Modular chains in infinite groups

Abstract

The aim of this paper is to investigate the behaviour of projective images of the groups which are finite over a term of their upper central series. In particular, we prove that for any positive integer k, the class of finitely generated groups in which the k-th term of the upper central series has finite index can be described in terms of lattice invariants, and so it is invariant under projectivities. In this context, we also study groups that have only finitely many maximal subgroups which are not permodular.

1 Introduction

One of the main problems in the study of subgroup lattices of groups is the translation of concepts and results of abstract group theory by means of purely lattice-theoretic objects, a crucial role in this context being played by modularity. For instance, a lattice translation of the famous result of Issai Schur [10] on the finiteness of the commutator subgroup of a group which is finite over the centre was obtained in [3] in terms of modularity. Recall here that a lattice $\mathfrak {L}$ is modular if the identity

$$\begin{aligned} (x\vee y)\wedge z=x\vee (y\wedge z) \end{aligned}$$

holds for all elements x, y, z of $\mathfrak {L}$ such that $x\le z$. The subgroup lattice of any abelian group is obviously modular and it is known that finite groups with a modular subgroup lattice are soluble, but unfortunately there exist also infinite simple groups with modular subgroup lattice, like for instance Tarski groups, i.e., infinite simple groups all of whose proper non-trivial subgroups have prime order. For this reason, modularity seems to be inadequate as a lattice translation of commutativity, at least in the case of infinite groups. A lattice $\mathfrak {L}$ is called permodular if it is modular and for all elements a, b of $\mathfrak {L}$ such that $a\le b$, the interval [b/a] is finite whenever it has finite length; of course modularity and permodularity are equivalent for finite lattices and it is known that a group G has a permodular subgroup lattice if and only if it is soluble and its subgroup lattice ${\mathfrak {L}}(G)$ is modular. Thus permodularity works much better than modularity and is naturally involved in the study of projectivities of abelian groups and their generalizations.

Schur’s theorem was generalized by R. Baer [1] by proving that if G is a group whose k-th centre $\zeta _k(G)$ has finite index for some positive integer k, then the corresponding term $\gamma _{k+1}(G)$ of the lower central series of G is finite. A partial converse of Baer’s theorem was obtained by Philip Hall [4], who showed that if G is any group such that $\gamma _{k+1}(G)$ is finite, then the index $|G:\zeta _{2k}(G)|$ is likewise finite, but the consideration of any infinite extraspecial group shows that a precise converse cannot hold in general.

The aim of this paper is to discuss the behaviour of groups which are finite over some term of their upper central series from a lattice point of view.

Let $\mathfrak {L}$ be a lattice with least element 0 and greatest element I. Recall that an element c of $\mathfrak {L}$ is called cyclic if the interval [c/0] is distributive and satisfies the maximal condition, while an element $p\in {\mathfrak {L}}$ is permodularly embedded in $\mathfrak {L}$ if the lattice $[p\vee c/0]$ is permodular for each cyclic element c of $\mathfrak {L}$. A chain

$$\begin{aligned} a_0\le a_1\le \cdots \le a_k \end{aligned}$$

of elements of $\mathfrak {L}$ is called a permodular chain if $a_{i+1}$ is permodularly embedded in $[I/a_i]$ for every non-negative integer $i<k$. We say that a lattice $\mathfrak {L}$ is nilmodular of class c if there exists in $\mathfrak {L}$ a permodular chain of length c from 0 to I and c is the smallest non-negative integer with such a property. The structure of finite groups with a nilmodular subgroup lattice has been described in [9]. Finally, if k is a positive integer, we say that an element $a\in {\mathfrak {L}}$ is k-permodularly embedded in $\mathfrak {L}$ if there is a permodular chain of length k from 0 to a; obviously, 1-permodularly embedded elements are precisely the permodularly embedded ones. Permodular chains were introduced by Kontorovič and Plotkin [6] in order to give a lattice characterization of torsion-free nilpotent groups.

If G is a group, it is known that the cyclic elements of ${\mathfrak {L}}(G)$ are precisely the cyclic subgroups of G, so that in particular the property of being finitely generated can be recognized in the lattice ${\mathfrak {L}}(G)$. Moreover, it is clear that all subgroups of $\zeta _k(G)$ are k-permodularly embedded in ${\mathfrak {L}}(G)$. Since the finiteness of the index of a subgroup may be described in a lattice-theoretic way (see [8]), our first main result is a lattice characterization of a relevant group class and so it proves in particular that such class of groups is invariant under projectivities.

Theorem A

  Let G be a finitely generated group and let k be a positive integer. Then the k-th centre $\zeta _k(G)$ has finite index in G if and only if G contains a k-permodularly embedded subgroup of finite index.

In the above statement, the condition that the group G is finitely generated cannot be omitted, as the following easy example shows. Let p be any prime number, and let be the semidirect product of an infinite abelian group P of exponent p by a cyclic subgroup $\langle x\rangle$ of prime order q such that $a^x=a^k$ for all $a\in P$, where $1<k<p$; then the lattice of subgroups of G is isomorphic to that of an abelian group and in particular, G has a permodular subgroup lattice. Therefore neither the class of central-by-finite groups nor the class of groups with a finite commutator subgroup are invariant under projectivities and hence they cannot be characterized in terms of lattice properties.

Recall that a subgroup P of a group G is called permodular if it satisfies the following conditions:

• (a) $\langle P,X\cap Y\rangle =\langle P,X\rangle \cap Y$ for all subgroups X, Y of G with $P\le Y$;

• (b) $\langle X,P\cap Y\rangle =\langle X,P\rangle \cap Y$ for all subgroups X, Y of G with $X\le Y$;

• (c) for each g in G and for each subgroup X of G such that $P\le X\le \langle P,g\rangle$ and $[\langle P,g\rangle /X]$ is a finite lattice, the index $|\langle P,g\rangle :X|$ is finite.

Then a group G has a permodular subgroup lattice if and only if all its subgroups are permodular. It is also clear that every normal subgroup is permodular and permodular subgroups were introduced by Zacher in [11] as the sharpest lattice translation of normal subgroups. Actually, permodular subgroups occur in most lattice characterizations of group classes which are invariant under projectivities.

It is almost obvious that if G is a group and the index $|G:\zeta _k(G)|$ is finite for some non-negative integer k, then G has only finitely many maximal subgroups that are not normal; our second main result deals with the behaviour of maximal subgroups in groups whose subgroup lattice is similar to that of a group which is finite over some term of its upper central series.

Theorem B

  Let G be a group containing a k-permodularly embedded subgroup of finite index for some positive integer k. Then G has only finitely many maximal subgroups which are not permodular.

Our notation is mostly standard and can be found in [7]. We refer to the monograph [8] for definitions and properties concerning arbitrary lattices and lattices of subgroups; in particular, if X is a subgroup of a group G, the symbol [G/X] denotes the set of all subgroups Y of G such that $X\le Y\le G$.

2 Preliminaries

This section is devoted to the description of properties of permodular subgroups which are necessary in our study of permodular chains. Notice first that in many relevant situations, permodular subgroups are close to being normal. As evidence of this fact, we mention the following two useful results, that were proved by Zacher [11] (see also [8, Lemma 6.2.3 and Lemma 6.2.8]).

Lemma 2.1

Let G be a group and let X be a permodular subgroup of G. If a is any element of infinite order of G such that $\langle a\rangle \cap X=\{1\}$, then $X^a=X$.

Lemma 2.2

Let G be a group and let X be a permodular subgroup of G. If there is a finitely generated subgroup E such that $G=\langle X,E\rangle$, then X has finite index in $X^G$.

One of the classical results on lattices of subgroups is Ore’s theorem stating that a group has a distributive subgroup lattice if and only if it is locally cyclic (see [8, Theorem 1.2.3]). It can be applied to obtain the following information on permodular subgroups.

Lemma 2.3

Let G be a group and let X be a permodular subgroup of G such that the interval [G/X] is an infinite distributive lattice satisfying the maximal condition. Then X is normal in G and G/X is infinite cyclic.

Proof

Since [G/X] satisfies the maximal condition, there exists a finitely generated subgroup E of G such that $G=\langle X,E\rangle$. Then X has finite index in $X^G$ by Lemma 2.2 and hence $G/X^G$ is infinite. Moreover, the subgroup lattice of the factor group $G/X^G$ is distributive and satisfies the maximal condition, so that $G/X^G$ is infinite cyclic. It follows that each element a of the set $G\setminus X^G$ has infinite order and $\langle a\rangle \cap X^G=\{1\}$, whence $X^a=X$ by Lemma 2.1. Therefore X is normal in $G=\langle G\setminus X^G\rangle$. $\square$

Corollary 2.4

Let G be a group and let X be a permodular subgroup of G such that the interval [G/X] is a distributive lattice satisfying the maximal condition. Then there exists an element g of G such that $G=\langle X,g\rangle$.

Proof

The statement is known if the lattice [G/X] is finite (see [9, Proposition 2.4]). On the other hand, if [G/X] is infinite, it follows from Lemma 2.3 that X is normal in G and G/X is cyclic, so that the statement holds also in this case. $\square$

Lemma 2.5

Let G be a group and let X and Y be subgroups of G such that $X\le Y$. If X is permodular in G and Y is permodularly embedded in the lattice [G/X], then $\langle Y,P\rangle$ is permodularly embedded in $\bigl [G/\langle X,P\rangle \bigr ]$ for each permodular subgroup P of G.

Proof

Let C be any cyclic element of the lattice $\bigl [G/\langle X,P\rangle \bigr ]$. Since $\langle X,P\rangle$ is a permodular subgroup of G (see [8, Theorem 6.2.21]), it follows from Corollary 2.4 that C contains an element g such that $C=\langle X,P,g\rangle$. Then

$$\begin{aligned} \bigl [\langle Y,P,C\rangle /\langle X,P\rangle \bigr ]=\bigl [\bigl \langle \langle Y,g\rangle ,\langle X,P\rangle /\langle X,P\rangle \bigr \rangle \bigr ]\simeq \bigl [\langle Y,g\rangle /\langle Y,g\rangle \cap \langle X,P\rangle \bigr ] \end{aligned}$$

and the latter is an interval of $\bigl [\langle Y,g\rangle /X\bigr ]$. Therefore $\bigl [\langle Y,P,C\rangle /\langle X,P\rangle \bigr ]$ is a permodular lattice and hence $\langle Y,P\rangle$ is permodularly embedded in $\bigl [G/\langle X,P\rangle \bigr ]$.

$\square$

Lemma 2.6

Let G be a group and let X and Y be subgroups of G such that $X\le Y$ and X is permodular in G. Then Y is permodularly embedded in [G/X] if and only if $\bigl [\langle Y,g\rangle /X\bigr ]$ is a permodular lattice for each element g of G.

Proof

Suppose first that Y is permodularly embedded in [G/X] and let g be any element of G. As X is permodular in G, we have $\bigl [\langle X,g\rangle /X\bigr ]\simeq \bigl [\langle g\rangle /\langle g\rangle \cap X\bigr ]$ and so the interval $\bigl [\langle X,g\rangle /X\bigr ]$ is distributive and satisfies the maximal condition. Thus $\langle X,g\rangle$ is a cyclic element of [G/X] and hence $\bigl [\langle Y,g\rangle /X\bigr ]$ is a permodular lattice.

Conversely, let C be any cyclic element of the lattice [G/X]. Then [C/X] is distributive and satisfies the maximal condition and so by Corollary 2.4, there exists an element g of G such that $C=\langle X,g\rangle$. Then $\bigl [\langle Y,C\rangle /X\bigr ]=\bigl [\langle Y,g\rangle /X\bigr ]$ is a permodular lattice and hence Y is permodularly embedded in [G/X]. $\square$

Corollary 2.7

Let G be a group and let X, Y be subgroups of G such that $X\le Y$. If X is permodular in G and Y is permodularly embedded in [G/X], then Y is permodular in G. In particular, if Y is permodularly embedded in G, then all its subgroups are permodular in G.

Proof

Since Y is permodularly embedded in [G/X], it follows from Lemma 2.6 that the lattice $\bigl [\langle Y,g\rangle /X\bigr ]$ is permodular for each element g of G and hence the subgroup Y is permodular in $\langle Y,g\rangle$ (see [8, Theorem 2.1.6]). Then Y is also permodular in G (see [11, Corollario B]). $\square$

It follows in particular from Corollary 2.7 that if G is any group and ${{\mathcal {C}}}$ is a permodular chain in ${\mathfrak {L}}(G)$ whose smallest term is a permodular subgroup of G, then all members of ${{\mathcal {C}}}$ are permodular in G. Therefore Lemma 2.5 has the following relevant consequence, which shows in particular that the class of nilmodular groups and the class of groups containing a k-permodularly embedded subgroup of finite index are closed with respect to homomorphic images.

Corollary 2.8

Let G be a group and let $\{1\}=X_0\le X_1\le \cdots \le X_k$ be a permodular chain in ${\mathfrak {L}}(G)$. If N is a normal subgroup of G, then $\{1\}=X_0N/N\le X_1N/N\le \cdots \le X_kN/N$ is a permodular chain in the lattice ${\mathfrak {L}}(G/N)$.

3 Proof of Theorem A

Recall that a group is hypercyclic if it has an ascending normal series with cyclic factors. It is known that all hypercyclic groups are locally supersoluble and in particular, a finite group is hypercyclic if and only if it is supersoluble. A normal subgroup N of a group G is hypercyclically embedded in G if N admits an ascending series with cyclic factors whose members are normal in G, or equivalently if for each proper G-invariant subgroup K of N the group N/K contains a cyclic non-trivial G-invariant subgroup.

Lemma 3.1

Let G be a group and let X be a cyclic permodular subgroup of G. Then the normal closure $X^G$ is hypercyclically embedded in G.

Proof

Obviously, $X_G$ is a cyclic normal subgroup of G and so we may suppose $X_G=\{1\}$. If X is finite, the statement follows directly from [8, Lemma 6.4.10]. Assume now that X is infinite. The arguments of the proofs of Lemma 2.1 and [2, Theorem 2.2] can be used to prove that the elements of finite order of $X^G$ form a subgroup T which is hypercyclically embedded in G and $X^G=XT$, so that also $X^G$ is hypercyclically embedded in G. $\square$

Corollary 3.2

Let G be a group and let N be a normal subgroup of G which is k-permodularly embedded in G for some positive integer k. Then N is hypercyclically embedded in G.

Proof

Let K be any proper G-invariant subgroup of N. Since N is k-permodularly embedded in G, we have that N/K contains a cyclic subgroup X/K which is permodular in G/K and it follows from Lemma 3.1 that $X^GK/K\le N/K$ is hypercyclically embedded in G/K. Therefore N/K contains a cyclic non-trivial G-invariant subgroup and hence N is hypercyclically embedded in G. $\square$

Corollary 3.3

Let G be a nilmodular group. Then G is hypercyclic and so locally supersoluble.

Our next result proves in particular that a torsion-free group is nilpotent of class at most k if and only if it contains a k-permodularly embedded subgroup of finite index and so it must be seen in relation to the result of Kontorovič and Plotkin mentioned in the introduction (see also [8, Theorem 7.2.3]).

Lemma 3.4

Let G be a group containing a k-permodularly embedded subgroup of finite index. Then the elements of finite order of G form a locally finite subgroup T and the factor group G/T is nilpotent of class at most k.

Proof

We prove the first part by induction on k, the statement being obvious if $k=0$, and we may obviously suppose $k>0$. Let $g_1,\ldots ,g_t$ be arbitrary elements of finite order of G. Then $E=\langle g_1,\ldots ,g_t\rangle$ admits a finite permodular chain $\{1\}=Y_0<Y_1<\cdots <Y_k$ such that the index $|E:Y_k|$ is finite. Since $Y_1$ is a permodular subgroup of E, the index $|Y_1^E:Y_1|$ is finite by Lemma 2.2. The induction assumption yields now that $E/Y_1^E$ is finite, so that $Y_1$ has finite index in E. Thus E contains a finite normal subgroup V such that E/V has modular subgroup lattice (see [3]) and so E is periodic. Moreover, the subgroup $Y_1$ is soluble because the lattice ${\mathfrak {L}}(Y_1)$ is permodular (see [8, Theorem 6.4.3]) and hence E is finite. Therefore the set T of all elements of finite order of G is a locally finite subgroup.

Since the hypothesis is inherited by homomorphic images, in order to prove the second part of the statement, we may assume without loss of generality that G is a torsion-free non-trivial group. Let $\{1\}=X_0<X_1<\cdots <X_k$ be a finite permodular chain of G such that the index $|G:X_k|$ is finite. Then $k>0$ and $X_1$ is contained in $\zeta (G)$ (see [8, Lemma 7.2.2]); in particular, if $k=1$, the group G is central-by-finite and so even abelian. The set $K/\zeta (G)$ of all elements of finite order of $G/\zeta (G)$ is a locally finite subgroup by the first part of the statement, so that it follows from Schur’s theorem that also $K'$ is locally finite and hence K is abelian. Let x be any element of K and let m be a positive integer such that $x^m\in \zeta (G)$. For each element g of G, we have $[x,g]^m=[x^m,g]=1$, so that $[x,g]=1$ and $K=\zeta (G)$. It follows that $G/\zeta (G)$ is torsion-free and so by induction on k, it is nilpotent of class at most $k-1$. Therefore G is nilpotent of class at most k. $\square$

Although a direct converse of Baer’s theorem does not hold in general (as mentioned above), a converse is known to hold for finitely generated groups (see for instance [5, Theorem 2.10]).

Lemma 3.5

Let G be a finitely generated group and let k be a positive integer. Then $\zeta _k(G)$ has finite index in G if and only if the subgroup $\gamma _{k+1}(G)$ is finite.

We are now in a position to prove our first main result.

Proof of Theorem A

The condition of the statement is clearly necessary since the subgroup $\zeta _k(G)$ is k-permodularly embedded in G.

Suppose conversely that there exists in G a finite permodular chain

$$\begin{aligned} \{1\}=X_0<X_1<\cdots <X_k \end{aligned}$$

such that the index $|G:X_k|$ is finite. By Lemma 3.5, we have to prove that the subgroup $\gamma _{k+1}(G)$ is finite. If $X_1$ has finite index in G, then G contains a finite normal subgroup N of G such that G/N has modular subgroup lattice (see [3]). Then G/N has a finite commutator subgroup ([8, Lemmas 2.4.8 and 2.4.9]) and hence $\gamma _2(G)=G'$ is likewise finite.

Assume now that the index $|G:X_1|$ is infinite, so that in particular $k>1$. Clearly, the factor group ${{\overline{G}}}=G/X_1^G$ contains a $(k-1)$-permodularly embedded subgroup of finite index and hence by induction on k, we may suppose that $\gamma _k({{\overline{G}}})$ is finite. In particular, ${{\overline{G}}}$ is polycyclic-by-finite and so there exists a finitely generated subgroup E of $X_1^G$ such that $X_1^G=E^G$ (see for instance [7, Part 1, Lemma 1.43]). Since $X_1$ has finite index in $X_1^G$, the group ${{\overline{G}}}$ is infinite and hence it is generated by elements of infinite order. On the other hand, if a is any element of infinite order of G such that $\langle a\rangle \cap X_1^G=\{1\}$, it follows from Lemma 2.1 that all subgroups of $X_1$ are normalized by a because the lattice ${\mathfrak {L}}\bigl (\langle a,X_1\rangle \bigr )$ is permodular. Thus every subgroup of $X_1$ is normal in G, so that $X_1^G=E^G=E$ is finitely generated and hence G is polycyclic-by-finite. On the other hand, the subgroup $\gamma _{k+1}(G)$ is periodic by Lemma 3.4 and so it is finite. The theorem is proved. $\square$

4 Maximal subgroups

This section contains the proof of Theorem B and a characterization of groups containing only finitely many maximal subgroups which are not permodular. We first discuss briefly groups in which all but finitely many maximal subgroups are normal.

If G is any group, we denote by $\delta (G)$ the intersection of all maximal subgroups of G which are not normal, with the convention that $\delta (G)=G$ whenever all maximal subgroups of G are normal. Obviously, $\delta (G)$ is a characteristic subgroup of G which contains both the centre $\zeta (G)$ and the Frattini subgroup $\Phi (G)$.

Lemma 4.1

If G is any group, then $\delta (G)/\Phi (G)=\zeta \bigl (G/\Phi (G)\bigr )$.

Proof

Since every maximal subgroup of G which is normal contains $G'$, we have $\bigl [\delta (G),G\bigr ]\le \delta (G)\cap G'\le \Phi (G)$ and so $\delta (G)/\Phi (G)\le Z/\Phi (G)=\zeta \bigl (G/\Phi (G)\bigr )$. On the other hand, every maximal non-normal subgroup of G obviously contains Z, so that  $Z\le \delta (G)$ and $\delta (G)/\Phi (G)=Z/\Phi (G)$. $\square$

Corollary 4.2

A group G has finitely many maximal non-normal subgroups if and only if the Frattini factor group $G/\Phi (G)$ is central-by-finite.

Proof

Suppose first that G has finitely many maximal non-normal subgroups. Then each of these subgroups has finitely many conjugates and hence has finite index in G. It follows that the index $|G:\delta (G)|$ is finite, so that $G/\Phi (G)$ is central-by-finite by Lemma 4.1. The converse statement is obvious since in this case again Lemma 4.1 yields that the group $G/\delta (G)$ is finite. $\square$

For our purposes, a special role is played by the intersection $\theta (G)$ of all maximal subgroups of a group G which are not permodular (again, we put $\theta (G)=G$ if all maximal subgroups of G are permodular). Clearly, $\theta (G)$ is a characteristic subgroup of G and $\Phi (G)\le \delta (G)\le \theta (G)$. Notice that, if all but finitely many maximal subgroups of the group G are permodular, then each maximal subgroup of G has finite index; it follows that a group G has only finitely maximal subgroups which are not permodular if and only if the factor group $G/\theta (G)$ is finite.

Lemma 4.3

If G is any group, then $\theta (G)/\Phi (G)$ is metabelian.

Proof

Let M be any maximal subgroup of G which is permodular. Then either M is normal in G or the factor group $G/M_G$ has order pq, where p and q are prime numbers; in particular, $G/M_G$ is metabelian and so $\bigl (\theta (G)\bigr )''\le G''\le M$. Therefore $\theta (G)''$ is contained in $\Phi (G)$ and hence $\theta (G)/\Phi (G)$ is metabelian. $\square$

We say that a normal subgroup N of a group G is $\theta$-embedded in G if every abelian complemented chief factor X/Y of G with $X\le N$ has prime order and $G/C_G(X/Y)$ is either trivial or of prime order.

Lemma 4.4

If G is any group, the normal subgroup $\theta (G)$ is $\theta$-embedded in G.

Proof

Let X/Y be an abelian complemented chief factor of G with $X\le \theta (G)$ and let M be a subgroup of G such that $G=MX$ and $M\cap X=Y$. Since X/Y is abelian, we have that M is a maximal subgroup of G. Of course, $\theta (G)$ is not contained in M, so that M is permodular in G and hence either M is normal in G or the order of $G/M_G$ is the product of two different primes. In both cases, M has prime index in G and so also the order of X/Y is a prime number. Moreover, we have

$$\begin{aligned}{}[XM_G,X]=X'[M_G,X]\le X'(M_G\cap X)\le Y, \end{aligned}$$

so that $XM_G\le C_G(X/Y)$ and hence $|G:C_G(X/Y)|\le |M/M_G|$ is either 1 or a prime number. Therefore $\theta (G)$ is $\theta$-embedded in G. $\square$

Lemma 4.5

Let G be a group and let N be a normal subgroup of G admitting an ascending G-invariant series with abelian factors. If M is a maximal subgroup of G which does not contain N, then there exists a chief factor X/Y of G such that $X\cap M\le Y<X\le N$.

Proof

Let $\{1\}=N_0<N_1<\cdots<N_\alpha<N_{\alpha +1}<\cdots <N_\tau =N$ be an ascending series consisting of normal subgroups of G such that $N_{\alpha +1}/N_\alpha$ is abelian for all ordinals $\alpha <\tau$. Consider the least ordinal $\mu \le \tau$ such that $N_\mu$ is not contained in M. Then $\mu$ is not a limit ordinal and $N_{\mu -1}\le M$. Since $N_\mu /N_{\mu -1}$ is abelian, it follows that $N_\mu \cap M$ is a normal subgroup of $G=N_\mu M$ and so there exists a chief factor X/Y of G such that $N_\mu \cap M\le Y<X\le N_\mu$. Clearly, $X\cap M=N_\mu \cap M$ and so the statement is proved. $\square$

Lemma 4.6

Let G be a group with $\Phi (G)=\{1\}$ and let N be a normal subgroup of G admitting an ascending G-invariant series with abelian factors. Then N is $\theta$-embedded in G if and only if $N\le \theta (G)$.

Proof

Of course, every G-invariant subgroup of $\theta (G)$ is $\theta$-embedded in G by Lemma 4.4. Suppose now that N is a $\theta$-embedded normal subgroup of G and assume for a contradiction that N is not contained in a maximal subgroup M of G which is not permodular. It follows from Lemma 4.5 that there exists a chief factor X/Y of G such that $X\cap M\le Y<X\le N$ and X/Y is abelian because N has an abelian G-invariant series with abelian factors. Then X is not contained in M, while $Y\le M$ by Dedekind’s modular law and hence $Y=X\cap M=X\cap M_G$. In particular, X/Y is complemented by M and hence it has prime order p, so that also $XM_G/M_G$ has order p. Moreover, since M is not normal, the centralizer $C=C_G(X/Y)$ is a proper subgroup of G and hence G/C has prime order q. The intersection $C\cap M$ is a normal subgroup of $G=XM$ and $M/C\cap M$ has order q, so that $C\cap M=M_G$ and hence has order pq, which is impossible as the subgroup lattice ${\mathfrak {L}}(G/M_G)$ cannot be modular. Therefore N is contained in all maximal subgroups of G that are not permodular and so also in $\theta (G)$.

$\square$

The following result can be proved by using the same argument as in the proof of [9, Theorem 2.11].

Lemma 4.7

Let G be a finite group and let X be a subgroup which is k-permodularly embedded in G for some positive integer k. Then there exists a normal subgroup N of G of prime order such that $N\le X$ and the factor group $G/C_G(N)$ is either trivial or of prime order.

Lemma 4.8

Let G be a finite group and let X be a subgroup which is k-permodularly embedded in G for some positive integer k. Then X is contained in $\theta (G)$.

Proof

Assume for a contradiction that the statement is false and choose a counterexample G of smallest possible order. Then G has a maximal subgroup M which is neither permodular nor contains X. By Lemma 4.7, the subgroup X contains a G-invariant subgroup N of prime order p such that $G/C_G(N)$ is either trivial or has prime order q. It follows from the minimal choice of G that X/N is a subgroup of $\theta (G/N)$, so that N is not contained in M and hence . Thus $G/C_G(N)$ has order q, so that $|M/C_M(N)|=q$ and $M_G=C_M(N)$ has index pq in G. Therefore the factor group $G/M_G$ has modular subgroup lattice, which is obviously a contradiction. $\square$

Lemma 4.9

Let G be a group with $\Phi (G)=\{1\}$ and let N be a k-permodularly embedded normal subgroup of G for some positive integer k. Then N is contained in $\theta (G)$.

Proof

Let X/Y be any abelian complemented chief factor of G with $X\le N$ and let M be a subgroup of G such that $G=XM$ and $X\cap M=Y$. Then M is maximal in G because X/Y is abelian. Moreover, since N is hypercyclically embedded in G by Corollary 3.2, we have that X/Y is finite of prime order and hence M has finite index in G. If M is normal in G, then $G/Y=X/Y\times M/Y$ and $X/Y\le \zeta (G/Y)$.

Suppose now that M is not normal in G, or equivalently that $C_G(X/Y)\ne G$. Then $X\cap M_G=Y$ and X/Y is G-isomorphic to $XM_G/M_G$; in particular, $XM_G/M_G$ is an abelian complemented chief factor of the finite group $G/M_G$. On the other hand, $NM_G/M_G$ is k-permodularly embedded in $G/M_G$, so that it is contained in $\theta (G/M_G)$ by Lemma 4.8. It follows now from Lemma 4.4 that $G/M_G$ induces on $XM_G/M_G$ a group of automorphisms of prime order and so also $G/C_G(X/Y)$ has prime order. Therefore N is $\theta$-embedded in G and hence $N\le \theta (G)$ by Lemma 4.6. $\square$

Proof of Theorem B

Clearly, there exists a normal subgroup of finite index N of G which is k-permodularly embedded in G. Put ${{\overline{G}}}=G/\Phi (G)$, so that ${{\overline{N}}}$ is k-permodularly embedded in ${{\overline{G}}}$ and hence ${{\overline{N}}}\le \theta ({{\overline{G}}})$ by Lemma 4.9. As $\theta ({{\overline{G}}})=\theta (G)/\Phi (G)$, it follows that $\theta (G)$ has finite index in G and so G has finitely many maximal subgroups which are not permodular. $\square$

Corollary 4.10

Let G be a locally nilmodular group. Then every maximal subgroup of G is permodular.

Proof

Let M be any maximal subgroup of G. Since nilmodular groups are hypercyclic by Corollary 3.3, the group G is locally supersoluble and so M has finite index in G. Then $G/M_G$ is a finite nilmodular group and hence $\theta (G/M_G)=G/M_G$ by Lemma 4.8. Therefore $M/M_G$ is permodular in $G/M_G$ and so M is permodular in G. $\square$

We can finally discuss the structure of groups in which all but finitely many maximal subgroups are permodular.

Theorem 4.11

A group G has only finitely many maximal sugroups which are not permodular if and only if it contains a normal subgroup L of finite index such that $L''\le \Phi (G)\le L$ and $L/\Phi (G)$ is $\theta$-embedded in $G/\Phi (G)$.

Proof

Suppose first that G has only finitely many maximal subgroups that are not permodular. Then each of these subgroups must have finite index in G and so also the index $|G:\theta (G)|$ is finite. Moreover, $\bigl (\theta (G)\bigr )''\le \Phi (G)\le \theta (G)$ by Lemma 4.3 and $\theta (G)$ is $\theta$-embedded in G by Lemma 4.4. Therefore it is enough to put $L=\theta (G)$.

Conversely let the group G contain a normal subgroup L of finite index such that $L''\le \Phi (G)\le L$ and $L/\Phi (G)$ is $\theta$-embedded in $G/\Phi (G)$. Of course, we may suppose without loss of generality that $\Phi (G)=\{1\}$, so that L is $\theta$-embedded in G. Assume for a contradiction that there exists a maximal subgroup M of G which is neither permodular nor contains L. By Lemma 4.5, we can find a chief factor X/Y of G such that $X\cap M\le Y<X\le L$. Clearly, X is not contained in M and so $G=XM$. On the other hand, Y lies in M since otherwise $G=YM$ and $X=YM\cap X=Y$. Thus $Y=X\cap M$, so that X/Y is a complemented chief factor and hence it has prime order p. Since M is not normal in G, the centralizer $C_G(X/Y)$ is a proper subgroup and so it has prime index q. Of course, $Y=X\cap M=X\cap M_G$ and so $|XM_G/M_G|=|X/Y|=p$. Finally, the subgroup $C=C_M(X/Y)$ is normal in G and hence it is contained in $M_G$; but M/C has order q, so that $C=M_G$ and $|M/M_G|=q$. It follows that $G/M_G$ has order pq and in particular, its subgroup lattice is modular, a contradiction. Therefore every maximal non-permodular subgroup of G contains L and so G has only finitely many maximal non-permodular subgroups. $\square$