2 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 we denote by
coreG(M):=⋂g∈GMg
the core of M in G. The soclesoc(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 Lk be the k-fold direct
product of L. The crown-based power of L of size k is the subgroup Lk of Lk defined by
Lk:={(l1,…,lk)∈Lk∣l1≡⋯≡lk(modA)}.
Equivalently, if we denote by diag(Lk) the diagonal subgroup of Lk, then Lk=Akdiag(Lk).
For the proof of the next lemma we need some basic terminology, which we borrow from [11, Sections 4.3 and 4.4]. Let κ be a positive integer and let A be a direct product S1×⋯×Sκ, where the Si are pairwise isomorphic non-abelian simple groups. We denote by πi:A→Si the natural projection onto Si. A subgroup X of A is said to be a strip if X≠1 and, for each i∈{1,…,κ}, either X∩Ker(πi)=1 or πi(X)=1. The support of the strip X is the set {i∈{1,…,κ}∣πi(X)≠1}. The strip X is said to be full if πi(X)=Si 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 πi(X)=Si for each i∈{1,…,κ}.
Scott’s lemma (see for instance [11, 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 Lk′ be a crown-based power of L of size k′ having non-abelian socle Nk′ and let H′ be a core-free subgroup of Lk′ contained in Nk′. Then |Nk′:H′|≥5k′.
Proof.
We argue by induction on k′. If k′=1, then the result is clear because Nk′=N has no proper subgroups having index less then 5. Suppose that k′≥2 and write N:=N1×⋯×Nk′, where N1,…,Nk′ are the minimal normal subgroups of Lk′ contained in Nk′. For each i∈{1,…,k′}, we denote by πi:Nk′→Ni the natural projection onto Ni.
Suppose that there exists i∈{1,…,k′} with πi(H′)<Ni. Then NiH′/Ni is a core-free subgroup of Lk′/Ni≅Lk′-1 and is contained in Nk′/Ni. Therefore, by induction, |Nk′:H′Ni|=|Nk′/Ni:H′Ni/Ni|≥5k′-1. Furthermore, |H′Ni:H′|=|Ni:H′∩Ni|≥5 because Ni has no proper subgroups having index less then 5. Therefore, |Nk′:H′|≥5k′.
Suppose that πi(H′)=Ni for every i∈{1,…,k′}. Since N is non-abelian, we may write Ni=Si,1×⋯×Si,ℓ for some pairwise isomorphic non-abelian simple groups Si,j of cardinality s. For each i∈{1,…,k′} and j∈{1,…,ℓ}, we denote by πi,j:Nk′→Si,j the natural projection onto Si,j. Since πi(H′)=Ni, we deduce πi,j(H′)=Si,j for every i∈{1,…,k′} and j∈{1,…,ℓ}. In particular, H′ is a subdirect subgroup of S1,1×⋯×Sk′,ℓ, and hence (by Scott’s lemma) H′ is a direct product of pairwise disjoint full strips. Since no Ni is contained in H′, there exist two distinct indices i1,i2∈{1,…,k′} and j1,j2∈{1,…,ℓ} such that (i1,j1) and (i2,j2) are involved in the same full strip of H′. If we now consider the projection πi1,i2:Nk′→Ni1×Ni2, we obtain |Ni1×Ni2:πi1,i2(H′)|≥s≥60≥52. The inductive hypothesis applied to Ker(πi1,i2)∩H′ yields |Ker(πi1,i2):Ker(πi1,i2)∩H′|≥5k′-2, and hence |Nk′:H′|≥5k′.
∎
In the proofs of Theorem 1.2 and 1.3, we use without mention the following basic fact.
Lemma 2.2.
Let M be a normal subgroup of a crown-based power Lk with socle Nk. Then either M≤Nk or Nk≤M.
Proof.
For each i∈{1,…,k}, we write Ni:={(n1,…,nk)∈Nk∣nj=1 for all j∈{1,…,k}∖{i}}. In particular, N=N1×⋯×Nk.
Let M be a normal subgroup of the crown-based power Lk with socle Nk and with M≰Nk. Let m∈M∖Nk. For each i∈{1,…,k}, since M does not centralize Ni, we deduce 1≠[M,Ni]≤M∩Ni. As Ni is one of the minimal normal subgroups of Lk, we must have Ni≤M. Therefore, Nk=N1×⋯×Nk≤M.
∎
2.2 Basic facts on G-groups
Given a group G, a G-groupA is a group A together with a group homomorphism θ:G→Aut(A) (for simplicity, we write ag for the image of a∈A under the automorphism θ(g)). Given a G-group A, we have the corresponding semi-direct productA⋊θG (or simply A⋊G when θ is clear from the context), where the multiplication is given by
g1a1⋅g2a2=g1g2a1g2a2
for every a1,a2∈A and for every g1,g2∈G.
A G-group A is said to be irreducible if G leaves no non-identity proper normal subgroup of A invariant.
Two G-groups A and B are said to be G-isomorphic (and we write A≅GB) if there exists an isomorphism φ:A→B such that
(ag)φ=(aφ)g
for every a∈A and for every g∈G. Similarly, we say that A and B are G-equivalent (and we write A∼GB) 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≇GB and A∼GB. However, when A and B are abelian, the converse is true, that is, if A and B are abelian, then A∼GB if and only if A≅GB; 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 with the structure of a 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, 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/coreG(M) has two minimal normal subgroups N1 and N2 that are
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 Lk are all Lk-equivalent.
2.3 Crowns of a finite group
Let X and Y be normal subgroups of G with A=X/Y being a chief factor of G. A complementU to A in G is a subgroup U of G such that
G=UX and Y=U∩X.
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 δG(A) of non-Frattini chief factors G-equivalent to A in any chief series of G does not depend on the series, and hence δG(A) is a well-defined integer depending only on the chief factor A.
We denote by LA the monolithic primitive group associated to A,
that is,
LA:={A⋊(G/CG(A)) if A is abelian,G/CG(A) otherwise.
If A is a non-Frattini chief factor of G, then LA is a homomorphic image of G.
More precisely, there exists
a normal subgroup N of G such that
G/N≅LA and soc(G/N)∼GA.
Consider now the collection 𝒩A of all normal subgroups N of G with G/N≅LA and soc(G/N)∼GA:
the intersection
RG(A):=⋂N∈𝒩AN
has the property that G/RG(A) is isomorphic to the crown-based power (LA)δG(A), that is, G/RG(A)≅(LA)δG(A).
The socle IG(A)/RG(A) of G/RG(A) is called the A-crown of G and it is a direct product of δG(A) minimal normal subgroups all G-equivalent to A.
We conclude this preliminary section with two technical lemmas and one of the main results from [9].
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 subgroup D of G with IG(A)=RG(A)×D.
Lemma 2.4 ([5, Proposition 11]).
Let G be a finite group with trivial Frattini subgroup, let IG(A),RG(A) and D be as in the statement of Lemma 2.3 and let K be a subgroup of G. If G=KD=KRG(A), then G=K.
Theorem 2.5 ([9, Theorem 1.4]).
There exists a constant c such that every finite group has at most cn3/2 core-free maximal subgroups of index n.
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 [9, Theorem 1.4]: to get Theorem 2.5 one should take into account [9, Theorem 1.4] and the remark following its statement.
3 Proofs of Theorems 1.2 and 1.3
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
a′:=∑u=1∞1u3/2.
Let c be the universal constant arising from Theorem 2.5. We define
a:=11ca′1-1/23/2.
Recall that max(H,G) is the number of maximal subgroups of G containing H. For the proofs 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~:=⋂H≤M<GM max. in GM.
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,
(3.1)H is an intersection of maximal subgroups of G.
Suppose that H contains a non-identity normal subgroup N of G. Since max(H,G)=max(H/N,G/N) and |G/N|<|G|, by induction, we have max(H,G)=max(H/N,G/N)≤a|G/N:H/N|3/2=a|G:H|3/2. Moreover, when G is soluble, we have max(H,G)=max(H/N,G/N)≤|G/N:H/N|-1=|G:H|-1. Therefore, we may suppose
(3.2)coreG(H)=1.
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.
By (3.1), we may write
H=X1∩⋯∩Xρ∩Y1∩⋯∩Yσ,
where X1,…,Xρ are the maximal subgroups of G not containing D
and Y1,…,Yσ are the maximal subgroups of G containing D.
We define
X:=X1∩⋯∩Xρ and Y:=Y1∩⋯∩Yσ.
Thus H=X∩Y.
For every i∈{1,…,ρ}, since D≰Xi, we have G=DXi, and hence Lemma 2.4 (applied with K:=Xi) yields R≤Xi. In particular,
(3.3)R≤X.
Since R=RG(A) for some chief factor A of G, Section 2.3 yields
G/R≅Lk
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 I/R=soc(G/R)≅soc(Lk)=Nk. Finally, let T:=X∩I. In particular,
TR=XR∩IR.
We have
H∩D=(X∩Y)∩D=X∩(Y∩D)=X∩D=X∩(I∩D)=(X∩I)∩D=T∩D.
It follows
|G:HD|=|G:H||HD:H|=|G:H||D:H∩D|=|G:H||D:T∩D|.
If D≤T, then D≤X, and hence D≤X∩Y=H because D≤Y. However, this is a contradiction because D≠1 and hence, from (3.2), D≰H. Therefore, D≰T and |D:T∩D|>1.
Applying our inductive hypothesis, we obtain
(3.4)σ=max(HD/D,G/D)≤a|G/D:HD/D|3/2=a|G:HD|3/2=a(|G:H||D:D∩T|)3/2≤a23/2|G:H|3/2.
Moreover, when G is soluble and HD is a proper subgroup of G, we obtain
(3.5)σ=max(HD/D,G/D)≤|G/D:HD/D|-1=|G:HD|-1=|G:H||D:D∩T|-1≤|G:H|2-1.
(Observe that, when G is soluble and G=HD, we have σ=0, and hence the inequality σ≤|G:H|/2-1 is valid also in this degenerate case.)
From (3.3) we deduce ρ≤max(HR,G). If R≰H, then |G:HR|<|G:H| and hence, applying our inductive hypothesis, we obtain
(3.6)ρ≤max(HR,G)≤a|G:HR|3/2=a(|G:H||HR:H|)3/2≤a23/2|G:H|3/2.
Moreover, when G is soluble and HR is a proper subgroup of G, we obtain
(3.7)ρ≤max(HR,G)≤|G:HR|-1=|G:H||HR:H|-1≤|G:H|2-1.
(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
max(H,G)=σ+ρ≤2a23/2⋅|G:H|3/2<a|G:H|3/2.
Similarly, when G is soluble, from (3.5) and (3.7) we have
max(H,G)=σ+ρ≤|G:H|2-1+|G:H|2-1<|G:H|-1.
In particular, for the rest of the proof, we may assume that R≤H. Now, (3.2) yields R=1, and hence G≅Lk and D=I. Therefore, we may identify G with Lk and D with Nk.
Set
𝒞:={coreG(Xi)∣i∈{1,…,ρ}}
and, for every C∈𝒞, set
ℳC:={Xi∣i∈{1,…,ρ},C=coreG(Xi)}.
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.
In this proof, we distinguish two cases.
Case 1: Suppose that N is non-abelian.
Since N is non-abelian, the group G=Lk has exactly k minimal normal subgroups. We denote by N1,…,Nk the minimal normal subgroups of G. In particular, I=Nk=N1×N2×⋯×Nk.
First, we claim that, for every i∈{1,…,ρ}, there exist x,y∈{1,…,k} such that Nℓ≤Xi for every ℓ∈{1,…,k}∖{x,y}, that is, Xi 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:=coreG(Xi). If C=1, then Xi is a maximal core-free subgroup of G, and hence the action of G on the right cosets of Xi 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 N1,…,Nk are the minimal normal subgroups of Lk, we deduce that there exists ℓ∈{1,…,k} with Nℓ≤C. Now, the proof of the claim follows applying the inductive hypothesis to G/Nℓ≅Lk-1 and to its maximal subgroup Xi/Nℓ.
The previous claim shows that, for every C∈𝒞, C contains all but possibly at most two minimal normal subgroups of Nk=I. Therefore,
|𝒞|≤k2.
Let C∈𝒞 and let M∈ℳC. The reader might find it useful to see Figure 1, where we have drawn a fragment of the subgroup lattice of G relevant to our argument.
Let k′ be the number of minimal normal subgroups of G contained in M. In particular, I∩M≅Nk′. Observe that I∩H is contained in I∩M and is core-free in G. Applying Lemma 2.1 (with H′ replaced by I∩H in a crowned-based group isomorphic to Lk′), we get |I∩M:I∩H|≥5k′. As k′≥k-2, we deduce t≥5k-2.
Now, M/C is a core-free maximal subgroup of G/C. From Theorem 2.5, when C=coreG(M) and z=|G:C| are fixed, we have at most cz3/2 choices for M. As t≥5k-2, we have z≤|G:H|/5k-2.
Thus
ρ=∑C∈𝒞|ℳC|≤∑C∈𝒞∑z∣|G:H|z≤|G:H|/5k-2cz3/2≤ck2∑z∣|G:H|z≤|G:H|/5k-2z3/2=ck2(|G:H|5k-2)3/2∑z∣|G:H|z≤|G:H|/5k-2(5k-2z|G:H|)3/2.
Therefore,
∑z∣|G:H|z≤|G:H|/5k-2(5k-2z|G:H|)3/2≤∑u=1∞1u3/2=a′.
Finally, it is easy to verify that k2/53(k-2)/2≤11 for every k. Summing up,
(3.8)ρ≤11ca′|G:H|3/2.
From (3.4), (3.8) and from the definition of a, we have
max(H,G)=σ+ρ≤a23/2|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 with the structure of an L-module. Since L is primitive, N is irreducible. Set q:=|EndL(N)|. Now, N is a vector space over the finite field 𝔽q with q elements, and hence |N|=qk′ for some positive integer k′.
Let C∈𝒞 and let M∈ℳC. By 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 (qk-1)/(q-1) choices for C. Moreover, |G:M|=|G/C:M/C|=|N|=qk′. From Theorem 2.5, when C is fixed, we have at most c|G:M|3/2=c(qk′)3/2 choices for M∈ℳC. This yields
(3.9)ρ≤|𝒞|⋅maxC∈𝒞|ℳC|≤qk-1q-1⋅cq3k′/2<cqk+3k′/2.
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
H∩I=(X1∩⋯∩Xρ)∩I
is an L-submodule of G and hence H∩I⊴G. Since H is core-free in G, we deduce H∩I=1, and hence |I|=|N|k=qkk′ divides |G:H|. In particular, |G:H|≥qkk′.
Therefore, from (3.9) we obtain
ρ≤c|G:H|k+3k′/2kk′.
When k≠1 or when (k,k′)≠(2,1), we have k+3k′/2kk′≤32. When k=1, by refining (3.9), we obtain the sharper bound ρ≤cq3k′/2≤c|G:H|3/2. When (k,k′)=(2,1), we may again refine (3.9):
ρ≤c(q+1)q3/2≤c⋅2q⋅q3/2=2cq5/2≤2c|G:H|5/4≤2c|G:H|3/2.
Summing up, in all cases we have
(3.10)ρ≤2c|G:H|3/2.
From (3.4) and (3.10) we have
max(H,G)=σ+ρ≤a23/2|G:H|3/2+2c|G:H|3/2<a|G:H|3/2,
as desired.
∎
The rest of the proof of Theorem 1.3 follows the same idea as in Case 2 above, but taking in account that the whole group G is soluble.
Proof of Theorem 1.3.
Since G=Lk and I=Nk, 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 with the structure of an L-module. Since L is primitive, N is irreducible. Set q:=|EndL(N)|. Now, N is a vector space over the finite field 𝔽q with q elements, and hence |N|=qk′ for some positive integer k′.
Let C∈𝒞 and let M∈ℳC. 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=qkk′ divides |G:H|. In particular, |G:H|=ℓqkk′ for some positive integer ℓ.
Since G is soluble and since M is a maximal subgroup of G supplementing I, we have M=C⋊Kx 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 (qk-1)/(q-1) choices for C. Moreover, we have at most |I/C|=|G:M|=|N|=qk′ choices for x.
This yields
ρ≤qk-1q-1qk′.
Now, (3.5) gives σ≤|G:H|/|D:D∩T|-1: recall that D=I=Nk and D∩T=D∩H=I∩H=1. Thus σ≤|G:H|/|D|-1=|G:H|/qkk′-1=ℓ-1. Therefore,
(3.11)max(H,G)=σ+ρ≤ℓ-1+qk-1q-1qk′.
When ℓ≥2, a computation shows that the right-hand side of (3.11) is less than or equal to
ℓqkk′-1=|G:H|-1.
In particular, we may suppose that ℓ=1. In this case, |G:H|=qkk′=|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 |EndL(N)|=q=|N|, we deduce that L/N is isomorphic to a subgroup of the multiplicative group of the field 𝔽q 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∈ℳC. 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 (qk-1)/(q-1) choices for C and hence we have at most (qk-1)/(q-1) choices for M. This yields
max(H,G)=σ+ρ=ρ≤qk-1q-1≤qk-1=|G:H|-1,
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 (3.11) is less than or equal to qkk′-1=|G:H|-1.
∎
