On the structure of some locally nilpotent groups without contranormal subgroups

Let 𝐺 be a group, and let L , K be normal subgroups of 𝐺 such that 𝐿 is periodic and locally nilpotent, L ≤ K and K / L is finite. Suppose that 𝐺 does not contain proper contranormal subgroups. Then 𝐾 is locally nilpotent.

Let 𝐺 be a group, and let 𝐴 be a normal nilpotent subgroup of 𝐺 such that G / A is a Chernikov 𝜋-group. Assume that 𝐴 has a finite series of 𝐺-invariant subgroups

such that every factor A j / A j + 1 , 0 ≤ j ≤ t - 1 , satisfies one of the following conditions:

• A j / A j + 1 is abelian, torsion-free and 𝐺-rationally irreducible;

• A j / A j + 1 is a periodic π ′ -group;

• A j / A j + 1 is a Chernikov 𝜋-group.

Let 𝐺 be a locally nilpotent group and 𝐴 a normal abelian subgroup of 𝐺 with G / A finite. Suppose that 𝐺 has a proper contranormal subgroup 𝐶. Then C = B ⁢ K , where B ≤ A is normal in 𝐺, 𝐾 is a finitely generated subgroup such that G = A ⁢ K , and A = B ⁢ [ K , A ] . In particular, the factor group G / B has the finite contranormal subgroup K ⁢ B / B .

Let 𝐺 be a hypercentral group. If 𝐺 contains a finite contranormal subgroup, then 𝐺 satisfies the following conditions:

1. G = V ⁢ C , where 𝑉 is a normal divisible abelian subgroup and 𝐶 is a finite contranormal subgroup of 𝐺;

2. Π ⁢ ( G ) = Π ⁢ ( C ) ; in particular, the set Π ⁢ ( G ) is finite;

3. 𝑉 has a family of 𝐺-invariant 𝐺-quasifinite subgroups { D μ ∣ μ ∈ M } such that V = ⟨ D μ ∣ μ ∈ M } ;

4. [ D μ , C ] = D μ for all μ ∈ M ; in particular, [ V , C ] = V .

Let 𝐺 be a group.

1. If 𝐶 is a contranormal subgroup of 𝐺 and 𝐻 is a normal subgroup of 𝐺, then C ⁢ H / H is a contranormal subgroup of G / H .

2. 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.

3. 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. ∎

Proof of Proposition A

First suppose that K / L is a minimal normal subgroup of G / L . Then either

Assume that the first holds. As K / L is finite, the factor group ( G / L ) / C G / L ⁢ ( K / L ) is finite. This factor group does not contain proper contranormal subgroups by Lemma 2.1. But a finite group which does not contain proper contranormal subgroups is nilpotent; thus K / L is abelian. Then K / L ≤ C G / L ⁢ ( K / L ) , and we obtain a contradiction. Then ( K / L ) ∩ C G / L ⁢ ( K / L ) = K / L , and K / L is a finite abelian 𝑝-subgroup. Suppose that 𝐾 is not locally nilpotent. Then 𝐾 is a not a 𝑝-subgroup; hence 𝐿 is a not a 𝑝-group. Let 𝑃 be the Sylow 𝑝-subgroup of 𝐿. Then 𝑃 is 𝐺-invariant, and we have K / P = ( L / P ) ⋊ S / P (see, for example, [5, Theorem 2.4.5]), where S / P is a finite Sylow 𝑝-subgroup of K / P . If S / P is 𝐺-invariant, then 𝑆 is a 𝐺-invariant Sylow 𝑝-subgroup of 𝐾. Then 𝑆 is locally nilpotent, and K = L ⁢ S is locally nilpotent since it is a product of two normal locally nilpotent subgroups. Therefore, S / P is not 𝐺-invariant. Then N G / P ⁢ ( S / P ) is a proper subgroup of G / P . Let 𝑔 be an element of 𝐺. Since 𝐾 is normal in 𝐺, ( S / P ) g ⁢ P ≤ K / P ; moreover, ( S / P ) g ⁢ P is a Sylow 𝑝-subgroup of K / P . Let U / P = ⟨ S / P , ( S / P ) g ⁢ P ⟩ . Clearly, 𝐾 is locally finite, so the subgroup U / P is finite. Therefore, the subgroups S / P and ( S / P ) g ⁢ P are conjugate in U / P . Thus the subgroup S / P is pronormal in G / P . But, in this case, the subgroup N G / P ⁢ ( S / P ) is abnormal in G / P (see, for example, [10, Proposition 2.6]). Since every proper abnormal subgroup is contranormal, we obtain that N G / P ⁢ ( S / P ) is contranormal in G / P , a contradiction since, by Lemma 2.1, G / P has no proper contranormal subgroups. This contradiction proves that 𝐾 is locally nilpotent.

Suppose now that K / L is not a minimal normal subgroup of G / L . Then there exists a series

of 𝐺-invariant subgroups such that L j + 1 / L j is a minimal normal subgroup of G / L j , 0 ≤ j ≤ n - 1 . Then, by induction, 𝐾 is locally nilpotent, as required. ∎

Let 𝐺 be a group, and let H , K be normal subgroups of 𝐺 such that 𝐻 is periodic and locally nilpotent, H ≤ K , and K / H is 𝐺-hyperfinite. Suppose that 𝐺 does not contain proper contranormal subgroups. Then 𝐾 is locally nilpotent. In particular, if G / H is hyperfinite, then 𝐺 is locally nilpotent.

Proof

Denote by 𝐿 the locally nilpotent radical of 𝐾, and assume that L < K . Since K / L is hyperfinite, there exists a non-trivial finite 𝐺-invariant subgroup F / L of K / L . Then 𝐹 is locally nilpotent by Proposition A; thus F ≤ L < F . This contradiction proves the result. ∎

Let 𝐺 be a periodic group, and let 𝐻 be a normal locally nilpotent subgroup of 𝐺 such that G / H is nilpotent. Suppose that 𝐺 does not contain proper contranormal subgroups. Then 𝐺 is locally nilpotent.

Proof

Every periodic nilpotent group is hyperfinite, and we can apply Corollary 2.2. ∎

Let 𝐺 be a periodic group, and let 𝐻 be a normal nilpotent subgroup of 𝐺 such that G / H is nilpotent and π ⁢ ( H ) ∩ π ⁢ ( G / H ) = ∅ . Suppose that 𝐺 does not contain proper contranormal subgroups. Then 𝐺 is nilpotent.

Proof

𝐺 is locally nilpotent by Corollary 2.3. 𝐻 is the Sylow π ⁢ ( H ) -subgroup of 𝐺; hence G = H × S , where 𝑆 is the Sylow π ⁢ ( G / H ) -subgroup of 𝐺. Therefore, S ≃ G / H implies that 𝑆 is nilpotent, and 𝐺 is nilpotent, as required. ∎

Let 𝐺 be a periodic group, and let 𝐻 be a normal locally nilpotent subgroup such that G / H is a Chernikov group. Suppose that 𝐺 does not contain proper contranormal subgroups. Then 𝐺 is locally nilpotent.

Let 𝐺 be a locally finite group, and let 𝐻 be a normal locally nilpotent subgroup such that the Sylow 𝑝-subgroups of G / H are Chernikov groups, for every prime 𝑝. Suppose that 𝐺 does not contain proper contranormal subgroups. Then 𝐺 is locally nilpotent.

Proof

A locally finite group without proper contranormal subgroups, whose Sylow 𝑝-subgroups are Chernikov groups, is Sylow-nilpotent (see, for example, [17, Corollary C3]). In particular, it is hyperfinite, and we can apply Corollary 2.2. ∎

Let 𝐺 be a hyperfinite group. Suppose that 𝐺 does not contain proper contranormal subgroups. Then 𝐺 is hypercentral.

Proof

𝐺 is locally nilpotent by Corollary 2.3. Then 𝐺 is hypercentral since it is hyperfinite. ∎

Let 𝐺 be a group, and let 𝐴 be a normal abelian 𝑝-subgroup of 𝐺 such that G / A is nilpotent and G / C G ⁢ ( A ) is periodic. Suppose that 𝐺 does not contain proper contranormal subgroups. Then G / C G ⁢ ( A ) is a 𝑝-group.

Proof

Assume that the factor group G / C G ⁢ ( A ) is a not a 𝑝-group. Then we have

where P / C G ⁢ ( A ) is a 𝑝-group and Q / C G ⁢ ( A ) is a non-trivial p ′ -group. Then the intersection Q / C G ⁢ ( A ) ∩ ζ ⁢ ( G / C G ⁢ ( A ) ) is non-trivial. Let g ⁢ C G ⁢ ( A ) be a non-trivial element of Q / C G ⁢ ( A ) ∩ ζ ⁢ ( G / C G ⁢ ( A ) ) . Then A = C A ⁢ ( g ) × [ g , A ] (see, for example, [1, Proposition 2.12]), where [ g , A ] is non-trivial since g ∉ C G ⁢ ( A ) . Since g ⁢ C G ⁢ ( A ) ∈ ζ ⁢ ( G / C G ⁢ ( A ) ) , the subgroup C A ⁢ ( g ) is 𝐺-invariant; then we can consider the factor group G / C A ⁢ ( g ) . Therefore, without loss of generality, we may assume that C A ⁢ ( g ) is trivial. Thus A = [ g , A ] and C A ⁢ ( g ) = ⟨ 1 ⟩ . Hence the subgroup 𝐴 has a complement 𝑆 in 𝐺 (see, for example, [6, Theorem 8.2.7]), and G = A ⋊ S . Hence 𝑆 is a proper subgroup of 𝐺 since 𝐴 is non-trivial. Then, from A = [ g , A ] , we obtain A = [ S , A ] . Obviously, [ S , A ] ≤ S G , and therefore G = A ⁢ S ≤ S G . Thus 𝑆 is a proper contranormal subgroup of 𝐺. By this contradiction, G / C G ⁢ ( A ) is a 𝑝-group. ∎

Let 𝐺 be a group and 𝐴 a normal nilpotent subgroup of 𝐺 such that G / A is a Chernikov group. Assume that 𝐵 and 𝐶 are 𝐺-invariant subgroups of 𝐴 with B ≤ C , C / B abelian, torsion-free and 𝐺-rationally irreducible, G / C nilpotent. Suppose that 𝐺 does not contain proper contranormal subgroups. Then C / B ≤ ζ ⁢ ( G / B ) .

Proof

Assume B = ⟨ 1 ⟩ without loss of generality. First notice that A ≤ C G ⁢ ( C ) . In fact, C ∩ ζ ⁢ ( A ) is a non-trivial 𝐺-invariant subgroup of 𝐶; thus C / ( C ∩ ζ ⁢ ( A ) ) is periodic. Let c ∈ C , a ∈ A ; then [ c s , a ] = 1 for a suitable positive integer 𝑠; then 1 = [ c s , a ] = [ c , a ] s since 𝐶 is abelian and 𝐺-invariant; hence [ c , a ] = 1 since 𝐶 is torsion-free. That holds for each a ∈ A , so C ≤ ζ ⁢ ( A ) and A ≤ C G ⁢ ( C ) .

Suppose G / C G ⁢ ( C ) ≠ ⟨ 1 ⟩ . Let e ∈ C ∖ ⟨ 1 ⟩ , and write E = ⟨ e ⟩ G . Then the factor group C / E is periodic. There exists a prime 𝑞 such that E ≠ E q = K and q ∉ π ⁢ ( G / A ) (see, for example, [11, Theorem 1.15]). Then the subgroup 𝐾 is 𝐺-invariant, the factor group C / K is periodic and its Sylow 𝑞-subgroup Q / K is non-trivial. We have C / K = Q / K × R / K , where R / K is the Sylow q ′ -subgroup of C / K . By Lemma 3.1, G / C G ⁢ ( C / R ) is a 𝑞-group. But A ≤ C G ⁢ ( C / R ) and G / A is a q ′ -group. Thus G = C G ⁢ ( C / R ) . Then the subgroup L = [ G , C ] ≤ R , and L ≠ C . Obviously, the subgroup L = [ G , C ] is 𝐺-invariant and non-trivial. The inclusion C / L ≤ ζ ⁢ ( G / L ) and the nilpotence of G / C imply that G / L is nilpotent. Since 𝐶 is 𝐺-rationally irreducible, the factor C / L is periodic. Choose in 𝐿 a non-trivial element 𝑣, and put V = ⟨ v ⟩ G . Then the factor C / V is periodic, and L / V is also periodic. Suppose that the set Π ⁢ ( L / V ) contains a prime 𝑝 such that p ∉ Π ⁢ ( G / A ) . Let S / V be the Sylow p ′ -subgroup of L / V ; then L / S is a non-trivial 𝑝-factor of 𝐺. Using again Lemma 3.1, we obtain that G = C G ⁢ ( L / S ) . Let c ∈ C , g ∈ G . Then ( c ⁢ S ) g ⁢ S = c ⁢ S ⁢ u ⁢ S for some element u ⁢ S ∈ L / S . Suppose that u ⁢ S ≠ S . We have g t ∈ A ≤ C G ⁢ ( C / S ) for some p ′ number 𝑡. It follows that ( u ⁢ S ) t = S . On the other hand, u ⁢ S is a 𝑝-element, and we obtain a contradiction. This contradiction shows that ( c ⁢ S ) g ⁢ S = c ⁢ S so that the factor C / S is 𝐺-central, and we obtain a contradiction. Suppose now that Π ⁢ ( L / V ) ⊆ Π ⁢ ( G / A ) . In this case, we can find a prime 𝑟 such that V ≠ V r = U and r ∉ Π ⁢ ( G / A ) (see, for example, [11, Theorem 1.15]). Then the factor L / U is periodic, and the set Π ⁢ ( L / U ) contains a prime 𝑟 such that r ∉ π ⁢ ( G / A ) . Now we can repeat the above arguments, and again, we obtain a contradiction. This contradiction shows that [ G , C ] is trivial. ∎

Let 𝐺 be a group, and let 𝐴 be a normal nilpotent subgroup of 𝐺 such that G / A is a Chernikov group. Suppose that 𝐴 contains 𝐺-invariant subgroups 𝐵 and 𝐶, B ≤ C , such that C / B is abelian, torsion-free and has finite 0-rank, and G / C is nilpotent. Suppose that 𝐺 does not contain proper contranormal subgroups. Then the factor C / B is 𝐺-central.

Proof

Since C / B has finite 0-rank, there exists a finite series

of 𝐺-invariant subgroups whose factors C j + 1 / C j are torsion-free and 𝐺-rationally irreducible, 0 ≤ j ≤ n - 1 . Using Lemma 3.2, we obtain that the factor C n / C n - 1 is 𝐺-central. Since G / C is nilpotent, G / C n - 1 is also nilpotent. Therefore, we can apply Lemma 3.2 to the factor C n - 1 / C n - 2 , and we get that this factor is 𝐺-central. By induction, we obtain that every factor C j + 1 / C j is 𝐺-central, 0 ≤ j ≤ n - 1 . Then G / B is nilpotent, and G / C G ⁢ ( C / B ) periodic implies that C / B is 𝐺-central, as required. ∎

Let 𝐺 be a group, and let 𝐴 be a normal abelian subgroup of 𝐺 such that G / A is a Chernikov 𝜋-group. Suppose that 𝐴 has a finite series of 𝐺-invariant subgroups

such that every factor A j / A j + 1 , 0 ≤ j ≤ t - 1 , satisfies one of the following conditions:

• A j / A j + 1 is torsion-free and 𝐺-rationally irreducible;

• A j / A j + 1 is a periodic π ′ -group;

• A j / A j + 1 is a Chernikov 𝜋-group.

Proof

G / A is nilpotent (see [17, Lemma 4.9]). Consider the factor group G / A 1 . If A / A 1 is torsion-free and 𝐺-rationally irreducible, then Lemma 3.2 implies that it is 𝐺-central. Hence G / A 1 is nilpotent. If the factor A / A 1 is a periodic π ′ -group, then Corollary 2.4 implies that G / A 1 is nilpotent (see also [20, Theorem 1]). Finally, suppose that A / A 1 is a Chernikov 𝜋-group. Then G / A 1 is a Chernikov group, and therefore it is nilpotent (see [17, Lemma 4.9]). Suppose that the factor group G / A j is nilpotent, and consider the factor G / A j + 1 . If A j / A j + 1 is torsion-free and 𝐺-rationally irreducible, then Lemma 3.2 implies that this factor is 𝐺-central. Hence G / A j + 1 is nilpotent. If A j / A j + 1 is a periodic π ′ -group, then Lemma 3.1 implies that it is 𝐺-central; thus G / A j + 1 is nilpotent. Finally, assume that A j / A j + 1 is a Chernikov 𝜋-group. Let D / A be the divisible part of G / A . Since A j / A j + 1 is union of 𝐺-invariant finite subgroups, then D ≤ C G ⁢ ( A j / A j + 1 ) . Since G / A j is nilpotent, we get that D / A j + 1 is nilpotent. Then G / A j + 1 is nilpotent-by-finite, and thus it is nilpotent (see [20, Theorem 1]). For j = t - 1 , we obtain the result. ∎

Proof of Theorem B

Let 𝐾 be a nilpotent normal subgroup of 𝐺 such that G / K is a Chernikov group. Suppose that G / K is a 𝜋-group and that 𝐾 has a finite series as in the hypothesis. We argue by induction on 𝑡.

First assume that t = 1 . In this case, 𝐾 is either a torsion-free abelian 𝐺-rationally irreducible group, or a periodic π ′ -group, or a Chernikov 𝜋-group. Write S = [ K , K ] . Lemma 2.1 implies that the factor group G / S does not contain proper contranormal subgroups. Moreover, G / S is abelian-by-Chernikov. If 𝐾 is abelian, then S = { 1 } ; in the other two cases, K / S is either a periodic π ′ -group, or a Chernikov 𝜋-group. Then Proposition 3.4 implies that G / S is nilpotent. Using now [8, Theorem 7], we obtain that 𝐺 is nilpotent, as required.

Now suppose t > 1 and, by induction, G / A t - 1 nilpotent. If A t - 1 is an abelian torsion-free 𝐺-rationally irreducible group, then Lemma 3.2 implies that this subgroup is 𝐺-central, and then 𝐺 is nilpotent.

Now suppose that A t - 1 is periodic. For every positive integer ℎ, write B h for the intersection of A t - 1 with the ℎ-center of 𝐾. Then B h is normal in 𝐺 for every ℎ, A t - 1 = B k for a suitable 𝑘, and B v = { 1 } for some v ≤ k . Moreover, K / B h + 1 ≤ C G / B h + 1 ⁢ ( B h / B h + 1 ) . We show that G / B h is nilpotent for every ℎ, and thus G / B v = G is nilpotent. Suppose, by induction, G / B h + 1 nilpotent. If A t - 1 / B h + 1 is a π ′ -group, then B h / B h + 1 is a π ′ -group, and Lemma 3.1 implies that B h / B h + 1 ≤ ζ ⁢ ( G / B h + 1 ) ; therefore G / B h is nilpotent. If A t - 1 is a Chernikov 𝜋-group, then, if D / K is the divisible part of G / K , arguing as in the proof of Proposition 3.4, we obtain that D / B h is nilpotent. Then G / B h is nilpotent-by-finite, and therefore it is nilpotent (see [20, Theorem 1]). ∎

Proof of Proposition C

Suppose A ⁢ C ≠ G . Then Lemma 2.1 implies that C ⁢ A / A is a proper contranormal subgroup of the finite nilpotent group G / A . But a nilpotent group does not contain a proper contranormal subgroup. Hence A ⁢ C = G . Choose in 𝐶 a finitely generated subgroup 𝐾 such that A ⁢ K = G ; then C = B ⁢ K , where B = C ∩ A . Since 𝐴 is normal in 𝐺, [ K , B ] ≤ A . On the other hand, [ K , B ] ≤ C so that [ K , B ] ≤ C ∩ A = B . Therefore, the subgroup 𝐵 is 𝐾-invariant. 𝐵 is also 𝐴-invariant since 𝐴 is abelian; thus, from G = A ⁢ K , we get that 𝐵 is 𝐺-invariant. The intersection K ∩ A is normal in 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. Since G = A ⁢ K , with 𝐴 normal in 𝐺, it follows that [ K , A ] is normal in 𝐺 and [ G , G ] = [ K , A ] ⁢ [ K , K ] ≤ K ⁢ [ A , K ] . Thus G / ( K ⁢ [ K , A ] ) is abelian. By Lemma 2.1, C ⁢ [ K , A ] / ( K ⁢ [ K , A ] ) is contranormal in G / ( K ⁢ [ K , A ] ) . It follows that C ⁢ [ K , A ] / ( K ⁢ [ K , A ] ) = G / ( K ⁢ [ K , A ] ) . Therefore, we have

In particular, we obtain that A = B ⁢ [ K , A ] . The subgroup 𝐵 is normal in 𝐺. Then we obtain that

It follows that G / B = ( K ⁢ B / B ) G / B ; hence K ⁢ B / B is contranormal in G / B . ∎

Let 𝐺 be an abelian-by-finite 𝑝-group, 𝑝 a prime. If 𝐺 contains a finite contranormal subgroup, then 𝐺 satisfies the following conditions:

1. G = V ⁢ C , where 𝑉 is a normal divisible abelian subgroup and 𝐶 is a finite contranormal subgroup of 𝐺;

2. 𝑉 has a family of 𝐺-invariant 𝐺-quasifinite subgroups { D μ ∣ μ ∈ M } such that V = ⟨ D μ ∣ μ ∈ M ⟩ ;

3. [ D μ , C ] = D μ for all μ ∈ M ; in particular, [ V , C ] = V .

Proof

Let 𝐴 be a normal abelian subgroup of 𝐺 having finite index, and let 𝐶 be a finite contranormal subgroup of 𝐺. By Lemma 2.1, C ⁢ A / A is contranormal in G / A . Since G / A is a finite 𝑝-group, it is nilpotent. The fact that a nilpotent group does not include proper contranormal subgroups implies that C ⁢ A / A = G / A , so G = C ⁢ A . If A = A p , then 𝐴 is divisible and (i) holds. Suppose B = A p ≠ A . Then 𝐵 is normal in 𝐺 and G / B is an extension of an elementary abelian 𝑝-subgroup by a finite 𝑝-group. Such groups are nilpotent [2]. On the other hand, Lemma 2.1 shows that C ⁢ B / B is a contranormal subgroup of G / B . The fact that a nilpotent group does not include a proper contranormal subgroup implies that C ⁢ B / B = G / B . It follows that A / B is finite. 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, [12, 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 C ⁢ V / V = G / V , so G = V ⁢ C , and 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, [7, 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 1 G is a divisible Chernikov subgroup. Since 𝑌 satisfies the minimal condition, 𝑌 contains an infinite 𝐺-invariant subgroup D 1 which is 𝐺-quasifinite. If D 1 p ≠ D 1 , then D 1 p is finite since D 1 is quasifinite, and D 1 / D 1 p is finite since it is an elementary abelian 𝑝-group with the minimal condition; hence D 1 is finite, a contradiction. Therefore, D 1 p = 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, [13, 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, [6, Corollary 3.2.10]). It follows that D 1 ≤ ζ ⁢ ( D 1 ⁢ C ) . Consider the factor group G / R . We have V / R = D 1 ⁢ R / R ≃ D 1 / ( D 1 ∩ R ) . The equality [ D 1 , C ] = ⟨ 1 ⟩ implies that [ V / R , C ⁢ R / R ] = [ D 1 ⁢ R / R , C ⁢ R / R ] = [ D 1 , C ] ⁢ R / R = ⟨ 1 ⟩ . It follows that V / R ≤ ζ ⁢ ( G / R ) . But, in this case, ( C ⁢ R / R ) G / R = C ⁢ R / R , and we obtain a contradiction with Lemma 2.1. This contradiction shows that [ D 1 , C ] = D 1 . Choose in the subgroup 𝑅 a Prüfer 𝑝-subgroup Q 2 . Again, Q 2 has only finitely many conjugates so that Q 2 G is a divisible Chernikov subgroup. As above, Q 2 G contains 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 [13, Corollary 5.11], 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 ∩ R ⁢ L 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 ) . ∎

Let 𝐺 be a periodic locally nilpotent abelian-by-finite group. If 𝐺 contains a finite contranormal subgroup, then 𝐺 satisfies the following conditions:

1. G = V ⁢ C , where 𝑉 is a normal divisible abelian subgroup and 𝐶 is a finite contranormal subgroup of 𝐺;

2. Π ⁢ ( G ) = Π ⁢ ( C ) ; in particular, the set Π ⁢ ( G ) is finite;

3. 𝑉 has a family of 𝐺-invariant 𝐺-quasifinite subgroups { D μ ∣ μ ∈ M } such that V = ⟨ D μ ∣ μ ∈ M } ;

4. [ D μ , C ] = D μ for all μ ∈ M ; in particular, [ V , C ] = V .

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 = C ⁢ A . Suppose that Π ⁢ ( G ) ≠ Π ⁢ ( C ) , and choose a prime q ∈ Π ⁢ ( G ) ∖ Π ⁢ ( C ) . The equality G = A ⁢ C 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

which shows that C ⁢ R / R cannot be a contranormal subgroup of G / R . Thus we obtain a contradiction with Lemma 2.1. This contradiction proves 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 4.1 prove the result. ∎

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 nilpotence class of 𝑆. If 𝐻 is a normal subgroup of 𝐺 such that G / H is finite, then Lemma 2.1 shows that S ⁢ H / 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 S ⁢ H / H = G / H . In particular, G / H has nilpotence 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 nilpotence 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 = S ⁢ L / 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. Thus 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. ∎

Let 𝐺 be a hypercentral periodic group. If 𝐺 contains a finite contranormal subgroup, then 𝐺 is abelian-by-finite.

Proof

Let 𝐿 be the ℱ-perfect part of 𝐺. Lemma 4.3 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]). ∎

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 Tor ⁡ ( G ) of all elements of 𝐺 having finite order is a characteristic subgroup of 𝐺. Since 𝐺 is not periodic, G ≠ Tor ⁡ ( G ) . Then the inclusion S ≤ Tor ⁡ ( G ) implies that S G ≠ G , and we obtain a contradiction which proves the result. ∎

Proof of Theorem D

Lemma 4.5 implies that the group 𝐺 must be periodic. By Corollary 4.4, 𝐺 is abelian-by-finite, and the result follows from Corollary 4.2. ∎