}}-\frac{\log _{2}17!}{176m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,11,m_{1},\alpha _2)\geqslant 0.266046446707\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}13+m_{1}\\ 13\end{array}}\right)$, and so

$$\begin{aligned} f(K,11,m_{1},\alpha _2)\geqslant 0.108746702344\geqslant b_{2}. \end{aligned}$$

Case 6  $m=10$. In this case, we know that $s(H)\geqslant 14$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 14^{m_{1}}/|P_{1}|$, and thus

$$\begin{aligned}f(K,10,m_{1},\alpha _2)\geqslant \frac{m_{1}\log _{2}14-\log _{2}|P_{1}|}{10m_{1}} -\frac{\log _{2}17!}{160m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,11,m_{1}, \alpha _2)\geqslant 0.292651091378\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}13+m_{1}\\ 13\end{array}}\right)$, and so

$$\begin{aligned} f(K,10,m_{1},\alpha _2)\geqslant 0.119621372578\geqslant b_{2}. \end{aligned}$$

Case 7  $m=6$. Then $s(H)\geqslant 8$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 8^{m_{1}}/|P_{1}|$, and consequently

$$\begin{aligned} f(K,6,m_{1},\alpha _2)\geqslant \frac{m_{1}\log _{2}8 -\log _{2}|P_{1}|}{6m_{1}}-\frac{\log _{2}17!}{96m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,6,m_{1}, \alpha _2)\geqslant 0.353192665287\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}7+m_{1}\\ 7\end{array}}\right)$, and so

$$\begin{aligned} f(K,6,m_{1},\alpha _2)\geqslant 0.127120125055\geqslant b_{2}. \end{aligned}$$

Case 8  $m=5$. In this case, we know that $s(H)\geqslant 6$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 6^{m_{1}}/|P_{1}|$, and thus

$$\begin{aligned} f(K,5,m_{1}, \alpha _2)\geqslant \frac{m_{1}\log _{2}6 -\log _{2}|P_{1}|}{5m_{1}}-\frac{\log _{2}17!}{80m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,5,m_{1},\alpha _2)\geqslant 0.340823698489\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}5+m_{1}\\ 5\end{array}}\right)$, and so

$$\begin{aligned}f(K,5,m_{1},\alpha _2)\geqslant 0.115304404373\geqslant b_{2}. \end{aligned}$$

Case 9  $m=4$. In this case, we know that $s(H)\geqslant 5$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 5^{m_{1}}/|P_{1}|$, and so

$$\begin{aligned} f(K,4,m_{1}, \alpha _2)\geqslant \frac{m_{1} \log _{2}5-\log _{2}|P_{1}|}{4m_{1}} -\frac{\log _{2}17!}{64m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,4,m_{1},\alpha _2)\geqslant 0.360271021652\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}4+m_{1}\\ 4\end{array}}\right)$, and thereby

$$\begin{aligned} f(K,4,m_{1},\alpha _2)\geqslant 0.117713287755\geqslant b_{2}. \end{aligned}$$

Case 10  $m=3$. Then $s(H)\geqslant 4$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 4^{m_{1}}/|P_{1}|$, and so

$$\begin{aligned} f(K,3,m_{1},\alpha _2)\geqslant \frac{m_{1}\log _{2}4 -\log _{2}|P_{1}|}{3m_{1}}-\frac{\log _{2}17!}{48m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,3,m_{1},\alpha _2)\geqslant 0.373051997241\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}3+m_{1}\\ 3\end{array}}\right)$, and so

$$\begin{aligned} f(K,3,m_{1},\alpha _2)\geqslant 0.117960009756\geqslant b_{2}. \end{aligned}$$

Case 11  $m=2$. Then $s(H)\geqslant 3$. By Lemma 2.3 (i), one has $s(K)\geqslant s(H)^{m_{1}}/|P_{1}|\geqslant 3^{m_{1}}/|P_{1}|$, and so

$$\begin{aligned} f(K,2,m_{1},\alpha _2)\geqslant \frac{m_{1}\log _{2}3 -\log _{2}|P_{1}|}{2m_{1}}-\frac{\log _{2}17!}{32m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned}f(K,2,m_{1},\alpha _2)\geqslant 0.352059246222\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}2+m_{1}\\ 2\end{array}}\right)$, and consequently,

$$\begin{aligned} f(K,2,m_{1},\alpha _2)\geqslant 0.110899910437\geqslant b_{2}. \end{aligned}$$

This completes the proof of this lemma. $\square$

Lemma 4.5

Let $G\lesssim \mathrm{S}_{17}\wr P_{1}\wr \cdots \wr P_k$ be a transitive permutation group of degree n which does not contain any $\mathrm{A}_\ell ~(\ell >17)$ as a composition factor, where all the $P_{i}$ are primitive groups. If $\deg (P_{1})\ne 17$, then $rs(G)\geqslant b_{2}$.

Proof

By Table 1, we know that $s(H)=18$. Let $K=\mathrm{S}_{17}\wr P_{1}$, where $\deg (P_{1})\ne 17$. By Lemma 4.3, it suffices to prove that

$$\begin{aligned}\frac{\log _{2}s(K)}{17m_{1}} -\frac{\log _{2}\alpha _2}{17m_{1}}\geqslant b_{2}. \end{aligned}$$

Denote by $f(K,m_{1},\alpha _2)$ the number $\frac{\log _{2}s(K)}{17m_{1}}- \frac{\log _{2}\alpha _2}{17m_{1}}$. By Lemma 2.3 (i), we have $s(K)\geqslant 18^{m_{1}}/|P_{1}|$, and so

$$\begin{aligned}f(K,m_{1},\alpha _2)\geqslant \frac{m_{1}\log _{2}18 -\log _{2}|P_{1}|}{17m_{1}}-\frac{\log _{2}17!}{272m_{1}}. \end{aligned}$$

If $m_{1}\geqslant 25$ and $m_{1}\ne 32$, then by Lemma 2.4 (iii), we have $|P_{1}|\leqslant 2^{0.76m_{1}}$, and so

$$\begin{aligned} f(K,m_{1}, \alpha _2)\geqslant 0.193475352539\geqslant b_{2}. \end{aligned}$$

If $m_{1}\leqslant 13$, then by Lemma 2.3 (ii), we have $s(K)\geqslant \left( {\begin{array}{c}17+m_{1}\\ 17\end{array}}\right)$, and so

$$\begin{aligned} f(K,m_{1},\alpha _2)\geqslant 0.107757777700\geqslant b_{2}. \end{aligned}$$

If $14\leqslant m_{1}\leqslant 24$ or $m_{1}=32$, then by Table 2, we calculate

$$\begin{aligned} f(K,m_{1},\alpha _2)\geqslant 0.166906219845\geqslant b_{2}. \end{aligned}$$

$\square$

Lemma 4.6

Let $H\cong \mathrm{S}_{17}\wr \mathrm{S}_{17}\wr \cdots \wr \mathrm{S}_{17}$ be a transitive permutation group of degree $17^{k}$ with $k\geqslant 1$. Let $P_{1},\dots , P_\ell$ be primitive permutation groups. If $\deg (P_{1})\ne 17$, then $rs(H\wr \mathrm{S}_{17}\wr \mathrm{S}_{17})\leqslant rs(H\wr P_{1}\wr \cdots \wr P_\ell )$.

Proof

Let $s(H)=A$ and $n=\deg (H)=17^{k}$. Then $A\geqslant s(\mathrm{S}_{17})=18$. Let $B=s(H\wr \mathrm{S}_{17})$. By Lemma 2.3 (ii), we have $B=\left( {\begin{array}{c}A+16\\ 17\end{array}}\right) =\frac{(A+16)(A+15) \cdots (A+2)(A+1)A}{17!}$, and so

$$\begin{aligned} B+16\leqslant \frac{(A+16)^{17}}{17!}. \end{aligned}$$

Again by Lemma 2.3 (ii), one has

$$\begin{aligned} s(H\wr \mathrm{S}_{17}\wr \mathrm{S}_{17})=\left( {\begin{array}{c}B+16\\ 17\end{array}}\right) =\frac{(B+16)(B+15) \cdots (B+2)(B+1)B}{17!}, \end{aligned}$$

and thereby

$$\begin{aligned} s(H\wr \mathrm{S}_{17}\wr \mathrm{S}_{17})\leqslant \frac{[\frac{(A+16)^{17}}{17!}]^{17}}{17!}. \end{aligned}$$

Then we obtain the following inequality

$$\begin{aligned}rs(H\wr \mathrm{S}_{17}\wr \mathrm{S}_{17})= & {} \frac{\log _{2}s(H\wr \mathrm{S}_{17}\wr \mathrm{S}_{17})}{n\cdot 17^{2}}\leqslant \frac{\log _{2}\frac{[\frac{(A+16)^ {17}}{17!}]^{17}}{17!}}{n\cdot 17^{2}}\\= & {} \frac{\log _{2}(A+16)}{n}-\frac{\log _{2}17!}{17n} -\frac{\log _{2}17!}{289n} \end{aligned}$$

On the other hand, let $\deg (P_{1})=m_{1}$. By Lemma 2.3 (i), we have $s(H\wr P_{1})\geqslant s(H)^{m_{1}}/|P_{1}|=A^{m_{1}}/|P_{1}|$, and so

$$\begin{aligned} \frac{\log _{2}s(H\wr P_{1})}{nm_{1}}-\frac{\log _{2}\alpha _2}{nm_{1}}\geqslant \frac{\log _{2}A}{n}-\frac{\log _{2}|P_{1}|}{nm_{1}} -\frac{\log _{2}\alpha _2}{nm_{1}} \end{aligned}$$

Let $G=H\wr P_{1}\wr \cdots \wr P_\ell$. It suffices to check the following inequality:

$$\begin{aligned} \frac{\log _{2}(A+16)}{n}-\frac{\log _{2}17!}{17n}-\frac{\log _{2}17!}{289n}\leqslant \frac{\log _{2}A}{n} -\frac{\log _{2}|P_{1}|}{nm_{1}}-\frac{\log _{2}\alpha _2}{nm_{1}} \end{aligned}$$

(6)

Consequently, (6) follows if and only if

$$\begin{aligned} \frac{\log _{2}|P_{1}|}{m_{1}}+\frac{\log _{2}\alpha _2}{m_{1}}\leqslant \frac{\log _{2}17!}{17}+\frac{\log _{2}17!}{289} -\log _{2}(\frac{A+16}{A}). \end{aligned}$$

Since $A\geqslant 18$, it follows that $\log _{2}(\frac{A+16}{A})\leqslant 0.917537839808$. Therefore, we only need to check the following inequality:

$$\begin{aligned} \frac{\log _{2}|P_{1}|}{m_{1}}+ \frac{\log _{2}\alpha _2}{m_{1}}\leqslant 2.093108733204. \end{aligned}$$

Denote by $f(P_{1},m_{1},\alpha _2)$ the number $\frac{\log _{2}|P_{1}|}{m_{1}}+\frac{\log _{2}\alpha _2}{m_{1}}$. If $m_{1}>24$, then by Lemma 2.4 (ii), we know that $|P_{1}|<2^{m_{1}}$, and so

$$\begin{aligned} f(P_{1},m_{1},\alpha _2)<1.120844008278<2.093108733204. \end{aligned}$$

If $m_{1}\leqslant 24$, then by Table 2, we calculate

$$\begin{aligned} f(P_{1},m_{1},\alpha _2)\leqslant 1.868687569222<2.093108733204. \end{aligned}$$

$\square$

We are now in a position to prove Proposition 4.1.

Proof of Proposition 4.1: Note that $b_{k}={\frac{\log _{2}s(\mathrm{S}_{17}\wr \mathrm{S}_{17}\wr \cdots \wr \mathrm{S}_{17})}{17^{k}}}$ and $M={\lim \limits _{k \rightarrow +\infty }b_{k}}$. Since the sequence $\{b_k\}_{k\geqslant 1}$ is strictly decreasing, it follows that $M<b_{2}$. If G is primitive, then $rs(G)\geqslant b_{2}$ by Lemma 4.2. Now, if G is not primitive, then G is induced from $\mathrm{S}_{17}$ by Lemma 4.4. Combining Lemma 4.5 and Lemma 4.6, one has

$$\begin{aligned} \inf \Big (\frac{\log _{2}s(G)}{n} \Big )={\lim _{k \rightarrow +\infty }\frac{\log _{2}s (\overbrace{\mathrm{S}_{17}\wr \mathrm{S}_{17}\wr \cdots \wr \mathrm{S}_{17}}^{\text{ k }\ terms})}{17^{k}}}. \end{aligned}$$

This completes the proof of this proposition. $\square$

5 Further Considerations

In this section, we first make some remarks on Proposition 3.1. For convenience, we use the notation defined in Proposition 3.1.

Remark 5.1

Using the proof of Lemma 3.6, we can obtain a good estimate of the limit. We first notice that

$$\begin{aligned} rs(\mathrm{M}_{24}\wr \mathrm{S}_{5}\wr \mathrm{S}_{5}\wr \cdots )\leqslant \frac{\log _{2}({{53}\atopwithdelims ()5}+4)}{24\cdot 5}-\frac{\log _{2}120}{24\cdot 5^{2}} -\frac{\log _{2}120}{24\cdot 5^{3}}-\cdots . \end{aligned}$$

This implies that

$$\begin{aligned} \inf \left( \frac{\log _2 s(G)}{n}\right) \leqslant \frac{\log _{2}({{53}\atopwithdelims ()5}+4)}{24\cdot 5}-\frac{\log _{2}120}{24\cdot 5}(\frac{1}{5}+\frac{1}{5^{2}}+\cdots ) \approx 0.164381169292. \end{aligned}$$

On the other hand, by Lemma 3.5 and the value $a_{2}$, we obtain

$$\begin{aligned} \inf \left( \frac{\log _2 s(G)}{n}\right) \geqslant 0.164381160912. \end{aligned}$$

Taking into consideration the possible mistakes in the last two digits, the following bound can be guaranteed,

$$\begin{aligned} 0.164381160900<\inf \left( \frac{\log _2 s(G)}{n}\right) <0.164381169300. \end{aligned}$$

We compare the bound of $\inf \left( \frac{\log _2 s(G)}{n}\right)$ in Proposition 3.1 with that of Proposition 4.1. Now, we make the following remarks. For convenience, we use the notation defined in Proposition 4.1.

Remark 5.2

By the proof of Lemma 4.6, one can obtain a good estimate of the limit. We first observe that

$$\begin{aligned} rs(\mathrm{S}_{17}\wr \mathrm{S}_{17}\wr \mathrm{S}_{17}\wr \cdots )\leqslant \frac{\log _{2} ({{34}\atopwithdelims ()17}+16)}{17^{2}}-\frac{\log _{2}17!}{17^3} -\frac{\log _{2}17!}{17^{4}}-\cdots . \end{aligned}$$

This implies that

$$\begin{aligned}\inf \left( \frac{\log _2 s(G)}{n}\right) \leqslant \frac{\log _{2}({{34} \atopwithdelims ()17}+16)}{17^{2}}-\frac{\log _{2}17!}{17^{2}}(\frac{1}{17}+\frac{1}{17^{2}}+\cdots ) \approx 0.097227729390. \end{aligned}$$

On the other hand, by Lemma 4.3 and the value $b_{2}$, we obtain

$$\begin{aligned} \inf \left( \frac{\log _2 s(G)}{n}\right) \geqslant 0.097227729356. \end{aligned}$$

Taking into consideration the possible mistakes in the last two digits, the bound can be guaranteed

$$\begin{aligned} 0.097227729300<\inf \left( \frac{\log _2 s(G)}{n}\right) <0.097227729395.\end{aligned}$$

Remark 5.3

One of the main difficulties of determining $\inf \left( \frac{\log _2 s(G)}{n}\right)$ is to identify the group that achieves the lower bound. We found a range of the bound, but we still do not know what the exact value of the above limit is. Although we conjecture this limit is a rational number, we cannot prove this. Therefore, one interesting problem is to determine whether the limit is a rational number or not.

Remark 5.4

A related problem would be to determine what is $\inf \left( \frac{\log _2 s(G)}{n}\right)$ where G does not contain any $\mathrm{A}_\ell$ with $\ell >t\geqslant 5$ for any positive integer t. The second author [10] answered this question for $t=4$, and the work in this paper has answered this question for $5\leqslant t \leqslant 166$. We believe the following is true.

Conjecture 5.5

Let G be a permutation group of degree n. If G does not contain any $\mathrm{A}_\ell$ with $\ell >t$ for $167 \leqslant t<\infty$ as a composition factor, then

$$\begin{aligned} \inf \left( \frac{\log _2 s(G)}{n}\right) =\lim _{k \mapsto +\infty } \frac{\log _2 s (\mathrm{S}_t\wr \mathrm{S}_t\wr \cdots \wr \mathrm{S}_t)}{t^k}. \end{aligned}$$

Although it might be possible to go beyond $t=166$ by tweaking the methods in this paper, it seems that our method is not strong enough for a general proof of this.