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

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


Introduction
Tim Wall in 1961 [12] has conjectured that the number of maximal subgroups of a finite group G is less than the group order |G|.Wall himself proved the conjecture under the additional hypothesis that G is soluble.The first remarkable progress towards a good understanding of Wall's conjecture is due to Liebeck, Pyber and Shalev [11]; they proved that all, but (possibly) finitely many, simple groups satisfy Wall's conjecture.Actually, Liebeck, Pyber and Shalev prove [11,Theorem 1.3] a polynomial version of Wall's conjecture: there exists an absolute constant c such that, every finite group G has at most c|G| 3/2 maximal subgroups.Based on the conjecture of Guralnick on the dimension of certain first cohomology groups [6] and on some computer computations of Frank Lübeck, Wall's conjecture was disproved in 2012 by the participants of an AIM workshop, see [7].
The question of Wall can be generalised in the context of finite permutation groups and this was done by Peter Cameron, see [3].(See [3] also for the motivation for this question.)Question 1.1 (Cameron [3]).Is the number of maximal blocks of imprimitivity through a point for a transitive group G of degree n bounded above by a polynomial of degree n? Find the best bound!To see that this question extends naturally the question of Wall we fix some notation.Given a finite group G and a subgroup H of G, we denote by max(H, G) := |{M | M maximal subgroup of G with H ≤ M }|, the number of maximal subgroups of G containing H. Now, if Ω is the domain of a transitive permutation group G and ω ∈ Ω, then there exists a one-to-one correspondence between the maximal systems of imprimitivity of G and the maximal subgroups of G containing the point stabiliser G ω and hence Question 1.1 asks for a polynomial upper bound for max(G ω , G) as a function of n = |G : G ω |.When n = |G|, that is, G acts regularly on itself, the question of Cameron reduces to the question of Wall and [11,Theorem 1.3] yields a positive solution in this case, with exponent 3/2.
The main result of this paper is a positive solution to Question 1.1.

Theorem 1.2.
There exists a constant a such that, for every finite group G and for every subgroup H of G, we have max(H, G) ≤ a|G : H| 3/2 .In particular, a transitive permutation group of degree n has at most an 3/2 maximal systems of imprimitivity.
In the case of soluble groups we actually obtain a much tighter bound, which extends the result of Wall [12, (8.6), page 58] for soluble groups on his own conjecture.
Theorem 1.3.If G is a finite soluble group and H is a proper subgroup of G, then max(H, G) ≤ |G : H|−1.In particular, a soluble transitive permutation group of degree n ≥ 2 has at most n − 1 maximal systems of imprimitivity.

Preliminaries
We start by reviewing some basic results on G-groups, on monolithic primitive groups and on crowns tailored to our proof of Theorem 1.2.For the first part we follow [5], for the second part we follow [8] and for the third part we follow [1, Chapter 1] and [5].This section will also help for setting some notation.All groups in this paper are finite.
2.1.Monolithic primitive groups and crown-based power.Recall that an abstract group L is said to be primitive if it has a maximal subgroup with trivial core.Incidentally, given a group G and a subgroup M be denote by core G (M ) := g∈G M g the core of M in G.The socle soc(L) of a primitive group L is either a minimal normal subgroup, or the direct product of two non-abelian minimal normal subgroups.A primitive group L is said to be monolithic if the first case occurs, that is, soc(L) is a minimal normal subgroup of L and hence (necessarily) L has a unique minimal normal subgroup.
Let L be a monolithic primitive group and let A := soc(L).For each positive integer k, let L k be the k-fold direct product of L. The crown-based power of L of size k is the subgroup L k of L k defined by Equivalently, if we denote by diag(L k ) the diagonal subgroup of L k , then For the proof of the next lemma we need some basic terminology, which we borrow from [9, Section 4.3 and 4.4].Let κ be a positive integer and let A be a direct product S 1 × • • • × S κ , where the S i s are pair-wise isomorphic non-abelian simple groups.We denote by π i : A → S i the natural projection onto S i .A subgroup X of A is said to be a strip, if X = 1 and, for each i ∈ {1, . . ., κ}, either X ∩ Ker(π i ) = 1 or π i (X) = 1.The support of the strip X is the set {i ∈ {1, . . ., κ} | π i (X) = 1}.The strip X is said to be full if π i (X) = S i , for all i in the support of X.Two strips X and Y are disjoint if their supports are disjoint.A subgroup X of A is said to be a subdirect subgroup if, for each i ∈ {1, . . ., κ}, π i (X) = S i .
Scott's lemma (see for instance [9,Theorem 4.16]) shows (among other things) that if X is a subdirect subgroup of A, then X is a direct product of pairwise disjoint full strips of A.
Lemma 2.1.Let L k ′ be a crown-based power of L of size k ′ having non-abelian socle N k ′ and let H ′ be a core-free subgroup of Proof.We argue by induction on k ′ .If k ′ = 1, then the result is clear because N k ′ = N has no proper subgroups having index less then 5. Suppose that k ′ ≥ 2 and write Suppose that there exists i ∈ {1, . . ., k because N i has no proper subgroups having index less then 5. Therefore, Suppose that, for every i ∈ {1, . . ., k ′ }, π i (H ′ ) = N i .Since N is non-abelian, we may write , for some pair-wise isomorphic non-abelian simple groups S i,j of cardinality s.For each i ∈ {1, . . ., k ′ } and j ∈ {1, . . ., ℓ}, we denote by π i,j : N k ′ → S i,j the natural projection onto S i,j .As π i (H ′ ) = N i , we deduce π i,j (H ′ ) = S i,j , for every i ∈ {1, . . ., k ′ } and j ∈ {1, . . ., ℓ}.In particular, H ′ is a subdirect subgroup of S 1,1 × • • • × S k ′ ,ℓ and hence (by Scott's lemma) H ′ is a direct product of pair-wise disjoint full strips.Since no N i is contained in H ′ , there exist two distinct indices i 1 , i 2 ∈ {1, . . ., k ′ } and j 1 , j 2 ∈ {1, . . ., ℓ} such that (i 1 , j 1 ) and (i 2 , j 2 ) are involved in the same full strip of H ′ .If we now consider the projection π i1,i2 : In the proof of Theorem 1.2 and 1.3, we use without mention the following basic fact.
Let M be a normal subgroup of the crown based power L k with socle N k and with

2.2.
Basic facts on G-groups.Given a group G, a G-group A is a group A together with a group homomorphism θ : G → Aut(A).(For simplicity, we write a g for the image of a ∈ A under the automorphism θ(g).)Given a G-group A, we have the corresponding semi-direct product A ⋊ θ G (or simply A ⋊ G when θ is clear from the context), where the multiplication is given by 1 a 2 , for every a 1 , a 2 ∈ A and for every g 1 , g 2 ∈ G.A G-group A is said to be irreducible if G leaves invariant no non-identity proper normal subgroup of A.
Two G-groups A and B are said to be G-isomorphic (and we write A ∼ =G B), if there exists an isomorphism ϕ : for every a ∈ A and for every g ∈ G. Similarly, we say that A and B are G-equivalent (and we write A ∼ G B), if there exist two isomorphisms ϕ : A → B and Φ : A ⋊ G → B ⋊ G such that the following diagram commutes.
Being "G-equivalent" is an equivalence relation among G-groups coarser than the "G-isomorphic" equivalence relation, that is, two G-isomorphic G-groups are necessarily G-equivalent.The converse is not necessarily true: for instance, if A and B are two isomorphic non-abelian simple groups and G := A × B acts on A and on B by conjugation, then A ≇ G B and A ∼ G B. However, when A and B are abelian, the converse is true, that is, if A and B are abelian, then A ∼ G B if and only if A ∼ =G B, see [8, page 178].
Let G be a group and let A := X/Y be a chief factor of G, where X and Y are normal subgroups of G. Clearly, the action by conjugation of G endows A of the structure of G-group and, in fact, A is an irreducible G-group.On the set of chief factors, the G-equivalence relation is easily described.Indeed, it is proved in [8,  G-isomorphic to A and B respectively.
(The example in the previous paragraph witnesses that the second possibility does arise.)From this, it follows that, for every monolithic primitive group L and for every k ∈ N, the minimal normal subgroups of the crown-based power L k are all L k -equivalent.

2.3.
Crowns of a finite group.Let X and Y be normal subgroups of We say that A = X/Y is a Frattini chief factor if X/Y is contained in the Frattini subgroup of G/Y ; this is equivalent to saying that A is abelian and there is no complement to A in G.The number δ G (A) of non-Frattini chief factors Gequivalent to A in any chief series of G does not depend on the series and hence δ G (A) is a well-defined integer depending only on the chief factor A.
We denote by L A the monolithic primitive group associated to A, that is, If A is a non-Frattini chief factor of G, then L A is a homomorphic image of G.More precisely, there exists a normal subgroup N of G such that G/N ∼ = L A and soc(G/N ) ∼ G A.

Consider now the collection
We conclude this preliminary section with two technical lemmas and one of the main results from [11].Theorem 2.5 is an improvement of [10, Corollary 2].We warn the reader that the statement of Theorem 2.5 is slightly different from that of Theorem 1.4 in [11]: to get Theorem 2.5 one should take into account Theorem 1.4 in [11] and the remark following its statement.In this section we prove Theorems 1.2 and 1.3.Our proofs are inspired from some ideas developed in [4].Moreover, our proofs have some similarities and hence we start by deducing some general facts holding for both.
We start by defining the universal constant a. Observe that the series ∞ u=1 u −3/2 converges.We write Let c be the universal constant arising from Theorem 2.5.We define Recall that max(H, G) is the number of maximal subgroups of G containing H.For the proof of Theorems 1.2 and 1.3 we argue by induction on |G : H| + |G|.The case |G : H| = 1 for the proof of Theorem 1.2 is clear because max(H, G) = 0. Similarly, the case that H is maximal in G for the proof of Theorem 1.3 is clear because max(H, G) = 1.In particular, for the proof of Theorem 1.2, we suppose |G : H| > 1 and, for the proof of Theorem 1.3, we suppose that H is not maximal in G.
Consider H := Observe that max(H, G) = max( H, G).In particular, when H < H, we have |G : H| < |G : H| and hence, by induction, we have max(H, G) = max( H, G) ≤ a|G : H| 3/2 < a|G : H| 3/2 ; moreover, when G is soluble, we have max(H, G) = max( H, G) ≤ |G : H| − 1 < |G : H| − 1.Therefore, we may suppose H = H, that is, ( H is an intersection of maximal subgroups of G. Let F be the Frattini subgroup of G. From (3.1), we have F ≤ H and hence, from (3.2), F = 1.In particular, we may now apply Lemma 2.3 to the group G.
Choose I, R and D as in Lemma 2.3.From (3.1), we may write where X 1 , . . ., X ρ are the maximal subgroups of G not containing D and Y 1 , . . ., Y σ are the maximal subgroups of G containing D. We define Thus H = X ∩ Y .For every i ∈ {1, . . ., ρ}, since D X i , we have G = DX i and hence Lemma 2.4 (applied with K := X i ) yields R ≤ X i .In particular, for some monolithic primitive group L and for some positive integer k.We let N denote the minimal normal subgroup (a.k.a. the socle) of L. From the definition of I and R, we have We have (As above, when G is soluble and G = HR, we have ρ = 0 and hence the inequality ρ ≤ |G : H|/2 − 1 is valid also in this degenerate case.)Now, from (3.4) and (3.6), we have • |G : H| 3/2 < a|G : H| 3/2 ; similarly, when G is soluble, from (3.5) and (3.7), we have In particular, for the rest of the proof, we may assume that R ≤ H. Now, ( For the rest of our argument for proving Theorems 1.2 and 1.3, we prefer to keep the proofs separate.
Proof of Theorem 1.2.Case 1: Suppose that N is non-abelian.
Since N is non-abelian, the group G = L k has exactly k minimal normal subgroups.We denote by N 1 , . . ., N k the minimal normal subgroups of G.In particular, We claim that, for every i ∈ {1, . . ., ρ}, there exist x, y ∈ {1, . . ., k} such that N ℓ ≤ X i , for every ℓ ∈ {1, . . ., k} \ {x, y}, that is, X i contains all but possibly at most two minimal normal subgroups of G.
We argue by induction on k.The statement is clearly true when k ≤ 2. Suppose then k ≥ 3 and let C := core G (X i ).If C = 1, then X i is a maximal core-free subgroup of G and hence the action of G on the right cosets of X i gives rise to a faithful primitive permutation representation.Since a primitive permutation group has at most two minimal normal subgroups [2, Theorem 4.4] and since G has exactly k minimal normal subgroups, we deduce that k ≤ 2, which is a contradiction.Therefore C = 1.
Since N 1 , . . ., N k are the minimal normal subgroups of L k , we deduce that there exists ℓ ∈ {1, . . ., k} with N ℓ ≤ C. Now, the proof of the claim follows applying the inductive hypothesis to G/N ℓ ∼ = L k−1 and to its maximal subgroup The previous claim shows that, for every C ∈ C, C contains all but possibly at most two minimal normal subgroups of Let C ∈ C and let M ∈ M C .The reader might find useful to see Figure 1, where we have drawn a fragment of the subgroup lattice of G relevant to our argument.
From (3.4), (3.8) and from the definition of a, we have max(H, G) = σ + ρ ≤ a 2 3/2 |G : H| 3/2 + 11ca ′ |G : H| 3/2 = a|G : H| 3/2 .Case 2: Suppose that N is abelian.As N is abelian, the action of L by conjugation on N endows N of the structure of an L-module.Since L is primitive, N is irreducible.Set q := |End L (N )|.Now, N is a vector space over the finite field F q with q elements, and hence |N | = q k ′ , for some positive integer k ′ .
Let C ∈ C and let M ∈ M C .From Lemma 2.2, C ≤ I. Now, the action of G/C on the right cosets of M/C is a primitive permutation group with point stabilizer M/C.Observe that in this primitive action, I/C is the socle of G/C.In particular, G/C acts irreducibly as a linear group on I/C and hence C is a maximal L-submodule of I. Since I is the direct sum of k pairwise isomorphic irreducible L-modules, we deduce that we have at most (q k − 1)/(q − 1) choices for C.Moreover, |G : As we have observed above, M ∩ I = C is an L-submodule of G. Since an intersection of L-submodules is an L-submodule, we deduce that is an L-submodule of I and hence H ∩ I G. Since H is core-free in I, we deduce H ∩ I = 1 and hence |I| = |N | k = q kk ′ divides |G : H|.In particular, |G : H| ≤ q kk ′ .Therefore, from (3.9), we obtain .
The rest of the proof of Theorem 1.3 follows the same idea as in the "Case 2" above, but taking in account that the whole group G is soluble.
Proof of Theorem 1.3.Since G = L k and I = N k , we may write G = I ⋊ K, where K is a complement of N in L. As in the proof of Theorem 1.2 for the case that N is abelian, we have that the action of L by conjugation on N endows N of the structure of an L-module.Since L is primitive, N is irreducible.Set q := |End L (N )|.Now, N is a vector space over the finite field F q with q elements, and hence |N | = q k ′ , for some positive integer k ′ .
Let C ∈ C and let M ∈ M C .As we have observed above (for the proof of "Case 2"), M ∩ I = C is a maximal L-submodule of G, H ∩ I = 1 and |I| = |N | k = q kk ′ divides |G : H|.In particular, |G : H| = ℓq kk ′ , for some positive integer ℓ.
Since G is soluble and since M is a maximal subgroup of G supplementing I, we have M = C ⋊ K x , for some maximal L-submodule C of I and some x ∈ I. Arguing as in the proof of Theorem 1.2 for the case that N is abelian, we deduce that we have at most (q k − 1)/(q − 1) choices for C.Moreover, we have at most |I/C| = |G : M | = |N | = q k ′ choices for x.This yields (3.11) ρ ≤ q k − 1 q − 1 q k ′ .max(H, G) = σ + ρ ≤ ℓ − 1 + q k − 1 q − 1 q k ′ .
When ℓ ≥ 2, a computation shows that the right hand side of (3.12) is less than or equal to ℓq kk ′ − 1 = |G : H| − 1.In particular, we may suppose that ℓ = 1.In this case, |G : H| = q kk ′ = |I| and hence G = IH = I ⋊ H.Moreover, σ = 0. Since H is not a maximal subgroup of G (recall the base case for our inductive argument), k ≥ 2.
Assume also k ′ = 1.Since |End L (N )| = q = |N |, we deduce that L/N is isomorphic to a subgroup of the multiplicative group of the field F q and hence |L : N | is relatively prime to q. Therefore |G : I| is relatively prime to q and hence so is |H|.Therefore, replacing H by a suitable G-conjugate, we may suppose that K = H.Using this information, we may now refine our earlier argument bounding ρ.Let C ∈ C and let M ∈ M C .Since G = I ⋊ H is soluble, M is a maximal subgroup of G supplementing I and H ≤ M , we have M = C ⋊ H, for some maximal L-submodule C of I. We deduce that we have at most (q k − 1)/(q − 1) choices for C and hence we have at most (q k − 1)/(q − 1) choices for M .This yields max(H, G) and the result is proved in this case.Assume k ′ ≥ 2. A computation (using ℓ = 1 and k, k ′ ≥ 2) shows that the right hand side of (3.12) is less than or equal to q kk ′ − 1 = |G : H| − 1.

Lemma 2 . 3 . [ 1 ,
Lemma 1.3.6]Let G be a finite group with trivial Frattini subgroup.There exists a chief factor A of G and a non-identity normal subgroupD of G with I G (A) = R G (A) × D. Lemma 2.4.[5,Proposition 11]  Let G be a finite group with trivial Frattini subgroup, let I G (A), R G (A) and D be as in the statement of Lemma 2.3 and let K be a subgroup of G.If G = KD = KR G (A), then G = K.Theorem 2.5.[11, Theorem 1.4]There exists a constant c such that every finite group has at most cn 3/2 core-free maximal subgroups of index n.
yields R = 1 and hence G ∼ = L k and D = I.Therefore, we may identify G with L k and D with N k .Set C := {core G (X i ) | i ∈ {1, . . ., ρ}} and, for every C ∈ C, set

Figure 1 .
Figure 1.Subgroup lattice for G

Now, ( 3 . 5 )
gives σ ≤ |G : H|/|D : D ∩ T | − 1: recall that D = I = N k and D ∩ T = D ∩ H = I ∩ H = 1.Thus σ ≤ |G : H|/|D| − 1 = |G : H|/q kk ′ − 1 = ℓ − 1.Therefore, (3.12) Proposition 1.4] that two chief factors A and B of G are G-equivalent if and only if either • A and B are G-isomorphic, or • there exists a maximal subgroup M of G such that G/core G (M ) has two minimal normal subgroups N 1 and N 2