ON THE LENGTH AND DEPTH OF FINITE GROUPS
TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

arXiv:1802.02194v1 [] 6 Feb 2018

With an appendix by D.R. Heath-Brown
Abstract. An unrefinable chain of a finite group G is a chain of subgroups G = G0 >
G1 > · · · > Gt = 1, where each Gi is a maximal subgroup of Gi−1 . The length (respectively, depth) of G is the maximal (respectively, minimal) length of such a chain. We
studied the depth of finite simple groups in a previous paper, which included a classification of the simple groups of depth 3. Here we go much further by determining the
finite groups of depth 3 and 4. We also obtain several new results on the lengths of finite
groups. For example, we classify the simple groups of length at most 9, which extends
earlier work of Janko and Harada from the 1960s, and we use this to describe the structure of arbitrary finite groups of small length. We also present a number-theoretic result
of Heath-Brown, which implies that there are infinitely many non-abelian simple groups
of length at most 9.
Finally we study the chain difference of G (namely the length minus the depth). We
obtain results on groups with chain difference 1 and 2, including a complete classification
of the simple groups with chain difference 2, extending earlier work of Brewster et al.
We also derive a best possible lower bound on the chain ratio (the length divided by the
depth) of simple groups, which yields an explicit linear bound on the length of G/R(G)
in terms of the chain difference of G, where R(G) is the soluble radical of G.

1. Introduction
An unrefinable chain of length t of a finite group G is a chain of subgroups
G = G0 > G1 > · · · > Gt−1 > Gt = 1,

(1)

where each Gi is a maximal subgroup of Gi−1 . The length of G, denoted by l(G), is the
maximal length of an unrefinable chain. This notion arises naturally in several different
contexts, finding a wide range of applications. For example, Babai [3] investigated the
length of symmetric groups in relation to the computational complexity of algorithms for
finite permutation groups. In a different direction, Seitz, Solomon and Turull studied the
length of finite groups of Lie type in a series of papers in the early 1990s [30, 32, 33],
motivated by applications to fixed-point-free automorphisms of finite soluble groups. In
fact, the notion predates both the work of Babai and Seitz et al. Indeed, Janko and
Harada studied the simple groups of small length in the 1960s, culminating in Harada’s
description of the finite simple groups of length at most 7 in [16].
Given the definition of l(G), it is also natural to consider the minimal length of an
unrefinable chain for G. Following [8], we call this number the depth of G, denoted by
λ(G). For example, if G is a cyclic group of order n > 2, then λ(G) = Ω(n), the number
of prime divisors of n (counting multiplicities). In particular, λ(G) = 1 if and only if G
has prime order. This notion appears in the earlier work of several authors. For example,
in [31] Shareshian and Woodroofe investigate the length of various chains of subgroups
The third author acknowledges the hospitality of Imperial College, London, while part of this work was
carried out. He also acknowledges the support of ISF grant 686/17 and the Vinik chair of mathematics
which he holds.
Date: February 8, 2018.
2010 Mathematics Subject Classification. Primary 20E32, 20E15; Secondary 20E28.
1

2

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

of finite groups G in the context of lattice theory (in their paper, the depth of G is
denoted by minmaxl(G)). There are also several papers on the so-called chain difference
cd(G) = l(G) − λ(G) of a finite group G. For example, a well known theorem of Iwasawa
[18] states that cd(G) = 0 if and only if G is supersoluble. The simple groups G with
cd(G) = 1 have been determined by Brewster et al. [7] (also see [17] and [28] for related
results).
In [8], we focus on the depth of finite simple groups. One of the main results is [8,
Theorem 1], which determines the simple groups of depth 3 (it is easy to see that λ(G) > 3
for every non-abelian simple group G); the groups that arise are recorded in Table 1. We
also show that alternating groups have bounded depth (indeed, λ(An ) 6 23 for all n,
whereas l(An ) tends to infinity with n) and we obtain upper bounds on the depth of each
simple group of Lie type. We refer the reader to [9] for results on analogous notions of
length and depth for connected algebraic groups over algebraically closed fields.
G
Ap
L2 (q)
Lǫn (q)
2B

2 (q)
M23 , B

Conditions
p and (p − 1)/2 prime, p 6∈ {7, 11, 23}

 (q + 1)/(2, q − 1) or (q − 1)/(2, q − 1) prime, q 6= 9; or
q prime and q ≡ ±3, ±13 (mod 40); or

q = 3k with k > 3 prime
n −ǫ
both prime, n > 3 and
n and (q−ǫ)q (n,q−ǫ)
(n, q, ǫ) 6= (3, 4, +), (3, 3, −), (3, 5, −), (5, 2, −)
q − 1 prime
Table 1. The simple groups G with λ(G) = 3

Our goal in this paper is to extend the depth results in [8] in several different directions,
both for simple groups, as well as arbitrary finite groups. We also revisit some of the
aforementioned results of Janko and Harada from the 1960s, providing a precise description
of the simple groups of small length. In turn, this allows us to describe the structure of
arbitrary finite groups of small length and we can use this to classify the simple groups G
with cd(G) = 2, which extends one of the main results in [7].
By a theorem of Shareshian and Woodroofe [31, Theorem 1.4], it follows that λ(G) > 3
for every insoluble finite group G. Our first main result determines all finite groups of
depth 3. In particular, notice that an obvious consequence of the theorem is that almost
simple groups of depth 3 are simple.
Theorem 1. A finite group G has depth 3 if and only if either G is soluble of chief length
3, or G is a simple group as in Table 1.
The next result classifies the finite groups of depth 4. In part (iv) of the statement, a
twisted wreath product T twrφ S for non-abelian simple groups S, T is as defined and studied
in [4]. The ingredients are a transitive action of S on k points with point stabiliser S1 , and
a homomorphism φ : S1 → Aut(T ) with image containing Inn(T ). Thus T is isomorphic
to a proper section of S; indeed, the subgroups C = φ−1 (Inn(T )) and D = ker(φ) ∩ C
satisfy C/D ∼
= T , and (C, D) forms an S-maximal section of S, as defined in [4, Definition
4.1]. Moreover T twrφ S is a semidirect product T k .S having S as a maximal subgroup.
Theorem 2. Suppose G is a finite group of depth 4. Then one of the following holds,
where p is a prime:
(i) G is soluble of chief length 4.

ON THE LENGTH AND DEPTH OF FINITE GROUPS

3

(ii) G = T × T , T × Cp or T p .Cp , where T is simple of depth 3 (as in Table 1).
(iii) G = (Cp )k .T (a possibly nonsplit extension), where T is simple of depth 3, and
acts irreducibly on (Cp )k .
(iv) G = T twrφ S, a twisted wreath product, where S, T are simple, T is a proper
section of S, and S has depth 3.
(v) G is quasisimple, and Z(G) is either 1 or Cp .
(vi) G is almost simple with socle T , and G/T is either 1 or Cp .
Remark 1. Let us make some comments on the statement of Theorem 2.
(a) Note that in part (iv), and also (ii) (apart from T p .Cp ), the group G has a maximal
subgroup that is simple of depth 3. In (iii), for split extensions this is also the case;
for nonsplit, G must have a maximal subgroup M = pk .M0 with M0 maximal in
T acting irreducibly on pk (so λ(M0 ) = 2 and λ(M ) = 3). See Remark 3.1 for
further comments on the nonsplit groups G = pk .T arising in part (iii).
(b) In (iv), being of depth 3, the possibilities for S can be found in Table 1. Every
proper non-abelian simple section T of S can occur. The simple sections of the
groups Ap and Lǫn (q) in Table 1 cannot be listed. However, the proper simple
sections of the other groups in the table can be listed: for L2 (q) they are A5 ,
L2 (q0 ) (with q = q0k and q0 > 4); for 2B2 (q) there are none (since q − 1 is prime);
and those for M23 and B can be listed using [12].
(c) Consider case (v), where G is quasisimple with Z(G) = Cp (p prime) and let
T = G/Z(G), a simple group. Here λ(G) = λ(T ) + 1, as explained in Remark 3.2,
so λ(G) = 4 if and only if T is one of the simple groups in Table 1. In particular, by
considering the Schur multipliers of the simple groups in Table 1 (see [21, Theorem
5.1.4], for example), we deduce that a quasisimple group with nontrivial centre has
depth 4 if and only if it appears in Table 2.

G
2.Ap
SLǫn (q)

SL2 (q)
2.2 B2 (8), 2.B

Conditions
p and (p − 1)/2 prime, p 6∈ {7, 11, 23}
q n −ǫ
both prime, n > 3 and
n = (n, q − ǫ) and n(q−ǫ)
(n, q, ǫ) 6= (3, 4, +), (3, 5, −)

 (q + 1)/2 or (q − 1)/2 prime, q 6= 9; or
q prime and q ≡ ±3, ±13 (mod 40); or

q = 3k with k > 3 prime

Table 2. The quasisimple groups G with Z(G) 6= 1 and λ(G) = 4
The next result sheds further light on the almost simple groups of depth 4 (case (vi) of
Theorem 2).
Theorem 3. Let G be an almost simple group with socle T and depth 4. Then one of the
following holds:
(i) G/T = Cp , p prime and λ(T ) = 3;
(ii) (G, T ) is one of the cases in Table 3 (in each case, λ(T ) = 4);
(iii) G = T has a soluble maximal subgroup M of chief length 3; the possibilities for
(G, M ) are listed in Table 4;
(iv) G = T has a simple maximal subgroup of depth 3.

4

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

In Table 4, the required conditions when G = L2 (q) and q is odd are rather complicated
to state (mainly due to the fact that we need λ(G) 6= 3). To simplify the presentation of
the table, we refer to the following conditions on q:

 Ω(q ± 1) > 3
q 6≡ ±3, ±13 (mod 40) if q is prime
(2)

q 6= 3k with k > 3 prime.

We refer the reader to Remark 3.3 for further details on the simple groups that arise in
part (iv) of Theorem 3.
T
A6
A7 , A11 , A23
L2 (q)
L3 (4)
U3 (5)

G
Conditions
PGL2 (9), M10
S7 , S11 , S23
PGL2 (q)
q prime, q ≡ ±11, ±19 (mod 40), Ω(q ± 1) > 3
PGL3 (4)
PGU3 (5)

Table 3. The almost simple groups G = T.p with λ(G) = λ(T ) = 4

G
Ap
A6
L2 (q)

Lǫ3 (q)
Lǫn (q)
2B

2 (q)

M
p:((p − 1)/2)
S4 , 32 :4
Fq :((q − 1)/2), Dq−1
Dq+1
S4
Fq :(q − 1), D2(q−1)
D2(q+1)
(Cq−ǫ )2 :S3 

q n −ǫ
(q−ǫ)(n,q−ǫ)

:n

D2(q−1)
√
(q ± 2q + 1):4
√
2 G (q) (q ± 3q + 1):6
2
3 D (q) (q 4 − q 2 + 1):4
4
J1
7:6, 11:10, 19:6, 23 :7:3
J4
43:14
Ly
67:22
Fi′24
29:14
Th
31:15

Conditions
p prime, Ω(p − 1) = 3
q
q
q
q
q
q

odd, Ω(q − 1) = 3, (2) holds
odd, Ω(q + 1) = 3, (2) holds
prime, q ≡ ±1 (mod 8), (2) holds
even, Ω(q − 1) = 2, Ω(q + 1) > 2
even, Ω(q + 1) = 2, Ω(q − 1) > 2
> 8 even, q − ǫ prime, Ω(q 2 + ǫq + 1) > 2
n

q −ǫ
n > 3 prime, Ω( (q−ǫ)(n,q−ǫ)
)=2

Ω(q − 1) = 2
√
q ± 2q + 1 prime, Ω(q − 1) > 2
√
q ± 3q + 1 prime, q > 3
q 4 − q 2 + 1 prime

Table 4. The simple groups G of depth 4 with a soluble maximal subgroup
M of depth 3
Next we turn to our main results on the lengths of finite groups. Recall that the finite
simple groups of small length were studied by Janko and Harada in the 1960s, beginning
with [19] which classifies the simple groups of length 4 (since λ(G) > 3, Iwasawa’s theorem
implies that l(G) > 4 for every non-abelian simple group G). In a second paper [20], Janko
describes the simple groups of length 5 and this was extended by Harada [16] to length at

ON THE LENGTH AND DEPTH OF FINITE GROUPS

5

most 7. In both papers, the main results state that either G = L2 (q) for some unspecified
prime powers q, or G belongs to a short list of specific groups.
The following result extends this earlier work by giving a precise classification of the
simple groups of length at most 9.
Theorem 4. Let G be a non-abelian finite simple group of length at most 9. Then G and
l(G) are recorded in Table 5, where p is a prime number.

l(G)
4
5
6
7

8

9

G
A5 , L2 (q)
L2 (q)
A7 , J1 , L2 (q)
M11 , U3 (3), U3 (5)
L2 (q)

Conditions
q = p > 5, max{Ω(q ± 1)} = 3 and q ≡ ±3, ±13 (mod 40)
q ∈ {7, 8, 9, 11, 19, 27, 29}, or q = p and max{Ω(q ± 1)} = 4
q ∈ {25, 125}, or q = p and max{Ω(q ± 1)} = 5
q ∈ {16, 32, 49, 121, 169}, or q = p and max{Ω(q ± 1)} = 6, or
q = p3 , Ω(q − 1) = 4 and Ω(q + 1) 6 6

M12 , 2 B2 (8), L3 (3)
L2 (q)
q
q
q
q
q
A8 , U4 (2), L3 (4)
U3 (q)
q
q
L2 (q)
q
q
q
q
q
q
q

= p and max{Ω(q ± 1)} = 7, or
= p2 , Ω(q − 1) = 6 and Ω(q + 1) 6 7, or
= p3 , Ω(q − 1) = 5 and Ω(q + 1) 6 7, or
= p3 , Ω(q − 1) 6 4 and Ω(q + 1) = 7, or
= p5 , Ω(q − 1) = 3 and Ω(q + 1) 6 7
∈ {4, 11, 13, 29}, or q = p, Ω(q ± 1) = 3, Ω(q 2 − q + 1) 6 8,
≡ 2 (mod 3) and q ≡ ±3, ±13 (mod 40)
∈ {81, 128, 2187}, or q = p and max{Ω(q ± 1)} = 8, or
= p2 , Ω(q − 1) = 7 and Ω(q + 1) 6 8, or
= p2 , Ω(q − 1) = 6 and Ω(q + 1) = 8, or
= p3 , Ω(q − 1) = 6 and Ω(q + 1) 6 8, or
= p3 , Ω(q − 1) 6 5 and Ω(q + 1) = 8, or
= p5 , Ω(q − 1) = 4 and Ω(q + 1) 6 8, or
= p5 , Ω(q − 1) = 3 and Ω(q + 1) = 8

Table 5. The simple groups G of length at most 9

The proof of Theorem 4 is given in Section 4.1, together with the proof of the following
corollary, which describes the structure of finite groups of small length. Recall that soluble
groups G have length Ω(|G|), so we focus on insoluble groups.
Corollary 5. Let G be a finite insoluble group, in which case l(G) > 4.
(i) l(G) = 4 if and only if G is simple as in line 1 of Table 5.
(ii) l(G) = 5 if and only if one of the following holds:
(a) G is simple as in line 2 of Table 5; or
(b) G = T × Cp with T simple of length 4 (as in Table 5) and p a prime; or
(c) G = SL2 (q) or PGL2 (q), and either q = 5, or q > 5 is a prime such that
max{Ω(q ± 1)} = 3 and q ≡ ±3, ±13 (mod 40).
(iii) l(G) = 6 if and only if one of the following holds:
(a) G is simple as in line 3 of Table 5; or

6

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

(b) G = T ×Cp , or a quasisimple group p.T , or an almost simple group T.p, where
T is simple of length 5 (as in Table 5) and p a prime; the quasisimple groups
occurring are SL2 (q), 3.L2 (9), and the almost simple groups are PGL2 (q),
M10 , S6 , L2 (8).3 and L2 (27).3; or
(c) G = L2 (q) × (p.r), (L2 (q) × p).2, SL2 (q) × p or 2.L2 (q).2, where p, r are primes
and L2 (q) has length 4, as in Table 5.
Let G = L2 (q), where q is a prime, and consider the conditions on q in the first row
of Table 5. One checks that the first ten primes that satisfy the given conditions are as
follows:
q ∈ {13, 43, 67, 173, 283, 317, 653, 787, 907, 1867},

but it is not known if there are infinitely many such primes. The following more general
problem is addressed in [1]: Does there exist an infinite set S of non-abelian finite simple
groups and a positive integer N such that l(G) 6 N for all G ∈ S? The main result of [1]
gives a positive answer to this question. The key ingredient is a purely number theoretic
result [1, Theorem C], which states that for each positive integer n, there is an infinite
set of primes P and a positive integer N such that Ω(pn − 1) 6 N for all p ∈ P. More
precisely, for n = 2 they show that the conclusion holds with N = 21, which immediately
implies that there are infinitely many primes p with l(L2 (p)) 6 20. The same problem
arises in work of Gamburd and Pak (see [14, p.416]), who state that l(L2 (p)) 6 13 for
infinitely many primes p (giving [15] as a reference).
We establish the following strengthening of the results in [1, 14].
Theorem 6. There are infinitely many finite non-abelian simple groups G with l(G) 6 9.
In fact we show that l(L2 (p)) 6 9 for infinitely many primes p. As explained in Section
4.2, this is easily deduced from the following number-theoretic result of Heath-Brown,
which is of independent interest.
Theorem 7 (Heath-Brown). There are infinitely many primes p ≡ 5 (mod 72) for which
Ω((p2 − 1)/24)) 6 7.

See Appendix A for the proof of this theorem, which implies that there are infinitely
many primes p for which max{Ω(p ± 1)} 6 8.
Next we consider the chain difference of simple groups. The following result determines
the simple groups of chain difference two (see Section 5.1 for the proof). This extends
earlier work of Brewster et al. [7, Theorem 3.3] (also see Theorem 5.1), who described the
simple groups of chain difference one.
Theorem 8. Let G be a finite simple group. Then cd(G) = 2 if and only if one of the
following holds:
(i) G = A7 , J1 or U3 (5).
(ii) G = L2 (q) and either q ∈ {7, 8, 11, 27, 125}, or q is a prime and one of the following
holds:
(a) max{Ω(q ± 1)} = 4 and either min{Ω(q ± 1)} = 2, or q ≡ ±3, ±13 (mod 40).
(b) max{Ω(q ± 1)} = 5, min{Ω(q ± 1)} > 3 and q 6≡ ±3, ±13 (mod 40).
The chain ratio of a finite group G is given by cr(G) = l(G)/λ(G) and the next result establishes a best possible lower bound on the chain ratio of simple groups. In the
statement, we define
G = {G : G simple, l(G) = 5 and λ(G) = 4}.

(3)

ON THE LENGTH AND DEPTH OF FINITE GROUPS

7

By combining Theorem 4 with [8, Theorem 1], it follows that the groups in G are of the form
L2 (q) and either q ∈ {9, 19, 29}, or q is a prime with max{Ω(q ±1)} = 4, min{Ω(q ±1)} > 3
and q 6≡ ±3, ±13 (mod 40).
Theorem 9. Let G be a non-abelian finite simple group. Then
5
cr(G) > ,
4
with equality if and only if G ∈ G.
It follows from [8, Corollary 9] that there exists an absolute constant a such that l(G) 6
a cd(G) for every non-abelian simple group G. As an immediate corollary of Theorem 9,
we deduce that a = 5 is the best possible constant.
Corollary 10. Let G be a non-abelian finite simple group. Then l(G) 6 5 cd(G), with
equality if and only if G ∈ G.
Our final result, which applies Theorem 9, relates the structure of an arbitrary finite
group G with its chain difference. We let R(G) denote the soluble radical of G.
Theorem 11. Let G be a finite group. Then
l(G/R(G)) 6 10 cd(G).
Combining this theorem with [1, Proposition 2.2], it follows that
Ω(|G/R(G)|) 6 100 cd(G)2 .
Note that the length of G itself need not be bounded in terms of cd(G); indeed, if G
is supersoluble then cd(G) = 0 while l(G) may be arbitrarily large. However, we show
in Proposition 5.10 below, that, if ss(G) denotes the direct product of the non-abelian
composition factors of G (with multiplicities), then
l(ss(G)) 6 5 cd(G).
This extends Corollary 10 dealing with simple groups, and serves as a useful tool in the
proof of Theorem 11 above.
2. Preliminaries
We begin by recording some preliminary results, which will be needed in the proofs of
our main theorems. Given a finite group G, we write chiefl(G) for the length of a chief
series of G. Recall that l(G) and λ(G) denote the length and depth of G, respectively, as
defined in the Introduction. Let cd(G) = l(G) − λ(G) be the chain difference of G.
Lemma 2.1. Let G be a finite group and let N be a normal subgroup of G.
(i) l(G) = l(N ) + l(G/N ).
(ii) λ(G/N ) 6 λ(G) 6 λ(N ) + λ(G/N ).
Proof. This is straightforward. For example, see [10, Lemma 2.1] for part (i).



Notice that Lemma 2.1(i) implies that the length of a finite group is equal to the sum
of the lengths of its composition factors. In particular, if G is soluble then l(G) = Ω(|G|),
which is the number of prime divisors of |G| (counting multiplicities).
The next result is [7, Lemma 1.3].
Lemma 2.2. If G is a finite group, B 6 G and A is a normal subgroup of B, then
cd(G) > cd(B/A) + cd(A).
In particular, cd(G) > cd(L) for every section L of G.

8

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

Lemma 2.3. Let G be a finite group.
(i) If G is soluble, then λ(G) = chiefl(G).
(ii) If G is insoluble, then λ(G) > chiefl(G) + 2.
Proof. Part (i) is [22, Theorem 2] and part (ii) is [31, Theorem 1.4].



Lemma 2.4. Let H be soluble, p a prime, and suppose G = H p hαi where αp ∈ H p and α
permutes the p factors transitively. Then λ(G) > λ(H) + 1.
Proof. We proceed by induction on |H|. The base case H = 1 is trivial. Now assume
H 6= 1, and let N be a minimal normal subgroup of H. By Lemma 2.3(i) we have
λ(H/N ) = λ(H) − 1.
Q
αi
Let M = p−1
i=0 N . Then 1 6= M ⊳ G, and λ(G) > λ(G/M ) + 1, again by Lemma 2.3(i).
Applying the induction hypothesis to G/M ∼
= (H/N )p .p, we have
λ(G/M ) > λ(H/N ) + 1.
It follows that λ(G) > λ(H/N ) + 2 = λ(H) + 1, as required.



The next lemma on the length of L2 (q) will be useful later.
Lemma 2.5. Let G = L2 (q), where q = pf > 5 and p is a prime.
(i) If q is even, then l(G) = Ω(q − 1) + f + 1.
(ii) If q is odd, then either
l(G) = max{Ω(q − 1) + f, Ω(q + 1) + 1},

(4)

or q ∈ {7, 11, 19, 29} and l(G) = 5, or q = 5 and l(G) = 4.
Proof. Part (i) is a special case of [32, Theorem 1], noting that 2f :(2f − 1) is a Borel
subgroup of G. Now assume q is odd. The case f = 1 follows from [10, Proposition 5.2],
so let us assume f > 2. We proceed by induction on Ω(f ).
First assume Ω(f ) = 1, so f is a prime, and let M be a maximal subgroup of G. By
inspecting [6, Tables 8.1, 8.2], either M = PGL2 (p) or A5 (for f = 2 only), or M =
pf :((pf − 1)/2), Dpf ±1 or L2 (p), which gives
l(G) = max{Ω(q − 1) + f, Ω(q + 1) + 1, l(L2 (p)) + 1 + δ2,f },

where δi,j is the familiar Kronecker delta. It is easy to check that (4) holds if p ∈
{3, 5, 7, 11, 19, 29}. For example, if p = 29 and f = 2, then Ω(q − 1) = 6 and l(L2 (p)) = 5.
For any other prime p,
l(L2 (p)) + 1 + δ2,f = max{Ω(p ± 1)} + 2 + δ2,f 6 max{Ω(q − 1) + f, Ω(q + 1) + 1}

and the result follows.
Similarly, if Ω(f ) > 2 then

l(G) = max{Ω(q − 1) + f, Ω(q + 1) + 1, l(L2 (q 1/r )) + 1 + δ2,r : r ∈ π(f )},

where π(f ) is the set of prime divisors of f , and induction gives
Therefore

l(L2 (q 1/r )) = max{Ω(q 1/r − 1) + f /r, Ω(q 1/r + 1) + 1}.

l(L2 (q 1/r )) + 1 + δ2,r 6 max{Ω(q − 1) + f, Ω(q + 1) + 1}
and we conclude that (4) holds.
3. Depth
In this section we prove our results on depths, namely Theorems 1, 2 and 3.



ON THE LENGTH AND DEPTH OF FINITE GROUPS

9

3.1. Proof of Theorem 1. Let G be a finite group. For soluble groups, the theorem is
an immediate corollary of Lemma 2.3(i), so let us assume G is insoluble. Here Lemma
2.3(ii) implies that λ(G) > chiefl(G) + 2. Hence if λ(G) = 3, then chiefl(G) = 1 and thus
G is simple, and is as in Table 1 by [8, Theorem 1]. Conversely, the groups in the table
indeed have depth 3.
3.2. Proof of Theorem 2. Suppose G is a finite group and λ(G) = 4. If G is soluble
then it has chief length 4 by Lemma 2.3(i), as in part (i) of Theorem 2. Now assume G is
insoluble. Since λ(G) > chiefl(G) + 2 by Lemma 2.3(ii), it follows that chiefl(G) 6 2. If
chiefl(G) = 1 then G is simple, so (v) holds.
Now assume that chiefl(G) = 2. Then G has a minimal normal subgroup N ∼
= Tk
for some simple (possibly abelian) group T , and G/N ∼
= S is simple (also possibly nonabelian).
Suppose first that k = 1 and S, T are both non-abelian. Then G ∼
= S ×T by the Schreier
hypothesis. When S ∼
6 T , any maximal subgroup of G is of the form S0 ×T or S ×T0 (with
=
S0 , T0 maximal in S, T respectively), and neither of these can have depth 3, by Theorem
1. Hence S ∼
= T . Now G has a maximal subgroup M of depth 3, and M cannot be of the
above form S0 × T or S × T0 . It follows that M is a diagonal subgroup isomorphic to T ,
and hence λ(T ) = 3 and G is as in conclusion (ii) of Theorem 2.
Next suppose k = 1 and S or T is an abelian simple group Cp . Then G is one of T × Cp ,
S ×Cp, a quasisimple group p.S or an almost simple group T.p. The latter two possibilities
are conclusions (v) and (vi). Now assume G = T × Cp , and let M be a maximal subgroup
of G of depth 3. Then M is either T or T0 × Cp , where T0 is maximal in T . In the
latter case Theorem 1 shows that T0 × Cp is soluble of chief length 3, hence chiefl(T0 ) = 2.
Therefore in both cases T has depth 3, and so G is as in conclusion (ii).
We may now assume that k > 1. Suppose T is non-abelian and S = Cp . Then k = p
and G = N hαi = T p hαi, where αp ∈ N and α permutes the p factors transitively. Let M
be a maximal subgroup of G of depth 3. Then M 6= N by Theorem 1, so, replacing α by
Q
αi
∼ p
another element in the coset N α if necessary, M is of the form p−1
i=0 H hαi = H .p for
some maximal subgroup H of T . Also M is soluble, again by Theorem 1. It now follows
from Lemma 2.4 that λ(M ) > λ(H) + 1, and hence λ(H) 6 2. Therefore λ(T ) 6 3, so
λ(T ) = 3 and conclusion (ii) of Theorem 2 holds.
Next consider the case where T = Cp and S is non-abelian. Here G = (Cp )k .S, where
S acts irreducibly on V := (Cp )k . Let M be a maximal subgroup of G of depth 3. If
V 66 M then M maps onto S, hence M ∼
= S by Theorem 1, and conclusion (iii) holds.
Now assume V 6 M , so that M is soluble, by Theorem 1. Then M/V = S0 , a maximal
subgroup of S, and by Lemma 2.3(i), λ(S0 ) < λ(M ) = 3. Hence λ(S) = 3 and again (iii)
holds.
It remains to handle the case where both T and S are non-abelian. Here G = T k .S
and S acts transitively on the k factors. Let M be a maximal subgroup of G of depth 3.
Then T k 66 M by Theorem 1. Hence M maps onto S, so M ∼
= S, again by Theorem 1. In
particular, λ(S) = 3. Write Ω = (G : M ), the coset space of M in G. As M is a core-free
maximal subgroup of G, it follows that G acts primitively on Ω and T k is a regular normal
subgroup. At this point the O’Nan-Scott theorem (see [24], for example) implies that G
is a twisted wreath product T twrφ S, as in conclusion (iv) of Theorem 2.
This completes the proof of Theorem 2.
Remark 3.1. Let us consider the nonsplit groups G = pk .T arising in part (iii) of Theorem
2. Here T is a simple group of depth 3 (so the possibilities for T are given in Table 1) and
V = (Cp )k is a nontrivial irreducible module for T over Fq with q = pf for a prime p. In

10

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

particular, dim V > 2, p divides |T | and the second cohomology H 2 (T, V ) is nontrivial.
As noted in Remark 1(a), G has a maximal subgroup M = pk .S with λ(M ) = 3, where
S < T is maximal and acts irreducibly on V . Note that S is soluble and has chief length
2. It will be difficult to give a complete classification of the depth 4 groups of this form,
but we can identify some genuine examples:
Example. Let T = M23 , so S = 23:11. Now T has an 11-dimensional irreducible module
V over F2 . Moreover, one checks that S acts irreducibly on V and H 2 (T, V ) 6= 0, hence
there is a nonsplit group G = 211 .M23 of depth 4.
Example. Take T = A5 and S = A4 . Set q = 5f with f > 1 and let V be a 3-dimensional
irreducible module for T over Fq . Then S acts irreducibly on V and H 2 (T, V ) 6= 0, so for
each f > 1 there is a nonsplit group 53f .A5 of depth 4.
 n 
−1
:n.
Example. Suppose T = Ln (r), where n > 3 is a prime and (n, r −1) = 1, so S = rr−1
Let V be the natural module for T = SLn (r). Then S acts irreducibly on V , and a theorem
of Bell [5] implies that H 2 (T, V ) 6= 0 if and only if
(n, r) ∈ {(3, 3a > 3), (3, 2), (3, 5), (4, 2), (5, 2)}.

In particular, there is a nonsplit group 39 .L3 (27) of depth 4 (note that we need λ(L3 (r)) =
3, which in this case means that r 2 + r + 1 is a prime). Thanks to Bell’s result, there are
also nonsplit groups 23 .L3 (2), 53 .L3 (5) and 25 .L5 (2), each of which has depth 4.
Remark 3.2. Let G be a quasisimple group with Z(G) = Cp (p prime) and G/Z(G) = T .
We claim that λ(G) = λ(T ) + 1, so λ(G) = 4 if and only if T is one of the simple groups
in Table 1. To see this, first note that λ(T ) 6 λ(G) 6 λ(T ) + 1 by Lemma 2.1(ii). Also
observe that if M is a maximal subgroup of a group of the form N.K, where N = Cp is
normal, then either M = N.L with L < K maximal, or M ∼
= K. Seeking a contradiction,
suppose λ(G) = λ(T ) = t and consider an unrefinable chain of G as in (1). Set Z = Z(G).
Since λ(G) = λ(T ), it follows that G1 ∼
6 T , so G1 = Z.T1 where T1 < T is maximal.
=
Similarly, G2 = Z.T2 with T2 < T1 maximal, and so on. In particular, Gt−1 = Z.Tt−1 with
Tt−1 < Tt−2 maximal. But Tt−1 6= 1 and thus |Gt−1 | is composite. This is a contradiction
and the claim follows.
3.3. Proof of Theorem 3. Let G be an almost simple group with λ(G) = 4 and socle
T . First assume G 6= T , so Theorem 2 implies that G/T = Cp for a prime p. If λ(T ) = 3,
then we are in case (i) of Theorem 3. Now assume λ(T ) > 4. We claim that (ii) holds, so
λ(T ) = 4 and (G, T ) is one of the cases in Table 3.
To see this, let M be a maximal subgroup of G of depth 3. Then M 6= T , so G = T M
and M ∩ T ⊳ M has index p. By Theorem 1, M is soluble and Lemma 2.3(i) implies that
λ(M ∩ T ) = 2, so M ∩ T is not maximal in T . Therefore, G has a novelty soluble maximal
subgroup M of chief length 3. This property is highly restrictive and we can determine
all the possibilities for G and M .
First assume T = An is an alternating group, so G = An .2. If n = 6 then λ(T ) = 4 and
one checks that λ(PGL2 (9)) = λ(M10 ) = 4, while λ(S6 ) = 5. Now assume n 6= 6, so that
G = Sn . By part (I) of the main theorem of [23], the novelty soluble maximal subgroups
of Sn are S2 ≀ S4 < S8 and Cp :Cp−1 < Sp for p ∈ {7, 11, 17, 23}. Of these, only Cp :Cp−1
for p ∈ {7, 11, 23} have depth 3, so S7 , S11 and S23 are the only depth 4 groups arising in
this case.
If T is a sporadic group then G = T.2 and one checks (by inspection of the Atlas [12])
that G does not have a maximal subgroup M with the required properties.
Now assume T is a simple group of Lie type over Fq . If T is an exceptional group of Lie
type, then all the maximal soluble subgroups of G are known (see [11, 25]) and one checks

ON THE LENGTH AND DEPTH OF FINITE GROUPS

11

that there are no relevant examples (it is helpful to note that if T = 2 B2 (q), 2 G2 (q), 2 F4 (q)
or 3 D4 (q), then G does not have any novelty maximal subgroups). Finally, suppose T is a
classical group. For the low-rank groups, it is convenient to consult the relevant tables in
[6]; in this way, one checks that the only cases that arise are the ones listed in Table 3 (in
each case, λ(T ) = 4). For example, if G = PGL2 (q), where q is a prime and q ≡ ±11, ±19
(mod 40), then G has a maximal subgroup M = S4 and M ∩ T = A4 is non-maximal in T
(note that the additional condition Ω(q ± 1) > 3 is needed to ensure that λ(T ) > 4, which
means that λ(T ) = 4 by [8, Lemma 3.1]). By inspecting [21], it is easy to check that no
examples arise when G is one of the remaining classical groups not covered by [6]. We
conclude that part (ii) of Theorem 3 holds.
To complete the proof, we may assume G = T has depth 4. Let M be a maximal
subgroup of G with λ(M ) = 3. By Theorem 1, either M is simple (and we are in part
(iv) of Theorem 3), or M is soluble of chief length 3. It remains to show that in the latter
case, the possibilities for G and M are given in Table 4. To do this, we essentially repeat
the above argument, but now there are more cases to consider because G = T and there
is no novelty condition.
First assume G = An . It is easy to verify the result for n 6 16 (with the aid of Magma,
for example), so let us assume n > 17. By the O’Nan-Scott theorem (see [24]), the only
soluble maximal subgroups of G are of the form M = AGL1 (p) ∩ G = Cp :C(p−1)/2 , with
n = p a prime. Here Lemma 2.3(i) implies that λ(M ) = Ω(p − 1), which explains the
condition Ω(p − 1) = 3 in Table 4.
Next assume G is a sporadic group. The groups with λ(G) = 4 can be read off from
[8, Lemma 3.3] and the cases appearing in Table 4 are obtained by inspecting the lists of
maximal subgroups of G in the Atlas [12].
Finally suppose G is a simple group of Lie type over Fq . As noted above, if G is an
exceptional group then all of the soluble maximal subgroups of G are known and it is
routine to read off the cases with such a subgroup of depth 3 (for G = 2 B2 (q), note that
we need the extra condition Ω(q −1) > 2 to ensure that λ(G) = 4). Similarly, the result for
classical groups is obtained by carefully inspecting [6] (for the low-rank groups) and [21]
(in the remaining cases). Once again, extra conditions on q are needed to get λ(G) = 4.
This completes the proof of Theorem 3.
Remark 3.3. It is not feasible to give a complete description of the depth 4 examples
(G, M ) arising in part (iv) of Theorem 3, but we can give some partial information:
Sporadic groups. By inspecting the Atlas [12], it is easy to show that if G 6= M is a
sporadic group, then (G, M ) is one of the following:
(M11 , L2 (11)) (M12 , L2 (11)) (M22 , L2 (11)) (M24 , L2 (7)) (M24 , L2 (23))
(M24 , M23 )
(J1 , L2 (11))
(J2 , A5 )
(Suz, L2 (25)) (Co2 , M23 )
(Co3 , M23 )
(Fi23 , L2 (23)) (Th, L3 (3))
At present, the maximal subgroups of the Monster group have not been completely determined; among the 44 conjugacy classes of known maximal subgroups, L2 (59) is the only
simple maximal group of depth 3. According to Wilson’s recent survey article [34], any
additional maximal subgroup of the Monster is almost simple with socle L2 (8), L2 (13),
L2 (16), U3 (4) and U3 (8). Note that each of these simple groups has depth 3. The latter
possibility is eliminated in [35], and Wilson reports that he has also ruled out L2 (8) and
U3 (4) in unpublished work (see the final paragraph in Section 3 of [34]). This leaves open
the two remaining possibilities.

12

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

Alternating groups. Let G = An be an alternating group. With the aid of Magma, it is
easy to check that for n 6 100, the possibilities for (M, n) are as follows:
(A5 , 6)
(L5 (2), 31)
(L2 (59), 60)
(L2 (13), 78)

(L2 (7), 7)
(L2 (37), 38)
(A59 , 60)
(L2 (83), 84)

(L3 (3), 13)
(L2 (43), 44)
(L2 (61), 62)
(A83 , 84)

(L2 (13), 14) (M23 , 23)
(L3 (5), 31)
(L2 (47), 48) (A47 , 48)
(L2 (53), 54)
(L2 (25), 65) (L2 (67), 68) (L2 (73), 74)
(A87 , 88)

The main theorem of [24] on the maximal subgroups of symmetric and alternating groups
provides some useful information in the general case, but it is not possible to state a precise
result.
Exceptional groups. Let G be an exceptional group of Lie type over Fq . If G = 2 B2 (q),
2 G (q), 2 F (q)′ , 3 D (q) or G (q), then the maximal subgroups of G are known and we can
2
4
4
2
read off the relevant examples: either G = 2 F4 (2)′ and M = L2 (25), or
• G = 2 B2 (q) and M = 2 B2 (q0 ) with q = q0k , q0 > 2 and both k and q0 − 1 are
primes; or
• G = G2 (q) and M = L2 (13) or L2 (8), with various conditions on q needed for the
maximality of M (see [6], for example).
In the remaining cases, by combining results of Liebeck and Seitz [27] with recent work
of Craven [13], we deduce that there are no examples with M an alternating or sporadic
group. Strong restrictions on the remaining possibilities when M is a group of Lie type can
be obtained by applying [26, Theorem 8] (defining characteristic) and the main theorem
of [27] (non-defining characteristic).
Classical groups. Finally, suppose G is a simple classical group with natural n-dimensional
module V . By Aschbacher’s subgroup structure theorem [2], either M belongs to a collection C(G) of geometric subgroups, or M ∈ S(G) is almost simple and acts irreducibly on
V . By inspecting [6, 21], it is possible to determine the relevant examples with M ∈ C(G)
(the precise list of cases will depend on some delicate number-theoretic conditions). For
example, suppose G = Un (q) is a unitary group, where n = q + 1 and q > 5 is a prime. If
(q n−1 + 1)/(q + 1) is also a prime, then G has a simple maximal subgroup M = Un−1 (q)
of depth 3 (here M is the stabiliser of a non-degenerate 1-space). For instance, G = U6 (5)
has a maximal subgroup M = U5 (5) of depth 3. It is not feasible to determine all the
cases that arise with M ∈ S(G), although this can be achieved for the low-dimensional
classical groups (that is, the groups with n 6 12) by inspecting the relevant tables in [6,
Chapter 8].
4. Length
In this section we prove our main results on length, namely Theorems 4 and 6, and
Corollary 5.
4.1. Proof of Theorem 4 and Corollary 5. Let G be a non-abelian finite simple group.
First recall that λ(G) > 3 and cd(G) > 1, so l(G) > 4. The simple groups of length 4 were
classified by Janko [19, Theorem 1]. In a second paper [20], he proves that every simple
group of length 5 is of the form L2 (q) for some prime power q (but he does not give any
further information on the prime powers that arise). In later work of Harada [16], this
result was extended to simple groups of length at most 7.
Theorem 4.1 (Harada, [16]). Let G be a finite simple group with l(G) 6 7. Then either
(i) G = U3 (3), U3 (5), A7 , M11 , J1 ; or
(ii) G = L2 (q) for some prime power q.

ON THE LENGTH AND DEPTH OF FINITE GROUPS

13

For the groups in part (i) of Theorem 4.1, it is easy to check that A7 and J1 have length
6, the others have length 7. We are now ready to prove Theorem 4.
Lemma 4.2. Theorem 4 holds if G ∼
= L2 (q).
Proof. Write q = pf , where p is a prime. In view of Lemma 2.5, the result is clear if p = 2
or f = 1, so let us assume p > 3 and f > 2, in which case
l(G) = max{Ω(q − 1) + f, Ω(q + 1) + 1}.
Since l(G) 6 9, it follows that f 6 7 and it is easy to verify the result when p ∈ {3, 5}.
Now assume p > 7, in which case Ω(p2 − 1) > 5 and f ∈ {2, 3, 5}.
Suppose f = 5, so Ω(q − 1) > 3 and we have l(G) ∈ {8, 9}. For l(G) = 8 we must have
Ω(q − 1) = 3 and Ω(q + 1) 6 7; one checks that there are primes p with these properties:
p ∈ {3, 7, 23, 83, 263, 1187, . . .}.
Similarly, for l(G) = 9 we need Ω(q − 1) = 4 and Ω(q + 1) 6 8, or Ω(q − 1) = 3 and
Ω(q + 1) = 8; in both cases, there are primes satisfying these conditions.
Next consider the case f = 3, so Ω(q − 1) > 3 and l(G) ∈ {6, 7, 8, 9}. First assume
l(G) = 6, so Ω(q − 1) = 3 and thus (p − 1)/2 and p2 + p + 1 are both primes. In particular,
p ≡ −1 (mod 12) and p > 11. Therefore, Ω(p + 1) > 4 and p2 − p + 1 is divisible by
3, hence Ω(q + 1) > 6 and we have reached a contradiction. Next assume l(G) = 7, so
Ω(q − 1) = 3 or 4. If Ω(q − 1) = 3 then we need Ω(q + 1) = 6, which forces Ω(p + 1) = 4
and p2 − p + 1 = 3r for some prime r. But 7 divides p6 − 1, so 7 must divide p + 1 and
thus p = 83 is the only possibility. But then p2 + p + 1 is composite, so this case does not
arise. However, there are primes
p ∈ {7, 11, 83, 1523, 20507, 28163, . . .}
satisfying the conditions Ω(q − 1) = 4 and Ω(q + 1) 6 6, so this case is recorded in Table
5. Similarly, if l(G) = m ∈ {8, 9} then either Ω(q − 1) = m − 3 and Ω(q + 1) 6 m − 1, or
Ω(q − 1) 6 m − 4 and Ω(q + 1) = m − 1. Moreover, one can check that there are primes p
satisfying these conditions.
Finally, let us assume f = 2. Here the condition p > 7 implies that Ω(q − 1) > 5, so
l(G) ∈ {7, 8, 9}. Suppose l(G) = 7. Here Ω(q − 1) = 5 and either (p − 1)/2 or (p + 1)/2
is a prime. Suppose (p − 1)/2 is a prime, so p ≡ 3 (mod 4) and (p + 1)/2 = 2r for some
prime r. Therefore, r, 2r − 1 and 4r − 1 are all primes. If r ∈ {2, 3} then p ∈ {7, 11} and
one checks that l(G) = 7. Now assume r > 5. If r ≡ 1 (mod 3) then 4r − 1 is divisible by
3. Similarly, if r ≡ 2 (mod 3) then 3 divides 2r − 1, so there are no examples with r > 5.
A similar argument applies if we assume (p + 1)/2 is a prime: here we need a prime r such
that 2r + 1 and 4r + 1 are also primes, and one checks that r = 3 is the only possibility,
which corresponds to the case G = L2 (169) with l(G) = 7.
Next assume l(G) = 8 and f = 2, so Ω(q − 1) = 5 or 6. The case Ω(q − 1) = 5 is
ruled out by arguing as in the previous paragraph. On the other hand, if Ω(q − 1) = 6
then we need Ω(q + 1) 6 7 and there are primes p with these properties. Finally, let us
assume l(G) = 9, so Ω(q − 1) ∈ {5, 6, 7}. The case Ω(q − 1) = 5 is ruled out as above,
whereas there are primes p such that Ω(q − 1) = 7 and Ω(q + 1) 6 8, or Ω(q − 1) = 6 and
Ω(q + 1) = 8.

In view of Theorem 4.1, it remains to determine the simple groups G ∼
6 L2 (q) with
=
l(G) = 8 or 9.
Lemma 4.3. Theorem 4 holds if l(G) = 8.

14

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

Proof. Let G be a simple group with l(G) = 8. By inspection, if G is a sporadic group then
G = M12 is the only example (also see [10, Tables III and IV]). For alternating groups,
[10, Theorem 1] gives


3n − 1
− bn − 1,
(5)
l(An ) =
2
where bn is the number of ones in the base 2 expansion of n. From the formula, it is easy
to check that no alternating group has length 8.
Now assume G is a simple group of Lie type over Fq , where q = pf with p a prime.
First we handle the exceptional groups. If G = 2 B2 (q) then p = 2, f > 3 is odd and
l(G) = Ω(q − 1) + 2f + 1

by [32, Theorem 1], hence l(G) = 8 if and only if f = 3. Next assume G = 2 G2 (q), so
p = 3 and f > 3 is odd. Since a Borel subgroup of G has order q 3 (q − 1), it follows that
l(G) > Ω(q − 1) + 3f + 1 > 8.

If G = G2 (q) then q > 3 and l(G) > l(GL2 (q)) + 5f + 1 > 8, so no examples arise. All of
the other exceptional groups can be eliminated in a similar fashion.
Finally, let us assume G is a classical group. If G = PΩǫn (q) is an orthogonal group with
n > 7, then it is clear that l(G) > 8 (indeed, a Sylow p-subgroup of G has length greater
than 8). Similarly, we can eliminate symplectic groups PSpn (q) with n > 6. Now assume
G = PSp4 (q) with q = pf > 3. Here l(G) > l(SL2 (q))+ 3f + 1, so we may assume f = 1, in
which case G has a maximal subgroup 24 .A5 or 24 .S5 (according to the value of q modulo
8) and thus l(G) > 1 + 4 + l(A5 ) = 9. Similarly, it is easy to show that l(Lǫn (q)) > 8 if
n > 4.
To complete the proof, we may assume that G = U3 (q) or L3 (q). Let B be a Borel
subgroup of G and first assume G = U3 (q), so q > 3. If q is even, then [32, Theorem 1]
gives
l(G) = Ω(|B|) + 1 = Ω(q 2 − 1) + 3f + 1 − Ω((3, q + 1))
(6)
and thus l(G) > 9. For q odd we have
l(G) > Ω(q 2 − 1) + 3f + 1 − Ω((3, q + 1))

(7)

l(G) = Ω(|B|) + 2 = 2Ω(q − 1) + 3f + 2 − Ω((3, q − 1))

(8)

l(G) > l(L2 (q)) + 2f + 2 + Ω(q − 1) − Ω((3, q − 1))

(9)

and we quickly deduce that f = 1, so q > 7 (since U3 (3) and U3 (5) have length 7). Now
Ω(q 2 − 1) > 5, so we must have Ω(q 2 − 1) = 5 and q ≡ 2 (mod 3), hence (q − 1)/2 and
(q + 1)/6 are both prime. This implies that q = 6r − 1, where r and 3r − 1 are primes, so
r = 2 is the only option and one checks that l(U3 (11)) = 9.
Finally, let us assume G = L3 (q). If q is even then
and it is easy to see that l(G) 6= 8. Now assume q is odd. If q = 3 then one can check that
l(G) = 8, so let us assume q > 5. Let H be a maximal parabolic subgroup of G. Then
l(G) > l(H) + 1, so
and we deduce that l(G) > 9.



Lemma 4.4. Theorem 4 holds if l(G) = 9.
Proof. This is very similar to the proof of the previous lemma. Let G be a simple group
with l(G) = 9. By inspection, G is not a sporadic group. In view of (5), G = A8 is the
only alternating group of length 9. Now assume G is a group of Lie type over Fq , where
q = pf with p a prime. The exceptional groups are easily eliminated by arguing as in the
proof of Lemma 4.3. Similarly, if G is a classical group then it is straightforward to reduce
to the cases G = PSp4 (q)′ and Lǫ3 (q) (note that L4 (2) ∼
= A8 and U4 (2) ∼
= PSp4 (3)).

ON THE LENGTH AND DEPTH OF FINITE GROUPS

15

Suppose G = PSp4 (q)′ . If q = 2 then G ∼
= A6 and l(G) = 5. Now assume q > 3. As
noted in the proof of the previous lemma, l(G) > l(SL2 (q)) + 3f + 1 and so we may assume
q = p is odd. Now G has a maximal subgroup H of type Sp2 (q) ≀ S2 , which implies that
l(G) > l(H) + 1 = 3 + 2l(L2 (p)).
If p = 3 then this lower bound is equal to 9 and one checks that l(PSp4 (3)) = 9. For p > 3
we get l(G) > 11.
Next assume G = L3 (q). If q is even, then (8) holds and one checks that l(G) = 9 if and
only if q = 4. Now assume q is odd. We have already noted that l(L3 (3)) = 8, so we may
assume q > 5. Moreover, in view of (9), we may assume that q = p. Since l(L2 (p)) > 4
and Ω(p − 1) > 2, it follows that l(L2 (p)) = 4, Ω(p − 1) = 2 and p ≡ 1 (mod 3). Clearly,
p = 7 is the only prime satisfying the latter two conditions, but l(L2 (7)) = 5.
To complete the proof of the lemma, we may assume G = U3 (q). If q is even then (6)
holds and we deduce that l(G) = 9 if and only if q = 4. Now suppose q is odd, so (7)
holds. If f > 2 then Ω(q 2 − 1) > 5 and thus l(G) > 11. Therefore, we may assume q = p
is odd. We have already noted that l(U3 (3)) = l(U3 (5)) = 7 and l(U3 (11)) = 9, and it is
straightforward to check that l(U3 (7)) = 10. Now assume q > 13. If Ω(q 2 − 1) > 7 then
(7) implies that l(G) > 10, so we must have Ω(q 2 − 1) = 5 or 6.
If Ω(q 2 − 1) = 5 then q = 13 is the only possibility (see the proof of Lemma 4.2) and
one checks that l(U3 (13)) = 9. Now assume Ω(q 2 − 1) = 6, so q ≡ 2 (mod 3) by (7). By
considering the maximal subgroups of G (see [6, Tables 8.5 and 8.6]), we see that
l(G) = max{9, Ω(q + 1) + l(L2 (q)) + 1, Ω(q 2 − q + 1) + 1}.
Note that Ω(q + 1) > 3 and l(L2 (q)) > 4 since q > 13 and q ≡ 2 (mod 3). If Ω(q + 1) > 4
then l(L2 (q)) > 5 and thus l(G) > 10. Therefore, Ω(q +1) = Ω(q −1) = 3 and l(L2 (q)) 6 5,
so either q = 29 or q ≡ ±3, ±13 (mod 40). In addition, we need Ω(q 2 − q + 1) 6 8 and one
checks there are primes p that satisfy these conditions:
p ∈ {173, 317, 653, 2693, 3413, 3677, . . .}.



This completes the proof of Theorem 4.
Proof of Corollary 5. First observe that part (i) is clear from the additivity of length (see
Lemma 2.1(i)) and the fact that every non-abelian simple group has length at least 4.
Next, assume G is a finite insoluble group with l(G) = 5. The simple groups of length 5
are given in Theorem 4, so we may assume that G is not simple. Therefore, G must have
exactly two composition factors: a non-abelian simple group T of length 4 and depth 3,
and a cyclic group Cp of prime order. In particular, λ(G) = 4 and so G is one of the groups
in Theorem 2. By inspecting the various possibilities, we see that either G = T × Cp , or
G is quasisimple with G/Z(G) = T and Z(G) = Cp , or G is almost simple with socle
T and G/T = Cp . Since l(T ) = 4, [19, Theorem 1] implies that T = L2 (q) and q is a
prime satisfying the conditions in the first row of Table 5. In particular, the only valid
quasisimple and almost simple groups are of the form SL2 (q) and PGL2 (q), respectively.
This completes the proof of part (ii) of Corollary 5.
Finally, suppose G is insoluble of length 6. Again, the simple groups of length 6 are
given by Theorem 4, so we may assume G is not simple. Then G has a unique non-abelian
composition factor T of length 4 or 5, and 6 − l(T ) abelian composition factors. It is
readily checked that the possibilities for G when l(T ) = 5 (resp. 4) are those in (iii)(b)
(resp. (c)) of Corollary 5.


16

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

4.2. Proof of Theorem 6. Set G = L2 (p), where p > 5 is a prime. By Lemma 2.5(ii),
l(G) 6 1 + max{4, Ω(p ± 1)}.

The result follows from Theorem 7 (see Appendix A), which implies that there are infinitely
many primes p with max{Ω(p ± 1)} 6 8.
5. Chain differences and ratios
In this section we prove our main results concerning chan differences and chain ratios,
namely Theorems 8, 9 and 11.
5.1. Proof of Theorem 8. For comparison, we start by recalling [7, Theorem 3.3], which
describes the simple groups of chain difference one.
Theorem 5.1 (Brewster et al. [7]). Let G be a finite simple group. Then cd(G) = 1 if
and only if G = L2 (q) and either q ∈ {4, 5, 9}, or q is a prime and one of the following
holds:
(i) 3 6 Ω(q ± 1) 6 4 and either q ≡ ±1 (mod 10) or q ≡ ±1 (mod 8).
(ii) Ω(q ± 1) 6 3 and q ≡ ±3, ±13 (mod 40).
We begin the proof of Theorem 8 by handling the alternating and sporadic groups.
Lemma 5.2. Let G be a simple alternating or sporadic group. Then cd(G) = 2 if and
only if G = A7 or J1 .
Proof. First assume G = An is an alternating group. A formula for l(G) is given in (5) and
it is easy to compute λ(An ) directly for small values of n: we get cd(A5 ) = cd(A6 ) = 1,
cd(A7 ) = 2 and cd(A8 ) = 4. By Lemma 2.2, it follows that cd(An ) > 4 for all n > 8.
The length and depth of each sporadic group G is given in [10, Tables III and IV] and [8,
Table 2], respectively, and we immediately deduce that cd(G) = 2 if and only if G = J1 .
(Note that the “possible values” for l(G) listed in [10, Table IV] are in fact the exact
values.)

Lemma 5.3. Let G be a simple group of Lie type over Fq with G ∼
6 L2 (q). Then cd(G) = 2
=
if and only if G = U3 (5).
Proof. We will follow a similar approach to the proof of [7, Theorem 3.3]. Set q = pf ,
where p is a prime and f > 1.
First assume G = L3 (q), in which case q > 3 since L3 (2) ∼
= L2 (7). If q = 3 then
l(G) = 8 and λ(G) = 3, so we may assume q > 4 and thus l(G) > 9 by Theorem 4.
If p is odd then L3 (p) has a maximal subgroup SO3 (p) ∼
= PGL2 (p), so λ(L3 (p)) 6 6
by [8, Corollary 3.4] and thus cd(L3 (p)) > 3. In view of Lemma 2.2, this implies that
cd(G) > 3 since L3 (p) 6 G. Now assume p = 2 and let H = QL be a maximal parabolic
subgroup of G, where Q is elementary abelian of order q 2 and L 6 GL2 (q) has index
d = (3, q − 1). Note that L acts irreducibly on Q, so L is a maximal subgroup of H.
Therefore, l(H) = Ω((q − 1)/d) + 2f + l(L2 (q)) and
λ(H) 6 λ(L) + 1 = Ω((q − 1)/d) + λ(L2 (q)) + 1,

so cd(H) > 2f − 1 + cd(L2 (q)) > 2f and the result follows since f > 2.
Now assume G = Ln (q) with n > 4. Since G has a subgroup L3 (q), the desired result
follows from Lemma 2.2 if q > 3. Similarly, if q = 2 then G has a subgroup L4 (2) and it
is easy to check that cd(L4 (2)) = 4.
Next suppose G is a classical group of rank r > 3, or an untwisted exceptional group,
or one of 3 D4 (q) or 2 E6 (q). It is easy to see that G has a section isomorphic to L3 (q) and

ON THE LENGTH AND DEPTH OF FINITE GROUPS

17

thus cd(G) > 3 if q > 3. Now assume q = 2. If G = G2 (2)′ ∼
= U3 (3) then cd(G) = 3. Since
−
U3 (3) < Sp6 (2) < Ω−
(2),
it
follows
that
cd(G)
>
3
if
G
=
Sp
6 (2) or Ω8 (2). In each of the
8
remaining cases (with q = 2), G has a section isomorphic to L4 (2) and thus cd(G) > 4.
Next assume G = U3 (q), so q > 3. Let H = QL be a Borel subgroup of G, where
d = (3, q + 1). Here Q = q 1+2 , L = (q 2 − 1)/d and Q/Z(Q) is elementary abelian of
order q 2 . Moreover, L acts irreducibly on Q/Z(Q). Therefore, l(H) = 3f + Ω(L) and
λ(H) 6 f + 1 + Ω(L), so cd(H) > 2f − 1 and we may assume q = p. If p ∈ {3, 7, 11}
then it is easy to check that cd(G) > 3, whereas cd(G) = 2 if p = 5. For p > 11,
Theorem 4 implies that l(G) > 9 and we get cd(G) > 3 since G has a maximal subgroup
SO3 (p) ∼
= PGL2 (p) of depth at most 5. Now assume G = Un (q) with n > 4. If q is even
then G has a section isomorphic to U4 (2) and one checks that cd(U4 (2)) = 4. Similarly,
if q is odd and q 6= 5 then G has a section U3 (q) with cd(U3 (q)) > 3. Finally, suppose
q = 5. One checks that λ(U4 (5)) = 5 and thus cd(U4 (5)) > 5 by Theorem 4. The result
now follows since G has a section isomorphic to U4 (5).
To complete the proof, we may assume G = PSp4 (q), 2 F4 (q)′ , 2 G2 (q) or 2 B2 (q). Suppose
G = PSp4 (q) with q > 3. If q is even then G has a maximal subgroup H = L2 (q) ≀ S2 .
Now l(H) = 2l(L2 (q)) + 1 and λ(H) 6 λ(L2 (q)) + 2, so cd(H) > l(L2 (q)) > 4. Similarly,
if q is odd then H = 24 .Ω−
4 (2) < G and the result follows since cd(H) = 4. Next assume
2
′
G = F4 (q) . One checks that the Tits group 2 F4 (2)′ has depth 4, so cd(2 F4 (2)′ ) > 6 by
Theorem 4 and thus cd(G) > 6 by Lemma 2.2.
Next assume G = 2 G2 (q), so q = 3f and f > 3 is odd. Let H be a Borel subgroup of G
and let K = 2 × L2 (q) be the centralizer of an involution. Then
l(G) > l(H) + 1 = Ω(q − 1) + 3f + 1

λ(G) 6 λ(K) + 1 6 λ(L2 (q)) + 2 6 Ω(q − 1) + 3
and thus cd(G) > 3f − 2 > 7. Finally, suppose G = 2 B2 (q), where q = 2f and f > 3 is
odd. Here l(G) = Ω(q − 1) + 2f + 1 by [32, Theorem 1] and λ(G) 6 Ω(q − 1) + 2 (since G
has a maximal subgroup D2(q−1) ). Therefore, cd(G) > 2f − 1 > 5.

The next result completes the proof of Theorem 8.
Lemma 5.4. If G = L2 (q) with q > 5, then cd(G) = 2 if and only if
(i) q ∈ {7, 8, 11, 27, 125}; or
(ii) q is a prime and one of the following holds:
(a) max{Ω(q ± 1)} = 4 and either min{Ω(q ± 1)} = 2, or q ≡ ±3, ±13 (mod 40).
(b) max{Ω(q ± 1)} = 5, min{Ω(q ± 1)} > 3 and q 6≡ ±3, ±13 (mod 40).
Proof. As before, write q = pf . First assume p = 2, so q > 4. Here l(G) = Ω(q − 1) + f + 1
by Lemma 2.5(i) and λ(G) 6 Ω(q−1)+2 (since D2(q−1) is a maximal subgroup). Therefore,
cd(G) > f − 1 and thus f ∈ {2, 3}. If f = 2 then cd(G) = 1, while cd(G) = 2 if f = 3
(the case q = 8 is recorded in part (ii)(a) of Theorem 8).
Now assume p > 3. For q 6 11, one checks that cd(G) = 2 if and only if q = 7 or 11,
so we may assume q > 13. This implies that Dq−1 is a maximal subgroup of G and thus
λ(G) 6 Ω(q − 1) + 1. Now l(G) > Ω(q − 1) + f by Lemma 2.5(ii), so cd(G) > f − 1 and
thus we may assume f ∈ {1, 2, 3}.
First assume q = p > 13. By [8, Corollary 3.4] we have

3 min{Ω(p ± 1)} = 2 or p ≡ ±3, ±13 (mod 40)
λ(G) =
4 otherwise.

18

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

If λ(G) = 3 then we need l(G) = 5, in which case Theorem 4 implies that max{Ω(p±1)} =
4. There are primes p that satisfy these conditions. For example, if
p ∈ {23, 59, 83, 227, 347, 563, . . .},

then Ω(p − 1) = 2 and Ω(p + 1) = 4. Similarly, we have λ(G) = 4 and l(G) = 6 if and
only if Ω(p ± 1) > 3, p 6≡ ±3, ±13 (mod 40) and max{Ω(p ± 1)} = 5. Once again, there
are primes p with these properties.
Next assume q = p2 with p > 5. Since PGL2 (p) is a maximal subgroup of G, it
follows that λ(G) 6 6 and thus l(G) 6 8. Now, if l(G) 6 7 then Theorem 4 implies
that p ∈ {5, 7, 11, 13} and in each case one checks that cd(G) > 3. Therefore, we may
assume λ(G) = 6 and l(G) = 8. By Theorem 4, l(G) = 8 if and only if Ω(q − 1) = 6
and Ω(q + 1) 6 7. Similarly, λ(G) = 6 if and only if Ω(q ± 1) > 5, p ≡ ±1 (mod 10) and
λ(L2 (p)) = 4. The latter constraint yields the additional condition Ω(p ± 1) > 3, so we
need Ω(p − 1) = Ω(p + 1) = 3 since Ω(q − 1) = 6. We claim that there are no primes p that
satisfy these conditions. For example, suppose p ≡ 1 (mod 10). Then (p − 1)/10 must be
a prime. If p ≡ 1 (mod 3) then p = 31 is the only possibility, but this gives Ω(p + 1) = 5.
On the other hand, if p ≡ 2 (mod 3) then (p + 1)/6 is a prime. But p2 ≡ 1 (mod 8) and
thus (p − 1)/10 = 2 or (p + 1)/6 = 2, which implies that p = 11 and Ω(p − 1) = 2. A very
similar argument handles the case p ≡ −1 (mod 10).
Finally, suppose q = p3 . If p = 3 then one checks that λ(G) = 3 and l(G) = 5, so
cd(G) = 2 in this case. Now assume p > 5. Since L2 (p) is a maximal subgroup of G, it
follows that

4 min{Ω(p ± 1)} = 2 or p ≡ ±3, ±13 (mod 40)
λ(G) =
5 otherwise
and thus l(G) 6 7. By applying Theorem 4, we see that λ(G) = 4 and l(G) = 6 if and
only if p = 5. Similarly, λ(G) = 5 and l(G) = 7 if and only if Ω(q − 1) = 4, Ω(q + 1) 6 6,
Ω(p ± 1) > 3 and p 6≡ ±3, ±13 (mod 40). Note that the conditions Ω(q − 1) = 4 and
Ω(p ± 1) > 3 imply that Ω(p − 1) = 3 and p2 + p + 1 is a prime, so p ≡ 2 (mod 3) and
p2 − p + 1 is divisible by 3, whence Ω(p2 − p + 1) > 2. There are primes p such that
Ω(p3 − 1) = 4, Ω(p3 + 1) 6 6 and Ω(p ± 1) > 3: the smallest one is 433373. However, we
claim that there is no prime p that also satisfies the condition p 6≡ ±3, ±13 (mod 40).
If p ≡ 1, 9, 17, 33 (mod 40) then p − 1 is divisible by 8 and thus Ω(p − 1) > 4. Similarly,
if p ≡ 7, 23, 31, 39 (mod 40) then p + 1 is divisible by 24, and p 6= 23 since we need
Ω(p − 1) = 3, so Ω(p + 1) > 5. But we have already noted that Ω(p2 − p + 1) > 2,
whence Ω(p3 + 1) > 7. Finally, suppose p ≡ 11, 19, 29 (mod 40). These cases are similar,
so let us assume p ≡ 11 (mod 40). Here (p − 1)/10 is a prime and p + 1 is divisible by
12, so Ω(p + 1) > 4 since p 6= 11. Since Ω(p3 + 1) 6 6, it follows that (p + 1)/12 and
(p2 − p + 1)/3 are both primes. Now p6 ≡ 1 (mod 7), so one of (p − 1)/10, p2 + p + 1,
(p + 1)/12 or (p2 − p + 1)/3 must be equal to 7, but it is easy to see that this is not
possible. For example, if (p − 1)/10 = 7 then p = 71 does not satisfy the required
congruence condition.

5.2. Proof of Theorem 9. Recall that cr(G) = l(G)/λ(G) is the chain ratio of G. In this
section we prove Theorem 9, which states that cr(G) > 5/4 for every finite non-abelian
simple group G, with equality if and only if G ∈ G (see (3)). We partition the proof into
several cases.
Lemma 5.5. Theorem 9 holds if G is a sporadic or alternating group.
Proof. First assume G is a sporadic group. The length and depth of G is given in [10,
Tables III and IV] and [8, Table 2], respectively, and we immediately deduce that cr(G) >
3/2, with equality if and only if G = J1 .

ON THE LENGTH AND DEPTH OF FINITE GROUPS

19

Now assume G = An is an alternating group. The length of G is given in (5) and [8,
Theorem 2] states that λ(G) 6 23. One checks that this bound is sufficient if n > 23. For
example, if n = 23 then l(G) = 34 − 4 − 1 = 29 and thus cr(G) > 29/23 > 5/4. For n < 23
we can use Magma to show that λ(G) 6 6, with equality if and only if n = 16. In view of
the above formula for l(G), we deduce that cr(G) > 5/4 if n > 8. For the smallest values
of n, we get cr(A5 ) = 4/3, cr(A6 ) = 5/4 and cr(A7 ) = 3/2.

Lemma 5.6. Theorem 9 holds if G is an exceptional group of Lie type.
Proof. Let G be a finite simple exceptional group of Lie type over Fq , where q = pk and p
is a prime. Let B be a Borel subgroup of G and let r be the twisted Lie rank of G. Then
l(G) > Ω(|B|) + r

(10)

and [8, Theorem 4] gives
λ(G) 6 3Ω(k) + 36
2B

if G 6= 2 (q).
First assume G = E8 (q). Here |B| = q 120 (q − 1)8 so

5
5
5
l(G) > 120k + 8 > (3 log2 k + 36) > (3Ω(k) + 36) > λ(G)
4
4
4

and the result follows. The case E7 (q) is handled in exactly the same way, and similarly
E6ǫ (q) and F4 (q) with k > 2. Suppose G = F4 (p), so l(G) > 28 by (10). If p = 2 then
2 F (2) < G is maximal and λ(2 F (2)) = 5, so λ(G) 6 6 and the result follows. Similarly,
4
4
if p is odd then
λ(F4 (p)) 6 λ(2.Ω9 (p)) + 1 6 λ(Ω9 (p)) + 2
and one of A10 , S10 or A11 is a maximal subgroup of Ω9 (p) (see [6, Table 8.59]). Therefore
λ(Ω9 (p)) 6 7, so λ(G) 6 9 and once again we deduce that cr(G) > 5/4. If G = E6ǫ (p) then
F4 (p) < G is maximal, so the previous argument yields λ(G) 6 10 and the result quickly
follows.
Next assume G = G2 (q)′ . If q = 2 then G ∼
= U3 (3) and one checks that λ(G) = 4 and
6
l(G) = 7. Now assume q > 2. Since |B| = q (q − 1)2 it follows that l(G) > 6k + 4 and
by considering a chain of subfield subgroups we deduce that λ(G) 6 Ω(k) + λ(G2 (p)). If
p > 5 then G2 (2) < G2 (p) is maximal, so λ(G2 (p)) 6 6 and it is easy to check that the
same bound holds if p = 2 or 3. Therefore λ(G) 6 Ω(k) + 6 and we deduce that
5
5
5
l(G) > 6k + 4 > (log2 k + 6) > (Ω(k) + 6) > λ(G).
4
4
4
If G = 3 D4 (q) then G2 (q) < G is maximal and thus λ(G) 6 Ω(k) + 7. In addition,
|B| = q 12 (q 3 − 1)(q − 1), so l(G) > 12k + 2 and the result follows.
To complete the proof of the lemma, we may assume G = 2 F4 (q)′ , 2 G2 (q) or 2 B2 (q).
Suppose G = 2 F4 (q)′ , so q = 2k with k odd. If k = 1 then λ(G) = 4 and l(G) > 13
since G has a soluble maximal subgroup of the form 2.[28 ].5.4. Similarly, if k > 1 then
λ(G) 6 Ω(k) + 5 (see the proof of [8, Theorem 4]), l(G) > 12k + 2 and these bounds are
sufficient. The case G = 2 G2 (q)′ , where q = 3k with k odd, is very similar. If k = 1 then
G∼
= L2 (8), so λ(G) = 3 and l(G) = 5. If k > 1, then the proof of [8, Theorem 4] gives
λ(G) 6 Ω(k) + 4 and we have l(G) > 3k + 2 since |B| = q 3 (q − 1). It is easy to check that
these bounds are sufficient.
Finally, let us assume G = 2 B2 (q), where q = 2k with k > 3 odd. Since |B| = q 2 (q − 1),
it follows that l(G) > 2k + 1 + Ω(q − 1) > 2k + 2. By [8, Theorem 4], we also have
λ(G) 6 Ω(k) + 1 + Ω(q − 1) < Ω(k) + k + 1.

20

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

Therefore,
l(G) > 2k + 2 >

5
5
(log2 k + k + 1) > λ(G)
4
4

as required.



Lemma 5.7. Theorem 9 holds if G is a classical group and G ∼
6 L2 (q).
=
Proof. Let G be a finite simple classical group over Fq and r be the twisted rank of G. As
before, write q = pk with p a prime. Let B be a Borel subgroup of G and recall that (10)
holds. Our initial aim is to reduce the problem to groups of small rank. To do this, we
will consider each family of classical groups in turn.
First assume G = Lr+1 (q). We claim that cr(G) > 5/4 if r > 9. To see this, first
observe that
1
l(G) > Ω(|B|) + r > kr(r + 1) + r
2
and λ(G) 6 3Ω(k) + 36 by [8, Theorem 4]. For r > 9, it is routine to check that
1
5
5
l(G) > kr(r + 1) + r > (3 log2 k + 36) > λ(G),
2
4
4
which justifies the claim. In a similar fashion, we can reduce the problem to r 6 6 when
−
G = PSp2r (q), Ω2r+1 (q) or PΩ+
2r (q); r 6 5 when G = PΩ2r+2 (q); and r 6 4 for G = U2r (q).
Finally, suppose G = U2r+1 (q). Here l(G) > kr(2r + 1) + r and [8, Theorem 4] states
that λ(G) 6 3Ω(k) + 36 if q or k is odd. If q = 2k and k is even, then the same theorem
gives
a
λ(G) 6 3Ω(k) + 35 + 2Ω(22 + 1),
a
where k = 2a b and b is odd. Since Ω(22 + 1) 6 k, it follows that λ(G) 6 3Ω(k) + 2k + 35
for all possible values of q and k, and one checks that
5
5
5
l(G) > kr(2r + 1) + r > (3 log2 k + 2k + 35) > (3Ω(k) + 2k + 35) > λ(G)
4
4
4
if r > 5.
Therefore, in order to complete the proof of the lemma, we may assume that we are in
one of the following cases:
(a)
(b)
(c)
(d)
(e)
(f)
(g)

G = Ω2r+1 (q) with 3 6 r 6 6 and q odd;
G = PSp2r (q) with 2 6 r 6 6;
G = PΩ+
2r (q) with 4 6 r 6 6;
−
G = PΩ2r+2 (q) with 3 6 r 6 5;
G = Lr+1 (q) with 2 6 r 6 8;
G = U2r (q) with 2 6 r 6 4;
G = U2r+1 (q) with 1 6 r 6 4.

Let us start by handling case (a). By considering a chain of subfield subgroups, we see
that λ(G) 6 2Ω(k) + λ(Ω2r+1 (p)). In addition, the proof of [8, Theorem 4] implies that
λ(Ω2r+1 (p)) 6 4 + λ(Sn ) for some n 6 2r + 3 = 15. One checks that λ(Sn ) 6 6 for n 6 15,
2
hence λ(G) 6 2Ω(k) + 10. Now |B| = 12 (q − 1)r q r , so (10) yields l(G) > kr 2 + 2r − 1 and
one checks that
5
kr 2 + 2r − 1 > (2 log2 k + 10)
4
for all possible values of k and r. The result follows.
Next consider (b). First assume p = 2, in which case λ(G) 6 Ω(k) + λ(Sp2r (2)) and
λ(Sp2r (2)) 6 4 + λ(Sn ) for some n 6 14 (see the proof of [8, Theorem 4]). Therefore,
λ(G) 6 Ω(k) + 10. Since l(G) > kr 2 + r by (10), the result follows unless (r, k) = (3, 1),

ON THE LENGTH AND DEPTH OF FINITE GROUPS

21

or if r = 2 and k 6 3. If (r, k) = (2, 1) then G ∼
= A6 and cr(G) = 5/4. In each of the
remaining cases we have λ(G) 6 5 and cd(G) > 2, which implies the desired bound. Now
assume p is odd, so l(G) > kr 2 + 2r − 1 and the proof of [8, Theorem 4] yields
λ(G) 6 2Ω(k) + λ(PSp2r (p)) 6 2Ω(k) + 8 + λ(Sr ) 6 2Ω(k) + 13.
This gives cr(G) > 5/4 unless (r, k) = (3, 1), or if r = 2 and k 6 4. If G = PSp6 (p) then
G > L2 (p3 ).3 > L2 (p3 ) > L2 (p)
is unrefinable, so λ(G) 6 7 and the result follows since cd(G) > 2. Now assume r = 2
and k 6 4. Note that |B| = 12 q 4 (q − 1)2 , so Ω(|B|) = 4k + 2Ω(q − 1) − 1. If k = 4 then
λ(G) 6 4 + λ(PSp4 (p)) and we note that one of A6 , S6 or S7 is a maximal subgroup of
PSp4 (p), so λ(PSp4 (p)) 6 7 and thus λ(G) 6 11. In addition, Ω(q − 1) > 5, so l(G) > 27
and the result follows. Similarly, if k = 2 or 3 then λ(G) 6 8 and l(G) > 15. Finally, if
k = 1 then λ(G) 6 7 and the result follows since cd(G) > 2.
Now let us turn to case (c), so G = PΩ+
2r (q) and r = 4, 5 or 6. First assume p = 2, in
which case l(G) > kr(r − 1) + r and λ(G) 6 Ω(k) + λ(Ω+
2r (2)). It is easy to check that
+
λ(Ω2r (2)) 6 9. For example, if r = 6 then there is an unrefinable chain
−
−
Ω+
12 (2) > Sp10 (2) > Ω10 (2).2 > Ω10 (2) > A12

and λ(A12 ) = 5, so λ(Ω+
12 (2)) 6 9. Therefore, l(G) > 12k + 4, λ(G) 6 Ω(k) + 9 and one
checks that these bounds are sufficient. Now assume p > 2. Here l(G) > kr(r − 1) + 2r − 2
+
and λ(G) 6 3Ω(k) + λ(PΩ+
2r (p)). One checks that λ(PΩ2r (p)) 6 9. For instance, if r = 6
then there is an unrefinable chain
PΩ+
12 (p) > PSO11 (p) > Ω11 (p) > H
with H = A12 , S12 or A13 , and the claim follows since λ(H) 6 6. Therefore, l(G) > 12k+6,
λ(G) 6 3Ω(k) + 9 and we conclude that cr(G) > 5/4. A very similar argument applies in
case (d) and we omit the details.
Next consider case (e), so G = Lr+1 (q) and λ(G) 6 2Ω(k) + λ(Lr+1 (p)). Note that
|B| =

q r(r+1)/2 (q − 1)r
.
(r + 1, q − 1)

Suppose r ∈ {3, 5, 7} is odd. Now Lr+1 (p) has a maximal subgroup of the form PSpr+1 (p)
or PSpr+1 (p).2, and we noted that λ(PSpr+1 (p)) 6 13 in the analysis of case (b), whence
λ(G) 6 2Ω(k) + 15. One now checks that the bound l(G) > kr(r + 1)/2 + r from (10)
is sufficient when r = 5 or 7. Now suppose r = 3. If q = 2 then cr(G) = 9/5 so we
can assume q > 2, in which case l(G) > 6k + 5. Now λ(L4 (p)) = 5 if p = 2 or 3, and
λ(L4 (p)) 6 9 if p > 5 (this follows from the fact that PSp4 (p).2 is a maximal subgroup
of L4 (p)). Therefore, λ(G) 6 2Ω(k) + 9 and the result follows if k > 1. Finally suppose
G = L4 (p) with p > 3. If p = 3 then λ(G) = 5 and l(G) > 11. Similarly, λ(G) 6 9 and
l(G) > 13 if p > 5. The result follows.
Now let us assume G = Lr+1 (q) and r ∈ {2, 4, 6, 8}. First assume p is odd. There
is an unrefinable chain Lr+1 (p) > PSOr+1 (p) > Ωr+1 (p), so λ(Lr+1 (p)) 6 12 and thus
λ(G) 6 2Ω(k) + 12. Now l(G) > kr(r + 1)/2 + 2r − 1 and the desired bound follows if
r > 2. Now assume G = L3 (q). Since Ω3 (p) ∼
= L2 (p) we deduce that λ(L3 (p)) 6 6, so
λ(G) 6 2Ω(k) + 6 and one checks that the bound l(G) > 3k + 3 is good enough if k > 2.
If G = L3 (p2 ) then l(G) > 11 and λ(G) 6 8 since there is an unrefinable chain
L3 (p2 ) > L3 (p).2 > L3 (p) > PSO3 (p) > Ω3 (p).
Similarly, if G = L3 (p) then λ(G) 6 6 and we note that l(G) > 10 if p > 5 (this follows
from Theorem 4). Finally, if G = L3 (3) then cr(G) = 8/3.

22

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

To complete the analysis of case (e), let us assume r ∈ {2, 4, 6, 8} and p = 2. Here
l(G) > kr(r + 1)/2 + r and λ(G) 6 2Ω(k) + λ(Lr+1 (2)). As noted in the proof of [8,
Theorem 4], there is an unrefinable chain Lr+1 (2) > 2r .Lr (2) > Lr (2) and one can check
that λ(Lr (2)) 6 5, so λ(G) 6 2Ω(k) + 7. This gives the desired bound unless r = 2 and
k 6 3. We can exclude the case k = 1 since L3 (2) ∼
= L2 (7). For k ∈ {2, 3} we get λ(G) 6 4,
l(G) > 9 and the result follows.
To complete the proof of the lemma, it remains to handle the unitary groups of dimension at most 9 arising in cases (f) and (g). First consider (f), so G = U2r (q), r ∈ {2, 3, 4}
and
q r(2r−1) (q 2 − 1)r
.
|B| =
(2r, q + 1)
By arguing as in the proof of [8, Theorem 4], we see that λ(G) 6 λ(PSp2r (q)) + 2. If
p = 2, it follows that
λ(G) 6 Ω(k) + 2 + λ(Sp2r (2)) 6 Ω(k) + 12
(recall that λ(Sp2r (2)) 6 10). In view of (10) we have l(G) > kr(2r − 1) + 2r − 1 and
one checks that these bounds are sufficient unless r = 2 and k = 1, 2. Here we compute
λ(U4 (4)) = λ(U4 (2)) = 5 and the result follows. Now assume p > 2. Here
λ(G) 6 2Ω(k) + 2 + λ(PSp2r (p)) 6 2Ω(k) + 15
and (10) gives l(G) > kr(2r − 1) + 4r − 2. These estimates give the result, unless r = 2
and k = 1, 2. There is an unrefinable chain
U4 (p2 ) > PSp4 (p2 ).2 > PSp4 (p2 ) > L2 (p2 ) > L2 (p).2 > L2 (p)
and thus λ(G) 6 9 for G = U4 (p2 ). Similarly, one checks that λ(G) 6 9 if G = U4 (p). In
both cases l(G) > 12 and the result follows.
Finally, let us consider case (g), where G = U2r+1 (q), r ∈ {1, 2, 3, 4} and
|B| =

q r(2r+1) (q 2 − 1)r
.
(2r + 1, q + 1)

First assume p > 2. Here
λ(G) 6 λ(Ω2r+1 (q)) + 2 6 2Ω(k) + 2 + λ(Ω2r+1 (p)) 6 2Ω(k) + 12
and (10) gives l(G) > kr(2r + 1) + 4r − 2. One checks that these bounds are sufficient
unless r = 1 and k 6 5. Suppose G = U3 (pk ) with k 6 5. If k = 3 or 5 then λ(G) 6
2 + λ(U3 (p)) 6 8 and the result follows since l(G) > 11. If k = 4 then l(G) > 14 and there
is an unrefinable chain
U3 (p4 ) > PSO3 (p4 ) > Ω3 (p4 ) > L2 (p2 ).2 > L2 (p2 ) > L2 (p).2 > L2 (p),
so λ(G) 6 10. Similarly, if k = 2 then λ(G) 6 8 and l(G) > 11. Finally, suppose
G = U3 (p). If p > 7 then U3 (p) > PSO3 (p) > Ω3 (p) is unrefinable, so λ(G) 6 6. It is easy
to check that the same bound holds when p = 3 or 5, and the desired result now follows
since cd(G) > 2.
Now suppose p = 2. If k is odd then by considering a chain of subfield subgroups we
get λ(G) 6 2Ω(k) + λ(U2r+1 (2)) and one checks that λ(U2r+1 (2)) 6 6. For example,
J3 < U9 (2) is maximal and λ(J3 ) = 5, so λ(U9 (2)) 6 6. Therefore, λ(G) 6 2Ω(k) + 6.
Since l(G) > kr(2r + 1) + 2r − 1, the result follows unless r = 1 and k = 3 (note that
(r, k) 6= (1, 1) since U3 (2) is soluble). A routine computation gives cr(U3 (8)) = 4.
Finally, suppose p = 2 and k is even. Here l(G) > kr(2r + 1) + 3r − 1 and we recall
that λ(G) 6 3Ω(k) + 2k + 35. These bounds are sufficient unless (r, k) = (3, 2), or r = 2
and k ∈ {2, 4, 6}, or if r = 1. If (r, k) = (3, 2) then G = U7 (4) has a maximal subgroup

ON THE LENGTH AND DEPTH OF FINITE GROUPS

23

3277:7 of depth 3, so λ(G) 6 4 and the result follows. Similarly, if r = 2 and k ∈ {2, 4, 6}
then by considering an unrefinable chain through the maximal subgroup
(q 5 + 1)
:5,
(q + 1)(5, q + 1)
we deduce that λ(G) 6 5 and the result follows since l(G) > 10k + 5. Finally, let us
assume G = U3 (2k ) with k even. Now G has a reducible maximal subgroup of the form
q+1
.L2 (q)
(3, q + 1)
and thus
λ(G) 6 λ(L2 (q)) + Ω(q + 1) − Ω((3, q + 1)) + 1 6 Ω(q 2 − 1) + Ω(k) + 2 − Ω((3, q + 1))

since λ(L2 (q)) 6 Ω(q − 1) + Ω(k) + 1 by [8, Theorem 4]. We also have
l(G) = Ω(q 2 − 1) + 3k + 1 − Ω((3, q + 1))

by [32, Theorem 1], and one checks that

5
(Ω(q 2 − 1) + Ω(k) + 2 − Ω((3, q + 1)))
4
if Ω(q 2 −1)+6 < 7k. Since q = 2k we have Ω(q 2 −1)+6 < 2k +6 and the result follows.
Ω(q 2 − 1) + 3k + 1 − Ω((3, q + 1)) >



Lemma 5.8. Theorem 9 holds if G ∼
= L2 (q).

Proof. Write q = pk with p a prime. If k = 1 then λ(G) ∈ {3, 4} by [8, Corollary 3.4],
and we have l(G) > λ(G) + 1, so cr(G) > 5/4 and equality holds if and only if λ(G) = 4
and l(G) = 5 (that is, if and only if G ∈ G; see (3)). The result now follows by combining
Theorems 4 and 5.1. For the remainder, we may assume k > 2.
Suppose p = 2. Here l(G) = Ω(q − 1) + k + 1 and λ(G) 6 Ω(q − 1) + Ω(k) + 1 by [32,
Theorem 1] and [8, Theorem 4(i)]. Now
5
Ω(q − 1) + k + 1 > (Ω(q − 1) + Ω(k) + 1)
4
if and only if
Ω(q − 1) + 5Ω(k) + 1 < 4k.
(11)
Since
Ω(q − 1) + 5Ω(k) + 1 < k + 5 log 2 k + 1,
it is routine to check that (11) holds for all k > 2.
Now suppose p > 2 and observe that l(G) > Ω(q − 1) + k by Lemma 2.5(ii). First
assume k > 3 is odd. By considering a chain of subfield subgroups (as in the proof of
[8, Theorem 4]), we deduce that λ(G) 6 Ω(k) + λ(L2 (p)). Therefore, λ(G) 6 Ω(k) + 2 if
p = 3, so
5
5
5
l(G) > Ω(q − 1) + k > k + 2 > (log3 k + 2) > (Ω(k) + 2) > λ(G)
4
4
4
as required. Similarly, if p > 5 then l(G) > k + 3, λ(G) 6 Ω(k) + 4 and for k > 3 the result
follows in the same way. If k = 3 then λ(G) 6 5 and Theorem 5.1 implies that cd(G) > 2,
so cr(G) > 5/4.
Finally, let us assume p > 2 and k is even. If q = 9 then G ∼
= A6 and we have
already noted that cr(G) = 5/4 in this case. Now assume q > 9, so Ω(q − 1) > 4 and
thus l(G) > k + 4. Also observe that λ(G) 6 2Ω(k) + λ(L2 (p)). If p = 3 then k > 4,
λ(G) 6 2Ω(k) + 2 and one checks that
5
k + 4 > (2 log 2 k + 2),
4

24

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

which gives the desired result. Now assume p > 5. Here λ(G) 6 2Ω(k) + 4 and we have
5
k + 4 > (2 log2 k + 4)
4
if k > 8. If k ∈ {4, 6, 8} then Ω(q − 1) > 6, so l(G) > k + 6 and the result follows. Finally,
if k = 2 then λ(G) 6 6 (since PGL2 (p) < G is maximal) and cd(G) > 2 by Theorem 5.1,
so cr(G) > 5/4 as required.

This completes the proof of Theorem 9. Notice that Corollary 10 follows immediately.
Indeed, we have l(G) 6 a cd(G) if and only if cr(G) > a/(a − 1), so Theorem 9 implies
that a = 5 is the best possible constant.
5.3. Proof of Theorem 11. We begin by recording some immediate consequences of
Lemma 2.2.
Proposition 5.9. Let G be a finite group.
(i) If 1 = GP
m ✁ Gm−1 ✁ · · · ✁ G1 ✁ G0 = G is a chain of subgroups of G, then
cd(G) > i cd(Gi−1 /Gi ).
P
(ii) If G = G1 × · · · × Gm , then cd(G) > i cd(Gi ).
(iii) If T1 , . . . P
, Tm are the composition factors of G, listed with multiplicities, then
cd(G) > i cd(Ti ).

Note that in part (iii) above we havePcd(Ti ) = 0 if Ti is abelian, so only the nonabelian composition factors contribute to i cd(Ti ). Now let T1 , . . . , Tm be the non-abelian
composition factors of G (listed with multiplicities), and let
ss(G) =

m
Y

Ti

i=1

be their direct product.
Proposition 5.10. We have l(ss(G)) 6 5 cd(G) for every finite group G.
Proof. With the above notation we have
l(ss(G)) =

m
X

l(Ti ).

i=1

By Corollary 10 we have l(Ti ) 6 5 cd(Ti ) for each i, and by combining this with Proposition
5.9(iii) we obtain
l(ss(G)) 6

m
X

5 cd(Ti ) = 5

m
X

cd(Ti ) 6 5 cd(G)

i=1

i=1

as required.



By a semisimple group we mean a direct product of (non-abelian) finite simple groups.
Lemma 5.11. If G is a finite semisimple group, then l(Aut(G)) 6 2l(G).
Q
ki
Proof. Write G = m
i=1 Ti where the Ti are pairwise non-isomorphic finite (non-abelian)
simple groups and ki > 1. Then
Aut(G) ∼
=

m
Y
i=1

Aut(Tiki )

∼
=

m
Y
i=1

Aut(Ti ) ≀ Ski .

ON THE LENGTH AND DEPTH OF FINITE GROUPS

Hence Out(G) ∼
=

Qm

i=1 Out(Ti )
m
X

≀ Ski , so

ki l(Out(Ti )) + l(Ski ) 6

l(Out(G)) =

m
X
i=1

i=1

25

ki (log2 |Out(Ti )| + 3/2),

where the last inequality follows from the main theorem of [10] on the length of the
symmetric group. Using the well known orders of Out(T ) for the finite simple groups T
(for example, see [21], pp. 170-171), it is easy to verify that log2 |Out(Ti )| + 3/2 6 l(Ti )
for all i. We conclude that
m
X
ki l(Ti ) = l(G)
l(Out(G)) 6
i=1

and thus l(Aut(G)) 6 2l(G) as required.



We are now ready to prove Theorem 11. Let R(G) be the soluble radical of G and
consider the semisimple group Soc(G/R(G)). Applying Proposition 5.10 to the group
G/R(G) we obtain
l(Soc(G/R(G))) 6 l(ss(G/R(G))) 6 5 cd(G/R(G)).
It is well known that G/R(G) 6 Aut(Soc(G/R(G))). Applying the inequality above with
Lemma 5.11 we obtain
l(G/R(G)) 6 l(Aut(Soc(G/R(G)))) 6 2l(Soc(G/R(G))) 6 10 cd(G/R(G)) 6 10 cd(G)
and the result follows.
This completes the proof of Theorem 11.
Appendix A. On the number of prime divisors of p ± 1
by D.R. Heath-Brown
In this appendix, we prove the following result.
Theorem A.1. There are infinitely many primes p ≡ 5 (mod 72) for which
Ω((p2 − 1)/24)) 6 7.

Hence there are infinitely many primes p for which

max{Ω(p ± 1)} 6 8.
We begin by showing how the second claim follows from the first. For any prime
p ≡ 5 (mod 72) one has 24|(p2 − 1). Indeed for such primes one has (p − 1, 72) = 4
and (p + 1, 72) = 6. One necessarily has Ω((p − 1)/4) > 1 when p > 5, so that if
Ω((p2 − 1)/24)) 6 7 one must have Ω((p + 1)/6)) 6 6. It then follows that Ω(p + 1) 6 8.
The proof that Ω(p − 1) 6 8 is similar.
To handle the first statement of the theorem we use sieve methods, as described in
the book by Halberstam and Richert [15], and in particular the weighted sieve, as in [15,
Chapter 10]. To be specific, we apply [15, Theorem 10.2] to the set
A = {(p2 − 1)/24 : p ≡ 5 (mod 72), p 6 x}

and the set P of all primes. The expected value of

|Ad | := #{n ∈ A : d|n}

is Xω(d)/d, with X = Li(x)/24, and where ω(d) is a multiplicative function satisfying

0 if p = 2, 3,
ω(p) =
2 if p > 5.

26

TIMOTHY C. BURNESS, MARTIN W. LIEBECK, AND ANER SHALEV

Condition (Ω1 ), see [15, p.29], is then satisfied with A1 = 2, while condition (Ω∗2 (κ)), see
[15, p.252], holds with κ = 2 and a suitable numerical constant A2 . Moreover |Ap2 | =
O(xp−2 ), which shows that condition (Ω3 ), see [15, p.253], also holds, for an appropriate
numerical constant A3 . Finally we consider the condition (Ω(R(2, α)) given in [15, p.219].
The primes p ≡ 5 (mod 72) for which d divides (p2 − 1)/24 fall into ω(d) residue classes
modulo 72d, so that (Ω(R(2, 12 )) holds by an appropriate form of the Bombieri–Vinogradov
theorem, as in [15, Lemma 3.5]. This verifies all the necessary conditions for Theorem 10.2
of [15], and the inequality (2.2) of [15, p.278] is satisfied (with α = 21 ) for any constant
µ > 4, if x is large enough.
Theorem 10.2 of [15] then tells us that there are ≫ X(log X)−2 elements n ∈ A which
are “Pr -numbers” (that is to say, one has Ω(n) 6 r), provided that

Rv
1
1 − ut dtt
2 u σ2 (v(α−1/t))
.
r > 2u − 1 +
1 − η2 (αv)

Finally, we refer to the calculations of Porter [29], and in particular the last 3 lines of [29,
p.420], according to which it will suffice to have r > 6.7 if one takes u = 2.2 and v = 22.
Since we then have α−1 < u < v and αv = 11 > ν2 = 4.42 . . ., by Porter [29, Table 2], the
final conditions (2.3) of [15, Theorem 10.2] are satisfied, and our theorem follows.
References

[1] K. Alladi, R. Solomon and A. Turull, Finite simple groups of bounded subgroup chain length, J. Algebra
231 (2000), 374–386.
[2] M. Aschbacher, On the maximal subgroups of the finite classical groups, Invent. Math. 76 (1984),
469–514.
[3] L. Babai, On the length of subgroup chains in the symmetric group, Comm. Algebra 14 (1986), 1729–
1736.
[4] R.W. Baddeley, Primitive permutation groups with a regular nonabelian normal subgroup, Proc. London Math. Soc. 67 (1993), 547–595.
[5] G.W. Bell, On the cohomology of the finite special linear groups, I, J. Algebra 54 (1978), 216–238.
[6] J.N. Bray, D.F. Holt and C.M. Roney-Dougal, The maximal subgroups of the low-dimensional finite
classical groups, London Math. Soc. Lecture Note Series, vol. 407, Cambridge University Press, 2013.
[7] B. Brewster, M. Ward and I. Zimmermann, Finite groups having chain difference one, J. Algebra 160
(1993), 179–191.
[8] T.C. Burness, M.W. Liebeck and A. Shalev, The depth of a finite simple group, Proc. Amer. Math.
Soc., to appear.
[9] T.C. Burness, M.W. Liebeck and A. Shalev, The length and depth of algebraic groups, submitted
(arXiv:1712.08214).
[10] P.J. Cameron, R. Solomon and A. Turull, Chains of subgroups in symmetric groups, J. Algebra 127
(1989), 340–352.
[11] A.M. Cohen, M.W. Liebeck, J. Saxl, and G.M. Seitz, The local maximal subgroups of exceptional
groups of Lie type, finite and algebraic, Proc. London Math. Soc. 64 (1992), 21–48.
[12] J.H. Conway, R.T. Curtis, S.P. Norton, R.A. Parker and R.A. Wilson, Atlas of Finite Groups, Oxford
University Press, 1985.
[13] D.A. Craven, Alternating subgroups of exceptional groups Lie type, Proc. Lond. Math. Soc. 115 (2017),
449–501.
[14] A. Gamburd and I. Pak, Expansion of product replacement graphs, Combinatorica 26 (2006), 411–429.
[15] H. Halberstam and H.-E. Richert, Sieve methods, LMS Monographs, Academic Press, London, 1974.
[16] K. Harada, Finite simple groups with short chains of subgroups, J. Math. Soc. Japan 20 (1968),
655-672.
[17] M.A. Hartenstein and R.M. Solomon, Finite groups of chain difference one, J. Algebra 229 (2000),
601–622.
[18] K. Iwasawa, Über die endlichen Gruppen und die Verbände ihrer Untergruppen, J. Fac. Sci. Imp. Univ.
Tokyo. Sect. I. 4 (1941), 171–199.
[19] Z. Janko, Finite groups with invariant fourth maximal subgroups, Math. Z. 82 (1963), 82–89.
[20] Z. Janko, Finite simple groups with short chains of subgroups, Math. Z. 84 (1964), 428–437.
[21] P. Kleidman and M.W. Liebeck, The subgroup structure of the finite classical groups, London Math.
Soc. Lecture Note Series, vol. 129, Cambridge University Press, 1990.

ON THE LENGTH AND DEPTH OF FINITE GROUPS

27

[22] J. Kohler, A note on solvable groups, J. London Math. Soc. 43 (1968), 235–236.
[23] M.W. Liebeck, C.E. Praeger and J. Saxl, A classification of the maximal subgroups of the finite
alternating and symmetric groups, J. Algebra 111 (1987), 365–383.
[24] M.W. Liebeck, C.E. Praeger and J. Saxl, On the O’Nan-Scott Theorem for finite primitive permutation
groups, J. Austral. Math. Soc. 44 (1988), 389–396.
[25] M.W. Liebeck, J. Saxl, and G.M. Seitz, Subgroups of maximal rank in finite exceptional groups of Lie
type, Proc. London Math. Soc. 65 (1992), 297–325.
[26] M.W. Liebeck and G.M. Seitz, A survey of of maximal subgroups of exceptional groups of Lie type,
in Groups, combinatorics & geometry (Durham, 2001), 139–146, World Sci. Publ., River Edge, NJ,
2003.
[27] M.W. Liebeck and G.M. Seitz, On finite subgroups of exceptional algebraic groups, J. reine angew.
Math. 515 (1999), 25–72.
[28] J. Petrillo, On the length of finite simple groups having chain difference one, Arch. Math. 88 (2007),
297–303.
[29] J.W. Porter, Some numerical results in the Selberg sieve method, Acta Arith. 20 (1972), 417–421.
[30] G.M. Seitz, R. Solomon and A. Turull, Chains of subgroups in groups of Lie type, II, J. London Math.
Soc. 42 (1990), 93–100.
[31] J. Shareshian and R. Woodroofe, A new subgroup lattice characterization of finite solvable groups, J.
Algebra 351 (2012), 448–458.
[32] R. Solomon and A. Turull, Chains of subgroups in groups of Lie type, I, J. Algebra 132 (1990),
174–184.
[33] R. Solomon and A. Turull, Chains of subgroups in groups of Lie type, III. J. London Math. Soc. 44
(1991), 437–444.
[34] R.A. Wilson, Maximal subgroups of sporadic groups, in Finite Simple Groups: Thirty Years of the
Atlas and Beyond, 57–72, Contemp. Math., 694, Amer. Math. Soc., Providence, RI, 2017.
[35] R.A. Wilson, The uniqueness of PSU3 (8) in the Monster, Bull. London Math. Soc. 49 (2017), 877–880.
T.C. Burness, School of Mathematics, University of Bristol, Bristol BS8 1TW, UK
E-mail address: t.burness@bristol.ac.uk
D.R. Heath-Brown, Mathematical Institute, Radcliffe Observatory Quarter, Woodstock
Road, Oxford OX2 6GG, UK
E-mail address: rhb@maths.ox.ac.uk
M.W. Liebeck, Department of Mathematics, Imperial College, London SW7 2BZ, UK
E-mail address: m.liebeck@imperial.ac.uk
A. Shalev, Institute of Mathematics, Hebrew University, Jerusalem 91904, Israel
E-mail address: shalev@math.huji.ac.il

