Relationships Between the Factors of the Upper and the Lower Central Series of a Group

Abstract

In the paper we study groups in which the factor-group by k-th hypercenter is locally finite and has finite exponent. We proved that in these groups the (k+1)-th term of lower central series is locally finite and has finite exponent. Moreover we are able to find bounds for the exponent of $\gamma _{k+1}(G)$ and for the exponent of the locally nilpotent residual of G.

1 Introduction

The aim of this paper is to establish a relationship between the factors of the upper and the lower central series of a group. Given a group G, we recall that the upper central series of G is the series:

$$\begin{aligned} \langle 1\rangle = \zeta _0(G)\le \zeta _1(G)\le \zeta _2(G)\le \cdots \le \zeta _\alpha (G)\le \zeta _{\alpha +1}(G)\le \cdots \zeta _\gamma (G), \end{aligned}$$

where $\zeta _1(G) = \zeta (G)$ is the center of G, $\zeta _{\alpha +1}(G)/\zeta _\alpha (G) = \zeta (G/\zeta _\alpha (G))$ for every ordinal $\alpha$, $\zeta _\lambda (G) = \bigcup _{\mu <\lambda }\zeta _\mu (G)$ for every limit ordinal $\lambda$, and $\zeta (G/\zeta _\gamma (G))=\langle 1\rangle$. The term $\zeta _\alpha (G)$ is said to be the $\alpha {\mathrm{th}}$–hypercenter of G, and the last term $\zeta _\gamma (G)$ of this series is said to be the upper hypercenter of G. The ordinal $\gamma$ is said to be the central length of G and is denoted by zl(G). On the other hand, the lower central series of G is the series

$$\begin{aligned} G = \gamma _1(G)\ge \gamma _2(G)\ge \cdots \gamma _\alpha (G)\ge \gamma _{\alpha +1}(G)\ge \cdots \gamma _\delta (G), \end{aligned}$$

where $\gamma _2(G) = [G,G]$ is the derived group of G, $\gamma _{\alpha +1}(G) = [\gamma _\alpha (G),G]$ for every ordinal $\alpha$, $\gamma _\lambda (G) = \bigcap _{\mu <\lambda }\gamma _\lambda (G)$ for every limit ordinal, and $\gamma _\delta (G)=[\gamma _\delta (G),G]$. The term $\gamma _\alpha (G)$ is said to be the $\alpha {\mathrm{th}}$–hypocenter of G, and the last term $\gamma _\delta (G)$ of this series is said to be the lower hypocenter of G.

Let G be a nilpotent group. Then there exists a positive integer k such that $G = \zeta _k(G)$. Equivalently $\gamma _{k+1}(G) = \langle 1\rangle$. Extending this well-known fact, Baer [1] has been able to show the following result:

Theorem. Given a group G, suppose that the factor-group $G/\zeta _k(G)$ is finite for some positive integer k. Then $\gamma _{k+1}(G)$ is likewise finite.

To express properly these results in a general and unified way, we introduce the following concept. A class of groups $\mathfrak {X}$ is said to be a Baer class if whenever G is a group and we have $G/\zeta _k(G)\in \mathfrak {X}$ for some positive integer k, then $\gamma _{k+1}(G)\in \mathfrak {X}$. A natural question here is Finding Baer classes of groups. Obviously the trivial class $\mathfrak {I}=\{\langle 1\rangle \}$ is a Baer class, and Baer’s theorem shows that the class $\mathfrak {F}$ of all finite groups is also a Baer class. Another important precedent appeared if one considers the case $k = 1$. I. Schur has studied the relationship between the central factor-group $G/\zeta (G)$ of a group G and the derived subgroup [G, G] of G [6]. In particular, from Schur’s results it follows that if $G/\zeta (G)$ is finite, then [G, G] is also finite. Inspired by this and related facts, in [2] a class of groups $\mathfrak {X}$ of groups are called a Schur class if for every group G such that $G/\zeta _1(G)\in \mathfrak {X}$ it follows that derived subgroup $\gamma _2(G)$ always belong to $\mathfrak {X}$; examples of Schur classes are related in the mentioned paper [2]. Therefore, $\mathfrak {I}$ and $\mathfrak {F}$ are Schur classes.

Obviously, every Baer class is a Schur class. This raises in a natural way the study of the converse: Which Schur classes are Baer classes?. Now we know many examples of Schur classes, most of them since a long time ago (see [2]). For example, the class $\mathfrak {F}$ of all finite groups, the class $L\mathfrak {F}_{\pi }$ of locally finite $\pi$–groups, for an arbitrary set $\pi$ of prime numbers, the class $\mathfrak {P}$ of polycyclic-by-finite groups, the class $\mathfrak {C}$ of Chernikov groups, the class $\mathfrak {S}_1$ of soluble-by-finite minimax groups, and many others. Many of these classes have been proved to be Baer classes. A few years ago, Mann [5] proved that the class $\mathfrak {L}$ of all locally finite groups having finite exponent is a Schur class. Moreover, there exists a function m such that the exponent of the derived subgroup of a locally finite of exponent e is bounded by m(e). Therefore, the question of deciding whether this is a Baer class or not naturally appears. The first main result of this paper gives a positive answer on this question.

Theorem A

Let G be a group and suppose that $G/\zeta _k(G)$ is a locally finite group, having finite exponent e. Then the subgroup $\gamma _{k+1}(G)$ is locally finite and has finite exponent. Moreover, there exists a function $\beta _1$ such that the exponent of $\gamma _{k+1}(G)$ is at most $\beta _1(e,k)$.

For the groups described in Theorem A, we may ask another related question. Given a group G, we recall that the locally nilpotent residual L of G is the intersection of all normal subgroups H of G such that G / H is locally nilpotent. It is well known that G / L need not to be locally nilpotent, and therefore, the case in which this factor-group is locally nilpotent is very interesting. In particular, such situation is obtained in our second main result.

Theorem B

Let G be a group and suppose that $G/\zeta _k(G)$ is a locally finite group having finite exponent e. Then the locally nilpotent residual L of G is locally finite having finite exponent, and G / L is locally nilpotent. Moreover, there exists a function $\beta _2$ such that the exponent of L is at most $\beta _2(e)$.

It is worth mentioning that in fact the exponent of the locally nilpotent residual depends on the exponent of $G/\zeta _k(G)$.

2 Proof of Theorem A

The proof relies on the following auxiliary results.

Lemma 2.1

Suppose that A is an abelian normal subgroup of a group G such that $G/C_G(A) = \langle x_1C_G(A),x_2C_G(A)\rangle$ for some elements $x_1, x_2\in G$. Then $[A,G] = [A,x_1][A,x_2]$.

Proof

Put $U = [A,x_1][A,x_2]$. If $a\in A$, then

$$\begin{aligned} \left[ a,x_j^2\right] = \left[ a,x_j\right] \left[ y,x_j\right] ^{x_j} = \left[ y,x_j\right] \left[ y^{x_j},x_j\right] \in \left[ A,x_j\right] \le U,\quad \ j\in \{1,2\}. \end{aligned}$$

It follows that $[a,x_j^n]\in U$ for each $n\in \mathbb {Z}$. Let $n, k\in \mathbb {Z}$ and put $u = x_1^n$ and $v = x_2^k$. Given $a\in A$, we have

$$\begin{aligned}{}[a,uv] = [a,v][a,u]^v\ \text{ and }\ [a,u]^v = [vcv^{-1},u]^v, \end{aligned}$$

where $c = v^{-1}av\in A$. Put $d = [vcv^{-1},u]$ so that

$$\begin{aligned}{}[vcv^{-1},u]^v = d^v = dd^{-1}v^{-1}dv = d[d,v]. \end{aligned}$$

Clearly $d\in [A,u] = [A,x_1^n]\le U$ and $[d,v]\in [A,v] = [A,x_2^k]\le U$ and then

$$\begin{aligned}{}[a,u]^v = [vcv^{-1},u]^v = d^v = d[d,v]\in U. \end{aligned}$$

It follows that $[a,uv]\in U$. Proceeding in this way and applying induction, we see that

$$\begin{aligned} \left[ a,x_1^{k_1}x_2^{t_1}\ldots x_1^{k_n}x_2^{t_n}\right] \in U,\ \text{ for }\ k_1, t_1,\ldots , k_n, t_n\in \mathbb {Z}. \end{aligned}$$

Let g be an arbitrary element of G. Then

$$\begin{aligned} g = x_1^{r_1}x_2^{s_1}\ldots x_1^{r_m}x_2^{s_m}c, \end{aligned}$$

for some element $c\in C_G(A)$ and integer numbers $r_1, s_1,\ldots , r_m, s_m\in \mathbb {Z}$. Then

$$\begin{aligned} \left[ a,x_1^{r_1}x_2^{s_1}\ldots x_1^{r_m}x_2^{s_m}c\right] = \left[ a,x_1^{r_1}x_2^{s_1}\ldots x_1^{r_m}x_2^{s_m}\right] \in U, \end{aligned}$$

and hence we obtain that U is a G–invariant subgroup of A. By the choice of G, we have $A/U\le \zeta (G/U)$ which gives that $[A,G]\le U$, as required. $\square$

Corollary 2.2

Let A be an abelian normal subgroup of a group G and suppose we have that $G = \langle C_G(A),M\rangle$ for a certain subset M of G. Then [A, G] is the product of all [A, x], when x runs M.

Proof

Put $V = \langle [A,x]\ |\ x\in M\rangle$. Clearly $V\le [A,G]$. Let $w\in [A,G]$ so that

$$\begin{aligned} w = [a_1,y_1]\ldots [a_n,y_n] \end{aligned}$$

for suitable elements $a_1,\ldots , a_n\in A$ and $y_1,\ldots , y_n\in G$. Then there exist elements $x_1,\ldots , x_m\in M$ such that

$$\begin{aligned} y_1,\ldots , y_n\in \langle x_1,\ldots , x_m, C_G(A)\rangle = H, \end{aligned}$$

and therefore, $w\in [A,H]$. Since the product $[A,x_j][A,x_k]$ is $\langle x_j,x_k\rangle$–invariant for any choice of $j,k\in \{1,\ldots ,m\}$ by Lemma 2.1, the subgroup $[A,x_1]\ldots [A,x_m] = U$ is H–invariant. Then the center of the section H / U includes A / U, that is $[A,H]\le U$. Since the converse inclusion is also true, we deduce that $[A,H] = U$. Therefore,

$$\begin{aligned} w\in [A,x_1]\ldots [A,x_m]\le V, \end{aligned}$$

and hence $[A,G] = V$, as required. $\square$

Lemma 2.3

Let A be an abelian normal subgroup of a group G and suppose that $A/(\zeta (G)\cap A)$ is locally finite and has finite exponent e. Then [A, G] is a locally finite subgroup having finite exponent at most e.

Proof

We pick a subset M of G such that $G = \langle C_G(A),M\rangle$. Given $g\in G$, we consider the mapping $\xi _g: a\rightarrow [a,g],\ a\in A$ so that $\xi _g$ is an endomorphism of A. Since $\zeta (G)\cap A\le C_A(g) = \text{ Ker }(\xi _g)$, $A/\text{ Ker }(\xi _g)$ is locally finite and has finite exponent at most e. Since

[A, g] is locally finite and has finite exponent at most e. By Corollary 2.2, [A, G] is the product of the subgroups [A, g], when g runs M. Since every subgroup [A, g] is locally finite and has finite exponent at most e, the same is true for [A, G]. $\square$

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

Proof of Theorem A

Let

$$\begin{aligned} \langle 1\rangle = Z_0\le Z_1\le \cdots \le Z_{k-1}\le Z_k = Z \end{aligned}$$

be the upper central series of G. We proceed by induction on k.

If $k = 1$, then $G/Z_1$ is a locally finite group having finite exponent e. Application of Mann’s theorem [5] shows that $\gamma _2(G) = [G,G]$ is locally finite, and there exists a function m such that the exponent of $\gamma _2(G)$ is bounded by m(e).

We now suppose that $k > 1$ and we have already proved that $\gamma _k(G/Z_1)$ is locally finite of finite exponent, and there exists a function $\beta _1$ such that the exponent of $\gamma _k(G/Z_1)$ is at most $\beta _1(e,k-1)$. Put $K/Z_1 = \gamma _k(G/Z_1)$ and $L = \gamma _k(G)$ so that $L\le K$. Applying Mann’s theorem [5] to K, we obtain that $D = [K,K]$ is locally finite and has finite exponent at most $m(\beta _1(e,k-1))$. Since the factor-group K / D is abelian, LD / D is also abelian. We have

$$\begin{aligned} (LD/D)(LD/D\cap Z_1D/D)= & {} (LD/D)((LD \cap Z_1D)/D)\cong LD/(LD\cap Z_1D)\\ \cong (LD)(Z_1D)/(Z_1D)= & {} (LZ_1D)/(Z_1D)\cong L/(L\cap Z_1D), \end{aligned}$$

which shows that $(LD/D)(LD/D\cap Z_1D/D)$ is an epimorphic image of $L/(L\cap Z_1)$. Since $L/(L\cap Z_1)\cong LZ_1/Z_1\le K/Z_1$, $L/(L\cap Z_1)$ is a locally finite group of finite exponent at most $\beta _1(e,k-1)$. Therefore, the same is true also for $(LD/D)(LD/D\cap Z_1D/D)$. Applying Lemma 2.3 to the factor-group G / D, we see that its subgroup $V/D = [LD/D,G/D]$ is locally finite and has finite exponent at most $\beta _1(e,k-1)$. Since the center of G / V includes LV / V and (G / V) / (LV / V) is nilpotent of class at most k, $\gamma _{k+1}(G)\le V$. It follows that $\gamma _{k+1}(G)$ is a locally finite subgroup having exponent at most $m(\beta _1(e,k-1))\beta _1(e,k-1) = \beta _1(e,k)$, and we are done. $\square$

It is worth mentioning that the function $\beta _1(t,k)$ constructed in this theorem is defined recursively by $\beta _1(e,1) = m(e)$, $\beta _1(e,2) = m(m(e))m(e)$, and

$$\begin{aligned} \beta _1(t,k) = m(\beta _1(e,k-1))\beta _1(e,k-1). \end{aligned}$$

3 Proof of Theorem B

To show the auxiliary results that lead to the proof of this theorem, we need the following module-theoretical concepts.

Let G be a group, R a ring, and A an RG–module. Then the set

$$\begin{aligned} \zeta _{RG}(A) = \{a\in A\ |\ a(g-1) = 0\ \text{ for } \text{ each } \text{ element }\ g\in G\} \end{aligned}$$

is a submodule called the RG–center of A. The upper RG–central series of A is,

$$\begin{aligned} \{0\} = A_0\le A_1\le \cdots \le A_\alpha \le A_{\alpha +1}\le \cdots A_\gamma , \end{aligned}$$

where $A_1 = \zeta _{RG}(A)$, $A_{\alpha +1}/A_\alpha = \zeta _{RG}(A/A_\alpha )$, $\alpha <\gamma$, and $\zeta _{RG}(A/A_\gamma ) = \{0\}$. The last term $A_\gamma$ of this series is called the upper RG–hypercenter of A and will be denoted by $\zeta _{RG}^\infty (A)$, while the ordinal $\gamma$ is said to be the RG–central length of A and will be denoted by $zl_{RG}(A)$. The RG–module A is said to be RG–hypercentral if $A=A_\gamma$ happens and RG–nilpotent if $\gamma$ is finite.

If B and C are RG–submodules of A and $B\le C$, then the factor C / B is said to be G–central if $G = C_G(C/B)$ and G-eccentric otherwise. An RG–submodule C of A is said to be RG–hypereccentric if C has an ascending series of RG–submodules

$$\begin{aligned} \{0\} = C_0\le C_1\le \cdots \le C_\alpha \le C_{\alpha +1}\le \cdots C_\gamma = C \end{aligned}$$

whose factors $C_{\alpha +1}/C_\alpha$ are G–eccentric simple RG–modules.

Following D.I. Zaitsev [7], an RG–module A is said to have the Z–decomposition if one has

$$\begin{aligned} A = \zeta _{RG}^\infty (A)\oplus E_{RG}^\infty (A), \end{aligned}$$

where $E_{RG}^\infty (A)$ is the unique maximal RG–hypereccentric RG–submodule of A. We actually note that a given maximal E includes every RG–hypereccentric RG–submodule B and, in particular, it is unique. For, if $(B + E)/E$ is non-zero, it has to include a non-zero simple RG–submodule U / E. Since $(B + E)/E \cong B/(B \cap E)$, U / E is RG–isomorphic to some simple RG–factor of B, and it follows that $G/C_G(U/E)\ne G$. On the other hand, $(B + E)/E \le A/E \le \zeta _{RG}^{\infty }(A)$, that is $G/C_G(U/E)=G$. This contradiction shows that $B \le E$, as claimed.

Lemma 3.1

Let G be a finite nilpotent group and A be a $\mathbb {Z}G$–module. Suppose that A includes a $\mathbb {Z}G$–nilpotent $\mathbb {Z}G$–submodule C such that A / C is a finite group of order t and exponent e. Then A includes a finite $\mathbb {Z}G$–submodule K such that $|K|\ |\ t$, the exponent of K is at most e, and A / K is $\mathbb {Z}G$–nilpotent.

Proof

We first remark that $A\zeta _{\mathbb {Z}G}^\infty (A)$ is a finite of order divisor of t and exponent e. Pick a finite subset M of elements of A such that

$$\begin{aligned} M\mathbb {Z}G+C = A/C. \end{aligned}$$

Put $V = M\mathbb {Z}G$ and $U = C\cap V$ so that U is clearly $\mathbb {Z}G$–nilpotent. Since

$$\begin{aligned} V/U = V/(V\cap C)\cong (V + C)/C = A/C, \end{aligned}$$

$|V/U| = t$ and V / U has exponent at most e. Since G is finite, the natural semidirect product $V\leftthreetimes G$ is a nilpotent-by-finite group. Being finitely generated, it satisfies the maximal condition on all subgroups, and it follows that U is a finitely generated subgroup. Therefore, the periodic part T of U is finite, and hence, $U = T\oplus W$, for some torsion-free subgroup W. Put $Y = U^{|T|}$ so that Y is a characteristic subgroup of U. In particular, Y is a $\mathbb {Z}G$–submodule, and U / Y is finite whence V / Y is finite too. Since G is nilpotent, the finite factor-module V / Y has the Z–decomposition [7], that is,

$$\begin{aligned} V/Y = Z/Y\bigoplus E/Y, \end{aligned}$$

where $Z/Y = \zeta _{ZG}^\infty (V/Y)$ and $E/Y = E_{ZG}^\infty (V/Y)$. Since U / Y is $\mathbb {Z}G$–nilpotent, $U/Y \le Z/Y$. Applying the latter, the isomorphisms

$$\begin{aligned} E/Y\cong (V/Y)/(Z/Y)\cong V/Z \end{aligned}$$

and the inclusion $(\zeta _{ZG}^\infty (A)+Y)/Y\le Z/Y$ at once, we obtain that E / Y is isomorphic to some factor-module of $A/\zeta _{\mathbb {Z}G}^\infty (A)$. In particular, E / Y is finite, $|E|\ |\ t$, and the exponent of E / Y is at most e.

The choice of E yields that E is a $\mathbb {Z}G$–submodule of V. Then the periodic part K of E is also a $\mathbb {Z}G$–submodule. Since Y is torsion-free, $K\cap Y=\{0\}$, and then K is isomorphic to some section of E / Y. Therefore, K is finite, $|K|\ |\ t$ and the exponent of K is at most e. The choice of E yields $|K|\ |\ |E/Y|\ |\ t$. The factor-module E / K is $\mathbb {Z}$–torsion-free and includes a $\mathbb {Z}G$–nilpotent submodule $(Y+K)/K$ having finite index. It follows that E / K is also $\mathbb {Z}G$–nilpotent. The isomorphisms

$$\begin{aligned} V/E\cong (V/Y)/(E/Y)\cong \zeta _{\mathbb {Z}G}^\infty (V/Y) \end{aligned}$$

give that V / K is $\mathbb {Z}G$–nilpotent. Since $A = V+C$ and $C\le \zeta _{\mathbb {Z}G}^\infty (A)$, A / K is $\mathbb {Z}G$–nilpotent, as required. $\square$

An RG–module is said to be locally RG–nilpotent if for every finitely generated subgroup F of G and every finite subset M of A,  the $\mathbb {Z}F$–submodule $M\mathbb {Z}F$ generated by M is $\mathbb {Z}F$–nilpotent.

Corollary 3.2

Let G be a periodic locally nilpotent group and A be a $\mathbb {Z}G$–module. Suppose that A includes a $\mathbb {Z}G$–nilpotent $\mathbb {Z}G$–submodule C such that the additive group of A / C is periodic and has finite exponent e. Then A includes a $\mathbb {Z}G$–submodule K, the additive group of K is periodic and has finite exponent at most e, and A / K is locally $\mathbb {Z}G$–nilpotent.

Proof

Let M be an arbitrary finite subset of A. If $\mathcal {L}$ is the local system of G consisting of all its finite subgroups and $F\in \mathcal {L}$, we consider the $\mathbb {Z}F$–submodule $M_F = C+M\mathbb {Z}F$. Since A / C is $\mathbb {Z}$–periodic and F is finite, $M_F/C$ is finite (perhaps trivial if $M\subseteq C$). By Lemma 3.1, $M_F$ includes a finite $\mathbb {Z}F$–submodule R such that $M_F/R$ is $\mathbb {Z}F$–nilpotent and the exponent of R is at most e. Then R includes a unique minimal finite $\mathbb {Z}F$–submodule $K_F$ such that $M_F/K_F$ is $\mathbb {Z}F$–nilpotent. Let $H\in \mathbb {L}$ be such that $F\le H$. Obviously $M_F\le M_H$. Since the factor-module $M_H/K_H$ is $\mathbb {Z}H$–nilpotent, it is clearly $\mathbb {Z}F$–nilpotent. It follows that $M_F/(K_H\cap M_F)$ is $\mathbb {Z}F$–nilpotent and then $K_F\le K_H\cap M_F$ whence $K_F\le K_H$ by the election of $K_F$. From the equation $G = \bigcup _{F\in \mathcal {L}}F$,

$$\begin{aligned} M_0 = \bigcup _{F\in \mathcal {L}}M_F\ \text{ and }\ K(M) = \bigcup _{F\in \mathcal {L}}K_F \end{aligned}$$

are $\mathbb {Z}G$–submodules. Let S be an arbitrary finite subset of $M_0$ and X be an arbitrary finite subgroup of G. Since $M_0$ is generated by M as $\mathbb {Z}G$–submodule, there exists a finite subgroup $F\in \mathcal {L}$ such that $S\le M_F$. Pick $H\in \mathcal {L}$ such that $X, F\le H$. Then $M\mathbb {Z}X\le M_H$. Since $M_H/K_H$ is $\mathbb {Z}F$–nilpotent, in particular, it is $\mathbb {Z}X$–nilpotent. Then, $(M\mathbb {Z}X+K_H)/K_H$ is $\mathbb {Z}X$–nilpotent, and therefore, $(M\mathbb {Z}X+K(M))/K(M)$ is $\mathbb {Z}X$–nilpotent. Hence, $M_0/K(M)$ is locally $\mathbb {Z}G$–nilpotent. Since $K_F$ has exponent at most e for each $F\in \mathcal {L}$, K(M) also has exponent at most e.

We now consider the local family $\mathcal {M}$ consisting of the finite subset of A. Let $M, S\in \mathcal {M}$ such that $M\subseteq S$ and pick $F\in \mathcal {L}$. Since $S_0/K(S)$ is locally $\mathbb {Z}G$–nilpotent, $(S\mathbb {Z}F+K(S))/K(S)$ is $\mathbb {Z}F$–nilpotent. It follows that $M\mathbb {Z}F/(M\mathbb {Z}F\cap K(S))$ is $\mathbb {Z}F$–nilpotent. Therefore, $K_F\le M\mathbb {Z}F\cap K(S)$ and then $K_F\le K(S)$. Thus, $\bigcup _{F\in \mathcal {L}}K_F\le K(S)$. Thus, $K(M)\le K(S)$. This means that the family $\{K(M)\ |\ M\in \mathcal {M}\}$ is local; hence, $K =\bigcup _{M\in \mathcal {M}}K(M)$ is a $\mathbb {Z}G$–submodule. Since $A=\bigcup _{M\in \mathcal {M}}M$, A / K is locally $\mathbb {Z}G$–nilpotent. By construction, K has exponent at most e. $\square$

Lemma 3.3

Let K be a locally finite normal subgroup of a group G such that G / K is locally nilpotent. Then the locally nilpotent residual L of G is locally finite. Moreover, if G satisfies locally the maximal condition on subgroups, then G / L is locally nilpotent.

Proof

Since G / K is locally nilpotent, $L\le K$, and it follows that L is locally finite. Replacing G by the factor-group G / L, we may suppose that $L = \langle 1\rangle$. Then the thesis is to prove that G is locally nilpotent. Pick a family $\{G_\lambda \ |\ \lambda \in \Lambda \}$ of normal subgroups of G such that $\bigcap _{\lambda \in \Lambda }G_\lambda = \langle 1\rangle$ and $G/G_\lambda$ is locally nilpotent for every $\lambda \in \Lambda$. Since the result is trivial if $\Lambda$ is finite, we suppose that the family is infinite. Put $K_\lambda = K\cap G_\lambda$ so that $\bigcap _{\lambda \in \Lambda }K_\lambda = \langle 1\rangle$, every subgroup $K_\lambda$ is G–invariant and $G/K_\lambda$ is locally nilpotent for every $\lambda \in \Lambda$. Let F be an arbitrary finitely generated subgroup of G. Then $F/(F\cap K)$ is a finitely generated nilpotent group, and the subgroup $F\cap K$ is locally finite. Since F satisfies the maximal condition on subgroups, $T = F \cap K$ have to be finite. Then there exists a finite subset M of $\Lambda$ such that $T\cap (\bigcap _{\lambda \in M}K_\lambda ) = \langle 1\rangle$. Put $V = \bigcap _{\lambda \in M}K_\lambda$ so that G / V is locally nilpotent. We have now

$$\begin{aligned} F\cap V = F\cap \left( \bigcap _{\lambda \in M}K_\lambda \right) = \bigcap _{\lambda \in M}(F\cap K_\lambda ) = \bigcap _{\lambda \in M}(F\cap (K\cap K_\lambda )) =\\ = \bigcap _{\lambda \in M}((F\cap K)\cap K_\lambda ) = \bigcap _{\lambda \in M}(T\cap K_\lambda ) = T\cap \left( \bigcap _{\lambda \in M}K_\lambda \right) = \langle 1\rangle . \end{aligned}$$

It follows that $F\cong F/(F\cap V)\cong FV/V$. Since G / V is locally nilpotent, FV / V is nilpotent. Therefore, an arbitrary finitely generated subgroup F of G is nilpotent, and hence, G is locally nilpotent, as required. $\square$

Lemma 3.4

Let Z be the upper hypercenter of a group G. If G / Z is locally finite, then every finitely generated subgroup of G is nilpotent-by-finite.

Proof

Let F be an arbitrary finitely generated subgroup of G. The factor-group FZ / Z is finite since it is finitely generated and locally finite. Since $FZ/Z\cong F/(F\cap Z)$, $F\cap Z$ has finite index in F. Then $F\cap Z$ is finitely generated too (see [3, Corollary7.2.1]), and being hypercentral, is nilpotent. $\square$

Proof of Theorem B

Let

$$\begin{aligned} \langle 1\rangle = Z_0\le Z_1\le \cdots \le Z_{k-1}\le Z_k = Z \end{aligned}$$

be the upper central series of G so that every term $Z_j$ is G–invariant and every factors $Z_j/Z_{j-1}$ is G–central. By Kaluzhnin’s theorem [4], the factor-group $G/C_G(Z)$ is nilpotent of nilpotency class at most $k-1$. Put $C = C_G(Z)$ so that $Z\le C_G(C)$. In particular, $G/C_G(C)$ is locally finite and has finite exponent at most e. Clearly $C\cap Z\le \zeta (C)$, and then $C/(Z\cap C)\cong CZ/Z$ is locally finite and has finite exponent at most e. By Mann’s theorem [5], the derived subgroup $D = [C,C]$ is locally finite, and there exists a function m such that the exponent of D is bounded by m(e). The subgroup D is G–invariant, and C / D is abelian. We think of C / D as a $\mathbb {Z}H$–module where $H = (G/D)/C_{G/D}(C/D)$. Since C / G is abelian, $C/D\le C_{G/D}(C/D)$ and then $(G/D)/C_{G/D}(C/D)$ is nilpotent. Since $G/C_G(C)$ is locally finite, $(G/D)/C_{G/D}(C/D)$ is also locally finite.

We have $(C\cap Z)D/D\le \zeta _{\mathbb {Z}H}^\infty (C/D)$ and $(C/D)/((C\le Z)D/D)\cong C/(C\cap Z)D$ is a locally finite group of finite exponent at most e. By Lemma 3.2, C / D includes a $\mathbb {Z}G$–submodule V / D such that the additive group of V / D is periodic and has finite exponent at most e. Moreover, the factor-module (C / D) / (V / D) is locally $\mathbb {Z}G$–nilpotent. Put $B = C/V$ and pick an arbitrary subset M of $E = G/V$ and put $F = \langle M\rangle$. By Lemma 3.4, F is nilpotent-by-finite, in particular, it is noetherian; that is, it satisfies the maximal condition on subgroups. Then its subgroup $K = F\cap B$ is finitely generated. In this case K is finitely generated as a $\mathbb {Z}F$–module. Since B is a $\mathbb {Z}G$–module locally $\mathbb {Z}G$–nilpotent, its finitely generated $\mathbb {Z}F$–submodule K is $\mathbb {Z}F$–nilpotent. In other words, the upper hypercenter of F includes K. Since $F/K = F/(F\cap B)\cong FB/B$ is nilpotent, F is likewise nilpotent. Thus, G / V is locally nilpotent, and hence, V includes the locally nilpotent residual L. Since D (respectively, V / D) is locally finite and has finite exponent at most m(e) (respectively, e), V is locally finite and has finite exponent at most em(e). In particular, L is locally finite and has finite exponent at most em(e).

Finally, Lemma 3.4 shows that G is locally noetherian, and it suffices to apply Lemma 3.3 to see that G / L is locally nilpotent, as required. $\square$