FINITE GROUPS WITH INDEPENDENT GENERATING SETS OF ONLY TWO SIZES

Abstract A generating set S for a group G is independent if the subgroup generated by 
$S\setminus \{s\}$
 is properly contained in G for all 
$s \in S.$
 We describe the structure of finite groups G such that there are precisely two numbers appearing as the cardinalities of independent generating sets for G.


Introduction
The minimal number of generators of a finite group G is denoted by d(G).A generating set S for a group G is independent (sometimes called irredundant) if S \ {s} < G for all s ∈ S.
Let m(G) denote the maximal size of an independent generating set for G.
The finite groups with m(G) = d(G) are classified by Apisa and Klopsch.
THEOREM 1.1 (Apisa-Klopsch, [1,Theorem 1.6]).If d(G) = m(G), then G is soluble.Moreover, either • G/ Frat(G) is an elementary abelian p-group for some prime p; or • G/ Frat(G) = PQ, where P is an elementary abelian p-group and Q is a nontrivial cyclic q-group for distinct primes p and q, such that Q acts by conjugation faithfully on P and P (viewed as a module for Q) is a direct sum of m(G) − 1 isomorphic copies of one simple Q-module.
In view of this result, Apisa and Klopsch suggest a natural 'classification problem': given a nonnegative integer c, characterise all finite groups G which satisfy m(G) − d(G) ≤ c.The particular case c = 1 has been recently highlighted by Glasby (see [7,Problem 2.3]).
A. Lucchini and P. Spiga [2] A nice result in universal algebra, due to Tarski and known as the Tarski irredundant basis theorem (see for example [3,Theorem 4.4]), implies that, for every positive integer k with d(G) ≤ k ≤ m(G), G contains an independent generating set of cardinality k.So the condition m(G) − d(G) = 1 is equivalent to the fact that there are only two possible cardinalities for an independent generating set of G.
Let G be a finite group.We recall that the socle of G, denoted soc(G), is the subgroup generated by the minimal normal subgroups of G; moreover, G is said to be monolithic primitive if G has a unique minimal normal subgroup and the Frattini subgroup Frat(G) of G is the identity.
In this paper, we prove the following two main results.
THEOREM 1.2.Let G be a finite group with G is a monolithic primitive group and G/ soc(G) is cyclic of prime power order.
It was proved by Whiston and Saxl [15] that m(PSL(2, p)) = 3 for any prime p with p not congruent to ±1 modulo 8 or 10.In particular, as d(S) = 2 for every nonabelian simple group, we deduce that there are infinitely many nonabelian simple groups In Section 4, we give examples of finite soluble groups G with m(G) = d(G) + 1 for each of the three possibilities arising in Theorem 1.3.

Preliminary results
Let L be a monolithic primitive group and let A be its unique minimal normal subgroup.For each positive integer k, let L k be the k-fold direct product of L. The crown-based power of L of size k is the subgroup L k of L k defined by In [4], it is proved that for every finite group G, there exists a monolithic group L and a homomorphic image A group L k with this property is called a generating crown-based power for G.
In [4], it is explained how d(L k ) can be explicitly computed in terms of k and the structure of L. A key ingredient (when one wants to determine d(G) from the behaviour of the crown-based power homomorphic images of G) is to evaluate, for each monolithic group L, the maximal k such that L k is a homomorphic image of G.This integer k arises from an equivalence relation among the chief factors of G.In what follows, we give some details.
Given groups G and A, we say that A is a G-group if G acts on A via automorphisms.In addition, A is irreducible if G does not stabilise any nontrivial proper subgroups of A. Two G-groups A and B are G-isomorphic if there exists a group isomorphism φ : A → B such that φ(g(a)) = g(φ(a)) for all a ∈ A and g ∈ G. Following [8], we say that two irreducible G-groups A and B are G-equivalent, denoted A ∼ G B, if there is an isomorphism Φ : A G → B G which restricts to a G-isomorphism φ : A → B and induces the identity G AG/A → BG/B G, in other words, such that the following diagram commutes: Observe that two G-isomorphic G-groups are G-equivalent, and the converse holds if A and B are abelian. Let this is equivalent to saying that A is abelian and there is no complement to A in G.The number δ G (A) of non-Frattini chief factors that are G-equivalent to A, in any chief series of G, does not depend on the particular choice of such a series.Now, we denote by L G (A) the monolithic primitive group associated to A, that is, More precisely, there exists a normal subgroup N such that G/N L G (A) and soc(G/N) ∼ G A. We identify soc(L G (A)) with A, as G-groups.
Consider now all the normal subgroups N of G with the property that G/N L G (A) and soc(G/N) ∼ G A. The intersection R G (A) of all these subgroups has the property Note that if L is monolithic primitive and L k is a homomorphic image of G for some k ≥ 1, then L L G (A) for some non-Frattini chief factor A of G and k ≤ δ G (A). Furthermore, if (L G (A)) k is a generating crown-based power, then so is (L G (A)) δ G (A) ; in this case, we say that A is a generating chief factor for G.
For an irreducible G-module M, set It can be seen that for any irreducible G-module M, and therefore The importance of h G (M) is clarified by the following proposition.When G admits a nonabelian generating chief factor A, a relation between δ G (A) and d(G) is provided by the following result.

PROPOSITION 2.2. If d(G) ≥ 3 and there exists a nonabelian generating chief factor A of G, then
PROOF.Suppose that d(G) ≥ 3 and let A be a nonabelian generating chief factor of G.
For a finite group X, let φ X (m) denote the number of ordered m-tuples (x 1 , . . ., x m ) of elements of X generating X. Define In [4], it is proved that if m ≥ d(L), then By the main result in [13], d(L) = max(2, d(L/A)).Since A is a generating chief factor, from the definition, we have

.4)
Moreover, A S n , where n is a positive integer and S is a nonabelian simple group.In the proof of Lemma 1 in [5], it is shown that Now, [9] shows that |Out(S)| ≤ log 2 (|S|) and hence From (2.3), (2.4) and (2.5), we obtain Recall that m(G) is the largest cardinality of an independent generating set of G. PROOF.Suppose first that G is simple.Let l be an element of G of order 2. Since G = l x | x ∈ G , the set {l x | x ∈ G} contains a minimal generating set of G. Since G cannot be generated by two involutions, this minimal generating set has cardinality at least three.Thus, m(G) ≥ 3. Suppose next that G is not simple.Let a and b be the number of non-Frattini and nonabelian factors in a chief series of G.As G is not simple, there exists a maximal normal subgroup N of G containing A and we have a chief series 1

Proof of the main results
Let G be a finite group, let d := d(G) and let m := m(G).Suppose that m = d + 1.Let A be a generating chief factor of G and let δ := δ G (A), L := L G (A).

3.2.
A is abelian.It follows from Proposition 2.1 and (2.1) that Now suppose that d = δ.By Theorem 2.3, G is soluble and contains only one non-Frattini chief factor which is not where P is a finite p-group, V is an irreducible P-module and d(P) = d.In particular, we obtain item (1) in Theorem 1.3.
Finally assume d = δ + 1.Notice that in this case, L = A H, where A is a faithful, nontrivial, irreducible H-module, and In particular, by Corollary 2.4, and N 1 / Frat(G) is an abelian minimal normal subgroup of G/ Frat(G).As m(H) = 1, H is cyclic of prime power order.In particular, we obtain item (3) in Theorem 1.3.[12].The group G acts by conjugation on the set {S 1 , . . ., S n } of the simple components of N.This produces a group homomorphism G → Sym(n) and the image K of G under this homomorphism is a transitive subgroup of Sym(n).Moreover, the subgroup X of Aut S induced by the conjugation action of N G (S 1 ) on the first factor S 1 is an almost simple group with socle S.
By [12,Proposition 4], μ(G) ≥ μ(X) = m(X) − m(X/S).Assume m(G) = 3. Observe that by Theorems 1.1 and 1.2, G/N is cyclic of prime power order.If X = S, then This implies that G/N = 1 and G = S is a simple group.If X S, then G N and Moreover, X/S is a nontrivial cyclic group of prime power order, so By Corollary 2.4, m(X) = 3.The groups PΣL 2 (9), M 10 , Aut(PSL 2 (7)) are currently the only known examples (to the best knowledge of the authors) of almost simple groups X with X soc(X) and m(X) = 3.We believe that there are other such examples, but our current computer codes are not efficient enough to carry out a thorough investigation.

THEOREM 2 . 3 [ 14 ,COROLLARY 2 . 4 .
Theorem 1.3].Let G be a finite group.Then m(G) ≥ a + b, where a and b are, respectively, the number of non-Frattini and nonabelian factors in a chief series of G.Moreover, if G is soluble, then m(G) = a.Assume that G is a finite group with a unique minimal normal subgroup A. If A is nonabelian, then m(G) ≥ 3.

there exist two normal subgroups N 1 , N 2 such that 1 N 1 ≤ N 2 , N 1 is an abelian
soluble, then one of the following occurs:(1) G V P, where P is a finite noncyclic p-group and V is an irreducible P-module, which is not a p-group; in this case, d(G) = d(P);