Groups in which each subnormal subgroup is commensurable with some normal subgroup

Let 𝐺 be a subsoluble T⁢[*]-group. Then 𝐺 is finite-by-metabelian; hence 𝐺 is soluble. Moreover, all finitely generated subgroups of 𝐺 are CF-groups; hence 𝐺 is locally (abelian-by-finite).

Let 𝐺 be a locally nilpotent T⁢[∗]-group. Then 𝐺 is hypercentral. Moreover, if 𝐺 is periodic, then 𝐺 is a CN-group with a nilpotent subgroup of index at most 2.

Let 𝐻 be a cn-subgroup of a group 𝐺. Then 𝐻 is cf if one of the following holds:

1. 𝐻 has a finite number of conjugates;

2. 𝐺 is abelian-by-finite;

3. 𝐻 is finitely generated;

4. 𝐻 has finite Prüfer rank.

Proof

To prove (1), note that 𝐻 is commensurable to each of its conjugates (see [1]), and therefore HG is the intersection of finitely many subgroups which are commensurable to 𝐻. Thus H/HG is finite.

To prove (2), let 𝐴 be a normal abelian subgroup with finite index of 𝐺. Then H1=H∩A has finite index in 𝐻, and therefore H1 is cn as well. Since we have A≤NG⁢(H1), H1 has a finite number of conjugates. Then, by (1), H1 is cf in 𝐺. Therefore, 𝐻 is cf in 𝐺.

Finally, (3) follows from [9, Lemma 2.1], while (4) follows from [6, Proposition 1]. ∎

For any prime 𝑝, there exists a nilpotent 𝑝-group G(p) with property CN such that G(p)=A⋊B, where 𝐴 and 𝐵 are abelian and 𝐵 is neither nn nor cf.

Proof

For the sake of completeness, we recall the construction of G(p) in [5, Proposition 1.2]. Take a sequence Pn of isomorphic groups of order p4,

Notice that Pn′=⟨xnp2⟩ has order 𝑝. Let P=Drn⁡Pn. Then P=X⋊Y with X=Drn⁡⟨xn⟩ and Y=Drn⁡⟨yn⟩. Consider the automorphism 𝛾 of 𝑃 such that xnγ=xn1+p and ynγ=yn for each n∈N. Then 𝛾 has order p2 if 𝑝 is odd, or 2 otherwise. Put G=P⋊⟨γ⟩, and note that G=X⋊H, where 𝑋 and

are abelian. Since N=⟨x0p2⁢xnp2∣n∈N⟩≤X is a 𝛾-invariant subgroup of P′ with index 𝑝, we can consider G¯=G/N. Then G(p)=G¯=X¯⋊H¯ is a nilpotent 𝑝-group with CN, which is not abelian-by-finite (see [5, Proposition 1.2]).

For each n∈N, we have [xn,γ]⁢N=xnp⁢N so that H¯G¯=H¯⁢[H¯,G¯]≥H¯⁢X¯p, and hence |H¯G¯:H¯|≥|H¯X¯p:H¯|=|X¯p| is infinite. On the other hand, we have that [H¯G¯,X¯]≤H¯G¯∩X¯={1} so that ⟨X¯,H¯G¯⟩=X¯×H¯G¯ is abelian and H¯ is not cf; otherwise, the index |G¯:X¯⁢H¯G¯| would be finite and G¯ would be abelian-by-finite, which is not possible. Therefore, the result is true with A=X¯ and B=H¯. ∎

Let 𝐺 be a group with a finite normal subgroup 𝐹 such that G/F is a CF-group (resp. T*-group). Then 𝐺 is a CN-group (resp. T⁢[*]-group).

Proof

If 𝐻 is any (resp. subnormal) subgroup of 𝐺, then H⁢F/F is a (resp. subnormal) subgroup of G/F; hence H⁢F/F is commensurable with some normal subgroup N/F of G/F. Then H⁢F is commensurable with 𝑁, which is a normal subgroup of 𝐺. Since 𝐻 has finite index in H⁢F, we have that 𝐻 commensurable with 𝑁 as wished. ∎

Let 𝐹 be any finite group and G0 a metabelian periodic 𝑇-group which is not finite-by-abelian-by-finite. Then F×G0 is a periodic soluble T*-group (which can have arbitrary derived length) and is not finite-by-abelian-by-finite.

For each prime 𝑝, there exists a metabelian periodic 𝑇-group G0 which is not finite-by-abelian-by-finite and such that p∉π⁢(G0).

Proof

Let p<q1<p1<⋯<qi<pi<⋯ be any sequence of primes such that qi divides pi-1 (recall that such a sequence exists by Dirichlet’s theorem). For every 𝑖, let Ai=⟨ai⟩ (resp. Bi=⟨bi⟩) be a cyclic group of order pi (resp. qi) and Gi=Ai⋊Bi, where bi acts on ai by means of an automorphism of Ai with order qi. Let G0=Dri⁢Gi, and denote A=Dri⁡Ai and B=Dri⁡Bi. Then G0=A⋊B is a 𝑇-group by [17, Lemma 5.2.2]. Moreover,

by [17, Lemma 2.2.2]. Finally, if by a contradiction G0 is finite-by-abelian-by-finite, then it is nilpotent-by-finite, that is A=Fit⁡(G0) has finite index, a contradiction. ∎

Let 𝐺 be a T⁢[∗]-group, and let 𝑋 be any subgroup of finite index of 𝐺. Then 𝑋 is a T⁢[∗]-group.

Proof

Let 𝐻 be a subnormal subgroup of 𝑋. Then K=H∩XG is a subnormal subgroup of the core XG of 𝑋 in 𝐺 so that 𝐾 is a subnormal subgroup of 𝐺, and hence 𝐾 is a cn-subgroup of 𝐺. Since the index |G:XG| is finite, the subgroup 𝐾 has finite index in 𝐻, and so it follows that 𝐻 is likewise a cn-subgroup (of 𝐺). ∎

For an abelian-by-finite group, the properties CF, CN, T*, T⁢[*] are all equivalent.

Proof

Since it is clear that in general

it is enough to prove that if 𝐺 is a T⁢[*]-group which has a normal abelian subgroup 𝐴 with finite index, then any subgroup 𝐻 of 𝐺 is cf in 𝐺. By Proposition 1 (2), 𝐺 is T*, and hence H1=H∩A is cf in 𝐺, being subnormal. Since H1 has finite index in 𝐻, the subgroup 𝐻 is likewise cf, as required. ∎

For a finitely generated soluble group 𝐺, the properties CF, CN, T*, T*, T⁢[*] are all equivalent to the fact that 𝐺 has a normal torsion-free abelian group 𝐴 with finite index on which 𝐺 acts by either the identity or the inversion map.

Proof

Let 𝐺 be a T⁢[*]-group; then we may apply [7, Corollary 2.35] to deduce that there is 𝐴 as in the statement. On the other hand, if there is such a subgroup 𝐴, then 𝐺 is clearly a CF-group; in particular, by Proposition 7 follows that the properties T*, T⁢[*], CF and CN are all equivalent for 𝐺.

Thus, since T*⊆T⁢[*], it remains to prove that such a group 𝐺 is T*. Now let 𝐻 be a subnormal subgroup of 𝐺 with defect 𝑛; then we have [G,nH]≤H. If [H,A]≠{1}, we have A2n=[A,nH]≤[G,nH]≤H; thus 𝐻 has finite index in 𝐺, and so 𝐻 is an nn-subgroup of 𝐺. Hence suppose that [H,A]={1}. In this case, we have that A≤NG⁢(H), and so 𝐻 has a finite number of conjugates in 𝐺. Since H/HG is finite, we have that HG/HG is finite by Dietzmann’s lemma. ∎

Let 𝐺 be a nilpotent-by-finite group. If 𝐺 is T⁢[∗]-group (resp. T∗-group), then 𝐺 is a CN-group (resp. CF-group).

Proof

Let 𝑁 be a nilpotent normal subgroup of finite index of 𝐺, and let 𝐻 be any subgroup of 𝐺. Then K=H∩N is a subnormal subgroup of 𝐺, and so 𝐾 is a cn-subgroup (resp. cf-subgroup) of 𝐺. Since 𝐾 has finite index in 𝐻, it follows that 𝐻 is likewise a cn-subgroup (resp. cf-subgroup) of 𝐺. ∎

Let Hi and Ki be subgroups of a group Gi for each 𝑖. Then

are commensurable subgroups of G=Dri⁡Gi if and only if Hi and Ki are commensurable for each 𝑖 and Hi=Ki for all but finitely many 𝑖.

Proof

Consider first the case when K≤H, that is Ki≤Hi for all 𝑖. Then recall that |H:K| is finite if and only if the group H/KH is finite, and a corresponding statements holds for each |Hi:Ki|. Since we have

the statement under consideration reduces to the elementary fact that a direct product of any family of groups is finite if and only if all but finitely many of them are trivial.

For the general case, apply the above to evaluate the indices H∩K in both 𝐻 and 𝐾. ∎

Let the group G=Drp⁡Gp be direct product of its primary components Gp. Then 𝐺 is a T⁢[*]-group if and only if each primary component Gp is a T⁢[*]-group and Gp is a 𝑇-group for all but finitely many primes 𝑝.

Proof

Recall that each subgroup 𝐻 is of the type Drp⁡Hp with Hp=H∩Gp. Moreover, 𝐻 is subnormal if and only if all the components Hp are subnormal with bounded defect.

For the sufficiency of the condition, note that if 𝐻 is subnormal, then each subgroup Hp is commensurable to some normal subgroup Np of Gp and Hp=Np for all but finitely many 𝑝. By Lemma 10, we have that 𝐻 is commensurable to N=Drp⁡Np◁G, as wished.

Concerning necessity, it is clear that all components Gp are T⁢[*]-groups as they are normal subgroups of the T⁢[*]-group 𝐺. Let then 𝜋 be the set of primes 𝑝 for which Gp is not a 𝑇-group, and assume π≠∅. Consider, for each p∈π, a subnormal subgroup Hp of Gp with defect exactly 2. Then we have that H=Drp⁡Hp is subnormal in 𝐺, and therefore it is commensurable with a normal subgroup N=Drp⁡(N∩Gp) of 𝐺. On the other hand, by Lemma 10, we have that

for all but finitely many primes 𝑝. Then 𝜋 must be finite. ∎

Let G1 and G2 be periodic groups such that π⁢(G1)∩π⁢(G2)=∅, and let 𝒳 be any of the following properties 𝑇, T*, T*, T⁢[*]. If G1 and G2 are 𝒳-groups, then G1×G2 is an 𝒳-group.

If G(p) is the 𝑝-group with CN of Example 2 and G0 is a group with p∉π⁢(G0) as in Example 5, then G(p)×G0 is a soluble periodic T⁢[*]-group which is neither T* nor T* nor finite-by-abelian-by-finite.

Let Γ be a group acting on an abelian group 𝐴. If each subgroup of 𝐴 is commensurable with a Γ-invariant subgroup, then there is a 𝐺-series

such that

1. A0 and A/A1 are finite, A1/X is periodic and X/A0 is torsion-free,

2. 𝐺 acts by means of power automorphisms on the factors of 𝒮 which are infinite groups,

3. each subgroup of A/A0 has a subgroup with finite index which is 𝐺-invariant,

4. G′ acts trivially on A1/X.

1. Σ is nilpotent of class at most 4;

2. Γ/Σ is finite-by-abelian;

3. if 𝐴 is a 𝑝-group and Γ is periodic, then Γ/Σ is finite;

4. if 𝐴 is torsion-free, then Γ consists at most of the identity and the inversion map.

Proof

Apply [5, Corollary 2.3] to get a finite Γ-invariant subgroup A0 as in (3). Then apply [11, Theorems 2.8 and 2.10] and get A1 and 𝑋 like in the statements (1) and (2) and such that A1/X=L/X×D/X×E/X, where 𝐺 acts by means of power automorphism on all direct factors. Since power automorphisms of an abelian group commute (with all other automorphisms, see above), we have that Γ′ acts trivially on all 3 direct factors and hence on A1/X. Thus (4) holds.

We have that (i) holds by a well-known fact (see [13]) as Σ is the stabilizer of a series of length 5. Moreover, Γ/Σ embeds canonically in the direct product of quotients G/C, where 𝐶 ranges over the centralizers in 𝐺 of the factors of 𝒮. By (2), these quotients are either finite or abelian. Thus (ii) holds as well. Further, (iii) follows from the fact that a periodic group of power automorphisms of an abelian 𝑝-group is finite (as recalled above). Finally, statement (iv) follows from [5, Lemma 2.5]. ∎

A soluble periodic T⁢[*]-group (resp. T∗-group) 𝐺 is a CN-group (resp. CF-group), provided π⁢(G′) is finite.

Proof

One may refine the derived series of 𝐺 to a finite normal series 𝒮 such that each factor, apart from G/G′ (which may well remain unrefined), is a 𝑝-group for some prime number 𝑝. Then, applying Proposition 14, one may further refine the series to a normal series such that 𝐺 acts by means of power automorphisms on each infinite factor. Then, on the one hand, the stability group 𝑆 of 𝒮 is nilpotent as in Proposition 14 (i). On the other hand, 𝑆 has finite index in 𝐺 by Proposition 14 (iii). Then 𝐺 is nilpotent-by-finite and hence CN by Lemma 9. ∎

Let 𝐺 be a soluble T⁢[∗]-group, and let 𝑇 be the largest periodic normal subgroup of 𝐺. Then G/T is either abelian or dihedral. In particular, G′′ is periodic. Moreover, if 𝐺 is torsion-free, then 𝐺 is abelian.

Proof

Replacing 𝐺 by G/T, we can assume that 𝐺 has no non-trivial normal periodic subgroups and that 𝐺 is not abelian. By [8, Theorem Ã], we have that 𝐺 has a self-centralizing normal torsion-free abelian subgroup A=CG⁢(A). Now, by Proposition 14 (iv), it follows that G/A has order 2 and there exists some element 𝑥 which acts on 𝐴 as the inversion map; since x2∈A, we obtain that x2=(x2)x=x-2. This implies that x2=1 and that G=A⋊⟨x⟩ is the dihedral group on 𝐴. ∎

Let 𝐺 be an hypercentral 𝑝-group. If 𝐺 is a T⁢[∗]-group, then 𝐺 is a nilpotent-by-finite CN-group.

Proof

By a well-known fact, there exists a self-centralizing abelian normal subgroup A=CG⁢(A) of 𝐺 (see for example [18, Part 1, Lemma 2.19.1]). Since all subgroups of 𝐴 are cn-subgroups of 𝐺, we may apply Proposition 14 to the group Γ=G/A=G/CG⁢(A) of automorphisms of 𝐴 to get that there is a 𝐺-series

on which Γ acts as a group of power automorphisms on infinite factors. Let

be the stabilizer in 𝐺 of the series. By Proposition 14 (iii), 𝐶 has finite index in 𝐺. On the other hand, A≤Z4⁢(C), and C/A is nilpotent because it stabilizes the above series; hence 𝐶 is nilpotent. The statement follows now from Lemma 9. ∎

Let 𝐺 be a locally nilpotent T⁢[∗]-group, and let 𝐴 be any abelian normal subgroup of 𝐺. Then 𝐴 is contained in the hypercenter of 𝐺.

Proof

All subgroups of 𝐴 are cn-subgroups of 𝐺 so that there is a 𝐺- series

like in Proposition 14. If A0≠X, then no element of 𝐺 can act on X/A0 as the inversion map as 𝐺 is locally nilpotent. Thus the non-periodic factor X/A0 of 𝒮 lies in the center of G/A0. On the other hand, chief factors of locally nilpotent groups are central (see [18, Part 1, Corollary 1, p. 154]) and 𝐺 acts as a group of power automorphisms on infinite factors of 𝒮, again by Proposition 14. Thus also each periodic factor of 𝒮 is an hypercentral factor of 𝐺. Hence 𝐴 is contained in the hypercenter of 𝐺. ∎

Let 𝐺 be a subsoluble locally nilpotent periodic T⁢[∗]-group whose primary components are nilpotent-by-finite. Then 𝐺 is a nilpotent-by-finite CN-group.

Proof

Note that G=Drp⁡Gp, where each Gp is the Sylow 𝑝-subgroup of 𝐺. According to Proposition 11, the set 𝜋 of odd primes for which Gp is not a 𝑇-group is finite. On the other hand, if the subsoluble group Gp has property 𝑇, then it is abelian since p≠2 (see [17, Lemma 4.2.1]). Thus both Gπ and Gπ′ are nilpotent-by-finite; hence G=Gπ×Gπ′ is itself nilpotent-by-finite. In particular, 𝐺 is a CN-group by Lemma 9. ∎

Let 𝐺 be a Baer T⁢[∗]-group; then 𝐺 is nilpotent.

Proof

Let 𝑥 be any element of infinite order of 𝐺. Then ⟨x⟩ is a subnormal subgroup of 𝐺 so that ⟨x⟩ is a cn-subgroup and hence cf by Proposition 1 (3); thus ⟨x⟩G≠{1}, and so ⟨x⟩G≤Z⁢(G) since 𝐺 is locally nilpotent. Therefore, G/Z⁢(G) is periodic, and hence we can assume that 𝐺 is periodic. Since a nilpotent-by-finite Baer group is nilpotent, Lemma 19 allow us to assume that 𝐺 is a 𝑝-group for some prime number 𝑝.

If 𝐺 is a Chernikov group (that is a finite extension of an abelian group satisfying the minimal condition on subgroups), then 𝐺 is nilpotent as an easy consequence of [18, Lemma 3.13]. Then assume that 𝐺 is not a Chernikov group so that 𝐺 contains a subnormal abelian infinite subgroup 𝐴 (see [18, Part 1, p. 176]). Since 𝐴 is a cn-subgroup, there exists a normal subgroup 𝑁 of 𝐺 such that A∩N has finite index in both 𝐴 and 𝑁. By the well-known fact that any abelian-by-finite group has a characteristic abelian subgroup of finite index, it follows that 𝑁 contains a 𝐺-invariant non-trivial subgroup N∗ of finite index. Replacing 𝐴 with N∗, it can be supposed that 𝐺 has a normal abelian subgroup A≠{1}; then we may apply Lemma 18 to deduce that 𝐴 is contained in the hypercenter of 𝐺. In particular, Z⁢(G) is not trivial. Since any factor of 𝐺 is a Baer 𝑝-group in T⁢[∗], it follows that any non-trivial quotient of 𝐺 has non-trivial center. Hence 𝐺 is hypercentral and we may apply Lemma 17 to deduce that 𝐺 is a nilpotent-by-finite Baer group. Therefore, 𝐺 is nilpotent.∎

Let 𝐺 be a subsoluble T⁢[∗]-group. Then 𝐺 is nilpotent-by-finite-by-abelian (hence soluble and metanilpotent-by-finite) and G′ is finite-by-abelian-by-finite. In particular, 𝐺 is soluble.

Proof

First note that 𝐺 is hyperabelian since 𝐺 is subsoluble and Proposition 20 yields that the Baer radical of a T⁢[∗]-group is nilpotent.

Let 𝐹 be the Fitting subgroup of 𝐺. Then again Proposition 20 yields that 𝐹 is a nilpotent T⁢[∗]-group so that 𝐹 is a CN-group, and hence it contains a finite-by-abelian subgroup 𝐴 of finite index (see [5]); moreover, 𝐴 can be chosen to be a normal subgroup of 𝐺.

Assume first that 𝐴 is abelian. By Proposition 14, there is a 𝐺-series

on which 𝐺 acts as a group of power automorphisms on infinite factors. Let

be the stabilizer of the series 𝒮. By Proposition 14 (ii), the factor G/C is finite-by-abelian. Moreover, A≤Z5⁢(C). Let now N/A be the Fitting subgroup of C/A. Then N/A is nilpotent by Proposition 20 so that 𝑁 is nilpotent, and hence 𝑁 is contained in 𝐹. Thus N/A is finite, and hence C/A is finite (since the Fitting subgroup of a hyperabelian group contains its centralizer; see [18, Part 1, Lemma 2.17]). It follows that G/A is finite-by-abelian.

In the general case, A′ is finite, and the previous argument applies to G/A′ and proves that G/A is finite-by-abelian. Thus G′⁢A/A is finite so that G′ is finite-by-abelian-by-finite, and being A≤F and 𝐹 nilpotent, the group 𝐺 is nilpotent-by-finite-by-abelian. ∎

Let 𝐺 be a subsoluble T⁢[∗]-group. Then 𝐺 is finite-by-metabelian.

Proof

By Proposition 21, the group 𝐺 is soluble and G′ contains a subgroup 𝐴 of finite index such that A′ is finite; in particular, it can be supposed that 𝐴 is a normal subgroup of 𝐺 (see [15]). Replacing 𝐺 by G/A′, it can be supposed even that 𝐴 is abelian. Then there is a 𝐺-series

as in Proposition 14. We can assume A0={1} as A0 is finite. Since G′ acts trivially on the factor A1/X, we have that [G′,A1]≤X is torsion-free. On the other hand, [G′,A1]≤G′′ is periodic by Proposition 16. Thus [G′,A1]={1} and G′ is central-by-finite; hence G′′ is finite. ∎

Let 𝐺 be a subsoluble T⁢[∗]-group. Then

1. either G/G′ or G′ is periodic;

2. each finitely generated subgroup of 𝐺 is a CF-group.

Proof

To prove (1), we can argue as in the proof of [11, Lemma 3.7] as 𝐺 acts on its abelian normal torsion-free sections by inducing either the inversion or the identity map by Proposition 14 (iv).

To prove (2), let 𝐸 be a finitely generated subgroup of 𝐺. Note that, by Proposition 3, in the sequel, one can factor out any finite normal subgroup of 𝐺. Thus, by Proposition 14 and Theorem 22, we may assume that G′ is abelian and each subgroup of G′ is a cf-subgroup of 𝐺.

If G/G′ is periodic and 𝐻 is any subgroup of 𝐸, then E⁢G′/G′ is finite. Hence H⁢G′/G′ is finite, and hence H∩G′ has finite index in 𝐻. On the other hand, H∩G′ is a cf-subgroup of 𝐺 by the above. Therefore, 𝐻 is a cf-subgroup of 𝐺, and 𝐸 is a CF-group.

In the case when G′ is periodic, we may apply [8, Lemma 3.3] to the group 𝐸 and its normal (abelian) periodic subgroup N=G′∩E and deduce that 𝑁 is finite. Thus E′≤N is finite. Since 𝐸 is finitely generated, it follows that 𝐸 is central-by-finite. Hence 𝐸 is a CF-group. ∎

Let 𝐺 be a soluble locally nilpotent T⁢[∗]-group. Then 𝐺 is hypercentral.

Proof

Let 𝑇 be the subgroup consisting of all elements of finite order of 𝐺; then Proposition 16 allow us to suppose that 𝑇 is not trivial since a dihedral group on a torsion-free group is not locally nilpotent. Since, in a locally nilpotent group, any finite normal subgroups is contained in some term of the upper central series and since the hypotheses are inherited by quotients, it is enough to prove that T∩Z⁢(G)≠{1}. To this aim, apply Lemma 18 to the last term of the derived series of 𝑇. ∎

Let 𝐺 be a periodic soluble locally nilpotent T⁢[∗]-group; then 𝐺 is a CN-group with a nilpotent subgroup of index at most 2.

Proof

By Proposition 24, the group 𝐺 is hypercentral so that Lemma 17 yields that each primary component of 𝐺 is nilpotent-by-finite, and hence Lemma 19 yields that 𝐺 is nilpotent-by-finite. Thus 𝐺 is a CN-group by Lemma 9 so that 𝐺 contains a finite normal subgroup 𝐹 such that G/F is a CF-group (see [5]). Since 𝐺 is locally nilpotent, there exists a positive integer 𝑛 such that F≤Zn⁢(G). Hence G/Zn⁢(G) is likewise a periodic soluble locally nilpotent CF-group, and so it contains a nilpotent subgroup of index at most 2 (see [11, Theorem 3.10]). Therefore, 𝐺 itself contains a nilpotent subgroup of index at most 2. ∎