Bounding the maximal size of independent generating sets of finite groups

Denote by $m(G)$ the largest size of a minimal generating set of a finite group $G$. We estimate $m(G)$ in terms of $\sum_{p\in \pi(G)}d_p(G),$ where we are denoting by $d_p(G)$ the minimal number of generators of a Sylow $p$-subgroup of $G$ and by $\pi(G)$ the set of prime numbers dividing the order of $G$.


Introduction
A generating set X of a finite group G is said to be minimal (or independent) if no proper subset of X generates G.We denote by m(G) the largest size of a minimal generating set of G. First steps toward investigating m(G) have been taken in the context of permutation groups.An exhaustive investigation has been done for finite symmetric groups [2,17], proving that m(Sym(n)) = n − 1 and giving a complete description of the independent generating sets of Sym(n) having cardinality n − 1. Partial results for some families of simple groups are in [16]: it turns out that already in the case G = PSL(2, q), the precise value of m(G) is quite difficult to obtain.Further Apisa and Klopsch [1] proposed a natural "classification problem": given a non-negative integer c, characterize all finite groups G such that m(G) − d(G) ≤ c, where d(G) is the minimal size of a generating set of G.In particular, they classified the finite groups for which the equality m(G) = d(G) holds.During the same period the first author started in [10,11] a systematic investigation of how m(G) can be estimated for an arbitrary finite group G.
In 1989, Guralnick [8] and the first author [9] independently proved that, if all the Sylow subgroups of a finite group G can be generated by d elements, then d(G) ≤ d + 1.One may ask, if minded so, whether a similar result holds also for m(G).More precisely, denote by d p (G) the minimal number of generators of a Sylow p-subgroup of G.
Is it possible to bound m(G) as a function of the numbers d p (G), with p running through the prime divisors of the order of G? As customary, we denote by π(G) the set of prime divisors of the order of G.It can be easily seen that, if G is a finite nilpotent group, then m(G) = p∈π(G) d p (G).For simplicity, we let In a private communication to the first author, Keith Dennis has conjectured that m(G) ≤ δ(G), for every finite group G.
This conjecture is true for soluble groups.
Proof.In [10], it is proved that m(G) = p∈π(G) α p (G), where α p (G) denotes the number of complemented factors of p-power order in a chief series of G. Now, an easy inductive argument on the order of G shows that α p (G) ≤ d p (G) (see for example [12,Lemma 4]).Therefore m(G) ≤ p∈π(G) d p (G) = δ(G).
Despite Theorem 1.1, Dennis' conjecture is false if G is a symmetric group.We study the asymptotic behaviour of the function δ(Sym(n)) in Section 5. We prove in Theorem 5. [17], the difference m(Sym(n)) − δ(Sym(n)) goes to infinity with n and the inequality m(Sym(n)) ≤ δ(Sym(n)) is satisfies only by finitely many values of n.Indeed, using the explicit upper bound on δ(Sym(n)) in Theorem 5.1 and some calculations, we have 3,4,5,8,10,11,16,17,18,19,25,30 The proof of Theorem 5.1 is rather technical and uses some explicit bounds on the prime counting function.However, in Lemma 4.4 we show by elementary means that, for every positive real number η > 1, there exists a constant c η such that m(Sym(n This motivates the following conjecture, which can be seen as a natural generalization of Dennis' conjecture. Conjecture 1.2.There exist two constants c and η such that m(G) ≤ c • δ(G) η for every finite group G.
Given a normal subgroup N of a finite group G, we let The main result of this paper is the following theorem.
Theorem 1.3.Let G be a finite group and assume that there exist two constants σ ≥ 1 and η ≥ 2 such that m(X, S) ≤ σ•|π(S)| η , for every composition factor S of G and for every almost simple group X with soc X = S. Then m(G) ≤ σ•δ(G) η .
Theorem 1.3 reduces Conjecture 1.2 to the following conjecture on finite almost simple groups.
Conjecture 1.4 holds true when soc X is an alternating group or a sporadic simple group.Therefore, we have the following corollary.
Corollary 1.5.There exists a constant σ such that, if G has no composition factor of Lie type, then m(G) ≤ σδ(G) 2 .
Very little is known about m(G), when G is an almost simple group with socle a simple group of Lie type.Whiston and Saxl proved that, if G = PSL(2, q) with q = p r and with p a prime number, then m(G) ≤ max(6, π(r) + 2) where π(r) is the number of distinct prime divisors of r.It follows from Zsigmondy's Theorem that π(r) ≤ π(q + 1) ≤ |π(PSL(2, q))|.Therefore Conjecture 1.4 holds true when G = PSL(2, q).In his PhD thesis [5], P. J. Keen found a good upper bound for m(SL(3, q)), when q = p r and p is odd.In preparation for this, he also investigated the sizes of independent sets in SO(3, q) and SU (3, q), getting in all the cases a linear bound in terms of π(r).These partial results lead to conjecture that, if soc(X) is a group of Lie type of rank n over the field with q = p r elements, then m(X, soc X) is polynomially bounded in terms of n and π(r).If this were true, then Conjecture 1.4 would also be true.
The proofs of Theorem 1.3 and Corollary 1.5 are in Section 4. These proofs require two preliminary results, one concerning the prime divisors of the order of a finite non-abelian simple group and the other about permutation groups, proved respectively in Sections 2 and 3.

A result on the order of a finite simple group
For later use we need to recall some definitions and some results concerning Zsigmondy primes.Definition 2.1.Let a and n be positive integers.A prime number p is called a primitive prime divisor of a n − 1 if p divides a n − 1 and p does not divide a e − 1 for every integer 1 ≤ e ≤ n − 1.We denote an arbitrary primitive prime divisors of a n − 1 by a n .Theorem 2.2 (Zsigmondy's Theorem [18]).Let a and n be integers greater than 1.There exists a primitive prime divisor of a n − 1 except in one of the following cases: (1) n = 2, a = 2 s − 1 (i.e. a is a Mersenne prime), and s ≥ 2.
( Proof.Let S = L(q) be a simple group of Lie type defined over the field with q elements, where q = p t and p is a prime number.From Burnside's theorem, |π(S)| ≥ 3. From [3], if |π(S)| = 3, then and for these groups the theorem holds by a direct inspection.Therefore, for the rest of the proof we may suppose In particular, the result immediately follows when |π(Out(S))| ≤ 2 and hence we may suppose |π(Out(S))| ≥ 3. The order of L(q) has the cyclotomic factorization in terms of q : where Φ m (q) is the m-th cyclotomic polynomial and Λ, d, h and r m are listed in Tables L.1, C.1 and C.2 of [7].Suppose that S = D 4 (q) and that S is untwisted.From [13, page 207], if r ≥ 2 and m ≥ 1 are integers such that r m•t is a primitive prime of (r t ) m − 1, then r m•t divides Φ m (r t ).From this and from Zsigmondy's theorem, we conclude that, except for the six cases listed below, there exist i, j ∈ Λ with 2 ≤ i < j such that x := p i•t and y := p j•t are distinct primitive prime divisors.In particular, x and y are odd divisors of |S| and are relatively prime to q − 1 because i ≥ 2.Moreover, by Lemma 2.3, x ≡ y ≡ 1 mod t and hence x and y are relatively prime to t.In particular, x and y are our required primes.(The case S = D 4 (q) is special in this argument because 3 is (potentially) an odd prime divisor of |Out(S)| not arising from field automorphisms.) We are going to analyze the groups for which the existence of x and y is not ensured from the previous argument.
It remains to deal with the case S = D 4 (q) and with the twisted groups of Lie type.Suppose S = D 4 (q).Since 3 divides |Out(S)|, the previous argument fails exactly when the primitive prime divisor x or y is 3.The existence of x = p 2•t , y = p 4•t and z = p 6•t is ensured when q / ∈ {2, 8} and when q is not a Mersenne prime.When q = 2, the result follows since |Out(S)| = 6; when q = 8, we have that t = 3 does not divide y and z; therefore y and z are prime numbers satisfying our statement .When q is a Mersenne prime, if q = 3, then q and z are prime numbers satisfying our statement; if q = 3, then 5 and 7 are prime numbers satisfying our statement.
Assume S ∈ { 2 B 2 (q), 2 G 2 (q), 2 F 4 (q)}.In these cases we have |Out(S)| = t, so we may assume that t is not a prime.Since the existence of x = p i•t and y = p j•t is ensured by Zsigmondy's Theorem, for two different elements i and j of Λ, we are done.
If S = 3 D 4 (q) and q / ∈ {2, 8} and q is not a Mersenne prime, then we can take x = p 2•t and y = p 6•t (notice that | Out S| divides 3 • t).When q = 2 or q = 8 or q is a Mersenne prime, then |Out(S)| is divisible only by 3, against our assumption.
If S = 2 D n (q), then |Out(S)| divides 8 • t.So, when q = 2 or when q is a Mersenne prime, the result holds since |Out(S)| has only one prime divisor.For the remaining cases, we can take x = p 4•t and y = p 6•t .
3. An auxiliary result Lemma 3.1.Let Q be a p-group, let P be a permutation p-group with domain ∆ and let n ∆ (P ) be the number of orbits of P on ∆.Then Proof.Let ∆ 1 , . . ., ∆ ℓ be the orbits of P on ∆.
Replacing Q by Q/ Frat(Q) if necessary, we may suppose that Q is an elementary abelian p-group.Let B be the base group of the wreath product W := Qwr ∆ P .
Using the fact that B is an abelian normal subgroup of W and standard commutator computations, we get [W, W ] = [B, P ][P, P ].Given σ ∈ P and f ∈ B, we have and hence (3.1) Frat(W ) = [B, P ] Frat(P ).
When (1) and ( 2) are satisfied (even if t is not necessarily the maximum), we say that U 1 , . . ., U t are indipendent subgroups of K.Moreover, let S be a finite non-abelian simple group and let us denote by π * (S) the set of primes dividing |S| but not |Out(S)|.
Theorem 3.2.Let K be a transitive permutation on Ω, let S be a non-abelian simple group and let G be a group with Proof.For every p ∈ π * (S), we have d p (G) = d p (Swr Ω K) and hence, without loss of generality, we may assume G = Swr Ω K.For simplicity, we write f (S, Ω, K) := p∈π * (S) We argue by induction on t := t Ω (K).When t = 1, from Theorem 2.4 we deduce Suppose then t > 1.Let ω ∈ Ω and let U 1 , . . ., U t be t independent subgroups of K with For each i ∈ {1, . . ., t}, we define Ūi to be the intersection j∈{1,...,t}\{i} U j ; (as K ω ≤ Ūi , the orbit ω Ūi := {ω x | x ∈ Ūi } is a block of imprimitivity for the action of K on Ω. ) Ω i to be the system of imprimitivity determined by the block of imprimitivity ω Ūi ; Ki to be the permutation group induced by K on Ω i ; (we also denote by σ i : K → Ki the natural projection, so Ki = σ i (K).) G i to be the wreath product G i := Swr Ωi Ki .
Let i ∈ {1, . . ., t}.Since the point stabilizer σ i ( Ūi ) of ω Ūi ∈ Ω i in Ki is defined as the intersection of the t − 1 independent subgroups {σ i (U j ) | j ∈ {1, . . ., t} \ {i}}, we have t Ωi ( Ki ) ≥ t − 1.Moreover, from our inductive argument, we have For each prime p ∈ π * (S), let Π p be a Sylow p-subgroup of S and let P be a Sylow p-subgroup of K.In particular, Pi := σ i (P ) is a Sylow p-subgroup of Ki .From Lemma 3.1, for every i ∈ {1, . . ., t}, we have where n Ωi ( Pi ) = n Ωi (P ) denotes the number of orbits of P on Ω i .Observe that d(P ) ≥ d( Pi ).
Since |π * (S)| ≥ 2, we may choose p ∈ π * (S) and i ∈ {1, . . ., ℓ} such that | Ūi : K ω | is not a power of p.Let δ1 , . . ., δs be a set of representatives of the orbits of P on Ω i , where s := n Ωi (P ).In other words, this means that and that this union is disjoint.For each j ∈ {1, . . ., s}, let δ j ∈ δj .As δj ⊆ Ω is a block of imprimitivity for the action of K on Ω, the union (3.5) is made by pairwise disjoint P -orbits and hence n Ω (P ) ≥ s = n Ωi (P ).Moreover, n Ω (P ) = n Ωi (P ) if and only if the equality in (3.5) is attained, which in turn happens, if and only if, for each j ∈ {1, . . ., s}, the points in δj ⊆ Ω are in the same P -orbit.Since we are assuming that n Ω (P ) = n Ωi (P ), the previous paragraph shows that the stabilizer P δj of the block δj is transitive on the points in δj .Since P is a p-group, we deduce | δj | = | Ūi : K ω | is a power of p, contradicting our choice of i and p.

Proofs of Theorem 1.3 and Corollary 1.5
If N is a normal subgroup of a finite group G, we denote by m(G, N ) the difference m(G) − m(G/N ).We recall in the first part of this section some results proved in [10,11], estimating the value of m(G, N ) when N is a minimal normal subgroup of G.   Proof.It suffice to prove that d p (G) > d p (G/N ) whenever p ∈ π(N ).This is clear when N is abelian.Assume that N is non-abelian.Let p ∈ π(N ) and let P be a Sylow p-subgroup of G.If P ∩ N ≤ Frat(P ), then Tate's Theorem [4, p. 431] shows that N has a normal p-complement.However, this is impossible because N is a direct product of non-abelian simple groups.Thus P ∩ N ≤ Frat(P ), and consequently d p (G/N ) + 1 ≤ d p (G).
Proof of Theorem 1.3.Clearly the statement is true if G is simple.Thus we suppose that S is not a simple group and we proceed by induction on the order of G.We may assume Frat(G) = 1.Let N be a minimal normal subgroup of G.If N is abelian, using Lemma 4.3 and the inductive hypotheses, we have (In the last inequality, we used the fact that σ ≥ 1 and η ≥ 2.) Assume that N is non-abelian.Let K, X and S be as in the statement of Lemma 4.2.By Theorem 3.2, we have Combining this with Lemma 4.3, we conclude that The last inequality follows from the fact that x η + y η + x + y ≤ (x + y) η , for every positive integers x and y and for every η ≥ 2.
In order to prove Corollary 1.5, we first need the following lemma.Lemma 4.4.For every positive real number η > 1, there exists a constant c η such that n ≤ c η π(n) η , where π(n) is the number of prime numbers less than or equal to n.
Lemma 4.5.There exists a constant ρ such that, if X is an almost simple group and S = soc(X) is not a simple group of Lie type, then m(X, S) Proof.First assume that S = Alt(n).By [17, Theorem 1], m(X, S) ≤ n − 1.By Lemma 4.4, there exists a constant c 2 such that m(X, S) Clearly there exists a constant c such that m(X, S) ≤ c • |π(S)| 2 , for every sporadic simple group S. Taking ρ = max{c, c 2 }, the result follows.
Proof of Corollary 1.5.It follows from Theorem 1.3 and Lemma 4.5.

Estimating δ(Sym(n))
In this section, we aim to bound, from above and from below, δ(Sym(n)) as a function of n.By [17, Theorem 1], m(Sym(n)) = n − 1 while, by Kalužnin's Theorem, if For not making the notation too cumbersome, we set As in the previous sections we denote by π : R → N the prime counting function, that is, π(x) is the number of prime numbers less than or equal to x.As d p (n) ≥ ℓ(p, n) = 0 for every prime p ≤ n, we have From the Prime Number Theorem, π(n) is asymptotic to n/ log n (that is, the ratio π(n)/(n/ log n) tends to 1 as n tends to infinity) and hence n/ log n ∈ O(d(n)).In this section, we actually prove that d(n) is asymptotic to a linear function.In particular, d(n) = n log 2 + O(n/ log n).
Proof.We start by collecting some basic inequalities that we use throughout this proof.From Theorem 1 and Theorem 2 in [15], we have Given a prime number p with p ≤ n, ℓ(p, n) ≤ ⌊log p n⌋ and hence We define the two auxiliary functions We aim to obtain explicit bounds on d ′ (n) and d ′′ (n) as functions of n.We start with d ′ (n).From (5.3), we get For every k ∈ N with k ≥ 1, we denote by p k the k th prime number.Using [15, Corollary, page 69], we have where the first inequality is valid for every k ≥ 1 and the second inequality is valid for every k ≥ 6.
Therefore, from (5.1), (5.5) and a computation (we are using n ≥ 11 in the last inequality), for every n ≥ 11, we have Actually, with a direct inspection, we see that this inequality holds true for every natural number n with 2 ≤ n ≤ 10.Therefore Arguing in a similar manner, for every k ≥ 6, we obtain An explicit computation yields that (5.7) is also valid when k ∈ {2, 3, 4, 5}.Therefore, using (5.7), we have For every t ∈ N with t ≥ 2, write f (t) := t k=2 k/(log k + log log k).When k > 2, we have k/(log k + log log k) ≤ k.Moreover, when k ≥ √ t, we have where the last inequality holds for t ≥ 8. Therefore, for every t ≥ 8, we have where the last inequality follows with some elementary computations.A direct computation with 2 ≤ t < 8 shows that the same upper bound for f (t) holds.Therefore, applying this upper bound with t := π( √ n), we get Now, for every n ≥ 67 2 , using (5.1) and (5.2), we see that the right hand side of (5.8) is bounded above by The second summand of (5.9) is at most Thus the second summand of (5.9) is at most where the last inequality follows with a computation using the fact that n ≥ 67 2 .For the first and third summand of (5.9), we have where this inequality follows again with some elementary computations using the fact that n ≥ 67

Lemma 4 . 1 .
If N is an abelian minimal normal subgroup of G, then m(G, N ) is either 0 or 1 depending on whether N ≤ Frat(G) or not.Proof.If follows from [10, Lemma 11 and Lemma 12].

Lemma 4 . 2 .
Assume that N is a non-abelian minimal normal subgroup of a finite group G.There exist a non-abelian simple group S and a positive integer r such that N = S 1 × • • • × S r , with S ∼ = S i for each 1 ≤ i ≤ r.Let K be the transitive subgroup of Sym(r) induced by the conjugacy action of G on the set {S 1 , . . ., S r } of the simple components of N .As in the previous section, let t(K) := t {1,••• ,r} (K) be the largest positive integer t such that the stabilizer in K of a point in {1, • • • , r} can be obtained as an intersection of t independent subgroups.Moreover let X be the subgroup of Aut S 1 induced by the conjugation action of N G (S 1 ) on the first factor S 1 .Then m(G, N ) ≤ m(X, soc X) + t(K).Proof.If follows from [10,Lemma 13] and[11, Theorem 1].