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

Abstract

A subgroup X of a group G is called transitively normal if X is normal in any subgroup Y of G such that $X\le Y$ 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 $\overline{T}$-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 ${\overline{T}}$-group.

1 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] 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 ${\overline{T}}$-group if all its subgroups have the T-property. It is known that every finite soluble T-group is a ${\overline{T}}$-group and that all finite groups with the ${\overline{T}}$-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 $X\le Y$ 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 $\langle X,X^g\rangle$. It is straightforward to show that pronormal subgroups are transitively normal, while there exists a group of order $2\cdot 3\cdot 7^2$ 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 ${\overline{T}}$-property if and only if all its subgroups are transitively normal; since there exist soluble ${\overline{T}}$-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 $\overline{T}_0$ 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 ${\overline{T}}_0$, 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 $G/\Phi (G)$ has the T-property.

2 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 $X\le Y$; 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 ${\overline{T}}_0$ is closed with respect to subgroups and homomorphic images.

Lemma 2.1

Let G be a group and let $(X_i)_{i\in I}$ be a collection of transitively normal subgroups of G. If $[X_i,X_j]\le X_i\cap X_j$ for all $i,j\in I$, then also the subgroup $X=\langle X_i\;|\; i\in I\rangle$ is transitively normal in G.

Proof

Let Y be any subgroup of G such that $X\le Y$ and X is subnormal in Y. For each $i\in I$, 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. $\square$

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 $\langle X,g\rangle$ is nilpotent, so that in particular X is subnormal in $\langle X,g\rangle$. Then $X^g=X$, and hence X is normal in G. $\square$

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 $\overline{T}$-property, the section $G''/M$ must be infinite. As a consequence, we have the following elementary result.

Lemma 2.4

Let G be a ${\overline{T}}$ -group with no infinite simple sections. Then G is metabelian.

Of course, every ${\overline{T}}_0$-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. $\square$

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 $G^{(n)}\ne G^{(n+1)}$ for each non-negative integer n. Then $G^{(n)}/G^{(n+3)}$ is a soluble group of derived length 3 and so it contains a subnormal non-normal subgroup $X_n/G^{(n+3)}$. Then

$$\begin{aligned} X_0>X_3>\cdots>X_{3n}>\cdots \end{aligned}$$

is an infinite descending chain of subnormal non-normal subgroup of G, contradicting the hypothesis. $\square$

Lemma 2.7

Let G be a ${\overline{T}}_0$ -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. $\square$

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 $\overline{T}_0$-group. Then G contains a subgroup X of finite index whose finite homomorphic images have the ${\overline{T}}$-property. Moreover, $X/X''$ is a T-group and $X''=X^{(3)}$.

Proof

It follows from the ${\overline{T}}_0$-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 ${\overline{T}}$-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. $\square$

Recall that a group class ${\mathfrak {X}}$ is said to be countably recognizable if, whenever all countable subgroups of a group G belong to ${\mathfrak {X}}$, then G itself is an ${\mathfrak {X}}$-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

$$\begin{aligned} X_1\cap C_G(X_1)=\zeta (X_1)=\{1\}. \end{aligned}$$

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

$$\begin{aligned} \langle X_1,X_2\rangle \cap C_G\bigl (\langle X_1,X_2\rangle \bigr )=\{1\} \end{aligned}$$

and $|G:C_G\bigl (\langle X_1,X_2\rangle \bigr )|$ is finite, so that the iteration of the argument allows us to construct a collection $(X_n)_{n\in {\mathbb {N}}}$ of finite non-trivial normal subgroups of G such that

$$\begin{aligned} X=\langle X_n\;|\; n\in {\mathbb {N}}\rangle =\mathop \mathrm{Dr}_{n\in {\mathbb {N}}}X_n. \end{aligned}$$

As the normal subgroup X of G is countably infinite, it has an abelian non-trivial normal subgroup A. Then the normal closure $A^G\le X$ 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. $\square$

3 Periodic ${\overline{T}}_0$-groups

The first result of this section shows that ${\overline{T}}_0$-groups with no infinite simple sections are close to be hyperabelian.

Lemma 3.1

Let G be a ${\overline{T}}_0$ -group with no infinite simple sections. Then G contains a hyperabelian subgroup of finite index.

Proof

As the class ${\overline{T}}_0$ 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 ${\mathfrak {L}}$ be any chain of infinite normal subgroups of G such that

$$\begin{aligned} L=\bigcap _{X\in {{\mathfrak {L}}}}X \end{aligned}$$

is finite, and put

$$\begin{aligned} G\setminus L=\{ x_n\;|\; n\in {\mathbb {N}}\}. \end{aligned}$$

Since ${\mathfrak {L}}$ cannot have a smallest member, there are elements $X_1,\ldots ,X_n,\ldots$ of ${\mathfrak {L}}$ such that

$$\begin{aligned} X_n\cap \{x_1,\ldots ,x_n\}=\emptyset \quad \text {and}\quad X_n>X_{n+1} \end{aligned}$$

for all n, and clearly

$$\begin{aligned} \bigcap _{n\in {\mathbb {N}}}X_n=L. \end{aligned}$$

It follows now from the ${\overline{T}}_0$-property that there exists a positive integer m such that for every $n\ge m$ the group $X_n/X_{n+1}$ has the ${\overline{T}}$-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 G-invariant subgroups are finite. Since all chief factors of G are finite by Lemma 2.5, it follows that M is generated by its proper G-invariant subgroups. Let N be any proper G-invariant subgroup of M. Since N is finite, the centralizer $C_M(N)$ has finite index in M and so $N\le \zeta (M)$ because M has no proper G-invariant subgroups of finite index. Therefore M is abelian and this last contradiction completes the proof. $\square$

We are now ready to prove the main theorem in the periodic case.

Theorem 3.2

Let G be a periodic $\overline{T}_0$-group with no infinite simple sections. Then G is either Černikov or a $\overline{T}$-group.

Proof

By Lemma 3.1 the group G contains a hyperabelian normal subgroup K of finite 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 $C_K(F)\le F$.

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

$$\begin{aligned} A=\mathop \mathrm{Dr}_{n\in {\mathbb {N}}}A_n, \end{aligned}$$

where each $A_n$ has prime order. For every positive integer k, put

$$\begin{aligned} B_k=\mathop \mathrm{Dr}_{n\ge k}A_n, \end{aligned}$$

so that

$$\begin{aligned} B_1>B_2>\cdots>B_n>B_{n+1}>\cdots \end{aligned}$$

and there exists a positive integer m such that $B_n$ is transitively normal in G for all $n\ge m$. 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 $n\ge m$.

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 $n\ge m$. Moreover, there is an integer $r\ge m$ such that $XB_n$ is transitively normal in G, and so normal in $EB_n$, for each $n\ge r$. On the other hand, we may obviously choose an integer $s\ge r$ such that $E\cap B_s=\{1\}$ and so $X=XB_s\cap E$ is normal in E. Therefore all finite subgroups of G have the T-property and hence G is a $\overline{T}$-group. $\square$

4 Non-periodic ${\overline{T}}_0$-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 ${\overline{T}}$-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 $\overline{T}_0$-property.

Lemma 4.1

Let G be a ${\overline{T}}_0$-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 $\langle a^{p^h}\rangle$ and $\langle a^{q^k}\rangle$ are transitively normal in G. Then $\langle a\rangle =\langle a^{p^h},a^{q^k}\rangle$ 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. $\square$

Lemma 4.2

Let G be a non-periodic locally nilpotent ${\overline{T}}_0$ -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. $\square$

The behaviour of the Hirsch-Plotkin radical plays an important role in the study of non-periodic ${\overline{T}}_0$-groups, as the following lemma shows.

Lemma 4.3

Let G be a ${\overline{T}}_0$-group whose Hirsch-Plotkin radical H is not periodic. Then H is contained in the centre $\zeta (G)$ of G.

Proof

Assume for a contradiction that the statement is false, so that there exists an element g of G such that $[H,g]\ne \{1\}$. Then $\langle g,H\rangle$ is a counterexample to the statement and hence we may suppose without loss of generality that $G=\langle g,H\rangle$. 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 $\langle b\rangle$ is obviously characteristic in $\langle a,b\rangle$ and $\langle a,b\rangle$ is normal in G by Lemma 4.1, and so $\langle b\rangle$ 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 $x\in H$. Since $g^2\in H$, we have $g^4=1$ and hence $\langle g\rangle \cap \langle a\rangle =\{1\}$. It follows that

$$\begin{aligned} \langle g,a\rangle>\langle g,a^2\rangle>\langle g,a^4\rangle>\cdots>\langle g,a^{2^n}\rangle>\langle g,a^{2^{n+1}}\rangle >\cdots \end{aligned}$$

is a strictly descending chain of subgroups, and so there exists a positive integer k such that $\langle g,a^{2^n}\rangle$ is a transitively normal subgroup of G for each $n\ge k$. On the other hand, every $\langle g,a\rangle /\langle a^{2^n}\rangle$ is a finite 2-group, so that $\langle g,a^{2^n}\rangle$ is subnormal in $\langle g,a\rangle$ and hence $\langle g,a^{2^n}\rangle$ is normal in $\langle g,a\rangle$ for all $n\ge k$. Therefore also

$$\begin{aligned} \langle g\rangle =\bigcap _{n\ge k}\langle g,a^{2^n}\rangle \end{aligned}$$

is a normal subgroup of $\langle g,a\rangle$, whence $[g,a]=1$. This contradiction proves the statement. $\square$

Our final result completes the proof of the main theorem.

Theorem 4.4

Let G be a non-periodic ${\overline{T}}_0$-group which has no infinite simple sections. Then G is abelian.

Proof

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

$$\begin{aligned} G=G_0>G_1>G_2>\cdots>G_n>G_{n+1}>\cdots \end{aligned}$$

such that the index $|G:G_n|$ is finite for each non-negative integer n. It follows from the ${\overline{T}}_0$-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 $n\ge k$. In particular, $G_n/G_{n+1}$ is a finite $\overline{T}$-group and so it is soluble. Thus $G_k/G_n$ is a finite soluble group for each $n\ge k$.

As $G_k$ is finitely generated, Lemma 2.8 yields that it contains a subgroup X of finite index whose finite homomorphic images are ${\overline{T}}$-groups, $X/X''$ has the T-property and $X''=X^{(3)}$. Then $XG_n/G_n$ is metabelian and hence $X''\le G_n$ for all $n\ge k$. Since X is finitely generated, the infinite metabelian ${\overline{T}}$-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 $X'\ne \{1\}$. 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 ${\overline{T}}_0$-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 $G/\zeta (G)$ is finite.

Among all finitely generated counterexamples choose one G for which the index $|G:\zeta (G)|$ is smallest possible. Then every proper subgroup of G containing $\zeta (G)$ is abelian, so that all proper subgroups of the finite group $G/\zeta (G)$ are abelian; in particular, $G/\zeta (G)$ is soluble and hence G itself is a soluble group. On the other hand, $\zeta (G)$ coincides with the Hirsch-Plotkin radical of G by Lemma 4.3, and so $G=C_G\bigl (\zeta (G)\bigr )=\zeta (G)$, which is of course impossible. This contradiction completes the proof. $\square$