On the structure of some locally nilpotent groups without contranormal subgroups

Following J.S. Rose, a subgroup H of a group G is said contranormal in G if G = H^G . In a certain sense, contranormal subgroups are antipodes to subnormal subgroups. It is well known that a finite group is nilpotent if and only if it has no proper contranormal subgroups. We prove that a nilpotent-by-finite group with no proper contranormal subgroup is nilpotent. There are locally nilpotent groups with a proper contranormal subgroup. We study the structure of hypercentral groups with a finite proper contranormal subgroup.

Abstract. Following J.S. Rose, a subgroup H of a group G is said contranormal in G if G = H G . In a certain sense, contranormal subgroups are antipodes to subnormal subgroups. It is well known that a finite group is nilpotent if and only if it has no proper contranormal subgroups. We prove that a nilpotent-by-finite group with no proper contranormal subgroup is nilpotent. There are locally nilpotent groups with a proper contranormal subgroup. We study the structure of hypercentral groups with a finite proper contranormal subgroup.

Introduction.
A subgroup H of a group G is called contranormal in G if H G = G, where H G = x −1 hx | h ∈ H, x ∈ G is the normal closure of H in G, the smallest normal subgroup of G containing H. For example G is contranormal in G, for any group G. The term "contranormal subgroup" has been introduced by J.S. Rose in the paper [14]. Contranormal subgroups have been studied for example in the paper [9]. If G is a group and H is a contranormal subgroup of G, then every subgroup K containing H is contranormal in G. In particular, if H and L are contranormal subgroups of G, then the subgroup H, L is also contranormal in G. However, the intersection of two contranormal subgroups is not always contranormal. For example, in the group A 4 every Sylow 3-subgroup is contranormal, but the intersection of every two Sylow 3-subgroups of A 4 is trivial, so that it is not contranormal. Notice also that if M is a maximal subgroup of G which is not normal, then clearly M is a contranormal subgroup of G. Moreover, every subgroup of a finite group G is a contranormal subgroup of a subnormal subgroup of G. As we can see by the definition, contranormal subgroups are in a certain sense, antipode of normal and subnormal subgroups: a contranormal subgroup H of a group G is normal (respectively subnormal) if and only if H = G. It follows that groups, whose subgroups are subnormal (in particular, nilpotent group), do not contain proper contranormal subgroups. For finite groups the converse is true.
A finite group G is nilpotent if and only if G does not have proper contranormal subgroups.
Indeed, suppose that there is a prime p such that G has a Sylow p-subgroup P which is not normal in G. Then N G (P ) = G. Since P is pronormal in G, N G (P ) is abnormal in G ( [13], 1.6). But every abnormal subgroup is contranormal, and we obtain a contradiction, which shows that Sylow q-subgroups of G are normal for each prime q. It follows that G is nilpotent.
There exist infinite non-nilpotent groups, whose subgroups are subnormal (it is possible to find examples of such groups in the survey [3]). Therefore the following question naturally appears: When a locally nilpotent group without proper contranormal subgroups is nilpotent?
We notice that there exist Chernikov locally nilpotent groups having proper contranormal subgroups, as the following example shows. Let D be a divisible abelian 2-group. Then D has an automorphism ϕ such that ϕ(d) = d −1 for each element d ∈ D. Define the semidirect product Let a be an arbitrary element of D. Since D is divisible, there exists an We note that the group G is not nilpotent, however the series 1 ≤ Ω 1 (D) ≤ · · · ≤ Ω n (D) ≤ Ω n+1 (D) ≤ · · · ≤ D ≤ G is central, so that G is a hypercentral abelian-by-finite group. Besides, the contranormal subgroup b is ascendant. This group is abelianby-finite, thus there exist hypercentral abelian-by-finite groups having proper contranormal subgroups, and also finite contranormal subgroups. This example raises the following question: What can we say about locally nilpotent abelian-by-finite groups having no proper contranormal subgroups?
Our first result gives an answer to this question. In fact we have the following Theorem.
Theorem A. Let G be a nilpotent-by-finite group. If G has no proper contranormal subgroups, then G is nilpotent.
Now the question appears about the structure of locally nilpotent abelian-by-finite groups having proper contranormal subgroups. We show here the following result.
Proposition B. Let G be a locally nilpotent group and A be a normal abelian subgroup of G with G/A finite. Suppose that G has a proper contranormal subgroup C, then C = BK where B ≤ A is normal in G, K is a finitely generated subgroup such that G = AK, and A = B[K, A]. In particular the factor group G/B has the finite contranormal subgroup KB/B. Therefore we naturally come to locally nilpotent abelian-by-finite groups having a finite contranormal subgroup. Our last result gives a description of hypercentral groups which include a finite contranormal subgroup.
Theorem C. Let G be a hypercentral group. If G contains a finite contranormal subgroup, then G satisfies the following conditions: where V is a normal divisible abelian subgroup and C is a finite contranormal subgroup of G; (ii) Π(G) = Π(C), in particular the set Π(G) is finite; Here an infinite normal abelian subgroup A of a group G is called G-quasifinite if every proper G-invariant subgroup of A is finite.

Nilpotent-by-finite groups without proper contranormal subgroups
We start our investigation with this easy and very useful Lemma. (ii) If H is a normal subgroup of G and C is a subgroup of G such that H ≤ C and C/H is a contranormal subgroup of G/H, then C is a contranormal subgroup of G.
(iii) If C is a contranormal subgroup of G and D is a contranormal subgroup of C, then D is a contranormal subgroup of G.
Proof. These assertions are obvious.
Let G be a nilpotent-by-finite group and assume that G has no contranormal subgroups. In order to prove Theorem A, we first assume that G is p-group, p a prime. Furthermore, we first suppose that G is abelian-by-finite, thus there exists a normal abelian subgroup A of G of finite index in G. We start stating three easy Lemmas, well known in the literature. We add the proofs for the sake of completeness. Proof. Write s the exponent of A and k = |G/C G (A)|. For each a ∈ A we have A ≤ C G (a) and |G : C G (a)| ≤ k. Thus a has at most k conjugates in G. Therefore a G is an abelian group, of exponent ≤ s, generated by at most k elements. Thus a G is a finite normal subgroup of order at most s k . Write m = s k . Since G is a soluble p-group, then G is locally nilpotent, hence a G is contained in the m − th term of the upper central series of G. That holds for each a ∈ A, therefore A ≤ ζ m (G). Proof. Suppose first that A is a direct product of cyclic groups. Then since A is not bounded, there exists a subgroup C of A such that A/C = Dr n∈N d n , where the element d n has order p n . Consider the subgroup B/C = d n d −p n+1 |n ∈ N . Then by this choice the factor group A/B is a Prüfer p-group.
Suppose now that A cannot be decomposed in a direct product of cyclic subgroups. Let D be a basic subgroup of A (see Theorem 32.3 of the book [6]). Then D is the direct product of cyclic subgroups, therefore D = A. Moreover A/D is a divisible group. Thus A/D is direct product of Prüfer p-groups and there exists a subgroup B/D of A/B such that A/B is a Prüfer p-group.
, · · · , n}, and H is closed under subgroups and finite direct products, it follows that A/C ∈ H.
Another general lemma we will use is the following: [9], Theoren 5.9). In particular, the intersection D ∩ C is finite. Now assume that G is a p-group, p a prime, and that G has no proper contranormal subgroups. Suppose that G has a normal abelian subgroup A of finite index in G. If A is bounded, then there exists a positive integer m such that A ≤ ζ m (G), the m − th term of the upper central series of G, by Lemma 2.2. Since G/A is a finite p-group, G/A is nilpotent. Therefore G is nilpotent and we have the result of Theorem A in this case. Then we can suppose that A is not bounded. Thus, by Lemma 2.3, there exists a subgroup B of A such that A/B is a divisible Chernikov group. By Lemma 2.4 we can also suppose that B is G-invariant. In this case we have. Lemma 2.6. Let G be a p-group, p a prime, and suppose that G contains a normal abelian subgroup A of finite index. Assume that A contains a G-invariant subgroup C such that A/C is a divisible Chernikov group. If G has no proper contranormal subgroups, then [G, A] ≤ C.

Proof.
A/C is a Chernikov group, thus G/C satisfies the minimal condition on subgroups. Then there exists a series This means that the subgroup K/C n−1 is contranormal in G/C n−1 . By Lemma 2.1, the subgroup K is contranormal in G, and we obtain a contradiction. This contradiction shows that [G/C n−1 , A/C n−1 ] is finite. In this case the factor group G/C n−1 is nilpotent. It follows that the center of G/C n−1 contains A/C n−1 (see, for example, [5] [9], Corollary 5.11). Then the factor A/D is divisible Chernikov and G-quasifinite. Using the result of the previous paragraph, we obtain that [G, A] ≤ D.
Since the last intersection is finite, the factor group G/C is nilpotent. It follows that the center of G/C contains A/C (see, for example, [5], Proposition 3.2.11). Hence [G, A] ≤ C, and the Lemma is proved.
From Lemma 2.6 we have the following lemma:   Next step is to prove the result of Theorem A for every locally nilpotent abelian-by-finite group. Corollary 2.9. Let G be a locally nilpotent group, and suppose that G contains a normal abelian subgroup A of finite index. If G has no proper contranormal subgroups, then G is nilpotent.
Proof. First, suppose that G is periodic. Let π = Π(G/A) and σ = Π(G) \ π, then the set π is finite and we have G = Dr p∈π G p × Dr p∈σ G p , where G p is a Sylow p-subgroup of G for all p ∈ Π(G). The isomorphism G p ≃ G/Dr q∈Π(G),q =p G q and Lemma 2.1 show that G p has no proper contranormal subgroups for every p ∈ π. Using Corollary 2.8 we obtain that G p is nilpotent for each p ∈ π. The finiteness of the set π implies that Dr p∈π G p is nilpotent. Obviously the subgroup G p is abelian for every p ∈ σ, hence Dr p∈σ G p is abelian. Therefore G is nilpotent. Now suppose that G is non-periodic. Then the set T or(G) of all elements of G having finite order, is a characteristic subgroup of G and the factor group S = G/T or(G) is torsion-free. On the other hand, S is abelian-by-finite. then S is a locally nilpotent torsion-free abelian-byfinite group, and then it is abelian (see, for example, [5], Corollary 1.2.8). Choose in the abelian subgroup A a maximal Z-independent subset M and let C be the subgroup of A generated by M. Then A/C is a periodic group. By Lemma 2.4 there exists a G-invariant subgroup E ≤ C such that A/E is periodic. Obviously E is torsionfree. Then E ∩ T or(G) = 1 . Using Remak's theorem, we obtain an embedding G G/E × G/T or(G). By Lemma 2.1 G/E does not include proper contranormal subgroups. Then G/E is nilpotent by Corollary 2.8, moreover G/T or(G) is abelian, therefore G is nilpotent and we have the result. Now we extend Corollary 2.9 to any abelian-by-finite group. We start with the following two results. If the abelian pgroup A/C is bounded, then it is the direct product of cyclic subgroups. In particular, A/C contains a proper subgroup having finite index. Then, by Lemma 2.4, A/C contains a proper G-invariant subgroup B/C, having finite index. By Lemma 2.1 the factor group G/C does not contain proper contranormal subgroups. Being finite, this factor group must be nilpotent. But in this case [A/C, S/C] = 1 , and we obtain a contradiction. If the abelian p-group A/C is not bounded, then by Lemma 2.3, A/C contains a subgroup D/C such that A/D is a divisible Chernikov group. By Lemma 2.4, A/C contains a proper Ginvariant subgroup E/C such that A/E is Chernikov. By Lemma 2.1 the factor group G/E does not contain proper contranormal subgroups. Being Chernikov, this factor group must be nilpotent ( [11], Lemma 4.9). But in this case [A/C, S/C] = 1 , and we again obtain a contradiction. This contradiction proves that A = C A (S). It follows that S = A × V where V is a finite p ′ -subgroup. Moreover, V is a Sylow p ′ -subgroup of S, so that V is normal in G. By Lemma 2.1 the factor group G/V does not contain proper contranormal subgroups. This factor group is an abelian-by-finite p-group, then it is nilpotent, by Corollary 2.8.
The equality A ∩ V = 1 and Remak's theorem imply the embedding G G/A × G/V , which implies that G is nilpotent.
Let G be a group and A be a normal subgroup of G. We put γ ( G, A) = A, γ 2 (G, A) = [G, A], and, recursively, γ α+1 (G, A) = [G, γ α (G, A)], for all ordinals α, moreover, if λ is a limit ordinal, we write γ λ (G, A) = µ<λ γ µ (G, A) Lemma 2.11. Let G be a group and suppose that G contains a normal abelian torsion-free subgroup A of finite index. If G has no proper contranormal subgroups, then G is nilpotent.
Proof. Let M be a finite subset of A and write B = M G . Since G/A is finite, the subgroup B is finitely generated. Being torsion-free, it is free abelian. Moreover, B is G-invariant. Put T /B = T or(A/B), then the subgroup T has finite 0-rank and it is G-invariant. Let r 0 (T ) = n, then T /B has special rank at most n. Let p be an arbitrary prime and consider the factor A/B p . Let S p /B p be the Sylow p-subgroup of A/B p , then S p /B p is a Chernikov group of special rank at most n. We have the direct decomposition A/B p = Sp/B p × Cp/B p (see, for example [6], Theorems 21.2 and 27.5). Thus A/C p is a Cernikov p-group of special rank at most n. By Lemma 2.4 there exists a Ginvariant subgroup D p , D p ≤ C p such that A/D p is a Chernikov pgroup, it is D p = g∈G C g p , thus A/D p has special rank at most kn where k = |G/A|. The inclusion D p ≤ C p implies that B ∩ D p = B p . It follows that (BD p )/D p ≃ B/(B∩D p ) = B/B p , in particular (BD p )/D p is an elementary abelian p-group, having finite order less or equal to p n . The factor-group G/D p is periodic, therefore, by Corollary 2.9, G/D p is nilpotent. Then (BD p )/D p ≤ γ n (G), the n − th term of the lower central series of G. It follows that γ n+1 (G, B) ≤ D p . On the other hand, since B is normal in G, γ n+1 (G, B) The last inclusion is true for each prime p, therefore γ n+1 (G, B) ≤ p∈P B p , where P is the set of all primes.. Since B is a free abelian subgroup, p∈P B p = 1 , thus γ n+1 (G, B) = 1 . It follows that B ≤ γ n (G). That holds for every finitely generated subgroup B of A, therefore A is contained in the hypercenter of G. By Lemma 2.1 the factor group G/A does not contain proper contranormal subgroups. Being finite, G/A is nilpotent. Then G is hypercentral. In particular, G is locally nilpotent, and, by Lemma 2.9, G is nilpotent.
Corollary 2.12. Let G be an abelian-by-finite group. If G has no proper contranormal subgroups, then G is nilpotent.
Proof. Let A be a normal abelian subgroup of G such that the factor group G/A is finite. First suppose that G is periodic. Let π = Π(G/A) and σ = Π(A) \ π, then the set π is finite and we have A = Dr p∈π A p × Dr p∈σ A p , where A p is the Sylow p-subgroup of A for all p ∈ Π(A). Put B p = Dr q∈Π(A),q =p A q , then the subgroup B p is G-invariant, A/B p ≃ A p and by Lemma 2.1 G/B p does not contain proper contranormal subgroups for every p ∈ Π(A). By Lemma 2.10 G/B p is nilpotent for each p ∈ Π(A). In particular, if p ∈ σ, then G/B p is abelian. Since p∈Π(A) B p = 1 , by Remak's theorem, we obtain an embedding G Dr p∈π G/B p × Cr p∈σ G/B p . Since the set π is finite Dr p∈π G/B p is nilpotent. Since G/B p is abelian for all p ∈ σ, then Cr p∈σ G/B p is abelian. Therefore G is nilpotent. Now suppose that G is not periodic. Since G is not periodic, A is also not periodic. write T = T ora(A). Then A = T . Obviously the subgroup T is G-invariant and A/T is torsion-free. Lemma 2.1 shows that G/T does not contain proper contranormal subgroups. Hence the factor group G/T is nilpotent, by Lemma 2.11. Choose in the abelian subgroup A a maximal Z-independent subset M and let C = M . Then A/C is a periodic group. By Lemma 2.4 there exists a G-invariant subgroup E such that E ≤ C and A/E is a periodic group. The inclusion E ≤ C implies that E is torsion-free. Thus E ∩ T = 1 . By Remaks theorem, we obtain an embedding G ≤ G/E × G/T . Lemma 2.1 shows that G/E does not contain proper contranormal subgroups. Being periodic, G/E is nilpotent, we know that G/T is nilpotent, hence G is nilpotent, as required.

Now we can prove Theorem A.
Proof of Theorem A. Let K be a nilpotent normal subgroup of G such that G/K is finite. Write D = [K, K]. Lemma 2.1 implies that the factor group G/D does not contain proper contranormal subgroups. Moreover, G/D is abelian-by-finite. Then Corollary 2.12 implies that G/D is nilpotent. Using now Theorem 7 of paper [7], we obtain that G is nilpotent, as required.

Locally nilpotent abelian-by-finite groups with a finite contranormal subgroup
We start this section by proving Proposition B.
Proof of Proposition B. Suppose that AC = G. Then Lemma 2.1 implies that CA/A is a proper contranormal subgroup of the finite nilpotent group G/A. But a nilpotent group does not contain a proper contranormal subgroups. Hence AC = G. Choose in C a finitely generated subgroup K such that AK = G, Considering the factor group G/(K ∩ A), without loss of generality we may assume that K ∩A is trivial. Then the subgroup K is finite. We start our investigation assuming that G is a p-group, p a prime. The finiteness of A/A p implies that A = F × V where V is a divisible subgroup and F is a finite subgroup (see, for example [8], Lemma 3). Clearly the subgroup V is G-invariant. Being a finite p-group, the factor group G/V is nilpotent. As above it follows that CV /V = G/V or G = V C, ad again (i) holds. Now suppose G = V C, where V is divisible, abelian and normal in G. Since V is an abelian divisible p-subgroup we have V = × λ∈Λ P λ , where P λ is a Prüfer p-subgroup for all λ ∈ Λ (see, for example [6], Theorem 23.1). Let Q 1 be a Prüfer p-subgroup of V . Since G/V is finite, Q 1 has only finitely many conjugates, so that Y = Q G 1 is a divisible Chernikov subgroup. Since Y satisfies the minimal condition, Y includes an infinite G-invariant subgroup D 1 which is G-quasifinite. If D p 1 = D 1 , then D p 1 is finite since D 1 is quasi finite, and D 1 /D p 1 is finite since it is an elementary abelian p-group with the minimal condition, hence D 1 is finite, a contradiction. Therefore D p 1 = D 1 and D 1 is divisible. Thus V = D 1 R for some subgroup R such that R is G-invariant, the intersection D 1 ∩ R is finite and (D 1 ∩ R) |C| = 1 (see, for example [9], Corollary 5.11]. Put |C| = p n , then D 1 ∩ R ≤ Ω n (V ). It is not hard to prove that the subgroup [D 1 , C] is G-invariant. If we suppose that [D 1 , C] is a proper subgroup of D 1 , then the fact that D 1 is G-quasifinite implies that [D 1 , C] must be finite. Then D 1 C is a finite-by-abelian p-group, so that D 1 C is nilpotent. Being Chernikov, D 1 C is central-by-finite (see, for example [5], Corollary 3.2.10). It follows that D 1 ≤ ζ(D 1 C). Consider the factor group G/R. We have Again Q 2 has only finitely many conjugates, so that Q G 2 is a divisible Chernikov subgroup. As above Q G 2 includes an infinite G-invariant subgroup D 2 , which is G-quasifinite. Arguing as before it is possible to prove that D 2 is divisible. Then, by Corollary 5.11 of [9], R = D 2 R 1 for some subgroup R 1 such that R 1 is G-invariant and the intersection D 2 ∩ R 1 is finite, moreover D 2 ∩ R 1 ≤ Ω n (V ). Using the above arguments, we obtain that [D 2 , C] = D 2 . Put L 1 = Ω n (D 1 ), then D 1 /L 1 ∩ RL 1 /L 1 = 1 and L 1 ≤ Ω n (V ). Similarly, put L 2 = Ω n (D 2 ), then D 2 /L 2 ∩ R 1 L 2 /L 2 = 1 and L 2 ≤ Ω n (V ). Repeating these arguments and using transfinite induction, we obtain that the subgroup V has a family of G-invariant G-quasifinite subgroups {D µ | µ ∈ M} such that V = D µ |µ ∈ M , [D µ , C] = D µ for all µ ∈ M, as required. Moreover we have V /Ω n (V ) = × µ∈M D µ Ω n (V )/Ω n (V ).

Now we can prove
Corollary 3.2. Let G be a periodic locally nilpotent abelian-by-finite group. If G contains a finite contranormal subgroup, then G satisfies the following conditions: (i) G = V C where V is a normal divisible abelian subgroup and C is a finite contranormal subgroup of G; (ii) Π(G) = Π(C), in particular the set Π(G) is finite; Proof. Let A be a normal abelian subgroup of G having finite index, and let C be a finite contranormal subgroup of G. Then, arguing as above we have G = CA. Suppose that Π(G) = Π(C) and choose a prime q ∈ Π(G) \ Π(C). The equality G = AC implies that A contains a Sylow q-subgroup Q of G. We have A = Q × R where R is a Sylow q ′ -subgroup of A. Then G/R = QR/R × CR/R, which shows that CR/R cannot be a contranormal subgroup of G/R. Thus we obtain a contradiction with Lemma 2.1. This contradiction proves that Π(G) = Π(C). We have G = × p∈Π(G) S p where S p is a Sylow p-subgroup of G. The isomorphism S p ≃ G/(× q∈Π(G),q =p S p ) and an application of Proposition 3.1 prove the result.
Recall that a group G is called F -perfect, if G does not contain a proper subgroup of finite index. In every group the subgroup F (G), generated by all F -perfect subgroups, is F -perfect. It is the greatest F -perfect subgroup of G. Clearly F (G) is a characteristic subgroup of G, and the factor group G/F (G) does not contain F -perfect subgroups. The subgroup F (G) is called the F -perfect part of G. Let X be a class of groups. If G is a group, then we denote by G X the intersection of all normal subgroups H of G such that G/H ∈ X . The subgroup G X is called the X -residual of the group G. If X = F is the class of all finite groups, then G F is called the finite residual of G. Lemma 3.3. Let G be a locally nilpotent periodic group. If G contains a finite contranormal subgroup, then the F -perfect part of G has finite index.
Proof. If G does not contain proper subgroups of finite index, then G is F -perfect and the result is proved. Therefore we suppose that G contains proper subgroups of finite index. Let S be a finite contranormal subgroup of G. Then S is nilpotent. Let k be the nilpotency class of S. If H is a normal subgroup of G such that G/H is finite, then Lemma 2.1 shows that SH/H is a contranormal subgroup of G/H. On the other hand, G/H is nilpotent, and a nilpotent group does not contain proper contranormal subgroups. It follows that SH/H = G/H. In particular, G/H has nilpotency class at most k. Let S be the family of all normal subgroups of G having finite index, and let L = H∈S H. By Remak's theorem there is an embedding G/L Cr H∈S G/H. Since G/H has nilpotency class at most k for every H ∈ S, this implies that G/L is a nilpotent group. It follows that G/L does not contain proper contranormal subgroups and we obtain the equality G/L = SL/L. This means that G/L is finite. If we suppose that L contains a proper subgroup K having finite index in L, then K has finite index in G. Then D = Core G (K) is normal in G and has finite index in G. Then D ∈ S, and therefore L ≤ D, a contradiction. This contradiction proves that L is F -perfect and L coincides with the F -perfect part of G. Proof. Let L be the F -perfect part of G. Lemma 3.3 implies that L has finite index in G. The result follows since a periodic hypercentral F -perfect group is abelian (see [4], Chapter 2, n. 2, Theorem 2.2). Lemma 3.5. Let G be a locally nilpotent group. If G is not periodic, then G does not contain finite contranormal subgroups.
Proof. Suppose the contrary, and let S be a finite contranormal subgroup of G. Since G is locally nilpotent, the set T or(G) of all elements of G having finite order is a characteristic subgroup of G. Since G is not periodic, G = T or(G). Then the inclusion S ≤ T or(G) implies that S G = G and we obtain a contradiction which proves the result. Now we can prove Theorem C.
Proof of Theorem C. Lemma 3.5 implies that a group G must be periodic. By Corollary 3.4 G is abelian-by-finite, and the result follows from Corollary 3.2.