An analogue of the Wielandt subgroup in infinite groups

Abstract

In this paper we define analogues of the Wielandt subgroup of a group. We say that a subgroup H of a group G is f-subnormal in G if there is a finite chain of subgroups $H=H_0\le H_1\le \cdots \le H_n=G$ such that either $|H_{i+1}: H_i|$ is finite or $H_i$ is normal in $H_{i+1}$, for $0\le i\le n-1$. We study in a group G the connection between the subgroups $\overline{\omega }(G)$ and $\overline{\omega }_i(G)$ which are, respectively, the sets of elements of G normalizing all f-subnormal subgroups of G and those normalizing all infinite f-subnormal subgroups of G. In particular we show that $\overline{\omega }_i(G)/\overline{\omega }(G)$ is always Dedekind and often abelian.

1 Introduction

A subgroup H of a group G is called f-subnormal in G if there is a finite chain of subgroups

$$\begin{aligned} H=H_0\le H_1\le \cdots \le H_n=G \end{aligned}$$

(1)

such that either $|H_{i+1}: H_i|$ is finite or $H_i$ is normal in $H_{i+1}$, for $0\le i\le n-1$. In this case we call the chain of subgroups (1) an f-subnormal series of length n for H in G. Clearly subnormal subgroups and subgroups of finite index in G are f-subnormal in G. In particular all subgroups of a finite group are f-subnormal in the group, so f-subnormal subgroups need not be subnormal, although it was shown in [4] that for locally nilpotent groups f-subnormality is equivalent to subnormality.

This generalization of subnormality was introduced by Phillips in [16], and groups all of whose subgroups are f-subnormal were shown to be finite-by-soluble in [4], while further results have been obtained in the recent papers [5, 8,9,10, 14].

In this paper we define and study the relationship between two subgroups of a group G. The first of these is the f-Wielandt subgroup of G, which we denote by ${\overline{\omega }}(G)$ and define by

$$\begin{aligned} {\overline{\omega }}(G)=\bigcap \{N_G(H)| H \text { is an f -subnormal subgroup of }G\}. \end{aligned}$$

This is an analogue of the Wielandt subgroup of a group G,

$$\begin{aligned} \omega (G)=\bigcap \{N_G(H)| H \text { is a subnormal subgroup of }G\}, \end{aligned}$$

first introduced by Wielandt in [23] and studied in various papers including [3] and [6]. It is always the case that ${\overline{\omega }}(G)\le \omega (G)$, but easy examples show that in general the two subgroups are different. The Wielandt subgroup is itself analogous to the kern or normN(G) of a group G, defined to be the intersection of the normalizers of all the subgroups of G, originally introduced by Baer in [1]. It was shown in [21] that $N(G)\le \zeta _2(G)$, the second centre of G, and that $G/C_G(N(G))$ is abelian. Some information concerning ${\overline{\omega }}(G)$ was obtained in [9]. We note that the intersection of the normalizers of the finite f-subnormal subgroups coincides with the intersection of the normalizers of the finite cyclic f-subnormal subgroups of prime power order, so we usually focus on these latter subgroups.

In [2] the generalized Wielandt subgroup of a group G, here denoted by $\omega _{i}(G)$, was introduced and defined to be

$$\begin{aligned} \omega _{i}(G)=\bigcap \{N_G(H)| H \text { is an infinite subnormal subgroup of }G\}. \end{aligned}$$

We remark that of course $\omega (G)\le \omega _{i}(G)$ and that $\omega _{i}(G)=G$ if there are no infinite subnormal subgroups. Here we introduce the generalized f-Wielandt subgroup

$$\begin{aligned} {\overline{\omega }}_{i}(G)=\bigcap \{N_G(H)| H \text { is an infinite f -subnormal subgroup of }G\}, \end{aligned}$$

as a natural generalization of $\omega _{i}(G)$ for f-subnormal subgroups. Clearly ${\overline{\omega }}_{i}(G)\le \omega _{i}(G)$ and ${\overline{\omega }}_{i}(G)=G$, if G has no infinite f-subnormal subgroups. Of course ${\overline{\omega }}(G)\le {\overline{\omega }}_{i}(G)$ in general.

One purpose of the article [2] was to discuss the situation when $\omega _{i}(G)\ne \omega (G)$; a further goal was to discuss the structure of the quotient group $\omega _{i}(G)/\omega (G)$. In this paper we discuss the situation when ${\overline{\omega }}_{i}(G)\ne {\overline{\omega }}(G)$ and also obtain structural results for ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$.

In Sect. 2 we obtain some interesting useful preliminary results. In Sect. 3 we obtain some of the main structural results concerning the quotient ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$. In particular Theorem 3.3 and Theorem 3.4 show that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is always residually finite and even a Dedekind group when ${\overline{\omega }}_{i}(G)$ is infinite. Section 4 discusses the case in which ${\overline{\omega }}_{i}(G)$ is finite and shows that in such circumstances ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is actually abelian. The results of Sects. 3 and 4 are combined in Theorem 4.7 to show that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is always Dedekind whenever G is infinite. In Sect. 5 we give some applications of our results in Sect. 3 to the class of soluble groups with finitely many normalizers of f-subnormal subgroups. Finally in Sect. 6 we discuss several examples and prove that every finite Dedekind group occurs as the quotient ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$.

2 Some preliminary results

In this section we collect some preliminary results which are of general interest and finish with specific structural results concerned with the situation when ${\overline{\omega }}_{i}(G)\ne {\overline{\omega }}(G)$.

Lemma 2.1

Let H be a finite f-subnormal subgroup of a group G and suppose that H has infinitely many conjugates in G. Then there are f-subnormal subgroups $H_1, H_2, H_3$ of G such that $H\le H_1\triangleleft \,H_2\triangleleft \,H_3$ satisfying

• (i) $H_1$ is finite;

• (ii) $H_2/H_1, H_3/H_2$ are infinite and H has infinitely many conjugates in $H_3$.

Proof

We consider an f-subnormal series

$$\begin{aligned} H=H_0\le H_1\le \cdots \le H_n=G \end{aligned}$$

for H in G of minimal length n. Clearly, there are consecutive terms of this series $H_1\triangleleft \,H_2$ such that $H_1$ is finite and $H_2/H_1$ is infinite. Since H has infinitely many conjugates in G it is also clear that $H_2\ne G$, so $H_2\lneqq H_3$. If $H_2$ has finite index in $H_3$, then Dietzmann’s lemma (see [18, Corollary 3 to Lemma 2.14]) implies that there is a finite normal subgroup L of $H_3$ containing $H_1$ and we then obtain a shorter f-subnormal series for H in G. Therefore, $H_2$ has infinite index in $H_3$, so $H_2\triangleleft \,H_3$. If H has finitely many conjugates in $H_3$, then Dietzmann’s lemma shows there is a finite normal subgroup of $H_3$ containing H again contradicting the minimality of n. Hence H has infinitely many conjugates in $H_3$. $\square$

The following corollary will be useful later.

Corollary 2.2

Let G be a group with the property that all finite subnormal subgroups of infinite f-subnormal subgroups have finitely many conjugates in them. Then all finite f-subnormal subgroups of G have finitely many conjugates.

Proof

If X is a finite f-subnormal subgroup with infinitely many conjugates, then we may apply Lemma 2.1 to obtain a contradiction. $\square$

Lemma 2.3

Let G be a group and let H be an f-subnormal subgroup of G.

• (i) If K is a subnormal subgroup of G and $K^H=K$, then HK is f-subnormal in G.

• (ii) If H is finite and G is infinite, then there is an infinite f-subnormal subgroup K of G containing H which normalizes H.

Proof

(i) Let $K=K_n\triangleleft \,K_{n-1}\triangleleft \,\cdots \triangleleft \,K_1=K^G\triangleleft \,K_0=G$ be the normal closure series for K in G. An easy induction shows that H normalizes $K_i$ for each i, so $K_{i+1}\triangleleft \,HK_i$ and $HK_{i+1}$ is an f-subnormal subgroup of $HK_i$. Hence HK is f-subnormal in G as required.

(ii) There are f-subnormal subgroups $H_1, H_2$ with $H_1\triangleleft \,H_2$ such that $H\le H_1$, $H_1$ is finite and $|H_2:H_1|$ is infinite. Clearly H has finitely many conjugates in $H_2$ so we may take $K=N_{H_2}(H)$. $\square$

If X, Y are subgroups of a group G and i is a natural number, then as usual we write

$$\begin{aligned}{}[X,\underbrace{Y,Y,\ldots , Y}_i] \end{aligned}$$

as $[X,_iY]$. The following result will often be useful.

Lemma 2.4

Let G be a group and let H be a finite f-subnormal subgroup of G.

• (i) If D is a divisible abelian normal subgroup of G, then $[D,H]=1$.

• (ii) If $G=PH$, where P is a Prüfer group, then H centralizes P.

• (iii) If K is an f-subnormal subgroup of G and P is a normal Prüfer subgroup of G, then $P\le K$ or $[P,K]=1$.

Proof

(i) Note that $[D,_m H]$ is a divisible abelian subgroup for each natural number m. There is an f-subnormal series

$$\begin{aligned} H=H_0\le H_1\le \cdots \le H_m=G \end{aligned}$$

for H in G of length $m>0$. If $H_{m-1}$ has finite index in $H_m$, then $D\le H_{m-1}$, whereas if $H_{m-1}$ has infinite index in $H_m$, then $H_{m-1}\triangleleft \,H_m$, so $[D,H]\le [D,H_{m-1}]\le H_{m-1}$. Thus in any case $[D,H]\le H_{m-1}$. Repeating this argument we see that $[D,H,H]\le H_{m-2}$ and inductively we have $[D,_m H]\le H.$ Since H is finite and $[D,_mH]$ is divisible we have $[D,_mH]=1$ and [18, Lemma 3.13] implies that $[D,H]=1$.

(ii) Since P has finite index in G and P has no subgroups of finite index, it follows that $P\triangleleft \,G$ and the result follows by (i).

(iii) The group [P, K] is divisible and is a subgroup of P, so we may assume that $[P,K]=P$ . By induction on the length of an f-subnormal series we may suppose that $K\triangleleft \,G$ or |G : K| is finite. In both cases $P\le K$. $\square$

This result enables us to immediately deduce the following corollary.

Corollary 2.5

Let G be a group containing a normal Prüfer subgroup P. Then ${\overline{\omega }}(G)$ contains P.

The next result is also very useful.

Lemma 2.6

Let G be a group and suppose that H is a finite f-subnormal subgroup of G. Suppose that G contains an infinite residually finite H-invariant subnormal subgroup K. If $g\in {\overline{\omega }}_{i}(G)$, then $g\in N_G(H)$.

Proof

By Lemma 2.3KH is an f-subnormal subgroup of G which is clearly also residually finite. Then K has a finite residual system $\{K_{\lambda }\}_{\lambda \in \Lambda }$ consisting of H-invariant subgroups with the property that if $\lambda , \mu \in \Lambda$ there is an element $\alpha \in \Lambda$ such that $K_{\alpha }\le K_{\lambda }\cap K_{\mu }$. It is well known that

$$\begin{aligned} H^g\le \bigcap _{\lambda \in \Lambda }(K_{\lambda }H)^g= \bigcap _{\lambda \in \Lambda }(K_{\lambda }H)=H \end{aligned}$$

and the result follows. $\square$

Our next result and proof is based on the corresponding result [2, Lemma 1], but there are some differences in the proof, so we give it for completeness. In this paper, as in [2], we use the term H is Prüfer-by-finite to include the possibility that the group H is finite. If G is an arbitrary group, then $\pi (G)$ denotes the set of primes dividing the orders of elements in G of finite order.

We make the (obvious) remark that if H is an f-subnormal subgroup of a group G and if K is a subgroup of H of finite index in H, then K is also an f-subnormal subgroup of G.

Proposition 2.7

Let G be a group with ${\overline{\omega }}_{i}(G)\ne {\overline{\omega }}(G)$. Then every abelian normal subgroup of G is Prüfer-by-finite.

Proof

Let A be an abelian normal subgroup of G and note that by hypothesis there is a finite f-subnormal subgroup H of G that is cyclic of prime power order and an element $g\in {\overline{\omega }}_{i}(G)$ such that $H^g\ne H$.

We first show that A is periodic and to this end let $a\in A$ be an element of infinite order. Let $B_0=\langle a^H\rangle$, a finitely generated abelian group. Then $B_0$ is a residually finite H-invariant subgroup of G, so Lemma 2.6 implies that $g\in N_G(H)$, giving a contradiction. Hence A is periodic.

Next we prove that $\pi (A)$ is finite, so suppose for a contradiction that $\pi (A)$ is infinite and let B denote the $\pi (H)'$-component of A. Note that B is infinite and a normal subgroup of G. As H is f-subnormal in HB, there is an infinite f-subnormal subgroup K of HB such that $H\triangleleft \,K$ by Lemma 2.3. Clearly $K=H\times (K\cap B)$, so H is the Sylow $\pi (H)$-subgroup of K. But $H^g(K\cap B)=H(K\cap B)$, so $H^g=H$, giving the required contradiction.

Next note that if $A[p]=\{a\in A|a^p=1\}$ is infinite for some prime $p\in \pi (A)$, then Lemma 2.6 implies that $H^g=H$, contrary to the choice of H and g. Hence A is a Chernikov group.

If $R_p\ne 1$ and $R_q\ne 1$ are divisible components of A with p, q distinct primes then $HR_p, HR_q$ are f-subnormal in G. Then

$$\begin{aligned} H^g\le HR_p\cap HR_q=H(R_p\cap HR_q). \end{aligned}$$

By Lemma 2.4$R_p\cap HR_q=H\cap R_p$ so $H^g\le H$ and we obtain the usual contradiction. Hence the divisible component D of A is trivial or $\pi (D)=\{p\}$. Clearly we may assume that D is nontrivial and Lemma 2.4 implies that $[D,H]=1$.

Suppose that $D=P_1\times P_2$, where $P_1,P_2$ are nontrivial and $H\cap D\le P_1$. Then $HP_i$ is f-subnormal in G by Lemma 2.3 and

$$\begin{aligned} H^g\le HP_1\cap HP_2=H(HP_1\cap P_2)=H((H\cap D)P_1\cap P_2)=H. \end{aligned}$$

Hence the minimal divisible subgroup containing $H\cap D$ is D itself and thus $D[p]\le H$, which means that D is locally cyclic and so Prüfer. The result follows. $\square$

We have the following useful observation.

Corollary 2.8

Let G be a group such that ${\overline{\omega }}_{i}(G)\ne {\overline{\omega }}(G)$ and suppose that H is a Chernikov f-subnormal subgroup of G. Then H is Prüfer-by-finite.

Proof

We may assume that H is infinite, so H contains a normal subgroup K of finite index such that K is subnormal in G by [4, Proposition 3.1]. The divisible part D of K is nontrivial and subnormal in G. It follows by [18, Lemma 4.46] that G has a nontrivial divisible normal periodic abelian subgroup, which is a Prüfer group by Proposition 2.7. The result follows. $\square$

Finally in this section we give an analogue of [2, Theorem 1] and parts of the proof are based on the proof of the corresponding result.

Theorem 2.9

Let G be a group satisfying ${\overline{\omega }}_{i}(G)\ne {\overline{\omega }}(G)$. Then the Baer radical of G is Prüfer-by-finite and hence is nilpotent.

Proof

Let B be the Baer radical of G. By assumption, there is a finite f-subnormal subgroup H and an element $g\in {\overline{\omega }}_{i}(G)$ such that $H^g\ne H$. Clearly we may assume that H is a cyclic q-group for some fixed prime q. We suppose for a contradiction that B is not Prüfer-by-finite.

First we note that B is periodic, for, if not, there is an element $b\in B$ of infinite order and clearly $L=\langle b^H\rangle$ is a finitely generated nilpotent subnormal subgroup of G. Then L is residually finite and H-invariant, so Lemma 2.6 implies that $g\in N_G(H)$ which is a contradiction. Therefore, B is periodic so it is the direct product of its p-components.

Suppose that $B=C\times D$, where C and D are both infinite, G-invariant subgroups, and $H\cap B\le D$. Then it is clear that HC, HD are infinite f-subnormal subgroups of G and

$$\begin{aligned} H^g\le HC\cap HD=H(C\cap B\cap HD)=H(C\cap D(H\cap B))=H, \end{aligned}$$

a contradiction. It follows that B is the direct product of only finitely many p-components at most one of which is infinite. If B has finite rank, then Proposition 2.7 implies that B is Prüfer-by-finite. Hence B has infinite rank.

Let $B_p$ be the infinite p-component of B for some prime p. Of course H is f-subnormal in $HB_p$ so there exist f-subnormal subgroups $H_1\triangleleft \,H_2\le HB_p$ such that $H_1$ is Chernikov, $H_2$ has infinite rank and $H\le H_1$. Let $H_2\cap B_p=X$. Then HX is f-subnormal in G.

Certainly $[X,H]\le H_1$, so [X, H, H] is finite by Lemma 2.4 and if $d=|H|$, then

$$\begin{aligned}{}[X,H]^d\le [X,H^d][X,H,H]=[X,H,H], \end{aligned}$$

so the Chernikov group [X, H] is finite. Then $M= C_{X}(H^{X})$ is of infinite rank and f-subnormal in B, so subnormal in G. Since M is a Baer group it does not satisfy the minimal condition on subnormal abelian subgroups by [2, Lemma 2]. Thus M contains an infinite elementary abelian subnormal subgroup A and Lemma 2.6 gives the contradiction sought. Finally, B is nilpotent by [2, Lemma 2]. $\square$

3 Some structural results concerning ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$

In this section we discuss the structure of ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ which is very tightly controlled, much more so than with the corresponding factor $\omega _{i}(G)/\omega (G)$.

Let $V_f(G)$ denote the subgroup generated by all the finite f-subnormal subgroups of G.

We begin with the following analogue of [2, Theorem 6]. Example 3 shows that Lemma 3.1 cannot really be improved further.

Lemma 3.1

Let G be a group in which $V_f(G)$ is Baer-by-finite. Then

• (i) ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is finite;

• (ii) There exists a finite normal subgroup N of G such that every f-subnormal subgroup of ${\overline{\omega }}_{i}(G)/N$ is a normal subgroup.

Proof

We may suppose that ${\overline{\omega }}_{i}(G)\ne {\overline{\omega }}(G)$; otherwise, (i) and (ii) hold trivially since ${\overline{\omega }}(G)$ is a T-group. Let $V=V_f(G)$.

(i) Since V is Baer-by-finite it contains a characteristic Baer subgroup of finite index and Theorem 2.9 implies that V is Prüfer-by-finite.

If V is finite, then G has only finitely many finite f-subnormal subgroups. If H is such a finite f-subnormal subgroup, then $|G:N_G(H)|$ is finite and as

$$\begin{aligned} {\overline{\omega }}(G)={\overline{\omega }}_{i}(G)\cap \bigcap \{N_G(H) | H \text { is finite f -subnormal in }G\} \end{aligned}$$

(i) follows in this case.

Hence we may suppose that V is infinite and let P be the Prüfer p-subgroup of V, for some prime p. Of course, $P\triangleleft \,G$. It follows by Lemma 2.4 that $V\le C_G(P)$. Furthermore, there is a finite subgroup F of G such that $V=FP$. Let $p^d$ denote the highest power of p dividing |F| and let Q denote the set of elements of P of order at most $p^d$. Then $K=QF$ is a finite characteristic subgroup of G, so $G/C_G(K)$ is finite. Let $C=C_G(K)$ and note that C acts as a group of power automorphisms on V. Hence C normalizes the finite f-subnormal subgroups of G, so $C\cap {\overline{\omega }}_{i}(G)\le {\overline{\omega }}(G)$. Then $|{\overline{\omega }}_{i}(G):C\cap {\overline{\omega }}_{i}(G)|$ is also finite, as is $|{\overline{\omega }}_{i}(G):{\overline{\omega }}(G)|$.

(ii) If V is infinite, then we choose K and P as above. In this case, if H / K is f-subnormal in ${\overline{\omega }}_{i}(G)K/K$ then of course H is normalized by ${\overline{\omega }}_{i}(G)$, if H is infinite. If H is finite, then $H\le V=KP$, so H / K is a characteristic subgroup of V / K and hence is normalized by G / K. It follows that ${\overline{\omega }}_{i}(G)K/K$ has the property stated. Hence we take $N={\overline{\omega }}_{i}(G)\cap K$.

If V is finite, then all nontrivial f-subnormal subgroups of ${\overline{\omega }}_{i}(G)V/V$ are infinite and hence are normalized by ${\overline{\omega }}_{i}(G)V/V$. Thus every f-subnormal subgroup of ${\overline{\omega }}_{i}(G)V/V$ is a normal subgroup and we take $N={\overline{\omega }}_{i}(G)\cap V$ in this case. This completes the proof. $\square$

It was shown in [2] that $\omega _{i}(G)/\omega (G)$ is always residually finite and we obtain the same result concerning ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$. However, its structure is even more restricted. First we prove the following preliminary result.

Lemma 3.2

Let G be a group and let F be a finite f-subnormal subgroup of G. Then $|{\overline{\omega }}_{i}(G):N_G(F)\cap {\overline{\omega }}_{i}(G)|$ is finite.

Proof

We may assume that ${\overline{\omega }}_{i}(G)$ is infinite. It is easy to see that we only need prove the result when F is abelian, so we assume this from now on. Let

$$\begin{aligned} {\bar{F}}=\bigcap \{K| F\le K, K \text { is an infinite f -subnormal of }G\}. \end{aligned}$$

Clearly $F\le {\bar{F}}$, whereas if $x\in {\overline{\omega }}_{i}(G)$, then $x\in N_G(K)$ for all infinite f-subnormal subgroups K of G, so $x\in N_G({\bar{F}})$. Hence ${\bar{F}}$ is normalized by ${\overline{\omega }}_{i}(G)F$. Since ${\overline{\omega }}_{i}(G)F$ is an infinite f-subnormal subgroup of G we have ${\bar{F}}\le {\overline{\omega }}_{i}(G)F$, so ${\bar{F}}\triangleleft \,{\overline{\omega }}_{i}(G)F$.

If ${\bar{F}}$ is finite, then F has only finitely many conjugates in ${\overline{\omega }}_{i}(G)F$, so that $|{\overline{\omega }}_{i}(G):N_{G}(F)\cap {\overline{\omega }}_{i}(G)|$ is finite in this case.

On the other hand, suppose that ${\bar{F}}$ is infinite. Since F is f-subnormal in G there is a finite subgroup L such that $F\le L\triangleleft \,{\bar{F}}$. Clearly $C_{{\bar{F}}}(L)F$ is an infinite f-subnormal subgroup of G containing F, so ${\bar{F}}=C_{{\bar{F}}}(L)F$. Since F is abelian we have $F\le Z({\bar{F}})$; also, $Z({\bar{F}})$ is a normal subgroup of ${\overline{\omega }}_{i}(G)F$, which is f-subnormal in G.

If $Z({\bar{F}})$ is finite, then F has finitely many conjugates in ${\overline{\omega }}_{i}(G)F$, and hence, $|{\overline{\omega }}_{i}(G):N_{G}(F)\cap {\overline{\omega }}_{i}(G)|$ is finite. If $Z({\bar{F}})$ is infinite, then it coincides with ${\bar{F}}$. Since ${\bar{F}}/F$ contains no proper infinite subgroups it is Prüfer. If $|F|=d$, then the subgroup of elements of ${\bar{F}}$ whose order divides d is a finite characteristic subgroup, M say, of ${\bar{F}}$ containing F and of course $M\triangleleft \,{\overline{\omega }}_{i}(G)F$. Hence, as before, $|{\overline{\omega }}_{i}(G):N_{G}(F)\cap {\overline{\omega }}_{i}(G)|$ is finite. $\square$

Theorem 3.3

For all groups G the quotient group ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is residually finite. Furthermore, ${\overline{\omega }}_{i}(G)$ is either finite or ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is Dedekind.

Proof

We may assume that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is infinite. Note that

$$\begin{aligned} {\overline{\omega }}(G)&={\overline{\omega }}_{i}(G)\cap X\text { where }\\ X&=\bigcap \{N_G(F)| F \text { is a finite f -subnormal subgroup of } G\}. \end{aligned}$$

Let F be a finite f-subnormal subgroup of G. By Lemma 3.2 it follows that $|{\overline{\omega }}_{i}(G):N_G(F)\cap {\overline{\omega }}_{i}(G)|$ is finite and it is immediate that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is residually finite.

Next note that if R is a normal subgroup of ${\overline{\omega }}_{i}(G)$ of finite index and if $R\le H\le {\overline{\omega }}_{i}(G)$, then H is f-subnormal in G so H is normalized by ${\overline{\omega }}_{i}(G)$. Hence ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is actually residually (finite Dedekind).

Factoring by ${\overline{\omega }}(G)$ we may assume that ${\overline{\omega }}_{i}(G)$ is an infinite residually (finite Dedekind) subgroup of G and suppose that $\{R_{\lambda }\}_{\lambda \in \Lambda }$ is a finite residual system for ${\overline{\omega }}_{i}(G)$ such that ${\overline{\omega }}_{i}(G)/R_{\lambda }$ is a finite Dedekind group and with the property that if $\lambda , \mu \in \Lambda$ there is an element $\alpha \in \Lambda$ such that $R_{\alpha }\le R_{\lambda }\cap R_{\mu }$. Then ${\overline{\omega }}_{i}(G)'R_{\lambda }/R_{\lambda }$ has order at most 2 and it is easy to see that this means that $({\overline{\omega }}_{i}(G)/{\overline{\omega }}(G))'$ itself has order at most 2. Lemma 2.6 shows immediately that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is Dedekind, as required. $\square$

If we add a generalized solubility condition, then we can further strengthen our results. We recall that a group G is subsoluble if it has an ascending subnormal series whose factors are abelian. We note that this class of groups is ${\mathbf {N}}$-closed and that if G is subsoluble then $C_G(B)\le B$, where B is the Baer radical of G. These results appear in [17].

Theorem 3.4

Let G be an infinite subsoluble group such that ${\overline{\omega }}_{i}(G)\ne {\overline{\omega }}(G)$. Then

• (i) G is a soluble group with a normal Prüfer p-subgroup P such that G / P is finite-by-(torsion-free abelian);

• (ii) $G/{\overline{\omega }}(G)$ has finite exponent;

• (iii) ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is a finite abelian $\{p, p-1\}$-group.

Proof

(i) Let B be the Baer radical of G and note that $C_G(B)\le B$. By Theorem 2.9 it follows that B is Prüfer-by-finite and nilpotent. Since G is infinite it is clear that we may write $B=PF$, where P is a Prüfer p-group and F is a finite characteristic subgroup of B.

Clearly $G/C_G(P)$ is abelian and has finite torsion subgroup. Also $G/C_G(F)$ is finite, so if $x\in C_G(P)$, then there exists a natural number n such that $x^n\in C_G(B)\le B$, and hence, x has finite order. Thus $C_G(P)$ is periodic. Also $C_G(P)/C_G(B)\cong C_G(P)C_G(F)/C_G(F)$ is finite, so $C_G(P)$ is Prüfer-by-finite. Thus (i) follows. Furthermore, $V_f(G)\le C_G(P)$ by Lemma 2.4.

(ii) By (i) G / P is centre-by-(finite exponent). Let Y be an infinite f-subnormal subgroup of G. If T / P is the torsion subgroup of G / P, then T is Prüfer-by-finite and $P\le Y\le T$ if Y is periodic. If Y contains an element of infinite order, then $Y\nleqq C_G(P)$, so Lemma 2.4 implies that $P\le Y$. In any case, if Z / P is the centre of G / P it follows that $[Z,Y]\le P\le Y$, so $Z\le {\overline{\omega }}_{i}(G)$. In particular (ii) follows since Lemma 3.1 implies that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is finite.

(iii) It remains to prove that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is an abelian $\{p,p-1\}$-group. Let q be a prime not dividing p or $p-1$ and suppose that there is a nontrivial element $a{\overline{\omega }}(G)$ of ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ of order q. There is a finite cyclic f-subnormal subgroup $E=\langle e\rangle$ of G of prime power order which is not normalized by a. Since EP is infinite, Lemma 2.4 implies that |e| is a power of p.

If aP has finite order, then we may choose a so that aP has order a power of q. Then the q-component of ${\overline{\omega }}_{i}(G)/P$ commutes with the subgroup EP / P, so $e^a=ec$ for some element $1\ne c\in P$.

On the other hand, if aP has infinite order we have $e^a=e^nc$ for some nontrivial element $c\in P$ and some integer n. Thus

$$\begin{aligned} (aeP)^{aP}= ae^nP. \end{aligned}$$

However, G / P is finite-by-abelian, so every subgroup of G / P is f-subnormal and in particular this is true of $\langle aeP\rangle$. But $\langle ae\rangle P$ is normalized by a, so it follows that

$$\begin{aligned} (aeP)^{aP}= aeP. \end{aligned}$$

Hence $e^a=ec'$ for some $c'\in P$.

Thus in either case, there is a natural number m such that $a^{p^m(p-1)}$ centralizes E. However, $a^q$ also normalizes E, and hence, a normalizes E, which is a contradiction. Thus ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is a $\{p,p-1\}$-group.

Clearly ${\overline{\omega }}_{i}(G)/P$ is a Dedekind group. Thus if G / P is not periodic, then ${\overline{\omega }}_{i}(G)/P$ is abelian as is ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ in this case.

Hence we may assume by (ii) that G is periodic, so it is Prüfer-by-finite. We suppose ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is nonabelian. Let Q / P be the 2-component of ${\overline{\omega }}_{i}(G)/P$. Let $Q'P/P=\langle aP\rangle$ and note that $Q'P/P\le Z(G/P)$. Clearly there is an f-subnormal subgroup $X=\langle x\rangle$ of prime power order that is not normalized by a. Hence $x^a=xc$ for some nontrivial $c\in P$. Of course $G/C_G(P)$ is cyclic, so $a\in C_G(P)$ and Lemma 2.4 shows that $x=x^{a^2}=xc^2$. This implies that P is a 2-group. By Lemma 2.4$[x,c]=1$, so $\pi (\langle x\rangle )=\{2\}$. Thus $H=QX$ is locally nilpotent and f-subnormal in G. In particular $Q\le {\overline{\omega }}_{i}(G)\le {\overline{\omega }}_{i}(H)=\omega _{i}(H)$, so $Q'\le \omega _{i}(H)={\overline{\omega }}_{i}(H)$ by [2, Theorem 8], which means that a normalizes X, a contradiction. The result follows. $\square$

We note that, in contrast, the authors of [2] exhibit an example of a periodic soluble group G in which $\omega _{i}(G)/\omega (G)$ is nonabelian.

Following [4] we shall call a group in which every subgroup is f-subnormal an S-group.

Corollary 3.5

Let G be a group.

• (i) Suppose that ${\overline{\omega }}_{i}(G)$ is contained in an infinite subsoluble normal subgroup of G. Then there is a prime p such that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is an abelian residually finite $\{p,p-1\}$-group.

• (ii) If G is an (infinite subsoluble)-by-S-group, then there is a prime p such that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is an abelian residually finite $\{p,p-1\}$-group.

• (iii) If G is an infinite subsoluble-by-finite group, then there is a prime p such that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is a finite abelian $\{p,p-1\}$-group.

Proof

We may suppose throughout that ${\overline{\omega }}_{i}(G)\ne {\overline{\omega }}(G)$.

(i) Let H be an infinite subsoluble normal subgroup of G containing ${\overline{\omega }}_{i}(G)$ and let X be a finite cyclic f-subnormal subgroup of G which is not normalized by some element of ${\overline{\omega }}_{i}(G)$. Then XH is an infinite subsoluble f-subnormal subgroup of G and it is easy to see that ${\overline{\omega }}_{i}(XH)\ne {\overline{\omega }}(XH)$. Since ${\overline{\omega }}_{i}(G)\le {\overline{\omega }}_{i}(HX)$, it follows that ${\overline{\omega }}_{i}(G)/({\overline{\omega }}_{i}(G)\cap N_G(X))$ is a finite abelian $\{p,p-1\}$-group, applying Theorem 3.4 to XH. This implies that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is abelian and a residually finite $\{p,p-1\}$-group.

(ii) Let H be the subsoluble radical of G. Since ${\overline{\omega }}_{i}(G)H/H$ is Dedekind we have ${\overline{\omega }}_{i}(G)\le H$ and the result follows by (i).

(iii) Let H be the subsoluble radical of G. Then ${\overline{\omega }}_{i}(G)\le H$, so by (ii) ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is an abelian residually finite $\{p,p-1\}$-group. Suppose that this quotient is infinite. Then there is an infinite family of finite cyclic f-subnormal subgroups $\{\langle x_{i}\rangle | i\in {\mathbb {N}}\}$ of G such that

$$\begin{aligned} {\overline{\omega }}_{i}(G)\cap H_{n+1}\lneqq {\overline{\omega }}_{i}(G)\cap H_n, \end{aligned}$$

where $H_n=\bigcap _{j\le n} N_G(\langle x_j\rangle )$. If T is a transversal to H in G, then we may write $x_{i}=t_{i}h_{i}$, where $t_{i}\in T$ and $h_{i}\in H$. It follows that there exists $t\in T$ and an infinite subset I of ${\mathbb {N}}$ such that $x_{i}=th_{i}$ for all $i\in I$. Then $H\langle t\rangle$ is subsoluble and f-subnormal in G and contains all these particular $\langle x_{i}\rangle$. By Theorem 3.4 applied to $H\langle t\rangle$ there exists $k\in I$ such that ${\overline{\omega }}_{i}(H\langle t\rangle )\cap H_k={\overline{\omega }}_{i}(H\langle t\rangle )\cap H_{n}$ for all $n\in I$ such that $n\ge k$. Then ${\overline{\omega }}_{i}(G)\cap H_k={\overline{\omega }}_{i}(G)\cap H_n$ for all such n, which is a contradiction. Thus ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is a finite abelian $\{p,p-1\}$-group as required. $\square$

There are many classes of soluble groups where ${\overline{\omega }}_{i}(G)={\overline{\omega }}(G)$; the following corollary exhibits two such.

Corollary 3.6

Let G be an infinite subsoluble group.

• (i) If G is finitely generated, then ${\overline{\omega }}_{i}(G)={\overline{\omega }}(G)$.

• (ii) If G contains no Prüfer subgroups, then ${\overline{\omega }}_{i}(G)={\overline{\omega }}(G)$.

Proof

(i) If we assume the contrary, then Theorem 3.4 implies that there is a Prüfer subgroup P such that G / P is finitely presented. Then P is finitely generated and this is a contradiction.

(ii) We assume the contrary, and in this case, the Baer radical is finite, and hence, G is finite. $\square$

We next consider the situation when ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is infinite and collect a certain amount of information concerning this situation. We note that Example 3 shows that every finite Dedekind group appears as the quotient ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ for some group G.

Lemma 3.7

Let G be a group such that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is infinite. Then $\gamma _3({\overline{\omega }}_{i}(G))$ is an infinite ${\mathfrak {F}}$-perfect subgroup, centralizing all finite f-subnormal subgroups of G. Moreover, $\gamma _3({\overline{\omega }}_{i}(G))$ is the nilpotent residual of ${\overline{\omega }}_{i}(G)$.

Proof

Let

$$\begin{aligned} W={\overline{\omega }}_{i}(G)/{\overline{\omega }}(G) \end{aligned}$$

and note that W is nilpotent of class 2, so $\gamma _3({\overline{\omega }}_{i}(G))\le {\overline{\omega }}(G)$. If $\gamma _k({\overline{\omega }}_{i}(G))$ is finite for some k, then ${\overline{\omega }}_{i}(G)$ is finite-by-nilpotent and hence nilpotent-by-finite. However, Theorem 2.9 shows that the Baer radical of G is Prüfer-by-finite, so ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ would be finite by Corollary 2.5. Hence $\gamma _k({\overline{\omega }}_{i}(G))$ is infinite for all k, and in particular, the nilpotent group ${\overline{\omega }}_{i}(G)/\gamma _k({\overline{\omega }}_{i}(G))$ is always Dedekind. Hence $\gamma _3({\overline{\omega }}_{i}(G))$ is infinite and also the nilpotent residual of ${\overline{\omega }}_{i}(G)$. Furthermore, if M is a subgroup of finite index in $\gamma _3({\overline{\omega }}_{i}(G))$, then M is an infinite f-subnormal subgroup of G, so M is normal in ${\overline{\omega }}_{i}(G)$. Then ${\overline{\omega }}_{i}(G)/M$ is also Dedekind and hence nilpotent of class at most 2. It follows that $M=\gamma _3({\overline{\omega }}_{i}(G))$, so $\gamma _3({\overline{\omega }}_{i}(G))$ has no proper subgroups of finite index. Furthermore, let $H=\langle h\rangle$ be a finite cyclic f-subnormal subgroup of G of prime power order. By Lemma 3.2 we know that $|{\overline{\omega }}_{i}(G):N_H|$ is finite, where $N_H=N_G(H)\cap {\overline{\omega }}_{i}(G)$. Since $N_H/C_{N_H}(H)$ is finite it follows that $\gamma _3({\overline{\omega }}_{i}(G))\le C_{{\overline{\omega }}_{i}(G)}(H)$. $\square$

Lemma 3.8

Let G be a group such that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is periodic. Then $\pi ({\overline{\omega }}_{i}(G)/{\overline{\omega }}(G))$ is finite.

Proof

We may assume that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is infinite. We let

$$\begin{aligned} Z=Z(\gamma _3({\overline{\omega }}_{i}(G))) \end{aligned}$$

and note that Z is Prüfer-by-finite by Theorem 2.9.

Let $H=\langle h\rangle$ be a finite cyclic f-subnormal subgroup of G of prime power order. Clearly $h\in Z(\gamma _3({\overline{\omega }}_{i}(G))H)$ and also $\gamma _3({\overline{\omega }}_{i}(G))H$ is normalized by ${\overline{\omega }}_{i}(G)$ by Lemma 3.7. Hence for each $g\in {\overline{\omega }}_{i}(G)$ we have

$$\begin{aligned} h^g=h^nl\in Z(\gamma _3({\overline{\omega }}_{i}(G))H), \end{aligned}$$

(2)

for some element $l\in Z$ and some natural number n. In particular the order of l must be a power of the only prime belonging to $\pi (H)$. We note that if H is not normalized by g, then $l\ne 1$.

Let ${\mathcal {L}}$ be the set of all finite cyclic f-subnormal subgroups of prime power order which are not normalized by ${\overline{\omega }}_{i}(G)$ and let $\pi ({\mathcal {L}})$ denote the set of all primes p such that there is a subgroup in ${\mathcal {L}}$ of order a power of the prime p. Our work above shows that $\pi ({\mathcal {L}})$ is finite since $\pi (Z)$ is finite.

Since Z is Prüfer-by-finite it follows that $|{\overline{\omega }}_{i}(G):C_{{\overline{\omega }}_{i}(G)}(Z)|$ is finite. Let

$$\begin{aligned} g\gamma _3({\overline{\omega }}_{i}(G))\in C_{{\overline{\omega }}_{i}(G)}(Z)/\gamma _3({\overline{\omega }}_{i}(G)) \end{aligned}$$

be an element of prime power order $p_1^c$ which does not normalize some subgroup $H=\langle h\rangle \in {\mathcal {L}}$ of prime power order $p_2^d$ and suppose that $p_1\ne p_2$. Then $H\gamma _3({\overline{\omega }}_{i}(G))/\gamma _3({\overline{\omega }}_{i}(G))$ is a $p_2$-group and it is centralized by the $p_2'$-component of ${\overline{\omega }}_{i}(G)/\gamma _3({\overline{\omega }}_{i}(G))$. In particular we have

$$\begin{aligned} h^g=hl, \text { where }l\in Z \text { is a }p_2\text {-element}. \end{aligned}$$

On the other hand,

$$\begin{aligned} h=h^{g^{p_1^c}}=hl^{p_1^c} \end{aligned}$$

since $g\in C_{{\overline{\omega }}_{i}(G)}(Z)$, so that l has order a power of $p_1$. Hence $p_1=p_2$. In particular we see that

$$\begin{aligned} \pi ({\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)) \text { is finite}, \end{aligned}$$

since $\pi ({\mathcal {L}})$ is finite and the result follows. $\square$

In particular we have the following.

Corollary 3.9

Let G be a group such that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is an infinite nonabelian group. Then $\pi ({\overline{\omega }}_{i}(G)/{\overline{\omega }}(G))$ is finite.

Proposition 3.10

Let G be a group such that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is an infinite nonabelian group. Let $Z=Z(\gamma _3({\overline{\omega }}_{i}(G)))$.

• (i) If Z contains a Prüfer p-subgroup P, then $p=2$ and ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is a 2-group.

• (ii) If Z is finite, then ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ has finite exponent.

Proof

(i) Suppose that Z is infinite and contains the Prüfer p-subgroup P. Let $g\gamma _3({\overline{\omega }}_{i}(G)) \in C_{{\overline{\omega }}_{i}(G)}(Z)/\gamma _3({\overline{\omega }}_{i}(G))$ be an element of order $q^m$ for some prime $q\ne p$. Then g does not normalize some finite cyclic f-subnormal subgroup H of prime power order and the proof of Lemma 3.7 shows that H also has order a power of the prime q. However, by Lemma 2.4, H is then a characteristic subgroup of $HP=H\times P$ and since HP is infinite, we obtain that g normalizes H, a contradiction. Hence $C_{{\overline{\omega }}_{i}(G)}(Z)/\gamma _3({\overline{\omega }}_{i}(G))$ is a p-group. Since $|{\overline{\omega }}_{i}(G):C_{{\overline{\omega }}_{i}(G)}(Z)|$ is finite the $p'$-component of ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is finite.

Suppose that $p>2$. Since $C_{{\overline{\omega }}_{i}(G)}(P)\ge \gamma _3({\overline{\omega }}_{i}(G))$ it follows that ${\overline{\omega }}_{i}(G)/C_{{\overline{\omega }}_{i}(G)}(P)$ is cyclic with 2-component of order at most 2. Let x be any element of ${\overline{\omega }}_{i}(G)$ which has order 4 modulo ${\overline{\omega }}(G)$. Then there is a cyclic subgroup E of prime power order which is not normalized by $x^2$. Now PE is an abelian f-subnormal p-subgroup (as above) and we may write $PE=P\times \langle a\rangle$ for some element a of order a power of p. Clearly,

$$\begin{aligned} a^x=a^nc_x \end{aligned}$$

for some element $c_x\in P$ and n coprime to p. Also $x^4\in C_{{\overline{\omega }}_{i}(G)}(P)\cap {\overline{\omega }}(G)$, and hence, $x^4$ acts as a group of power automorphisms on the abelian p-group PE. Then [22, Lemma 1.5.4] shows that this automorphism is universal, so $x^4$ centralizes PE. Therefore, x acts on $\langle a\rangle P/P$ as an automorphism of order precisely 4; if not $a^{x^2}=ac$, with $1\ne c\in P$, so $p=2$, or $a^{x^2}=a$ in which case E would be normalized by $x^2$. In particular we have $a^{x^2}=a^{-1}c_{x^2}$ for some $c_{x^2}\in P$.

If y is another element of ${\overline{\omega }}_{i}(G)$ which has order 4 modulo ${\overline{\omega }}(G)$, then $y^2=x^2c$ for some $c\in {\overline{\omega }}(G)$, but $c=x^{-2}y^2\in C_{{\overline{\omega }}_{i}(G)}(P)$, so c centralizes PE as above. It follows that $a^{x^2}=a^{y^2}$. We choose $x,y, w=xy$ to be distinct elements of ${\overline{\omega }}_{i}(G)$ having order 4 modulo ${\overline{\omega }}(G)$ and such that x, y act in the opposite way on $\langle a\rangle P/P$. Then we have

$$\begin{aligned} a^w=ac_w \end{aligned}$$

for some $c_w\in P$. But then

$$\begin{aligned} a^{w^2}=ac_wc_w^w=a^{-1}c_{w^2}, \end{aligned}$$

a contradiction.

Consequently $p=2$.

Suppose next that $g\gamma _3({\overline{\omega }}(G))\in {\overline{\omega }}_{i}(G)/\gamma _3({\overline{\omega }}(G))$ is a nontrivial element of order some prime $q\ne 2$. Then there is a cyclic f-subnormal 2-subgroup E which is not normalized by g, so $PE=P\times \langle a\rangle$ for some element a of order a power of 2. Now $a^g=a^nc$ for some $c\in P$ and integer n coprime to 2. Hence there is a power of 2, say l such that $g_1=g^l$ centralizes P and $\langle a\rangle P/P$. Thus $a^{g_1}=ac$ for some $c\ne 1$ and this implies the contradiction $c^q=1$. It follows that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is actually a 2-group.

(ii) Suppose now that Z is finite of order s say and that, for a contradiction, there is a prime q such that the exponent of the q-component of ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is infinite. For each $m\in {\mathbb {N}}$ there is some $g_m\gamma _3({\overline{\omega }}_{i}(G))\in C_{{\overline{w}}_i(G)}(Z)/\gamma _3({\overline{w}}_i(G))$ for which $g_m^{q^m}$ does not normalize some cyclic f-subnormal subgroup $H_m=\langle h_m\rangle$ of order some power of q as in the proof of Lemma 3.8; we set

$$\begin{aligned} h_m^{g_m}=h_m^{n_m}l_m \end{aligned}$$

(3)

for some $n_m\in {\mathbb {N}}$ and $1\ne l_m\in Z$. Let ${\mathcal {L}}_q$ be the set of all these $H_m$.

Suppose that the subgroups in ${\mathcal {L}}_q$ have orders at most k and set $g=g_t$ where $t>ks$. Then (3) implies

$$\begin{aligned} h_t^{g^r}=h_t^{n_t^r}l_t^{1+n_t+\cdots +n^{r-1}_t} \end{aligned}$$

for $r\le q^t$. Hence there exist $u,\, v\in {\mathbb {N}}$ such that $u<v\le q^t$ and $h_t^{g^u}=h_t^{g^v}$, so $g^{v-u}$ normalizes $h_t$, a contradiction to the choice of g.

Thus the orders of the elements of ${\mathcal {L}}_q$ are unbounded and we take a subset ${\mathcal {L}}_q'$ of ${\mathcal {L}}_q$ having infinitely many, and all distinct, orders. Let $u\in {\mathbb {N}}$. If q divides $n_u$, then (3) shows that the order of $l_u$ is the same as that of $h_u$. Since Z is finite there are only finitely many $m\in {\mathbb {N}}$ for which $H_m\in {\mathcal {L}}_q'$ and q divides $n_m$. We remove the corresponding $H_m$ from ${\mathcal {L}}_q'$ and denote the resulting subset by ${\mathcal {L}}_q''$. Let $H_m\in {\mathcal {L}}_q''$. Then $1\ne l_m^{1+n_m+\cdots +n_m^{r-1}}\in Z$ for all $r\le q^m$. Moreover, if $i<j\le q^m$ and

$$\begin{aligned} l_m^{1+n_m+\ldots +n_m^{i-1}}= l_m^{1+n_m+\cdots +n_m^{j-1}} \end{aligned}$$

then

$$\begin{aligned} l_m^{n^i(1+n_m+\cdots +n_m^{j-i-1})}=1 \end{aligned}$$

and so also

$$\begin{aligned} l_m^{1+n_m+\cdots +n_m^{j-i-1}}=1, \end{aligned}$$

which is a contradiction. Thus there are infinitely many elements in Z, which is the final contradiction. $\square$

A Dedekind 2-group of course has exponent at most 4, so we state the following immediate corollary.

Corollary 3.11

Let G be a group such that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is an infinite nonabelian group. Then ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ has finite exponent.

4 Groups in which ${\overline{\omega }}_{i}(G)$ is finite

In this section we discuss the situation when ${\overline{\omega }}_{i}(G)$ is finite. It is easy to see that if G is the direct product of a quaternion group $Q_8$ of order 8 with infinitely many copies of the alternating group $A_5$, then ${\overline{\omega }}_{i}(G)\cong Q_8\cong {\overline{\omega }}(G)$. On the other hand, in this case we have $\omega _{i}(G)=G=\omega (G)$.

Lemma 4.1

Let G be a group such that $C_G({\overline{\omega }}_{i}(G))$ is infinite. Suppose that either

• (i) all cyclic subgroups of $G/C_G({\overline{\omega }}_{i}(G))$ are f-subnormal or

• (ii) all cyclic subgroups of ${\overline{\omega }}_{i}(G)$ are f-subnormal.

Then ${\overline{\omega }}_{i}(G)$ is nilpotent of class at most 2.

Proof

Let F be a cyclic subgroup of prime power order in ${\overline{\omega }}_{i}(G)$ and let $X=C_G({\overline{\omega }}_{i}(G))$. Since X is infinite, either of (i) or (ii) imply that FX is an infinite f-subnormal subgroup of G, so

$$\begin{aligned} {\overline{\omega }}_{i}(G)\le N_G(FX)\le N_G(Z(FX)) = N_G(FZ(X)). \end{aligned}$$

It follows that FZ(X) is an abelian normal subgroup of ${\overline{\omega }}_{i}(G)X$. This implies that ${\overline{\omega }}_{i}(G)Z(X)/Z(X)\cong {\overline{\omega }}_{i}(G)/Z({\overline{\omega }}_{i}(G))$ is a Dedekind group, so ${\overline{\omega }}_{i}(G)$ is also nilpotent of class at most 3.

Suppose that ${\overline{\omega }}_{i}(G)/Z({\overline{\omega }}_{i}(G))$ is nonabelian and hence periodic. Let p be an odd prime and let Y be the Sylow p-subgroup of ${\overline{\omega }}_{i}(G)$. Then $YZ({\overline{\omega }}_{i}(G))/Z({\overline{\omega }}_{i}(G))$ is abelian, so Y is nilpotent of class at most 2. Let L be the Sylow 2-subgroup of ${\overline{\omega }}_{i}(G)$. Then L / Z(L) is Dedekind and nonabelian, so

$$\begin{aligned} L/Z(L)= \langle aZ(L),bZ(L)\rangle /Z(L) \times C/Z(L), \end{aligned}$$

where

$$\begin{aligned} \langle aZ(L),bZ(L)\rangle /Z(L) \end{aligned}$$

is a quaternion group of order 8 and C / Z(L) is an elementary abelian 2-group generated by elements $c_iZ(L)$ for $i\in I$ an index set.

We shall prove that $a^2\in Z(L)$, giving a contradiction. To see this, observe that $a^2$ commutes with a and Z(L). Also $a^2=b^2z$ for some $z\in Z(L)$, so $a^2$ also commutes with b. However, $(ac_i)^2=a^2z_i$ for some $z_i\in Z(L)$ and $(ac_i)^2$ commutes with $ac_i$ for all $i\in I$. Thus

$$\begin{aligned} ac_i=(ac_i)^{a^2z_i}=a(c_i^{a^2}) \end{aligned}$$

which implies that $c_i=c_i^{a^2}$, so that $a^2$ commutes with each $c_i$ and it follows that $a^2\in Z(L)$, giving the contradiction sought. Hence L / Z(L) is also abelian, so L is nilpotent of class at most 2, as is ${\overline{\omega }}_{i}(G)$, giving the result. $\square$

We may immediately deduce the following result.

Corollary 4.2

Let G be an infinite group.

• (i) If ${\overline{\omega }}_{i}(G)$ is finite, then ${\overline{\omega }}_{i}(G)$ is nilpotent of class at most 2.

• (ii) If ${\overline{\omega }}_{i}(G)$ is a subgroup of the Baer radical of G, then ${\overline{\omega }}_{i}(G)$ is nilpotent of class at most 2.

Proof

(i) Clear from Lemma 4.1.

(ii) By (i) we may suppose that ${\overline{\omega }}_{i}(G)$ is infinite. If ${\overline{\omega }}_{i}(G)={\overline{\omega }}(G)$, then ${\overline{\omega }}_{i}(G)$ is Dedekind and the result follows. Hence we may suppose that ${\overline{\omega }}_{i}(G)\ne {\overline{\omega }}(G)$, so Theorem 2.9 implies that ${\overline{\omega }}_{i}(G)$ is nilpotent and Prüfer-by-finite. Therefore, the Prüfer subgroup of ${\overline{\omega }}_{i}(G)$ is central in ${\overline{\omega }}_{i}(G)$, and hence, $C_G({\overline{\omega }}_{i}(G))$ is infinite. We may now apply Lemma 4.1(ii) to deduce the result. $\square$

Lemma 4.3

Let G be an infinite group such that ${\overline{\omega }}_{i}(G)$ is finite and let $\langle g\rangle$ be a finite f-subnormal subgroup of G of prime power order. Then ${\overline{\omega }}_{i}(G)$ contains a normal subgroup Z such that $Z\le N_G(\langle g\rangle )$ and ${\overline{\omega }}_{i}(G)/Z$ is Dedekind. Furthermore,

$$\begin{aligned} {\overline{\omega }}_{i}(G)/({\overline{\omega }}_{i}(G)\cap N_G(\langle g\rangle )) \end{aligned}$$

is abelian.

Proof

Let $W={\overline{\omega }}_{i}(G)$. There are f-subnormal subgroups F and H such that

$$\begin{aligned} \langle g\rangle \le F\triangleleft \,H \end{aligned}$$

with F finite and H infinite. Let $H_1=C_H(FW)$. Then $H_1W\triangleleft \,HW$, so $H_1W\langle g\rangle$ is f-subnormal in G. Now $T=H_1\cap W\langle g\rangle$ is clearly central in $H_1W\langle g\rangle$, so $T\le N_G(\langle g\rangle )$. Moreover, as $H_1K$ is an infinite f-subnormal subgroup of G for each subgroup $K/T\le W\langle g\rangle /T$ it follows that each subgroup of $W\langle g\rangle /T$ is normalized by W, so WT / T is Dedekind. Let $Z=T\cap W$.

We may clearly suppose that W / Z is nonabelian. Then W contains a subgroup Q such that Q / Z is quaternion. The structure of Q in this situation is given in [12, Theorem 1]. In this case $Q=ZR$, where R is a 2-group generated by elements a, b such that $Q/Z=\langle aZ,bZ\rangle$. We shall prove that $[a,b]\in N_g=N_G(\langle g\rangle )$ so that then $W/(N_g\cap W)$ is abelian.

If $QT/T\cap \langle g\rangle T/T\ne 1$, then $[a,b]=g^ix$ for some $x\in T$ and $i\in {\mathbb {N}}$. Then $[a,b]\in N_g$ since T centralizes g. Hence we may suppose that $QT/T\cap \langle g\rangle T/T =1$. In this case, if the order of g is a power of an odd prime, then as $\langle g\rangle$ is subnormal in $W\langle g\rangle$ and W is nilpotent by Corollary 4.2, it follows that $W\langle g\rangle$ is nilpotent and $R\le N_g$ in this case. Thus we may assume that g has order a power of 2.

Every subgroup of $W\langle g\rangle /T$ is W-invariant, so if gT has order 2, then $g^a=gx, g^b=gy$, for some $x,y\in T$ and clearly $[a,b]\in N_g$. If gT has order at least 4 and if hT is an element of order precisely 4 in $\langle gT\rangle$, then it is easy to check that if aT, bT centralize hT, then $\langle ah\rangle T/T$ is not normalized by b. If say $h^aT=h^{-1}T$, then $(ah)^aT=ah^{-1}T\ne T, ahT, (ahT)^2=a^2T, (ah)^3T=a^{-1}hT$, so in this case we obtain a contradiction. The required result follows. $\square$

The following result is now immediate.

Theorem 4.4

If G is an infinite group in which ${\overline{\omega }}_{i}(G)$ is finite, then ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is abelian.

Lemma 4.5

Let G be a group. Then ${\overline{\omega }}_{i}(G)$ is finite if and only if ${\overline{\omega }}(G)$ is finite.

Proof

Let $W={\overline{\omega }}_{i}(G)$ and let $V={\overline{\omega }}(G)$. Clearly, if W is finite, then so is V, so suppose that V is finite, but that W / V is infinite. Then Theorem 3.3 implies that W / V is nilpotent. Let S be a subgroup of W and note that |SV : S| is finite. However, SV is subnormal in W, so S is f-subnormal in G. Hence all subgroups of W are f-subnormal in G. Thus if $X\le W$, then $V\le N_W(X)$ and by a result of Schenkman [21] $V\le Z_2(W)$, the second centre of W. Thus W is itself nilpotent. Theorem 2.9 implies that W contains a G-invariant Prüfer subgroup P and Corollary 2.5 yields the contradiction that $P\le V$. It follows that W must be finite. $\square$

We may therefore state Theorem 4.4 slightly differently.

Corollary 4.6

If G is an infinite group in which ${\overline{\omega }}(G)$ is finite, then ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is a finite abelian group.

We observe that in any case, using Theorems 3.3 and 4.4:

Theorem 4.7

If G is an infinite group, then ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is Dedekind.

We close this section by exhibiting a further class of groups in which ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is abelian for all groups G in the class.

Corollary 4.8

Let G be an infinite FC-group. Then ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is abelian.

Proof

It is easy to see that for an FC-group, ${\overline{\omega }}(G)$ coincides with the norm of G, so ${\overline{\omega }}(G)$ is Dedekind in this case. By Theorem 4.7${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is also Dedekind, so ${\overline{\omega }}_{i}(G)$ is subsoluble. If ${\overline{\omega }}_{i}(G)$ is finite, then the result follows by Corollary 4.6; otherwise, ${\overline{\omega }}_{i}(G)$ is infinite and Corollary 3.5 yields the result in this case. $\square$

5 Some applications of the results

In this short section we give an application of the results obtained to groups with finitely many normalizers of f-subnormal subgroups. Note that groups with finitely many normalizers of subnormal subgroups were studied in [7]. Let $FC_{sn}$ be the class of groups such that all subnormal subgroups have finitely many conjugates. Let ${\overline{FC}}_{sn}$ be the largest subclass of $FC_{sn}$ closed with respect to taking subgroups of finite index, normal subgroups and factor groups.

Theorem 5.1

Let G be an ${\overline{FC}}_{sn}$-group having finitely many normalizers of infinite f-subnormal subgroups. Then every f-subnormal subgroup of G has finitely many conjugates.

Proof

We use induction on the number of proper normalizers, k, of infinite f-subnormal subgroups. If $k=0$, then all infinite f-subnormal subgroups of G are normal in G. The result follows in this case using Lemma 3.2.

Suppose that $k\ge 1$. Let X be an infinite f-subnormal subgroup of G. By [4, Proposition 3.1] there is a subnormal subgroup S of G with finite index in X and $X\le N_G(S)$. Then $|G:N_G(S)|$ is finite since $G\in {\overline{FC}}_{sn}$. Let $Z=N_G(S)$ and suppose that $Z\ne G$. Since ${\overline{FC}}_{sn}$ is closed under taking subgroups of finite index we have $Z\in {\overline{FC}}_{sn}$. Then Z has one fewer proper normalizer of infinite f-subnormal subgroups than G so by the induction hypothesis $|Z:N_Z(X)|$ is finite. Hence $|G:N_G(X)|$ is likewise finite.

If $S\triangleleft \,G$, then X / S is a finite f-subnormal subgroup of G / S. However, $G/S\in {\overline{FC}}_{sn}$ so Corollary 2.2 implies that $|G/S:N_{G/S}(X/S)|$ is finite, and hence, $|G:N_G(X)|$ is likewise finite.

On the other hand, if X is a finite f-subnormal subgroup of G, then X has finitely many conjugates by Corollary 2.2. $\square$

There are some very easy consequences of this result which enable us to prove results concerning the quotient group $G/{\overline{\omega }}_{i}(G)$. The proof of the next corollary uses [7, Theorem 3.6] and Theorem 5.1.

Corollary 5.2

Let G be a soluble group with finitely many normalizers of f-subnormal subgroups. If G is locally polycyclic, then $|G:{\overline{\omega }}(G)|$ is finite.

By [7, Theorem 4.4] the class of periodic soluble groups with finitely many normalizers of infinite f-subnormal subgroups lies in the class ${\overline{FC}}_{sn}$. We note also that in [7, p.385] it is shown that if G is a periodic soluble IT-group, then $G/\omega (G)$ is finite.

The next result follows using Theorem 3.4 and Theorem 5.1.

Corollary 5.3

Let G be a periodic soluble group with finitely many normalizers of infinite f-subnormal subgroups. Then $|G:{\overline{\omega }}(G)|$ is finite, and in particular, G has finitely many normalizers of f-subnormal subgroups.

Corollary 5.4

Let G be a soluble group with finitely many normalizers of infinite f-subnormal subgroups. If G / Z(G) is periodic, then $|G:{\overline{\omega }}(G)|$ is finite.

Proof

Using Corollary 5.3 we may assume Z(G) is not periodic and hence contains an infinite cyclic subgroup J. If F is a finite f-subnormal subgroup of G, then FJ is f-subnormal in G and F is characteristic in FJ which implies that ${\overline{\omega }}_{i}(G)\le N_G(FJ)\le N_G(F)$ so ${\overline{\omega }}_{i}(G)={\overline{\omega }}(G)$. This completes the proof. $\square$

Corollary 5.5

Let G be a finitely generated soluble-by-finite group with finitely many normalizers of infinite f-subnormal subgroups. Then $|G:{\overline{\omega }}(G)|$ is finite. Furthermore, if G is soluble, then G is nilpotent and centre-by-finite. In particular if G is torsion-free and soluble, then G is abelian.

Proof

It follows from [7, Theorem 3.2] that G is abelian-by-finite so is a polycyclic-by-finite group. This implies in particular that G is an ${\overline{FC}}_{sn}$-group. From Theorem 3.4 and Theorem 5.1 we see that $|G:{\overline{\omega }}(G)|$ is finite. The last part of the statement follows from [7, Theorem 3.2]. $\square$

6 Examples and closing remarks

In this final section we collect together some examples which further exhibit the relationships between ${\overline{\omega }}(G), {\overline{\omega }}_{i}(G), \omega (G)$ and $\omega _{i}(G)$. If G is the dihedral group of order 6, then it is easy to see that $\omega (G)=\omega _{i}(G)$ but that ${\overline{\omega }}_{i}(G)\ne {\overline{\omega }}(G)$. On the other hand, if G is finite and if ${\overline{\omega }}_{i}(G)={\overline{\omega }}(G)$, then ${\overline{\omega }}(G)=G$, so G is a Dedekind group, and hence, $\omega (G)=\omega _{i}(G)$. If G is a finite non-Dedekind p-group then ${\overline{\omega }}_{i}(G)=\omega _{i}(G)=G\ne {\overline{\omega }}(G)=\omega (G)$.

For infinite groups the following example shows that if ${\overline{\omega }}(G)={\overline{\omega }}_{i}(G)$ we need not have $\omega (G)=\omega _{i}(G)$.

Example 1

There is an infinite group G such that ${\overline{\omega }}_{i}(G)= {\overline{\omega }}(G)$, but $\omega (G)\ne \omega _{i}(G)$. Furthermore, $\omega _{i}(G)/\omega (G)$ is not Dedekind.

Proof

Let $p>2$ be prime and for $i\ge 1$ let $X_i$ be a finite perfect group such that $X_i/Z(X_i)$ is a nonabelian simple group, where $Z(X_i)=\langle x_i\rangle$ has order p for all such i. Let $Y=\langle a,b\rangle$ be an elementary abelian p-group of order $p^2$ and define automorphism s, t of Y by $a^s=a^t=a, b^s=ab, b^t=b^{-1}$, so that $Z=\langle s,t\rangle$ is dihedral of order 2p. Let $S=Y\rtimes Z$ be the semidirect product of Y and Z, so S has order $2p^3$ and $Z(S)=\langle a\rangle$. We let G be the central product of S and the groups $X_i$.

Then $Z(G)=\langle a\rangle$. If R is an infinite subnormal subgroup of G, then there is some i for which $X_i\cap R$ is nontrivial. Thus we have $Z(G)\le X_i\le X_i\cap R$ since $X_i$ is perfect. However, using similar arguments it is easy to show that all the infinite subnormal subgroups of G / Z(G) are normal and so $\omega _{i}(G)=G$. On the other hand, $\omega (G)=\langle X_1,X_2,\ldots \rangle$. Since every finite subgroup of G is f-subnormal in G we have ${\overline{\omega }}_{i}(G)={\overline{\omega }}(G)=\langle a\rangle$. $\square$

The next example provides in particular an infinite finitely generated group G in which $\omega _{i}(G)\ne \omega (G)$, answering a question posed in [2]. We know of no example of a finitely generated group G in which ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ (or $\omega _{i}(G)/\omega (G)$) is nonabelian.

Example 2

Let A be a nontrivial finite abelian group. Then there is an infinite group G such that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)\cong A\times A$.

Proof

Let p be a sufficiently large prime number and let $H=G(\infty )$ be the group constructed in [15, 27.2], where we take $n_0=p^i$. Then H is an infinite 2-generator group whose maximal subgroups are cyclic of order $p^i$ and any two distinct subgroups of order $p^i$ have trivial intersection in H, by [15, Theorem 28.2]. The group H has a (graded) presentation F / N where F is the free group of rank 2 and N is a normal subgroup of F. Furthermore, this presentation satisfies Condition R by [15, Lemma 27.2]. It follows by [15, Corollary 31.1] that N / [F, N] is a free abelian group of countable rank.

The Schur multiplier $(F'\cap N)/[F,N]$ is then a free abelian group of at most countable rank. However, $N/(N\cap F')\cong NF'/F'=F/F'$ is 2-generator, so the Schur multiplier must have countable rank. By the universal coefficient theorem (see [19, 11.4.18], for example) there is a nonsplitting central extension M of an arbitrary countable abelian group by H. Moreover, using [19, Exercise 10, p.354] and the fact that M / Z(M) is a nonabelian simple group, we may arrange that M is perfect.

Write $A=\text {Dr}\,\langle a_i\rangle$ and define $B=\text {Dr}\,\langle b_i\rangle , C=\text {Dr}\,\langle c_i\rangle$ with $|a_i|=|b_i|=|c_i|$. Now use the above construction to form a group M with $Z(M)=A$ and set $G=(M\times B)\rtimes C$. We define an action of A on $G\times B$ by $b_i^{c_i}=b_ia_i, b_i^{c_j}=b_i$ for $j\ne i$ and $g^{c_i}=g$ for all $g\in M$, for all $i\ge 1$. Clearly G is a finitely generated group. If S is an infinite subnormal subgroup of G, then it is easy to see that, since M is perfect, $M\le S$. Thus $\omega _{i}(G)={\overline{\omega }}_{i}(G)=G$. It is also easy to see that $\omega (G)={\overline{\omega }}(G)=M$ and this completes the proof. $\square$

We saw above that ${\overline{\omega }}_{i}(G)$ is finite if and only if ${\overline{\omega }}(G)$ is finite. We note that the results of [2] can be used to show that $\omega _{i}(G)$ is finite if and only if $\omega (G)$ is finite. Furthermore, if $\omega _{i}(G)$ is finite, then so is ${\overline{\omega }}_{i}(G)$. However, it is easy to construct an infinite periodic soluble group G in which $G/\omega (G)$ is finite, while ${\overline{\omega }}(G)$ is trivial. For example, let $p_i,q_i$ for $i\ge 1$ be infinite sequences of distinct primes such that $p_i\equiv 1\pmod {q_i}$ which can be constructed by Dirichlet’s theorem on the number of primes in arithmetic progressions and for $i\ge 1$ let $G_i$ be the natural semidirect product of the group of order $p_i$ by the group of order $q_i$. In this case it is easy to see that ${\overline{\omega }}_{i}(G)={\overline{\omega }}(G)=1$, but that $\omega _{i}(G)=\omega (G)=G$.

In general the group ${\overline{\omega }}_{i}(G)$ need not be a Dedekind group. We know from [2, Example 1] that there is a finite perfect group X such that X / Z(X) is a nonabelian simple group such that Z(X) has order p. For each $i\ge 1$, let $X_i\cong X$ and let H be the central product of these groups $X_i$ with amalgamated centres. Let $K=H\times \langle a\rangle$, where $\langle a\rangle$ is an element of order p. Let $\langle b\rangle$ denote the centre of H and let $G=K\rtimes \langle c\rangle$, where c is an element of order p, $h^c=h$ and $a^c=ab$. If Y is an infinite f-subnormal subgroup of G, then Y contains b and it follows easily that ${\overline{\omega }}_{i}(G)=\langle a,c\rangle$, which is nilpotent of class 2, but not Dedekind.

Next we adapt a construction due to Heineken (see [13, Example 3]) which shows that every finite Dedekind group can be realized as the quotient ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ for some group G.

Example 3

If G is a finite Dedekind group, then there is a group R such that ${\overline{\omega }}_{i}(R)/{\overline{\omega }}(R)\cong G\cong \omega _{i}(R)/\omega (R)$.

Proof

The group G is a subgroup of the Galois group of some finite Galois extension F of ${\mathbb {Q}}$, the field of rational numbers. Let $L=GL(V,F)$ be the group of invertible linear mappings of the F-vector space V of countably infinite dimension and let $L^*$ denote the set of those mappings which fix elementwise some subspace of finite codimension. Rosenberg [20] showed that $L^+=L/L^*$ is simple modulo its centre, that $L=L'$ and $Z(L^+)\cong F^*$, the multiplicative group of F. By [11, Theorem 127.2] $F^*$ is isomorphic to the direct product of a finite cyclic group and a free abelian group of countably infinite rank. Let $K=L^+\rtimes G$ where the action of G is as follows. We fix a basis for V, say $\{x_1,x_2,\ldots \}$, let $\sigma \in G$ and define $(fx_i)^{\sigma }=f^{\sigma }x_i$; then the action of G on L is conjugation, namely if $\alpha \in L$ and $\sigma \in G$ then we define $\alpha ^{\sigma }$ to be the composite map $\sigma ^{-1}\alpha \sigma$ and under this action $L^*$ is G-invariant, so there is an induced action of G on $L^+$. We note that if $f\in F^*=Z(L^+)$, then $f^{\sigma }=\sigma (f)\in F^*$, for all $\sigma \in G$.

Let $Z(L^+)=S\times U$, where S is a finite cyclic group and U is freely generated by elements $u_1,u_2,\ldots$. Let $1\ne g\in G$. We shall define a subgroup $H_g$ with finite index in $Z(L^+)$ and an element $f_g\in Z(L^+)$ such that $(f_g)^gH_g$ is not a power of $f_gH_g$.

First, suppose that, for some i, $u_i^g=u_i^ntu_{j_1}^{m_1}\ldots u_{j_k}^{m_k}$, where $t\in S$, $i, j_1,\ldots j_k$ are pairwise distinct, $k\in {\mathbb {N}}$ and $m_1,\ldots , m_k,n\in {\mathbb {Z}}$. If $t\ne 1$, take $H_g=U$. Then $u_i^gH_g=u_i^ntH_g\ne u_i^rH_g$, for any integer r.

If $t=1$ and there is some index, $j_1$ say, such that $u_{j_1}^{m_1}\ne 1$, then we let

$$\begin{aligned} H_g =\langle u_j, u_{j_1}^p|j\ne i, j_1 \text { and }p \text { is a prime greater than }|m_1|\rangle . \end{aligned}$$

Again we can easily see that $u_i^gH_g\ne u_i^rH_g$, for any integer r. In the case when $u_i$ is mapped to some power $u_i^{m_i}$ for all i it follows that $m_i=\pm 1$, since g has finite order. If g fixes U, then there is some element $s\in S$ such that s has prime power order $q^m$, say, and $s^g\ne s$. Then $(su_1)^g\ne (su_1)^k$ for any integer k and we take

$$\begin{aligned} H_g=\langle u_j, u_{1}^{q^m}|j \ge 2\rangle . \end{aligned}$$

Suppose that there is an index l such that $u_l^g=u_l^{-1}$. If there is an index $m\ne l$ such that $u_m$ is fixed by g, then we replace $u_m$ by $u_m'=u_lu_m$. Then $(u_m')^g=u_m'u_l^{-2}$ and we can use the argument above to define $H_g$. Hence we may assume that g acts by inversion on U. Now if $n\in {\mathbb {Q}}$, then n is fixed by g, but $n=su$ with $s\in S, u\in U$, so $su=n=n^g=s^gu^{-1}$ which gives a contradiction. This completes the construction of the group $H_g$ and we let

$$\begin{aligned} J=\bigcap _{g\in G}H_g, T=J_G, R= K/T. \end{aligned}$$

If there were an element $g\in G$ acting as a power automorphism on $Z(L^+)/T$, then this would also be the case for $Z(L^+)/H_g$, which is a contradiction to the choice we made. Suppose that H / T is an infinite f-subnormal subgroup of R, so that there is an infinite subnormal subgroup M / T of R such that $M\triangleleft \,H$ and H / M is finite. Since $L^+/Z(L^+)$ is simple and $L^+$ is perfect, it is easy to see that $L^+\le M\le H$ and as G is Dedekind we have $H\triangleleft \,K$. Thus ${\overline{\omega }}_{i}(R)=R$.

However, every subgroup of $Z(L^+)/T$ is subnormal in K / T, so ${\overline{\omega }}(R)=\omega (R)=L^+/T$. Thus ${\overline{\omega }}_{i}(R)/{\overline{\omega }}(R)\cong G$ as required. $\square$

In fact we know no example of a group G for which ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is infinite.

Let N(G) denote the norm of the group G. The final result shows in particular that, by contrast, there is no finite group G such that ${\overline{\omega }}_{i}(G)/{\overline{\omega }}(G)$ is isomorphic to the quaternion group of order 8.

Proposition 6.1

There is no group G such that G / N(G) is isomorphic to the quaternion group of order 8.

Proof

Let $N=N(G)$ and suppose, for a contradiction, that G is a group such that $G/N=\langle xN,yN\rangle$ is quaternion of order 8. Then there is a cyclic subgroup H of G which is not normalized by $y^2$, and hence, x, y, xy do not normalize H either. By symmetry we may assume that H is one of the subgroups $\langle n\rangle ,\langle nx\rangle , \langle ny\rangle , \langle ny^2\rangle$, for some $n\in N$. Since $y^2=n_1x^2$ for some $n_1\in N$ indeed we may assume that H is one of the three possibilities

$$\begin{aligned} \langle n\rangle ,\langle ny\rangle , \langle ny^2\rangle . \end{aligned}$$

It is easy to see that the norm of $N\langle y\rangle$ is N and it follows from [1, Zusatz, p. 267] that N must be abelian. Let $H=\langle ny\rangle$. Since N normalizes $\langle y\rangle$ we have

$$\begin{aligned} z_y=[ny,y]=[n,y] \in N\cap \langle y\rangle \le Z(N\langle y\rangle ) \text { and } y^n= yz_y^{-1}. \end{aligned}$$

Furthermore, $z_y\notin \langle ny\rangle$ and since $(ny)^{y^2}=nyz_y^2\notin \langle ny\rangle$, but $ny=(ny)^{y^4}=nyz_y^4$ we see that $z_y$ has order exactly 4. Thus $\langle z_y\rangle \cap \langle ny\rangle =1$. Then $(ny)^n=ny^n=nyz_y^{-1}$ and since $n\in N$, the norm of G, we obtain the contradiction that $z_y\in \langle ny\rangle$. We obtain a similar contradiction if $H=\langle ny^2\rangle$.

On the other hand, let $H=\langle n\rangle$. Using the notation above, we see that $z_x^2=z_y^2=z_{xy}^2$, as $x^2N=y^2N=(xy)^2N$. Also $n^{xy}=nz_yz_x^y=nz_{xy}$ and $z_{xy}$ commutes with xy so that

$$\begin{aligned} n^{(xy)(yx)^{-1}}=n^{(xy)^2}=n(z_yz_x^y)^2=nz_y^2. \end{aligned}$$

From this we obtain

$$\begin{aligned} z_yz_x^yz_yz_x^y=z_{y}^2, \text { so } z_x^yz_yz_x^y=z_y \text { and } z_x^2=1, \end{aligned}$$

since N is abelian. This further contradiction gives the result. $\square$