Maximal Intersections in Finite Groups

For a finite group G, we investigate the behavior of four invariants, MaxDim(G),\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\text {MaxDim}(G),$$\end{document}MinDim(G),\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\text {MinDim}(G),$$\end{document}MaxInt(G)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\text {MaxInt}(G)$$\end{document} and MinInt(G),\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\text {MinInt}(G),$$\end{document} measuring in some way the width and the height of the lattice M(G)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {M}}(G)$$\end{document} consisting of the intersections of the maximal subgroups of G.

We will say that a subgroup H of a finite group G is a maximal intersection in G if there exists a family M 1 , . . . , M t of maximal subgroups of G with H = M 1 ∩ · · · ∩ M t . We will denote by M(G) the subposet of the subgroup lattice of G consisting of G and all the maximal intersections in G.
Let X be a set of maximal subgroups of the finite group G. We say that X is irredundant if the intersection of the subgroups in X is not equal to the intersection of any proper subset of X . The maximal dimension MaxDim(G) of G is defined as the maximal size of an irredundant set of maximal subgroups of G. This definition arises from the study of the maximum size m(G) of an irredundant generating set for G (that is, a generating set that does not properly contain any other generating set). Indeed, it is easy to see that m(G) ≤ MaxDim(G). However, in [6], it was proved that the difference MaxDim(G) − m(G) can be arbitrarily large. Independent of this motivation, MaxDim(G) can be viewed as a measure of the width of the poset M(G). The study of the maximal dimension of an arbitrary finite group is quite complicated and it is difficult to find good strategies to investigate this invariant. For example, one could ask the following question.

Negative Answers to Questions 1 and 2
Our first aim is to give a negative answer to question 1. Let F be the field with 11 elements and let C = c be the subgroup of order 5 of the multiplicative group of F. Let V = F 5 be a five-dimensional vector space over F and let σ = (1, 2, 3, 4, 5) ∈ Sym (5). The wreath group H = C σ has an irreducible action on V defined as follows: if v = (f 1 , . . . , f 5 ) ∈ V and h = (c 1 , . . . , c 5 )σ ∈ H, then v h = (f 1σ −1 c 1σ −1 , . . . , f 5σ −1 c 5σ −1 ). We will concentrate our attention on the semidirect product G := V H (notice that G = G 11,5 in the notations of [6,Section 3] [6,Lemma 10], the subset I of {1, . . . , 5} consisting of the indices i with x i = 0 contains at least 3 elements and give a negative answer to question 2. Let G = AGL (2,5), N = soc(G) and F/N = Frat(G/N ). We have N ∼ = C 5 × C 5 , F/N ∼ = Frat(GL (2,5)) ∼ = C 4 and G/F ∼ = Sym (5 Although question 2 has a negative answer, a weaker result holds. Proof. It is not restrictive to assume Frat(G) = 1.

An Example
Let H be a cyclic group of order m := p a q b , where p and q are two different primes and 2 ≤ a ≤ b. For any divisor r of m, denote by H r the unique subgroup of H of order r.
The maximal subgroups of G are the following: Now, we study the intersections of these maximal subgroups. Notice that if t ≥ 2 and x 1 , . . . , x t are distinct elements of X p r , then Let Y be a family of maximal subgroups of G. For any i ∈ I, let Y i be the set of subgroups in Y of the form B i,x , for some x ∈ X i . Moreover, define For Finally, set In this case, Let y 1 , y 2 be two different elements in X q . If follows from the fact that Let y 1 , y 2 be two different elements in X p . If follows from the fact that Moreover, if Y is a maximal irredundant family, then by Proposition 8, This implies |Y| = a + b. From the previous discussion, it follows: Clearly, we have an unrefinable chain in M(G) of length |I| = a + b − 2 from H to G and a chain of length 2 from H p a−1 q b−1 to H. Moreover, we have the following two unrefinable chains from 1 to H p a−1 q b−1 : In particular, we may easily conclude:

Proof of Theorem 3 and Further Considerations
Proof. By induction on n.
We substitute the original family with the subfamily {M j | j ∈ I \ {k}} and we conclude by induction.
Proof. We may assume Frat(G) = 1. Let t = MinInt(G) and assume that is a chain in M of length t, and this implies MaxInt(G) ≥ t.
hence H e1 , . . . , H eq is an irredundant family of maximal subgroups of G q,p,r and therefore MaxDim(G p,q,r ) ≥ q. Now assume that p has order q − 1 mod q : in that case C q is the direct sum of the irreducible σ -module, C 1 and C 2 , of dimension, respectively, 1 and q − 1. If we consider We are going to prove that the previous question has an affirmative answer if the Fitting length of G is at most 2. But the question remains open in the general case.

Lemma 16. Let G be a finite nilpotent group and let F be a family of subgroups of G which contains G and all the maximal subgroups of G and is closed under taking intersections. Let
Proof. Clearly, since C is not refinable, we must have X 0 = G. Let F = Frat(G). If X t ≤ F, then there exists a maximal subgroup M of G not containing X t . But then X t ∩ M < X t . However X t ∩ M ∈ F, and this would contradict the assumption that C cannot be refined in F. For any H ≤ G, let H := HF/F. We have a chain C : F = X t ≤ · · · ≤ X 0 .
Assume that X i = X i+1 . This implies that there exists a maximal subgroup is a maximal subgroup of X i (and therefore X i+1 is a maximal subgroup of X i ). This implies that the length of C is precisely u, hence t ≥ u.

MinInt(G) ≥ m(G).
Proof. We may assume Frat(G) = 1. In this case Fit(G) = soc(G) has a complement, say X, in G and Z(G) has a complement, say W , in Fit G, which is normal in G. Let H = Z(G)X. We have Consider the family F consisting of H and all the possible intersections of elements of D and M. Now let 1 = Y ρ < · · · < Y 0 = G be a chain of maximal intersections in G that cannot be refined. By an iterated application of [7,Theorem 15], there exists w ∈ W such that for Assume |J| = a and order the elements of J so that j 1 < j 2 < · · · < j a : we have that 0 < U ja < U ja−1 · · · < U j2 < U j1 = W is an H-composition series of W and this implies a = δ 1 + · · · + δ t . Now let J * := {j | U j+1 = U j }. Assume |J * | = b and order the elements of J so that j 1 < j 2 < · · · < j b : we have that Remark. One could hope to generalize Lemma 16 as follows: let G be a finite soluble group and let F be a family of subgroups of G which contains G and all the maximal subgroups of G and is closed under taking intersections. Let C = X t < X t−1 < . . . X 1 < X 0 be a chain in F. If C cannot be refined in F, then t ≥ m(G). This would allow to prove Theorem 17 for arbitrary finite soluble groups. However, this more general statement is false. (3) and let K i be the Sylow 3-subgroup of H i . Let 1 = k i ∈ K i and let X = k 1 , k 2 , k 3 . Let F be the family of subgroups of G consisting of G, the maximal intersections in G and X. A maximal subgroup of G containing X, contains also K :

Strongly and Weakly Minmax Finite Groups
are two unrefinable chains in M(G), so, since G is weakly minmax, they have the same length, but then t = u and therefore MinInt(G/N ) = MaxInt(G/N ). Now assume that X 1 , . . . , X a is a family of maximal subgroups of G with minimal size with respect to the property X 1 ∩ · · · ∩ X a = N and that Y 1 , . . . , Y b is a family of maximal subgroups of G containing N and with the property that the chain N ). We want to prove that a = b. First, notice that, since N is a maximal intersection and C is not refinable, Y 1 ∩ · · · ∩ Y b = N. There exists Z 1 , . . . , Z c such that X 1 , . . . , X a , Z 1 , . . . , Z c is a maximal irredundant family of maximal subgroups of G. Consider the chain of maximal intersections: Since G is weakly minmax, this chain cannot be refined (and in particular a + c = MinInt(G) = MaxInt(G)). On the other hand, is a chain of maximal intersections and we must have c + b ≤ MaxInt(G) = a + c, hence b ≤ a (and consequently b = a).

Theorem 19. A finite weakly minmax group is soluble.
Before proving this theorem, we need to introduce a couple of definitions and related lemmas.
Definition 20. Let X be an almost simple group and S = soc X.
1. We define σ(X) as the largest positive integer σ for which there exists a core-free maximal subgroup Y of X and s 1 , s 2 , . . . , s σ in S such that 2. We define τ (X) as the minimal size of a family of core-free maximal subgroups of X with trivial intersection.
Proof. Let Y be a core-free maximal subgroup of X and let T = S ∩ Y . We have T = 1 (see for example the last paragraph of the proof of the main theorem in [13]). If suffices to prove that there exists s ∈ S such that 1 < T ∩ T s < T. We have Y = N G (T ), so T = N S (T ). Assume by contradiction 1 = T ∩ T s for every s ∈ S\T : this means that S is a Frobenius group and T is a Frobenius complement, but this is in contradiction with the fact that S is a non-abelian simple group. Proof of Theorem 19. Let G a finite weakly minmax group. If G is not soluble, then it admits a non-abelian chief factor H/K. Let C = C G (H/K). Then, G/C is a monolithic group (with socle isomorphic to H/K) and is weakly minmax by Lemma 18. So, to complete the proof, it would suffice to prove that a finite monolithic group with non-abelian socle cannot be weakly minmax.
Let G be a monolithic primitive group, and assume N = soc(G) ∼ = S n , with S a non-abelian simple group.
Let ψ be the map from N G (S 1 ) to Aut(S) induced by the conjugacy action on S 1 . Set X = ψ(N G (S 1 )) and note that X is an almost simple group with socle S = Inn(S) = ψ(S 1 ). Let T := {t 1 , . . . , t n } be a right transversal of N G (S 1 ) in G; the map is an injective homomorphism. So we may identify G with its image in X T , where T = {π g | g ∈ G} is a transitive subgroup of Sym(n). In this identification, N is contained in the base subgroup X n and S i is a subgroup of the i-th component of X n .
Let F/N = Frat(G/N ) and assume that Y 1 , . . . , Y t is a family of maximal subgroups of G of minimal size with respect to the property F = Y 1 ∩· · ·∩Y t . Now, choose a core-free maximal subgroup Y of X and s 1 , . . . , s σ as in the definition of σ = σ(X) and let M = G ∩ (Y T ). By [1] Proposition 1.1.44, M is a maximal subgroup of G. For 1 ≤ i ≤ n and 1 ≤ j ≤ σ, let τ i,j = (1, . . . , 1, s j , 1 . . . , 1) ∈ S n , where s j is in the i-th position of τ i,j and let M i,j = M τi,j . We order lexicologically the pairs (i, j). Let Σ k,l = ∩ 1≤i≤k,1≤j≤l M i,j ∩ F. We have Σ k,l ∩ M k,l < Σ k,l and this implies MaxInt(G) ≥ t + n · σ(X). (4.1) Now let τ = τ (X) and suppose that R 1 , . . . , R τ are core-free maximal subgroups of X with trivial intersection. Again by [1] Proposition 1.1.44, ∩ Z τ and let π : G → T the epimorphism sending g to π g . Since W ∩ X n = 1, we have W ∩ ker π = 1, so W is isomorphic to a (proper) subgroup of T. Moreover since W ∩ N = 1, we have W ∼ = W N/N ≤ F/N, hence W is a nilpotent subgroup of Sym(n). With the same arguments used by Cameron, Solomon and Turull in [5], it can be easily proved that the maximal length l(K) of a chain of subgroups in a nilpotent permutation group of degree n is at most n − 1. It follows l(W ) ≤ n − 1. Since G has trivial Frattini subgroup, there exist at most n − 1 maximal subgroups of G whose total intersection with W is trivial, hence 34 Page 12 of 15 Since G is weakly minmax, combining (4.1) and (4.2), we get n · (σ(X) − 1) < τ(X).  Table 1]) and we deduce that MaxInt(M 22 ) ≥ 5. If G is an exceptional group of Lie type, then [3, Theorem 1] implies that G = G 2 (2) ∼ = U 3 (3), however, by a theorem of Wagner [16], G can be generated by 4 involutions and no fewer, so m(G) ≥ 4.
So we may assume S < G. Let H be a core-free maximal subgroup of G and let b = b(G, H) be the minimal size of a set of conjugates of H with trivial intersection (i.e., the base size of the primitive action of G on the set of the right cosets of H). This set of conjugates of H is an irredundant family of maximal subgroups with maximal size, so α(G) ≤ b(G, H) ≤ MaxDim(G) ≤ MaxInt(G). In particular, if G is weakly minmax, then b(G, H) is the same for any choice of a core-free maximal subgroup H of G. It follows from [2], that if soc G is a sporadic simple group, then G has faithful primitive actions with different base sizes, hence G is not weakly minmax. If n ≥ 5, then b(Sym(n), Sym(n − 1)) = n − 1 ≥ 4 > 3, and therefore Sym(n) is not weakly minmax. Finally, assume that G is an almost simple with a socle S of Lie type and S < G. Let B be Borel subgroup of S and let u be the number of the nodes of the associated Dynkin diagram, or the number of the orbits for a suitable groups of symmetries of this diagram if S is of twisted type or G involves a graph automorphism of G). There exists a family Y 1 , . . . , Y u of maximal parabolic subgroups of G such that Moreover, as in the proof of Lemma 21, since N S (B) = B and S is not a Frobenius group, there exists x ∈ S with 1 < B ∩ B x < B. Let m = m(G/S). There exists an irredundant family X 1 , . . . , X m of maximal subgroup of G containing S. Let X = X 1 ∩ · · · ∩ X m . Then, 1 < (X ∩ Y 1 ∩ · · · ∩ Y t ) ∩ (X ∩ Y 1 ∩ · · · ∩ Y t ) x < X ∩Y 1 ∩· · ·∩Y t < · · · < X ∩Y 2 ∩Y 1 < Y 1 ∩X < X < · · · < X 2 ∩X 1 < X 1 < G is a chain in M(G), so τ (G) ≥ MinInt(G) ≥ m+u+2 ≥ 3+u, a contradiction.
Proposition 23. Let G be a primitive monolithic soluble group. If G is weakly minmax, then the derived length of G is at most 3.