Groups with many abelian or self-normalizing subgroups

Abstract

Groups in which every non-abelian subgroup is equal to its normalizer have been recently completely described. This paper investigates locally soluble groups with restrictions on subgroups which are neither abelian nor self-normalizing.

1 Introduction

The structure of groups in which the set of all non-normal subgroups is small in some sense has been investigated in many papers and in several different situations. In particular, Romalis and Sesekin [19,20,21] studied groups whose non-abelian subgroups are normal. A dual problem has been considered recently in [7] where a group G is defined as an ${\mathcal {H}}$-group if any non-abelian subgroup H of G is self-normalizing (i.e. H is equal to the normalizer $N_{G}(H)$ of H in G). That paper deals with soluble ${\mathcal {H}}$-groups while finite non-soluble ${\mathcal {H}}$-groups were fully described later in [5]. Clearly, soluble ${\mathcal {H}}$-groups are metabelian and it turns out that any finite ${\mathcal {H}}$-group is either soluble or simple (see [5, Theorem 2.14]) and also that any infinite locally finite ${\mathcal {H}}$-group is soluble (see [5, Theorem 3.4]). The only finite non-abelian simple ${\mathcal {H}}$-groups are the alternating group $\text {Alt}(5)$ of degree 5 or the projective special linear group $\text {PSL}(2,2^{n})$, where $2^{n}-1$ is a prime (see [5, Theorem 2.17]). A soluble non-nilpotent group G is an ${\mathcal {H}}$-group if and only if $G=\left\langle x\right\rangle \ltimes G^{\prime }$ where x has order a power of a prime p, the derived subgroup $G^{\prime }$ is a periodic abelian $p^{\prime }$-group, and $x^{p}$ belongs to $C_{G}(G^{\prime })$ (see [7, Theorem 1.7]). Of course, any nilpotent group does not have self-normalizing subgroups so that a nilpotent ${\mathcal {H}}$-group which is not abelian is a minimal non-abelian group and hence its structure is well-known.

In [7], it was also proved that any soluble group whose infinite non-abelian subgroups are self-normalizing is either a Chernikov group or an ${\mathcal {H}}$-group. Here in Sect. 2, we will consider minimal conditions related to ${\mathcal {H}}$-groups and the just quoted result will be generalized as follows.

Theorem A1

Let G be a locally soluble group satisfying the minimal condition on subgroups which are neither abelian nor self-normalizing. Then G is either a Chernikov group or an ${\mathcal {H}}$-group.

Theorem A2

Let G be a locally soluble group satisfying the minimal condition on non-${\mathcal {H}}$-subgroups. Then G is either a Chernikov group or an ${\mathcal {H}}$-group.

Recall that a group G is said to have finite (Prüfer) rank r if every finitely generated subgroup of G can be generated by r elements and r is the least such integer. In recent years, the influence on a locally soluble group of the behavior of the subgroups of infinite rank has been investigated (see for instance [3, 4, 6, 8, 9, 12, 13]). This point of view will be adopted here in Sect. 3 were the following results will be obtained.

Theorem B1

Let G be a locally soluble group of infinite rank whose non-abelian subgroups of infinite rank are self-normalizing. Then G is an ${\mathcal {H}}$-group.

Theorem B2

Let G be a locally (soluble-by-finite) group of infinite rank whose subgroups of infinite rank are ${\mathcal {H}}$-groups. Then G is an ${\mathcal {H}}$-group.

For notation and basic facts, we refer to [18].

2 Minimal conditions

The first three lemmas deal with a property of ${\mathcal {H}}$-groups that we need in what follows. Recall that a group G is said to be locally graded if every finitely generated non-trivial subgroup of G contains a proper subgroup of finite index. The class of locally graded groups is quite large and contains, in particular, all locally (soluble-by-finite) groups.

Lemma 2.1

Let G be a locally graded group whose finitely generated non-abelian subgroups are self-normalizing. Then G is a locally (abelian-by-finite) ${\mathcal {H}}$-group.

Proof

Let E be any finitely generated subgroup of G and assume, by contradiction, that E is not abelian-by-finite. Since G is locally graded, E contains a proper normal subgroup $E_{1}$ of finite index. Then $E_{1}$ is finitely generated and non-abelian, so that $E_{1}=N_{G}(E_{1})$; hence $E_{1}=E$. This contradiction proves that E is abelian-by-finite. In particular, G locally satisfies the maximal condition.

Let H be any non-abelian subgroup of G which is not finitely generated and let g be any element of the normalizer $N_{G}(H)$. Consider any non-abelian finitely generated subgroup X of H. Then $\left\langle X,g\right\rangle$ satisfies the maximal condition, so that the subgroup $Y=X^{\left\langle g\right\rangle }$ is a finitely generated non-abelian subgroup of $H^{\left\langle g\right\rangle }=H$ which is normalized by g. Since $N_{G}(Y)=Y$, it follows that g belongs to Y and so also to H. Thus $N_{G}(H)=H$. Therefore G is an ${\mathcal {H}}$-group. $\square$

Lemma 2.2

Let G be a locally graded group. If every finitely generated subgroup of G is an ${\mathcal {H}}$-group, then G is an ${\mathcal {H}}$-group.

Proof

Let H be any finitely generated subgroup of G and let $g\in N_{G} (H)\smallsetminus H$. Then the finitely generated subgroup $K=\left\langle H,g\right\rangle$ is an ${\mathcal {H}}$-group and g belongs to $N_{K}(H)\smallsetminus H$. Thus H is abelian. Therefore all finitely generated subgroups of G are either abelian or self-normalizing, hence application of Lemma 2.1 proves that G is an ${\mathcal {H}}$-group. $\square$

Lemma 2.3

Let G be a locally graded ${\mathcal {H}}$-group. Then G is either finite or metabelian.

Proof

The group G is locally (abelian-by-finite) by Lemma 2.1. If G is periodic, then G is locally finite and so it is either finite or metabelian (see [5, Theorem 3.4]). Hence assume that G is not periodic. In order to prove that G is abelian, it can be assumed that G is finitely generated. Then G contains a normal abelian subgroup A of finite index; in particular, A is not periodic. If $g\in G$, then $A\left\langle g\right\rangle$ is a soluble non-periodic ${\mathcal {H}}$-group and hence it is abelian (see [5, Theorem 3.2] or [7, Lemma 1.2]). Therefore $A\le Z(G)$, so that G/Z(G) is finite. Thus $G^{\prime }$ is finite (see [18, Part 1, Theorem 4.12]) and so G is abelian (see [7, Lemma 1.2]). $\square$

Lemma 2.4

Let G be a nilpotent group satisfying the minimal condition on subgroups which are neither abelian nor self-normalizing. Then G is either abelian or a Chernikov group.

Proof

Since a nilpotent group has no self-normalizing subgroups, G satisfies the minimal condition on non-abelian subgroups and hence G is either abelian or a Chernikov group (see [2]). $\square$

Lemma 2.5

Let G be a group satisfying the minimal condition on subgroups which are neither abelian nor self-normalizing. Then either G is abelian or $G/G^{\prime }$ is periodic.

Proof

Let x be any element of G such that the coset $xG^{\prime }$ has infinite order. Since G has the minimal condition on subgroups which are neither abelian nor self-normalizing, from the consideration of the infinite descending chain of G-invariant subgroups

$$\begin{aligned} \left\langle x^{2},G^{\prime }\right\rangle>\left\langle x^{4},G^{\prime }\right\rangle>\cdots>\left\langle x^{2^{n}},G^{\prime }\right\rangle >\cdots , \end{aligned}$$

it follows that there exists $h\in {\mathbb {N}}$ such that $\langle x^{2^{h} },G^{\prime }\rangle$ is abelian. Similarly also $\langle x^{3^{k}},G^{\prime }\rangle$ is abelian for some $k\in {\mathbb {N}}$. Therefore Fitting’s theorem yields that

$$\begin{aligned} \langle x,G^{\prime }\rangle =\langle x^{2^{h}},G^{\prime }\rangle \langle x^{3^{k}},G^{\prime }\rangle \end{aligned}$$

is nilpotent. It follows that if $x_{1}G^{\prime },\ldots ,x_{t}G^{\prime }$ are elements of infinite order of $G/G^{\prime }$, the subgroup $\left\langle x_{1},G^{\prime }\right\rangle \cdots \left\langle x_{t},G^{\prime }\right\rangle$ is nilpotent by Fitting’s theorem and hence even abelian by Lemma 2.4. Therefore, if $G/G^{\prime }$ is not periodic and ${\mathcal {I}}$ is the set of all elements x of G such that the coset $xG^{\prime }$ has infinite order, the group $G=\left\langle \langle x,G^{\prime }\rangle :x\in {\mathcal {I}}\right\rangle$ itself is abelian. $\square$

Lemma 2.6

Let $G=\left\langle x,A\right\rangle$ where A is a non-periodic finitely generated abelian normal subgroup whose index in G is a power of a prime. If G satisfies the minimal condition on subgroups which are neither abelian nor self-normalizing, then G is abelian.

Proof

Consider first the case in which $\left\langle x\right\rangle \cap A=\{1\}$. Assume further that A is torsion-free, and let the index of A in G be a power of the prime p. If $\left\langle x\right\rangle A^{p^{n}}=\left\langle x\right\rangle A^{p^{n+1}}$ for some positive integer n, then

$$\begin{aligned} A^{p^{n}}=\left\langle x\right\rangle A^{p^{n+1}}\cap A^{p^{n}} =A^{p^{n+1}}(\left\langle x\right\rangle \cap A^{p^{n}})=A^{p^{n+1}} \end{aligned}$$

and this is not possible since A is a free abelian group of finite rank. Thus $\left\langle x\right\rangle A^{p^{n}}\ne \left\langle x\right\rangle A^{p^{n+1}}$ for every positive integer n. On the other hand, $G/A^{p^{n}}$ is a finite p-group, so that it is nilpotent and this implies that $\left\langle x\right\rangle A^{p^{n}}$ is a subnormal subgroup of G for any positive integer n. Therefore the consideration of the descending chain

$$\begin{aligned} \left\langle x\right\rangle A^{p}>\left\langle x\right\rangle A^{p^{2}}>\cdots>\left\langle x\right\rangle A^{p^{n}}>\cdots \end{aligned}$$

and the minimal condition on subgroups which are neither abelian nor self-normalizing gives that $\left\langle x\right\rangle A^{p^{n}}$ is abelian for some positive integer n. Hence $G=\left( \left\langle x\right\rangle A^{p^{n}}\right) A$ is nilpotent. Therefore application of Lemma 2.4 gives that G is abelian when A is torsion-free. If A is not torsion-free and T is the subgroup consisting of all elements of finite order of A, the previous argument gives that G/T is abelian. Therefore $G^{\prime }$ is periodic and, since G is not periodic, it follows from Lemma 2.5 that G is abelian.

Assume now that $\left\langle x\right\rangle \cap A\ne \{1\}$; notice that $\left\langle x\right\rangle \cap A\le Z(G)$. The first part of this proof yields that either $A/\left\langle x\right\rangle \cap A$ is periodic or $G/\left\langle x\right\rangle \cap A$ is abelian. If $A/\left\langle x\right\rangle \cap A$ is periodic, then $G/\left\langle x\right\rangle \cap A$ is finite; thus $G^{\prime }$ is finite (see [18, Part 1, Theorem 4.12]) and so, since G is not periodic, it follows from Lemma 2.5 that G must be abelian. On the other hand, if $G/\left\langle x\right\rangle \cap A$ is abelian, we have that G is nilpotent and so even abelian by Lemma 2.4. The lemma is proved.

$\square$

The next result also solves the non-periodic case of Theorem A1.

Lemma 2.7

Let G be a locally graded group satisfying the minimal condition on subgroups which are neither abelian nor self-normalizing. If G is not periodic, then G is abelian.

Proof

In order to prove that G is abelian, without loss of generality, it can be supposed that G is finitely generated.

Assume first that G is soluble. Suppose by contradiction that G is not abelian and let G be a counterexample with minimal derived length. Application of Lemma 2.5 yields that $G/G^{\prime }$ is finite, hence $G^{\prime }$ is likewise a non-periodic finitely generated group with the minimal condition on subgroups which are neither abelian nor self-normalizing, and so $G^{\prime }$ is abelian by the minimality of derived length of G. Let $G/G^{\prime }=\left\langle x_{1}G^{\prime },\ldots , x_{t}G^{\prime }\right\rangle$ with every $x_{i}G^{\prime }$ of prime-power order. Lemma 2.6 yields that every $\left\langle x_{i},G^{\prime }\right\rangle$ is abelian, so that $G=\left\langle x_{1},G^{\prime }\right\rangle \cdots \left\langle x_{t},G^{\prime }\right\rangle$ is nilpotent and hence even abelian by Lemma 2.4. This contradiction proves the statement for non-periodic soluble groups with the minimal condition on subgroups which are neither abelian nor self-normalizing.

In the general case, let G be locally graded. Since G is finitely generated, it contains a descending chain $G_{1}>\cdots>G_{n}>\cdots$ of proper normal subgroups of finite index; in particular, each $G_{i}$ is not periodic. Since G satisfies the minimal condition on subgroups which are neither abelian nor self-normalizing, there exists a positive integer n such that $G_{n}$ is abelian. If $g\in G$, then $\left\langle g\right\rangle G_{n}$ is a soluble non-periodic group with the minimal condition on subgroups which are neither abelian nor self-normalizing and hence it is abelian by the first part of the proof. Therefore $G_{n}\le Z(G)$ and so G/Z(G) is finite; thus $G^{\prime }$ is finite (see [18, Part 1, Theorem 4.12]). Hence $G\ne G^{\prime }$ and Lemma 2.5 yields that either G is abelian or $G/G^{\prime }$ is periodic. Since G is not periodic, it follows that G is abelian. $\square$

Proof of Theorem A1

Lemma 2.7 allows us to suppose that G is periodic. Assume that G is not a Chernikov group, and let H be an arbitrary finite non-abelian subgroup of G. Let $g\in N_{G}(H)$, and consider the finite subgroup $K=\left\langle H,g\right\rangle$. Then G contains an abelian subgroup A which does not satisfy the minimal condition such that $A^{K}=A$ (see [22]). Hence the socle S of A is infinite and so it is possible to find finite K-invariant subgroups $A_{1},A_{2},\ldots$ of S such that

$$\begin{aligned} \left\langle A_{n}:n\in {\mathbb {N}}\right\rangle =\underset{n\in {\mathbb {N}}}{\text {Dr}}A_{n} \end{aligned}$$

(see [11, Lemma 3.8]). Clearly, the subgroup $\left\langle A_{n}:n\in {\mathbb {N}}\right\rangle \cap K$ is contained in a direct product of finitely many $A_{n}$’s, so that, by replacing $\left\langle A_{n}:n\in {\mathbb {N}}\right\rangle$ with a suitable subgroup of finite index, it can be assumed that

$$\begin{aligned} \left\langle A_{n}:n\in {\mathbb {N}}\right\rangle \cap K=\{1\}. \end{aligned}$$

For any $n\in {\mathbb {N}}$, consider the K-invariant subgroup

$$\begin{aligned} B_{n}=\underset{i\ge n}{\text {Dr}}A_{i}. \end{aligned}$$

Then $B_{1}H>\cdots>B_{n}H>\cdots$ is an infinite descending chain of non-abelian subgroups of G and so there exists a positive integer k such that $N_{G}(B_{k}H)=B_{k}H$. Since $g\in N_{G}(B_{k}H)$, it follows that

$$\begin{aligned} g\in B_{k}H\cap K=H(B_{k}\cap K)=H. \end{aligned}$$

Thus $H=N_{G}(H)$. Therefore all finite non-abelian subgroups of G are self-normalizing and so G is an ${\mathcal {H}}$-group by Lemma 2.1. $\square$

Next we consider groups with the minimal condition on subgroups which do not have the property ${\mathcal {H}}$. The proof of Theorem A2 will be essentially the same as Theorem A1, but we need a corresponding version of Lemma 2.4.

Lemma 2.8

Let G be a nilpotent group satisfying the minimal conditions on non-${\mathcal {H}}$-subgroups. Then G is either abelian or a Chernikov group.

Proof

Let $G_{1}>\cdots>G_{n}>\cdots$ an infinite descending chain of non-abelian subgroups. Then there exists a positive integer n such that $G_{n}$ is an ${\mathcal {H}}$-group. Since G is nilpotent, it follows that $G_{n}$ is a minimal non-abelian group and hence it is finite. This contradiction proves that G satisfies the minimal condition on non-abelian subgroups, thus G is either abelian or a Chernikov group (see [2]). $\square$

Lemma 2.9

Let G be a locally graded group satisfying the minimal conditions on non-${\mathcal {H}}$-subgroups. If G is not periodic, then G is abelian.

Proof

Since any locally graded non periodic ${\mathcal {H}}$-group is abelian, the proof can be obtained in a similar way as in the proofs of Lemmas 2.5, 2.6, and 2.7, replacing in the argument Lemma 2.4 with Lemma 2.8. $\square$

Proof of Theorem A2

Lemma 2.9 allows us to suppose that G is periodic. Assume that G is not a Chernikov group. Consider any finite subgroup H of G. In order to prove the statement, by Lemma 2.2, it is enough to prove that H is an ${\mathcal {H}}$-group. The group G contains an abelian subgroup A which does not satisfy the minimal condition such that $A^{H}=A$ (see [22]). As in the proof of Theorem A1, H-invariant subgroups $B_{1},B_{2},\ldots$ of A can be found such that $B_{1}H>\cdots>B_{n}H>\cdots$ is an infinite descending chain. Hence there exists a positive integer k such that $B_{k}H$ is an ${\mathcal {H}}$-group. Thus H is likewise an ${\mathcal {H}}$-group and the theorem is proved. $\square$

Finally we conclude this section with a result concerning groups in which non-${\mathcal {H}}$-subgroups fall into finitely many conjugacy classes.

Proposition 2.10

Let G be an infinite locally graded group with finitely many conjugacy classes of non-${\mathcal {H}}$-subgroups. Then G is an ${\mathcal {H}}$-group.

Proof

Since any locally graded ${\mathcal {H}}$-group is either finite or soluble by Lemma 2.3 and hence also abelian-by-finite (see [5, Lemma 2.3]), the group G is locally (abelian-by-finite) (see [15, Proposition 3.3]). Then the elements of any chain of non-${\mathcal {H}}$-subgroups are pairwise not conjugate (see [1, Lemma 4.6.3]) and hence, since G has finitely many conjugacy classes on non-${\mathcal {H}}$-subgroups, any chain of non-${\mathcal {H}}$-subgroups subgroups of G is finite. In particular, G satisfies both the minimal and the maximal condition on non-${\mathcal {H}}$-subgroups. Assume that G is not an ${\mathcal {H}}$-group, so that G is periodic by Lemma 2.9. Let H be any minimal element of the set of all non-${\mathcal {H}}$-subgroups of G. Then all proper subgroups of H are ${\mathcal {H}}$-subgroups and hence it follows from Lemma 2.2 that H is finitely generated. Thus H is finite. Since G is infinite and locally finite, it follows that there are infinitely many elements $g_{1},g_{2},\ldots$ of G which give a strictly infinite ascending chain of subgroups

$$\begin{aligned} H<\left\langle H,g_{1}\right\rangle<\cdots< \left\langle H,g_{1},\ldots ,g_{n}\right\rangle <\cdots . \end{aligned}$$

Then $\left\langle H,g_{1},...,g_{n}\right\rangle$ is an ${\mathcal {H}}$-group for some $n\in {\mathbb {N}}$, a contradiction which proves the statement. $\square$

3 Restrictions on subgroups of infinite rank

Lemma 3.1

Let G be a group of infinite rank whose non-abelian subgroups of infinite rank are self-normalizing. If $G^{\prime }$ has finite rank, then G is abelian.

Proof

Let X be any finitely generated subgroup of G. Then $XG^{\prime } /G^{\prime }$ is an abelian finitely generated group and hence, as $G^{\prime }$ has finite rank, the subgroup $XG^{\prime }$ has likewise finite rank. Thus $G/XG^{\prime }$ has infinite rank and so there exists a proper subgroup of infinite rank Y of G containing $XG^{\prime }$; clearly Y is a normal subgroup and so it is abelian. Hence X is itself abelian and so the lemma is proved. $\square$

Lemma 3.2

Let G be a group of infinite rank whose non-abelian subgroups of infinite rank are self-normalizing. If G is abelian-by-finite, then G is an ${\mathcal {H}}$-group.

Proof

Let A be an abelian normal subgroup of finite index of G. Then A has infinite rank and there exists a sequence $(X_{n})_{n\in {\mathbb {N}}}$ of finitely generated G-invariant subgroups of A such that

$$\begin{aligned} X=\left\langle X_{n}:n\in {\mathbb {N}}\right\rangle =\underset{n\in {\mathbb {N}}}{\text {Dr}}X_{n}, \end{aligned}$$

$r(X_{n})<r(X_{n+1})$ for every $n\in {\mathbb {N}}$ (see [8, Lemma 6]). Let H be any finitely generated non-abelian subgroup of G, and let $g\in N_{G}(H)$. Then $K=\left\langle H,g\right\rangle$ is abelian-by-finite and finitely generated, hence $K\cap X$ is likewise finitely generated and so, by replacing X with a suitable subgroup (which is a direct product of infinitely many of the $X_{n}$’s), it can be assumed that $X\cap K=\{1\}$. Then $Y=\left\langle X_{n}:n\ge 2\right\rangle$ is a proper G-invariant subgroup of infinite rank of X, hence HY is a proper non-abelian subgroup of infinite rank of G and so $HY=N_{G}(HY)$. Since $g\in N_{G}(HY)$, it follows that

$$\begin{aligned} g\in HY\cap K=H(Y\cap K)=H. \end{aligned}$$

Thus $H=N_{G}(H)$. Therefore every finitely generated non-abelian subgroup of G is self-normalizing and hence G is an ${\mathcal {H}}$-group by Lemma 2.1. $\square$

As in many problems concerning groups of infinite rank, the existence of a proper normal subgroup of infinite rank plays a crucial role. We have in fact the following result on which the proof of Theorem B1 will depend.

Proposition 3.3

Let G be a locally graded group of infinite rank whose non-abelian subgroups of infinite rank are self-normalizing. If G contains a proper normal subgroup of infinite rank, then G is an ${\mathcal {H}}$-group.

Proof

By Lemma 3.2, it is enough to prove that G is abelian-by-finite. Let N be a proper normal subgroup of infinite rank. Then N is abelian, and so G/N is locally graded (see [17]). Clearly G/N is an ${\mathcal {H}}$-group and hence Lemma 2.3 yields that either G/N is finite or $G^{\prime \prime }\le N$. If G/N is finite, G is abelian-by-finite and so we have done. Assume that $G^{\prime \prime }\le N$, so $G\ne G^{\prime }$. Lemma 3.1 allows us to suppose that $G^{\prime }$ has infinite rank, so that $G^{\prime }$ is abelian. Clearly it can be assumed that G is not abelian, hence G cannot be nilpotent (see [14, Theorem A]) and so it follows from Fitting’s theorem that $G/G^{\prime }$ cannot be the product of two proper subgroups. On the other hand, if $a,b\in G$ are such that $[a,b]\ne 1$, then $\left\langle a,b,G^{\prime }\right\rangle$ is a non-abelian normal subgroup of infinite rank and hence $G=\left\langle a,b,G^{\prime }\right\rangle$; therefore $G/G^{\prime }$ is finitely generated. Hence $G/G^{\prime }$ is a cyclic group of prime-power order and so G is abelian-by-finite. The proof is completed. $\square$

Corollary 3.4

Let G be a locally (soluble-by-finite) group of infinite rank whose non-abelian subgroups of infinite rank are self-normalizing. If G has no simple homomorphic images of infinite rank, then G is an ${\mathcal {H}}$-group.

Proof

Assume, by contradiction, that G is not an ${\mathcal {H}}$-group. Then any proper normal subgroup of G has finite rank by Proposition 3.3; in particular, it follows from Lemma 3.1 that $G=G^{\prime }$. Let $N\$be any proper normal subgroup of G and assume that N is not abelian. Then N has finite rank and G/N is a locally (soluble-by-finite) group of infinite rank whose subgroups of infinite rank are self-normalizing; hence G/N is soluble (see [6]) which is a contradiction because G is perfect. Therefore any proper normal subgroup of G is abelian. Since any proper normal subgroup of G has finite rank and the product of finitely many normal subgroups of finite rank has likewise finite rank, it follows that the join L of all proper normal subgroups of G is abelian. Hence $G\ne L$ and so L has finite rank. Thus G/L has infinite rank and so, since clearly G/L is simple, we obtain the contradiction which concludes the proof.

$\square$

Proof of Theorem B1

Any simple locally soluble group has prime order (see [18, Part 1, Corollary 1 on page 154]), hence the statement follows immediately from Corollary 3.4. $\square$

Corollary 3.5

Let G be a group of infinite rank whose non-abelian subgroups of infinite rank are self-normalizing. If $G\ne G^{\prime }$, then G is an ${\mathcal {H}}$-group.

Proof

If $G^{\prime }$ has finite rank, then G is abelian by Lemma 3.1. If $G^{\prime }$ has infinite rank, then $G^{\prime }$ is abelian and the result follows from Theorem B1. $\square$

Finally we consider groups in which all proper subgroups of infinite rank are ${\mathcal {H}}$-groups.

Lemma 3.6

Let G be a locally graded group of infinite rank whose proper subgroups of infinite rank are ${\mathcal {H}}$-groups. If $G^{\prime }$ has finite rank, then G is an ${\mathcal {H}}$-group.

Proof

Let X be any finitely generated subgroup of G. Then $XG^{\prime } /G^{\prime }$ is abelian and finitely generated and so, since $G^{\prime }$ has finite rank, $XG^{\prime }$ has finite rank. Hence $G/XG^{\prime }$ is an abelian group of infinite rank and so there exists a subgroup of infinite rank Y such that $XG^{\prime }<Y<G$. Thus Y is an ${\mathcal {H}}$-group and so X is likewise an ${\mathcal {H}}$-group. Therefore G itself is an ${\mathcal {H}}$-group by Lemma 2.2.

$\square$

Lemma 3.7

Let G be a non-periodic locally (soluble-by-finite) group of infinite rank whose proper subgroups of infinite rank are ${\mathcal {H}}$-groups. Then G is abelian.

Proof

Assume, by contradiction, that G is not abelian. Hence G is not an ${\mathcal {H}}$-group, so that Lemma 3.6 yields that $G^{\prime }$ is a subgroup of infinite rank and hence $G/G^{\prime }$ is finitely generated (see [10, Lemma 2.8] and Lemma 2.2). Since every soluble ${\mathcal {H}}$-group is either abelian or periodic, the derived subgroup $G^{\prime }$ is periodic (see [10, Theorem A]). Hence $G/G^{\prime }$ is not periodic and so we may write $G/G^{\prime }=\left\langle x_{1}G^{\prime }, \ldots ,x_{t}G^{\prime }\right\rangle$ with any $x_{i}G^{\prime }$ of infinite order. If $i\in \{1,\ldots ,t\}$, p and q are distinct primes, the subgroups $\left\langle x_{i}^{p}, G^{\prime }\right\rangle$ and $\left\langle x_{i} ^{q},G^{\prime }\right\rangle$ are proper normal non-periodic subgroups of G of infinite rank, so that they are both ${\mathcal {H}}$-groups and hence even abelian. Thus

$$\begin{aligned} G=\left\langle x_{1}^{p},G^{\prime }\right\rangle \left\langle x_{1} ^{q},G^{\prime }\right\rangle \cdots \left\langle x_{t}^{p},G^{\prime }\right\rangle \left\langle x_{t}^{q},G^{\prime }\right\rangle \end{aligned}$$

is nilpotent. Therefore, since $G/G^{\prime }$ is finitely generated, G itself is finitely generated (see [18, Part 1, Theorem 2.26]) and hence G has finite rank. A contradiction which concludes the proof. $\square$

Proof of Theorem B2

Lemma 2.3 yields that all subgroups of infinite rank of G are soluble, and so G itself is soluble (see [9, Theorem 9]). Moreover, Lemma 3.7 allows us to assume that G is periodic. Let H be any finite subgroup of G, $g\in N_{G}(H)\smallsetminus H$, and $K=\left\langle H,g\right\rangle$. Then K is finite and there exist abelian K-invariant subgroups of infinite rank $A_{1}$ and $A_{2}$ such that $A_{2}<A_{1}$ and $K\cap A_{1}=\{1\}$ (see [16]). Then $A_{2}K$ is a proper subgroup of infinite rank of G, so that it is an ${\mathcal {H}}$-group. Since $g\in N_{A_{2} K}(H)\smallsetminus H$, it follows that H is abelian. Therefore all finite non-abelian subgroups of G are self-normalizing and hence G in an ${\mathcal {H}}$-group by Lemma 2.1. $\square$