Groups satisfying the minimal condition on subgroups which are not transitively normal

A subgroup X of a group G is called transitively normal if X is normal in any subgroup Y of G such that X≤Y\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$X\le Y$$\end{document} and X is subnormal in Y. Thus all subgroups of a group G are transitively normal if and only if normality is a transitive relation in every subgroup of G (i.e. G is a T¯\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\overline{T}$$\end{document}-group). It is proved that a group G with no infinite simple sections satisfies the minimal condition on subgroups that are not transitively normal if and only if either G is Černikov or a T¯\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\overline{T}}$$\end{document}-group.


Introduction
A group G is said to be a T-group (or to have the T-property) if normality in G is a transitive relation, i.e. if all subnormal subgroups of G are normal. The structure of soluble T-groups was described by W. Gaschütz [4] in the finite case and by D.J.S. Robinson [10] The first and the third authors are supported by GNSAGA (INdAM) and are members of AGTA -Advances in Group Theory and Applications (www.advgrouptheory.com). The second author is supported by National Research Foundation of Ukraine (Grant n. 2020.02/0066). Part of this work was carried out within the "VALERE: VAnviteLli pEr la RicErca" project. for arbitrary groups. In particular, it was proved that all soluble groups with the T-property are metabelian and locally supersoluble, and that a finitely generated soluble T-group either is finite or abelian. Obviously, any simple group has the T-property, so that the class of T-groups is not subgroup closed; a group is called a T-group if all its subgroups have the T-property. It is known that every finite soluble T-group is a T-group and that all finite groups with the T-property are soluble.
A subgroup X of a group G is said to be transitively normal if X is normal in any subgroup Y of G such that X ≤ Y and X is subnormal in Y (see [8]). Obviously, an ascendant subgroup which is transitively normal is normal and all self-normalizing subgroups of an arbitrary group are transitively normal. Transitively normal subgroups may be considered as the most natural generalization of the so-called pronormal subgroups: recall here that a subgroup X of a group G is pronormal if for each element g of G the subgroups X and X g are conjugate in ⟨X, X g ⟩ . It is straightforward to show that pronormal subgroups are transitively normal, while there exists a group of order 2 ⋅ 3 ⋅ 7 2 containing a subgroup of order 21 which is transitively normal but not pronormal (see for instance [3] and [7], where transitively normal subgroups were considered under different denominations).
Of course, a group G has the T-property if and only if all its subgroups are transitively normal; since there exist soluble T-groups containing non-pronormal subgroups, it follows that even a soluble group with only transitively normal subgroups may contain subgroups which are not pronormal. The aim of this paper is to give a further contribution to this topic of the theory by studying groups which have many transitively normal subgroups. More precisely, we investigate here the class T 0 consisting of all groups satisfying min − ntn , the minimal condition on subgroups which are not transitively normal. In order to avoid Tarski groups (i.e. infinite simple groups whose proper non-trivial subgroups have prime order) and similar pathological groups belonging to the class T 0 , in our main result the attention is restricted to a suitable class of generalized soluble groups.
Theorem Let G be a group satisfying the minimal condition on subgroups which are not transitively normal. If G has no infinite simple sections, then either G is Černikov or all subgroups of G are transitively normal. In particular, if G is not periodic, then it is abelian.
The proof of the theorem will be treated separately in the periodic and in the non-periodic case.
Our notation is mostly standard and can be found in [11]. Notice that the symbol T 0 was used in [13], with a completely different meaning, to denote the class of groups G whose Frattini factor group G∕Φ(G) has the T-property.

Some preliminaries
If X is any transitively normal subgroup of a group G, it is clear that X is transitively normal in every subgroup Y of G such that X ≤ Y ; moreover, if N is a normal subgroup of G, a subgroup X/N of G/N is transitively normal if and only if X is transitively normal in G. It follows that the class T 0 is closed with respect to subgroups and homomorphic images. Lemma 2.1 Let G be a group and let (X i ) i∈I be a collection of transitively normal sub- Proof Let Y be any subgroup of G such that X ≤ Y and X is subnormal in Y. For each i ∈ I , the subgroup X i is obviously normal in X and so subnormal in Y. Thus every X i is normal in Y, so that X itself is normal in Y. Therefore X is a transitively normal subgroup of G. ◻

Corollary 2.2 Let G be a group, and let A be an abelian subgroup which is generated by transitively normal subgroups of G. Then A itself is transitively normal in G.
Our next easy result shows in particular that if all cyclic subgroups of a locally nilpotent group G are transitively normal, then G is a Dedekind group.

Lemma 2.3 Let G be a locally nilpotent group, and let X be a finitely generated transitively normal subgroup of G. Then X is normal in G.
Proof If g is any element of G, the subgroup ⟨X, g⟩ is nilpotent, so that in particular X is subnormal in ⟨X, g⟩ . Then X g = X , and hence X is normal in G. ◻ Let G be a finitely generated T-group which is not soluble. Since G∕G �� is either finite or abelian, the subgroup G ′′ contains a maximal G-invariant subgroup M and G �� ∕M is a simple non-abelian group. In particular, if G has the T-property, the section G �� ∕M must be infinite. As a consequence, we have the following elementary result.

Lemma 2.4 Let G be a T-group with no infinite simple sections. Then G is metabelian.
Of course, every T 0 -group satisfies the minimal condition on subnormal non-normal subgroups and our next two lemmas describe the behaviour of chief factors and derived series in groups with such property.

Lemma 2.5 Let G be a group with no infinite simple sections. If G satisfies the minimal condition on subnormal non-normal subgroups, then every chief factor of G is finite.
Proof Let X/Y be a chief factor of G and let Z be any normal subgroup of X properly containing Y. Then Z is a subnormal non-normal subgroup of G and so X/Y satisfies the minimal condition on normal subgroups. It follows that X/Y is the direct product of finitely many simple subgroups (see [11] Part 1, Corollary 2 to Lemma 5.23) and hence it is finite. ◻

Lemma 2.6 Let G be a group satisfying the minimal condition on subnormal non-normal
subgroups. Then the derived series of G stops after finitely many steps.
Proof Assume for a contradiction that G (n) ≠ G (n+1) for each non-negative integer n. Then is a soluble group of derived length 3 and so it contains a subnormal non-normal subgroup X n ∕G (n+3) . Then is an infinite descending chain of subnormal non-normal subgroup of G, contradicting the hypothesis. ◻

Lemma 2.7 Let G be a T 0 -group. Then the Fitting subgroup F of G is hypercentral.
Proof Obviously, all nilpotent normal subgroups of G satisfying the minimal condition on subgroups are contained in the hypercentre of F, so that without loss of generality we may suppose that G contains at least one nilpotent normal subgroup K which is not Černikov.
Let N be any nilpotent normal subgroup of G. As the product KN is likewise nilpotent, it satisfies the minimal condition on non-normal subgroups and hence it is a Dedekind group by a result of Černikov (see [2] or [9]). It follows that in this case F itself is a Dedekind group and so the statement is proved. ◻ Since every finitely generated non-trivial group contains a maximal normal subgroups, it is clear that all groups with no infinite simple sections are locally graded, i.e. all their finitely generated non-trivial subgroups contain proper subgroups of finite index.

Lemma 2.8
Let G be a finitely generated locally graded T 0 -group. Then G contains a subgroup X of finite index whose finite homomorphic images have the T-property. Moreover, X∕X �� is a T-group and X �� = X (3) .
Proof It follows from the T 0 -property that G contains a subgroup X of finite index such that all proper subgroups of finite index of X are transitively normal in G, and hence all finite homomorphic images of X are T-groups. Clearly, X∕X (3) is a finitely generated soluble group whose finite homomorphic images have the T-property, so that also X∕X (3) is a T-group (see [12], Theorem 2), and hence it is metabelian. Therefore X �� = X (3) and X∕X �� has the T-property. ◻ Recall that a group class is said to be countably recognizable if, whenever all countable subgroups of a group G belong to , then G itself is an -group. Countably recognizable group classes have been introduced by Baer [1] more than fifty years ago and recently many relevant group classes have been proved to be detectable by the behaviour of countable groups (see for instance [5] and [6]). In particular, Baer proved that hyperabelian groups form a countably recognizable class; for our purposes we need to show that also the class of hyperabelian-by-finite groups is countably recognizable.

Theorem 2.9 The class of groups containing a hyperabelian subgroup of finite index is countably recognizable.
Proof Let G be an uncountable group whose countable subgroups have a hyperabelian subgroup of finite index and assume that G has no abelian non-trivial normal subgroups. Since the class of hyper-(abelian or finite) groups is countably recognizable (see [5], Lemma 4.3), G contains a finite non-trivial normal subgroup X 1 and Obviously, the index |G ∶ C G (X 1 )| is finite and so C G (X 1 ) contains a finite non-trivial G-invariant subgroup X 2 . Again we have and �G ∶ C G � ⟨X 1 , X 2 ⟩ � � is finite, so that the iteration of the argument allows us to construct a collection (X n ) n∈ℕ of finite non-trivial normal subgroups of G such that As the normal subgroup X of G is countably infinite, it has an abelian non-trivial normal subgroup A. Then the normal closure A G ≤ X is hyperabelian and hence it contains an abelian non-trivial G-invariant subgroup. This contradiction shows that every uncountable homomorphic image of G has an abelian non-trivial normal subgroup. On the other hand, if G/N is a countably infinite homomorphic image of G, we have G = NL for a suitable countable subgroup L of G, so that G/N has a hyperabelian subgroup of finite index and hence also an abelian non-trivial normal subgroup. Therefore all infinite homomorphic images of G have an abelian non-trivial normal subgroup and so G contains a hyperabelian subgroup of finite index and the statement is proved. ◻

Periodic T 0 -groups
The first result of this section shows that T 0 -groups with no infinite simple sections are close to be hyperabelian.

Lemma 3.1 Let G be a T 0 -group with no infinite simple sections. Then G contains a hyperabelian subgroup of finite index.
Proof As the class T 0 is subgroup closed, by Theorem 2.9 it is enough to prove the statement when G is countable. Assume for a contradiction that G is infinite but has no abelian non-trivial normal subgroups. Let be any chain of infinite normal subgroups of G such that is finite, and put Since cannot have a smallest member, there are elements X 1 , … , X n , … of such that for all n, and clearly It follows now from the T 0 -property that there exists a positive integer m such that for every n ≥ m the group X n ∕X n+1 has the T-property and hence is metabelian by Lemma 2.4. Then X m ∕L is soluble by Lemma 2.6 and so X m is soluble-by-finite. This contradiction shows that the intersection of any chain of infinite normal subgroups of G is infinite and so an application of Zorn's Lemma yields that G contains an infinite normal subgroup M whose proper index. Let F be the Fitting subgroup of K and assume first that F is Černikov. Then also K∕C K (F) is a Černikov group (see [11] Part 1, Theorem 3.29) and hence G is Černikov because C K (F) ≤ F. Suppose now that F is not a Černikov group. Then F does not satisfy the minimal condition on abelian subgroups and so it contains a subgroup of the form where each A n has prime order. For every positive integer k, put so that and there exists a positive integer m such that B n is transitively normal in G for all n ≥ m . Since F is hypercentral by Lemma 2.7, all its subgroups are ascendant in G and so B n is normal in G for every n ≥ m.
Consider an arbitrary finite subgroup E of G and let X be any subnormal subgroup of E. Then the product XB n is a subnormal subgroup of EB n for each n ≥ m . Moreover, there is an integer r ≥ m such that XB n is transitively normal in G, and so normal in EB n , for each n ≥ r . On the other hand, we may obviously choose an integer s ≥ r such that E ∩ B s = {1} and so X = XB s ∩ E is normal in E. Therefore all finite subgroups of G have the T-property and hence G is a T-group. ◻

Non-periodic T 0 -groups
As we mentioned in the introduction, a finitely generated soluble T-group is either finite or abelian and hence it follows from Lemma 2.4 that non-periodic T-groups with no infinite simple sections are abelian. Thus, in order to complete the proof of our main theorem, we need to show that such a conclusion holds in the case of groups with the T 0 -property.

Lemma 4.1 Let G be a T 0 -group, and let A be an abelian non-periodic subgroup of G.
Then A is transitively normal in G.
Proof Let a be any element of infinite order of A, and let p and q be different prime numbers. As G satisfies min − ntn , there exist positive integers h and k such that both subgroups ⟨a p h ⟩ and ⟨a q k ⟩ are transitively normal in G. Then ⟨a⟩ = ⟨a p h , a q k ⟩ is likewise transitively normal in G by Corollary 2.2. Since A is generated by its elements of infinite order, a further application of Corollary 2.2 yields that also A is transitively normal in G. ◻ Lemma 4.2 Let G be a non-periodic locally nilpotent T 0 -group. Then G is abelian.
Proof By Lemma 4.1 every infinite cyclic subgroup of G is transitively normal, and so even normal in G by Lemma 2.3. As G is locally nilpotent, it follows that all elements of infinite order of G belong to the centre, and hence G is abelian because it is generated by elements of infinite order. ◻ The behaviour of the Hirsch-Plotkin radical plays an important role in the study of non-periodic T 0 -groups, as the following lemma shows.

Lemma 4.3 Let G be a T 0 -group whose Hirsch-Plotkin radical H is not periodic. Then H is contained in the centre (G) of G.
Proof Assume for a contradiction that the statement is false, so that there exists an ele- . Then ⟨g, H⟩ is a counterexample to the statement and hence we may suppose without loss of generality that G = ⟨g, H⟩ . The subgroup H is abelian by Lemma 4.2, so that in particular G is soluble and C G (H) = H . Of course, all subgroups of H are subnormal in G and it follows from Lemma 4.1 that every infinite cyclic subgroup of H is normal in G. Let a be an element of infinite order of H. If b is any element of finite order of H, the subgroup ⟨b⟩ is obviously characteristic in ⟨a, b⟩ and ⟨a, b⟩ is normal in G by Lemma 4.1, and so ⟨b⟩ is normal in G. Therefore all subgroups of H are normal in G, so that G/H has order 2 and x g = x −1 for all x ∈ H . Since g 2 ∈ H , we have g 4 = 1 and hence ⟨g⟩ ∩ ⟨a⟩ = {1} . It follows that is a strictly descending chain of subgroups, and so there exists a positive integer k such that ⟨g, a 2 n ⟩ is a transitively normal subgroup of G for each n ≥ k . On the other hand, every ⟨g, a⟩∕⟨a 2 n ⟩ is a finite 2-group, so that ⟨g, a 2 n ⟩ is subnormal in ⟨g, a⟩ and hence ⟨g, a 2 n ⟩ is normal in ⟨g, a⟩ for all n ≥ k . Therefore also is a normal subgroup of ⟨g, a⟩ , whence [g, a] = 1 . This contradiction proves the statement. ◻ Our final result completes the proof of the main theorem.

Theorem 4.4 Let G be a non-periodic T 0 -group which has no infinite simple sections.
Then G is abelian.
Assume for a contradiction that the statement is false. Clearly, we may suppose that G is finitely generated, so that there exists in G an infinite descending chain of normal subgroups such that the index |G ∶ G n | is finite for each non-negative integer n. It follows from the T 0 -property that there is a positive integer k such that each subgroup of G n ∕G n+1 is transitively normal in G∕G n+1 for all n ≥ k . In particular, G n ∕G n+1 is a finite T-group and so it is soluble. Thus G k ∕G n is a finite soluble group for each n ≥ k.
As G k is finitely generated, Lemma 2.8 yields that it contains a subgroup X of finite index whose finite homomorphic images are T-groups, X∕X �� has the T-property and X �� = X (3) . Then XG n ∕G n is metabelian and hence X ′′ ≤ G n for all n ≥ k . Since X is finitely generated, the infinite metabelian T-group X∕X �� cannot be periodic, so that it is abelian and X � = X �� is the last term of the derived series of X. Assume that X � ≠ {1} . Clearly, X ′ is the normal closure in X of a finite subset and hence it contains a maximal proper X-invariant subgroup Y. Then X � ∕Y is a minimal normal subgroup of X/Y, and the T 0 -property yields that X � ∕Y satisfies the minimal condition on normal subgroups, so that it is the direct product of finitely many simple groups (see [11] Part 1, p.150). On the other hand, G has no infinite simple sections, so that X � ∕Y is finite and hence X/Y is soluble-by-finite and then even soluble. This contradiction shows that X is abelian. Then G is abelian-by-finite, and it follows from Lemma 4.3 that G∕ (G) is finite.
Among all finitely generated counterexamples choose one G for which the index |G ∶ (G)| is smallest possible. Then every proper subgroup of G containing (G) is abelian, so that all proper subgroups of the finite group G∕ (G) are abelian; in particular, G∕ (G) is soluble and hence G itself is a soluble group. On the other hand, (G) coincides with the Hirsch-Plotkin radical of G by Lemma 4.3, and so G = C G (G) = (G) , which is of course impossible. This contradiction completes the proof. ◻ Funding Open access funding provided by Università degli Studi di Napoli Federico II within the CRUI-CARE Agreement.
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