Nilpotence relations in products of groups

Problem.

Let G be a factorized group, G=A⁢B, with A and B subgroups, and assume that A and B satisfy some group-theoretical property P. What can be said about the structure of the whole group G?

Theorem A.

Let G be a group and A,B≤G=A⁢B, where A and B are supersolvable N-connected subgroups. Then G is supersolvable.

Corollary 1.1.

Let G be a group and A,B≤G=A⁢B, where A and B are finitely generated nilpotent N-connected subgroups. Then G is a finitely generated nilpotent group.

Corollary 1.2.

Let G be a group, G=G1⁢G2⁢…⁢Gn, where G1,G2,…,Gn are pairwise permutable supersolvable (finitely generated nilpotent) subgroups of G. If, for all i,j:1,…,n such that i≠j, we have that Gi and Gj are N-connected, then G is supersolvable (finitely generated nilpotent).

Theorem B.

Let G be a group, and let A,B be two subgroups of G such that G=A⁢B. Assume that A and B are N-connected and Černikov groups. Then G is a Černikov group.

Corollary 1.3.

Let G be a group, and let A,B be two subgroups of G such that G=A⁢B. Assume that A and B are N-connected solvable groups with Min. Then G is a solvable group with Min.

Corollary 1.4.

Let G be a group, G=G1⁢G2⁢…⁢Gn, where G1,G2,…,Gn are pairwise permutable Černikov (solvable with Min) subgroups of G. If, for all i,j:1,…,n such that i≠j, we have that Gi and Gj are N-connected, then G is Černikov (solvable with Min).

Lemma 2.1 ([2, 10]).

Let the group G=A⁢B be the product of two subgroups A and B. For a subgroup S of G, the following conditions are equivalent:

1. if a⁢b∈S, with a∈A and b∈B, then a∈S;

2. S=(A∩S)⁢(B∩S) and A∩B≤S.

Definition 1 ([2, 10]).

A subgroup S of a factorized group G is said to be factorized if it satisfies one of the equivalent conditions of Lemma 2.1.

Lemma 2.2.

Let the group G=A⁢B be factorized by A,B≤G. Let the group D be such that B≤D≤G, and let a subgroup Y≤G be factorized in AD. Then it is also factorized in AB.

Proof.

By Definition 1, we have Y=(A∩Y)⁢(D∩Y) and Y≥A∩D. Applying the Dedekind modular law twice, we get

and therefore

Definition 2 ([8, Chapter 3, p. 64]).

A group G is said to satisfy Max, the maximal condition on subgroups, if, for all chains H0≤H1≤H2≤⋯, there exists n∈ℕ such that Hn=Hn+1=⋯ or, in other words, every ascending chain of subgroups is finite. Analogously, a group G is said to satisfy Min, the minimal condition on subgroups, if, for all chains H0≥H1≥H2≥⋯, there exists n∈ℕ such that Hn=Hn+1=⋯ or, in other words, every descending chain of subgroups is finite.

Definition 3 ([8, Chapter 5, p. 147]).

A group is said to be polycyclic if it has a finite cyclic series.

Proposition 2.3 ([8, Chapter 5, Proposition 5.4.12, p. 147]).

A group is polycyclic if and only if it is solvable and satisfies the maximal condition on subgroups.

Definition 4 ([8, Chapter 5, p. 145]).

A group is said to be supersolvable if it has a finite normal cyclic series.

Proposition 2.4 ([8, 9]).

Let G be a polycyclic group. Then

1. G is nilpotent if and only if every finite quotient of G is nilpotent,

2. G is supersolvable if and only if every finite quotient of G is supersolvable;

moreover, if G is supersolvable, then

1. Fit⁢(G) is nilpotent and G/Fit⁢(G) is finite abelian,

2. 1≠N⊲G implies 1≠〈x〉⊲G, for some x∈N.

Definition 5 ([8, Chapter 5, Proposition 5.4.13, p. 147]).

Let G be a polycyclic group. The number of infinite factors in a cyclic series is independent of the series; thus it is an invariant, called the Hirsch length of G, h⁢(G).

Definition 6 ([8, Chapter 5, p. 151]).

A group is said to be Černikov if it is an extension of a finite direct product of quasicyclic groups by a finite group.

Proposition 2.5 ([8, Chapter 5, Proposition 5.4.23, p. 151]).

A solvable group satisfies the minimal condition on subgroups if and only if it is a solvable Černikov group.

Definition 7 ([6, Chapter 5, p. 86]).

A group is said to be minimax if it has a series of finite length for which each factor satisfies Max or Min.

Definition 8 ([6, Chapter 5, p. 87]).

Let G be a solvable-by-finite minimax group. We know that there exists a series 1=G0⊲G1⊲⋯⊲Gn=G in which each factor satisfies Max or Min. Clearly, it is possible to refine this series and obtain a series with cyclic factors (finite or infinite) and quasicyclic factors. By a routine application of the Schreier refinement theorem, it is possible to show that the number of infinite factors of that series is an invariant, and we call it the minimax length or minimality of G, m⁢(G). It can be thought of as a generalization of Hirsch length for polycyclic groups.

Definition 9 ([8, Chapter 9, p. 269]).

A class of finite groups ℱ is said to be a formation if

1. every homomorphic image of an ℱ-group is an ℱ-group,

2. if GN1,GN2∈ℱ, then GN1∩N2∈ℱ.

Moreover, a formation ℱ is said to be saturated if G∈ℱ whenever FΦ⁢(G)∈ℱ

Theorem 2.6 ([3]).

Let G=G1⁢G2⁢…⁢Gr be a finite group such that G1,…,Gr are pairwise permutable subgroups of G. Let F be a saturated formation such that N⊆F. If, for every pair i,j∈{1,…,r}, i≠j, the subgroups Gi and Gj are N-connected F-groups, then G∈F.

Lemma 3.1.

Let G be a group, and let A,B≤G be such that G=A⁢B and A,B are N-connected. If A and B are supersolvable (f.g. nilpotent), then G is polycyclic if and only if it is supersolvable (f.g. nilpotent).

Proof.

If G is supersolvable (f.g. nilpotent), the result is clear. Suppose that G is polycyclic and A and B are supersolvable. Consider a normal subgroup N of finite index in G. We have

that is, a product of finite supersolvable 𝒩-connected subgroups. The class of finite supersolvable groups is a saturated formation that contains the class of finite nilpotent groups. So we can apply Carocca’s theorem, Theorem 2.6, and we get that G/N is supersolvable. Thus, we have a polycyclic group in which every finite homomorphic image is supersolvable. Applying Proposition 2.4 (2), G is supersolvable. The same holds when A and B are finitely generated nilpotent subgroups, using, in this case, Proposition 2.4 (1). ∎

Lemma 3.2.

Let G be a group and H,K≤G such that G=〈H,K〉. If we have |G:K|<∞, then |H:H∩K|<∞

Lemma 3.3.

Let G be a finitely generated nilpotent group and H≤G such that NG⁢(H)/H is a finite group. Then |G:H|<∞

Corollary 3.4.

Let G be a supersolvable group, F=Fit⁢(G) and H≤F such that NG⁢(H)/H is a finite group. Then |G:H|<∞

Proof.

Consider G and H as in the hypotheses. Clearly, |NF(H):H|<∞. Applying Lemma 3.3, |F:H|<∞, and together with the fact that |G:F|<∞ by Proposition 2.4, we have the thesis. ∎

Lemma 3.5.

Let G be a group, A,B≤G such that G=A⁢B. If A satisfies Max, then G satisfies the maximal condition on the subgroups containing B.

Proof.

Consider S1≤S2≤⋯ an ascending chain of subgroups containing B. Each Si is factorized, with the following factorization:

The chain S1∩A≤S2∩A≤⋯ is an ascending chain in A, which is a group that satisfies Max. Thus, there exists t∈ℕ such that St=St+j for all j≥0. Hence,

and the thesis follows. ∎

Proof of Theorem A.

We proceed by induction on h:=h(A)+h(B), where h⁢(X) is the Hirsch length of the group X. If h⁢(A)=h⁢(B)=0, then A and B are finite, and the result holds by Theorem 2.6. If h⁢(A)=0, then A is finite, so B is a subgroup of finite index in G; hence, since BG is a normal supersolvable subgroup and the factor G/BG is a supersolvable finite group, we conclude using Lemma 3.1. Thus, we can assume that both h⁢(A),h⁢(B)≥1. We define the sets

We prove that either 𝒜 or ℬ is non-empty. In fact, if h⁢(A∩B)≥1, both 𝒜 and ℬ are non-empty. Suppose that h⁢(A∩B)=0, i.e. A∩B is a finite group; by Proposition 2.4, we can choose a∈A, b∈B such that |a|=∞=|b| and 〈a〉⊲A, 〈b〉⊲B. By 𝒩-connection, we know that 〈a,b〉 is an infinite nilpotent group, and for this reason, we have that

It is then easy to prove that at least one of NA⁢(〈b〉) and NB⁢(〈a〉) is infinite. Suppose that NA⁢(〈b〉) is infinite, and let T=NG⁢(〈b〉).

If |A:NA(〈b〉)|=∞, then observing that

we get that T is supersolvable by the inductive hypothesis, and ℬ≠∅. Otherwise, if |A:NA(〈b〉)|<∞, consider

Observing that h⁢(〈b〉)=1, by the inductive hypothesis we have that T/〈b〉 is supersolvable, and by Lemma 3.1, T is supersolvable. Hence, ℬ≠∅.

Clearly, by Lemma 3.5, ℬ admits a maximal element D. Stressing that every subgroup containing A or B factorizes by Lemma 2.1, we get D=(D∩A)⁢B.

We know by the definition of Hirsch length that h⁢(A∩D)≤h⁢(A). Suppose by contradiction that h⁢(A∩D)<h⁢(A), i.e. |A:D∩A|=∞. Set FA=Fit⁢(A) and FD=Fit⁢(D).

Define the following chain of subgroups:

We know that FD is nilpotent and normal in D, so there exists t∈ℕ such that Mt=NFD⁢(Mt-1)=FD. Thus, we show by induction on i that the following factorization holds for all i:1⁢…⁢t:

If i=0, consider g∈NG⁢(M0); then g=a⁢b for certain a∈A and b∈D, and from M0g=M0, we deduce M0a=M0b-1≤FA∩FD=M0. Knowing that M0 is nilpotent, we get a∈NA⁢(M0), which is the thesis.

Suppose that i≥1, and consider g∈NG⁢(Mi). As before, g=a⁢b with a∈A, b∈D, and then Mia=Mib-1. It is clear that

that is, M0a≤M0, i.e. a∈NA⁢(M0). Suppose that 0≤j≤i is the greatest index such that a∈NA⁢(Mj), and suppose that j≤i-1. Knowing that

and 〈a,Mj+1〉≤NG⁢(Mj), we deduce

that is, Mj+1a≤Mj+1. By inductive hypothesis, NG⁢(Mj)=NA⁢(Mj)⁢ND⁢(Mj), and by Lemma 2.2, NG⁢(Mj)=NA⁢(Mj)⁢NB⁢(Mj). Passing to the quotient,

These two factors are 𝒩-connected supersolvable, and using the properties of Hirsch length and the fact that h⁢(Mj)≥h⁢(M0)≥1, we can say that NG⁢(Mj)/Mj is supersolvable, and hence, by Lemma 3.1, NG⁢(Mj) is supersolvable. So we have that 〈a,Mj+1〉≤NG⁢(Mj) satisfies Max, and then Mj+1a=Mj+1. This contradicts the assumption on j. Thus, i=j, and the statement is proved. Therefore, NG⁢(FD)=NA⁢(FD)⁢ND⁢(FD)=NA⁢(FD)⁢D.

Now we prove that NG⁢(FD) is supersolvable. In fact, we know that

holds; by Lemma 2.2, we have

Passing to the quotient, we see that

is a product of supersolvable 𝒩-connected subgroups. Since h⁢(FD)≥1, we can apply the inductive hypothesis and use Lemma 3.1 to deduce that NG⁢(FD) is supersolvable.

The last fact we prove is that NG⁢(FD)>D. Actually, we demonstrate a stronger fact, that is, |NG(Mi):Mi+1|=∞ for all i:0,…,t-1. Proceed by induction on i. If i=0, we know by Corollary 3.4 that |NA(M0):M0|=∞. By Lemma 3.2, |NG(M0):ND(M0)|=∞, and then |NG(M0):M1|=∞. Suppose that i≥1. By the inductive hypothesis, we know that

Keeping in mind that Mi is nilpotent, we want to prove that

Let W=NG⁢(Mi-1), FW=Fit⁢(W) and K=FW∩FD. Clearly, K≤Mi. If K=Mi, then by Corollary 3.4, we get the thesis. Otherwise, consider NW⁢(K). By Corollary 3.4, we have that |NW(K):K|=∞. Since K⊲ND⁢(Mi-1), we have that NW⁢(K)≥ND⁢(Mi-1). For this reason, the factorization

holds. Consider R=NA⁢(K)⁢K, S=RNW⁢(K) (the core of R in NW⁢(K)) and T=S∩FW. By construction, K≤T⊲NW⁢(K). Define L=T⁢ND⁢(Mi-1). Considering the quotient

we have that T/K is nilpotent, and by Proposition 2.4, ND⁢(Mi-1)/K is abelian.

In particular, L/K is supersolvable, and it is the product of two nilpotent 𝒩-connected subgroups; hence, by Lemma 3.1, L/K is nilpotent. We know that |L/K:Mi/K|=∞, so by Lemma 3.3, we get

which implies

and then

Thus, by Lemma 3.2,

and then the thesis |NG(Mi):Mi+1|=∞.

Gathering all these facts, we obtain NG⁢(FD)∈ℬ and NG⁢(FD)>D. This fact contradicts the maximality of D. So h⁢(A)=h⁢(A∩D), or in other words, |G:D|<∞. Thus, DG is a supersolvable normal subgroup of G. By Theorem 2.6, G/DG is supersolvable, and by Lemma 3.1, G is supersolvable. ∎

Problem 3.6.

Let G be a group and A,B≤G=A⁢B, where A and B are polycyclic N-connected subgroups. Is it true that G is a polycyclic group?

Lemma 4.1.

Let G=A⁢B be a Černikov group. If we denote by A0, B0 and G0 the finite residuals of, respectively, A, B and G, then G0=〈A0,B0〉.

Proof.

By a lemma of Amberg (see for instance [1, Lemma 1.2.5]), the group 〈A0,B0〉 has finite index in G, and therefore G0≤〈A0,B0〉. The other inclusion is clear. ∎

Lemma 4.2.

Let G be a periodic group and 1≠A≤G a divisible abelian subgroup. Let a∈A and u∈CG⁢(x) for all x∈A such that a∈〈x〉. Then u∈CG⁢(A).

Proof.

Let h∈A, and let |h|=m. Consider a1∈A such that a1m=a. Therefore, (a1⁢h)m=a1m=a. Hence, the following holds:

which is the thesis. ∎

Notation 4.3.

Given a group G such that G=A⁢B, where A,B≤G, and an element g∈G, we set

and for any subset X of G, we set

Theorem 4.4.

Let G be a group, and let A,B be two subgroups of G such that G=A⁢B. Assume that A and B are N-connected and Černikov p-groups, where p is a prime number. Then G is a Černikov p-group.

Proof.

Since A and B are Černikov p-groups, denote by A0 and B0 the finite residuals of, respectively, A and B. We know that they are both a direct product of a finite number of copies of the Prüfer p-group. We divide the proof into 4 steps. In the first three, we assume, in addition, that A0∩B0=1.

Step 1. Consider a∈A0 and b∈B0, and let H=〈a,b〉. We prove that if CG⁢(H) is infinite, then the set CA0⁢(H)∪CB0⁢(H) is infinite.

We know that H is nilpotent by 𝒩-connection. Using the notation introduced above, let A∗=ΠA⁢(CG⁢(H)) and B∗=ΠB⁢(CG⁢(H)). Since CG⁢(H)⊆A∗⁢B∗, we have that A∗∪B∗ is infinite. Without loss of generality, we can suppose that A∗ is infinite. For all x∈A∗, there exists a y∈B such that x⁢y∈CG⁢(H). In particular, bx⁢y=b, from which bx=by-1∈bB, where bB is a conjugacy class in B. Since b∈B0, we have that bB is finite; thus there exists an infinite subset X⊆A∗ such that bx=bx0 for all x∈X and for a fixed x0∈X. Consequently, {x0⁢x-1∣x∈X}⊆CA⁢(b) is infinite. Therefore, CA0⁢(b) is infinite too, and since a∈A0 is abelian, we can conclude by observing that

Step 2. Suppose that Z⁢(G)=1; we claim that there exist a∈A0 and b∈B0 such that CB0⁢(a)=1=CA0⁢(b).

In fact, consider the set {CB0⁢(a)∣a∈A0}; it clearly admits a minimal element. Let C=CB0⁢(a) be one such minimal element. If C≠1, then by divisibility and minimality, we have that, for all x such that a∈〈x〉, also CB0⁢(〈x〉)=CB0⁢(〈a〉). Therefore, we can apply Lemma 4.2 and conclude that CB0⁢(a)=CB0⁢(A0). So CG⁢(C)≥〈A0,B0〉, and hence it has finite index. Since G is a p-group, this implies that Z⁢(G)≠1. Thus, if Z⁢(G)=1, then the minimal element of the set {CB0⁢(a)∣a∈A0} is exactly {1}. The same holds for {CA0⁢(b)∣b∈B0}.

Step 3. Now our objective is to prove that Z⁢(G)≠1.

Suppose that Z⁢(G)=1. Then, by step 2, there exist a∈A0 and b∈B0 such that CB0⁢(a)=1=CA0⁢(b). By step 1, CG⁢(〈a,b〉) is finite. Moreover, by 𝒩-connection, we can deduce that 1≠Z⁢(〈a,b〉)≤CG⁢(〈a,b〉), namely CG⁢(〈a,b〉) is nontrivial and finite. We can choose a and b such that |CG⁢(〈a,b〉)| is minimal. Let 1≠u∈CG⁢(〈a,b〉) and x∈A0, where a∈〈x〉; thus CG⁢(〈x,b〉)≤CG⁢(〈a,b〉); hence equality holds, and in particular, u∈CG⁢(〈x,b〉)≤CG⁢(〈x〉). By Lemma 4.2, we have u∈CG⁢(A0). Analogously, we can show that u∈CG⁢(B0), and then CG⁢(u)≥〈A0,B0〉, which implies Z⁢(G)≠1.

Step 4. We claim that G is a Černikov group.

First of all, we show that G is hypercentral proving that every nontrivial quotient group of G has nontrivial center. In fact, since all the hypotheses pass to quotient groups, it is sufficient to show that Z⁢(G)≠1. So if A0∩B0=1, then we obtain our thesis by step 3. Otherwise, denoting K:=A0∩B0≠1, it is true that CG⁢(K)≥〈A0,B0〉 has finite index. Hence, as we observed in the previous steps, Z⁢(G)≠1. In conclusion, since a hypercentral p-group with Min-n is a Černikov p-group (see [7]), the theorem is proved. ∎

Proof of Theorem B.

Since A and B are Černikov groups, we know that the finite residual A0 and B0 have the following structure:

where, in A0, each Pi is a direct product of hi copies of Prüfer pi-groups, in which pi are distinct primes for i:1,…,s; analogously, in B0, each Qj is the direct product of kj copies of Prüfer qj-groups, in which qj are distinct primes for j:1,…,t.

In terms of minimality, this means hi=m⁢(Pi) and kj=m⁢(Qj). So we have

We proceed by induction on m⁢(A)+m⁢(B). If m⁢(A)=0 (or m⁢(B)=0), then B is a subgroup of finite index in G; hence BG is a Černikov normal subgroup of finite index in G, but clearly, a virtually Černikov group is a Černikov group, so we have the thesis. Otherwise, we can assume that m⁢(A)≥1 and m⁢(B)≥1. For the previous theorem, we can suppose that A0 and B0 are not both p-groups. Hence, without loss of generality, we can also assume that h1,h2≥1 and k1≥1. If pi≠qj for all i and j, then 〈A0,B0〉 is abelian by 𝒩-connection, so we assume p1=q1.

Let P:=P1. Since char⁡PA0, we have that P⊲A. Consider NG⁢(P). By the Wielandt lemma (see [10]), we know that the following factorization holds:

Moreover, we know

Consider two cases.

Case (1) If Q1≤NB⁢(P), then |G:NG(P)|<∞, so we can reduce ourselves to proving that NG⁢(P) is Černikov. Now, P is Černikov, and NG⁢(P)/P is Černikov by the inductive hypothesis; hence NG⁢(P) is Černikov.

Case (2) If Q1≰NB⁢(P), denote B∗:=NB(P). We have that B0∩B∗<B0 and |B∗:B0∩B∗|<∞, so if B0∗ is the finite residual of B∗, we deduce that

Hence, B0∗ and B0 are both direct products of a finite number of Prüfer groups and B0∗<B0, and B0 has no proper subgroups of finite index; then this fact allows us to apply the inductive hypothesis on NG⁢(P) and to see that it is Černikov. By Lemma 4.1, we deduce that the finite residual of NG⁢(P) is exactly 〈A0,B0∗〉, from which we have [Pi,Qj]=1 for all i:1,…,s and for all j:2,…,t. Now, considering P2 and repeating the same argument, if case (1) holds, the proof is done; otherwise, [Pi,Qj]=1 must hold for all i:1,…,s and for all j:1,3,…,t. Gathering all these facts, we conclude that [A0,B0]=1, from which 〈A0,B0〉 is abelian and it has finite index in G.

Moreover, G is the product of two subgroups with Min; thus it satisfies Min-n, i.e. the minimal condition on normal subgroups (see [1]).

Finally, every subgroup of finite index of G satisfies Min-n (see [11]); hence 〈A0,B0〉 is abelian Černikov, and then G is Černikov. ∎

Proof of Corollary 1.3.

By Proposition 2.5, A and B are Černikov groups, and by Theorem B, G is also Černikov. The solvability of G follows from the fact that if G0 is the finite residual of G, then G0 is abelian and G/G0 is solvable by Theorem 2.6. ∎