Groups whose non-permutable subgroups are metaquasihamiltonian

Lemma 2.1.

Let X be a local group class. Then, for each positive integer k, the class Xk is also local.

Proof.

Since 𝔛1=𝔛, the statement is obvious if k=1. Suppose now that the class 𝔛k is local for some positive integer k. As 𝔛k+1 is subgroup closed, it is enough to prove that a group G belongs to 𝔛k+1 provided that all its finitely generated subgroups are 𝔛k+1-groups. Let X be any subgroup of G which is not in 𝔛k, and let 𝒲X be the set of all finitely generated subgroups of X which are not contained in an 𝔛k-subgroup of X. Then 𝒲X is not empty because 𝔛k is local. If g is any element of G and U∈𝒲X, the subgroup 〈g,U〉 belongs to 𝔛k+1, whence 〈g〉⁢U=U⁢〈g〉. It follows that all elements of 𝒲X are permutable in G. Moreover, 〈x,U〉∈𝒲X for all x∈X, and so X=〈V∣V∈𝒲X〉 is likewise permutable in G. Therefore, G belongs to 𝔛k+1, and hence 𝔛k+1 is a local class. ∎

Corollary 2.2.

For each positive integer k, the class of k-quasihamiltonian groups is local.

Lemma 2.3.

Let X be a class of locally (soluble-by-finite) groups. Then all locally graded, finitely generated Xk-groups are soluble-by-finite for all k≥1.

Proof.

Working by induction on k, we may assume k≥2. Let G be a locally graded, finitely generated 𝔛k-group. If all proper subgroups of G′′ have the 𝔛k-1-property, then G′′ is either 𝔛k-1 itself or finitely generated (see Lemma 2.1). In any case, being locally graded, it is soluble-by-finite by the induction hypothesis and hence even G is such.

By Lemma 2.1, it is now possible to take a finitely generated proper subgroup X of G′′ which is not 𝔛k-1 (use Lemma 2.1). Then X is permutable in G, and so |XG:X| is finite. Clearly, G/XG is a quasihamiltonian group so that G′′=XG for all such subgroups X; in particular, G′′′=G′′, and G′′ is perfect. Thus the finite group XG/XXG cannot be non-trivial soluble, and hence XXG has the 𝔛k-1-property. By induction, it is soluble-by-finite, along with G′′ and G. ∎

Lemma 2.4.

Let X be a group class, and let X,Y be subgroups of a group G such that X is not permutable in G and Y≤X. If Y is compressed in G by the class Xk for some integer k>1, then Y is compressed in X by Xk-1.

Proof.

Let N be an 𝔛k-compressor for Y in G. Then X/N is a non-permutable subgroup of the 𝔛k-group G/N, and hence it belongs to 𝔛k-1, which means that Y is compressed by 𝔛k-1 in X. ∎

Theorem 2.5.

Let X be a Robinson class of groups. Then, for each positive integer k, also the class Xk has the Robinson property.

Proof.

The statement is obvious if k=1. Suppose now by induction on k that 𝔛k is a Robinson class for some positive integer k, and let G be any finitely generated hyper-(abelian or finite) group in which all subgroups of finite index are compressed by 𝔛k+1. If all subgroups of finite index of G are permutable, then every finite homomorphic image of G is nilpotent, and so it follows from Robinson’s theorem that G itself is nilpotent, and hence also polycyclic. On the other hand, if G contains a non-permutable subgroup X of finite index, Lemma 2.4 yields that every subgroup of finite index of X is compressed in X by the Robinson class 𝔛k; then, by definition, X must contain a polycyclic subgroup of finite index, and so G is polycyclic-by-finite in any case.

Let H be a subgroup of G which is not 𝔛k. Then, by induction, there is a subgroup K of finite index in H that is not 𝔛k-compressed in H. Now K=H∩L for some subgroup L of finite index in G. If N is any normal subgroup of finite index of G, then H/H∩LG∩N is not 𝔛k, and hence H⁢(LG∩N)/(LG∩N) is permutable in G/(LG∩N). Now [10, Theorem A*] shows that H is permutable in G. This shows that G is 𝔛k+1 and completes the proof of the theorem. ∎

Theorem 2.6.

Let X be a Robinson class made of locally (soluble-by-finite) groups which is local and closed with respect to homomorphic images. If X is accessible, then the class Xk is accessible for all positive integers k.

Proof.

Assume for a contradiction that the statement is false, and let k be the smallest positive integer such that there exists an infinite locally graded group G which is not an 𝔛k-group while all its proper subgroups belong to 𝔛k. Clearly, k>1 since 𝔛 is accessible. Moreover, G is finitely generated because 𝔛k is a local class by Lemma 2.1, and hence Lemma 2.3 shows that G is actually soluble-by-finite.

Now [9] implies that the Frattini subgroup Φ⁢(G) of G has infinite index in G. Thus if N is any normal subgroup of finite index in G, then there is some maximal subgroup M such that M⁢N=G, and hence all finite quotients of G are 𝔛k. Since 𝔛k is a Robinson class (see Theorem 2.5), we get a contradiction. ∎

Corollary 2.7.

For each positive integer k, the class of k-quasihamiltonian is accessible.

Lemma 2.8.

Let X be a local group class which is closed by subgroups, homomorphic images and is made of soluble-by-finite groups. Then every locally graded Xk-group is soluble-by-finite for all k≥1.

Proof.

Assume for a contradiction that the statement is false, and let k be the smallest positive integer such that the class 𝔛k contains a locally graded group G which is not soluble-by-finite; clearly k>1.

If G′′ were not perfect, then G′′′ would either have the 𝔛k-1-property, and hence G would be soluble-by-finite by our choice of k, or G/G′′′ would be quasihamiltonian and hence metabelian, a contradiction. Thus G′′=G′′′.

Let N be any proper normal subgroup of G′′. Then it must be an 𝔛k-1-group; otherwise, G′′ would not be perfect, and by the choice of k, it is therefore soluble-by-finite. Let SN be the largest normal soluble subgroup of N. Then CG′′⁢(N/SN) has finite index in G′′, which means that N/SN is abelian, and so N=SN is soluble.

By Lemma 2.1 and the fact that G′′ is not an 𝔛k-1-group, it follows that there is a finitely generated subgroup F of G′′ not having the 𝔛k-1-property. Since every proper normal subgroup of G′′ is 𝔛k-1, it follows that F is not contained in any proper normal subgroup of G′′ and hence that G′′ contains a maximal proper normal subgroup S (being also the product of two proper normal subgroups of G′′ still proper), which is therefore soluble.

Being simple, G′′/S does not contain any proper non-trivial permutable subgroup, and this yields that all its proper subgroups have the 𝔛k-1-property. However, Lemma 2.3 shows that G′′/S is locally graded, and hence it must be either finitely generated or 𝔛k-1 by Lemma 2.1. In any case, the group G is soluble-by-finite. ∎

Lemma 2.9.

Let G be a soluble group in the class (An)k, where n and k are positive integers. Then G has derived length at most n+3⁢(k-1).

Proof.

Working by induction on k, we may assume k≥2. Obviously, if G′′′ were not in (𝔄n)k-1, then G′′=G′′′={1} and we would be done. Thus G′′′ is in (𝔄n)k-1, and by induction, it is soluble of derived length at most n+3⁢(k-2). It follows that the derived length of G is at most n+3⁢(k-2)+3=n+3⁢(k-1), and the result is proved. ∎

Corollary 2.10.

Let k be a positive integer, and let G be a locally soluble k-quasihamiltonian group. Then G is soluble of derived length at most 3⁢k-2.

Lemma 2.11.

Let X be a class of locally (nilpotent-by-finite) groups. Then all locally graded, finitely generated Xk-groups are nilpotent-by-finite for all positive integers k.

Proof.

Working by induction on k, let G be any finitely generated locally graded 𝔛k-group for some k>1; clearly G is soluble-by-finite by Lemma 2.3. If all subgroups of G having finite index are permutable, then Robinson’s theorem shows that G is nilpotent. Thus let H be a subgroup of finite index which is not permutable in G. Then H has the 𝔛k-1-property, and it is therefore nilpotent-by-finite, as well as G. Induction completes the proof. ∎

Lemma 2.12.

Let G be a periodic locally graded Ak for some positive integer k. Then there is a finite normal subgroup N such that the factor group G/N is quasihamiltonian.

Proof.

Suppose the result is false, and let k be the smallest positive integer for which there is a periodic locally graded 𝔄k-group G not satisfying the statement; clearly k>1. By Lemma 2.1, there is a finitely generated subgroup E which is not 𝔄k-1, and hence it is permutable in G; Lemma 2.8 yields that E is finite.

Let X be any subgroup of G; then the product XE is permutable in G, and the index |X⁢E:X| is finite. Therefore, every subgroup of G has finite index in a permutable subgroup, and hence G contains a finite normal subgroup N such that G/N is quasihamiltonian (see [3]). ∎

Lemma 2.13.

Let G be a torsion-free locally graded k-quasihamiltonian group for some positive integer k. Then G is abelian.

Proof.

Let k be the smallest positive integer for which there is a torsion-free 𝔄k-group G which is not abelian. Then all non-permutable subgroups are abelian so that G is metaquasihamiltonian, and being torsion-free, it is abelian by [2, Lemma 2.5]. ∎

Lemma 2.14.

Let G be a locally graded k-quasihamiltonian group for some positive integer k. Then the set of all periodic elements is a subgroup.

Proof.

Suppose the result is false, and let k be the smallest positive integer such that there exists a counterexample G in 𝔄k; clearly k≥2. By Lemma 2.11, G is locally (nilpotent-by-finite), so we can find two periodic elements x,y in G such that 〈x,y〉 is not finite and the 0-rank of 〈x,y〉 is smallest possible. If T is the periodic radical of 〈x,y〉, then we can clearly assume G (=〈x,y〉/T) has a normal torsion-free nilpotent subgroup N of finite index, and Lemma 2.13 yields that N is abelian.

If [Nm,x]=1=[Nn,y] for two positive integers n,m, then Nn⁢m≤Z⁢(G) so that G would be center-by-finite and hence finite-by-abelian, a contradiction. We may thus assume [Nm,x]≠1 for all integers m. If Nm⁢〈x〉 belonged to 𝔄k-1 for some m, then we would get [Nm,x]=1, a contradiction. Therefore, 〈x〉⁢Nm/Nm is permutable in 〈x〉⁢N/Nm for all m, and if we chose m such that (m,o⁡(x))=1, then 〈x〉⁢N/Nm would even be abelian, being locally nilpotent. As 〈x〉⁢N is residually abelian, it is also abelian, the final contradiction. ∎

Lemma 2.15.

Let p be a prime, and let G=〈g〉⋉H be a semidirect product of an infinite cyclic subgroup 〈g〉 and a quasihamiltonian p-group H with infinite commutator subgroup. Then G∉A∞.

Proof.

Suppose G∈𝔄k for the minimal possible integer k (clearly k≥2). By the structure of locally finite quasihamiltonian groups, the subgroup H has finite exponent; moreover, there exist an element h∈H and an abelian subgroup K such that H=K⁢〈h〉 and h acts like a universal power automorphism on K. By our choice of k, there is a finitely generated subgroup F which is not (k-1)-quasihamiltonian and hence is permutable. Since G locally satisfies the maximal condition, we can assume F=〈gm〉⁢S for some integer m and some finite subgroup S of H. Since H has finite exponent, there exists an infinite subgroup T of K such that T∩S={1} and 〈h〉⁢T has infinite commutator subgroup. Let t∈T, and note that, since F⁢〈t〉=〈gm〉⁢S⁢〈t〉 and S⁢〈t〉 are subgroups (actually 〈t〉 is normal in S⁢〈t〉), then g1=gm normalizes S⁢〈t〉. But it normalizes S, and hence tg1=s⁢tm1 for some s∈S and some positive integer m1 which is prime to the order of t. Now there is a power g2 of g1 centralizing S, and since the exponent of H is finite, it is clear by the above argument that there is a power g3 of g2 centralizing T. Finally, as G locally satisfies the maximal condition on subgroups, there is a power of g3 centralizing h. All we have proved so far allows us to assume G to be of the form 〈g〉×〈h〉⁢K.

Let A and B be subgroups of K such that both 〈h〉⁢A and 〈h〉⁢B have infinite commutator subgroups and A∩B={1}=〈h〉∩A⁢B.

Take a in A not commuting with h. We show that the subgroup

does not permute with 〈a⁢g〉. Indeed, the only way in which the element

could possibly belong to 〈a⁢g〉⁢L is by taking a power a1=am of a such that a1h=a and m<o⁡(a), and looking at products of the type

for some integers k1,k2 and b∈B; but m≢1(modo⁡(a)), and so the exponent m+o⁡(a)⋅(k1+k2) of g can never be equal to 1.

The subgroup L is subject to the same conditions as G, and this allows us to construct non-permutability chains of arbitrarily large length, a contradiction proving the statement. ∎

Proposition 2.16.

Let G be a locally graded k-quasihamiltonian group for some positive integer k. If the 0-rank of G is 1, then G is finite-by-quasihamiltonian.

Proof.

Let k be the smallest positive integer for which there is a counterexample G to the statement; clearly k≥3. Lemma 2.14 implies that the set of all periodic elements of G is a subgroup, say T, while G/T is abelian by Lemma 2.13. Furthermore, there is a finite normal subgroup N of T such that T/N is quasihamiltonian (see Lemma 2.12). In particular, T/N is locally nilpotent, and its primary components are either abelian or of finite exponent. The same can then be said about T/NG so that we may suppose T has those properties, replacing G with G/NG (however, note that quasihamiltonianity could not be preserved).

Our next aim is to show that T′ is finite. Having this in mind, suppose first that there is some p-primary component Tp of T with infinite commutator subgroup. Let g be any aperiodic element of G, and consider the section 〈g,Tp〉⁢N〈g〉/N〈g〉 to which an application of Lemma 2.15 gives a contradiction.

Now suppose there are infinitely many non-abelian primary components in T, and let Tp1,Tp2,…,Tpn be different non-abelian primary components of T such that pi does not divide |N| for all i=1,…,n. Let gn be some aperiodic element, and put S0={1}, Si=Tp1⁢…⁢Tpi for all i=1,…,n. Suppose an aperiodic element gi+1 has been defined for some i in {0,1,…,n-1}. The group 〈gi+1〉⁢Si+1/Si contains an infinite cyclic subgroup 〈gi+1mi+1⁢si+1⁢Si〉 (si+1∈Si+1) which is not permutable in 〈gi+1〉⁢Si+1/Si, and we let gi=gi+1mi+1⁢si+1. In particular, the subgroup 〈gi〉⁢Si is not permutable in 〈gi+1〉⁢Si+1. This construction leads to a non-permutability chain of arbitrary length, which contradicts G being a k-quasihamiltonian group.

Thus we have reduced so far to the case in which the non-abelian primary components of T have finite commutator subgroup and are finite in number so that we can factor T′ out and get T abelian.

Let F=〈g〉⁢S be a non-(k-1)-quasihamiltonian (and hence permutable) subgroup of G, for some g∈G∖T and some finite subgroup S of T; moreover, write

where h1=g and 〈hn〉⁢T≤〈hn+1〉⁢T for all n∈ℕ. If n is any positive integer, we have g=hnkn⁢un, where kn is an integer and un∈T, so that the subgroup 〈hnkn〉⁢S is permutable in G, being isomorphic to F. Therefore, 〈hn〉⁢S is a subgroup so that hn normalizes S. The arbitrariness of n shows that S is normal in G. Being finite, we may assume S={1}, and hence all subgroups 〈hn〉 are permutable in G. In particular, each element of infinite order acts as a power automorphism on T, and all elements of prime order and all elements of order 4 of T are central. Theorem 2.4.11 of [13] now implies that G is quasihamiltonian. ∎

Lemma 2.17.

Let G be a group and P a normal subgroup of G such that G/P is free abelian of finite 0-rank r≥2 and P is an abelian p-group of infinite exponent for some prime p. If there is some element h∈G acting as an automorphism of infinite order on P, then G∉A∞.

Proof.

It is enough to show the result in the case r=2 and G=〈g〉⁢〈h〉⁢P for some element g∈G of infinite order such that 〈g〉∩P⁢〈h〉={1}. Suppose by way of contradiction that G is k-quasihamiltonian for some positive integer k. Then Proposition 2.16 shows that P⁢〈h〉 is finite-by-quasihamiltonian, so there is a finite normal subgroup N such that P⁢〈h〉/N is quasihamiltonian. Factoring out NG, which is finite, we reduce ourselves to the case in which h acts like a (locally universal) power automorphism on P. In a similar way, we may assume g acting like a (locally universal) power automorphism on P. Finally, hg=h⁢t for some t∈P, and we even factor out 〈t〉G so that [h,g]=1.

Let a be an element of P not commuting with h, and let ga be a power of g centralizing a. Then the subgroup 〈gao⁡(a),h〉 does not permute with 〈a⁢ga〉 since the element h⋅a⁢ga∈〈gao⁡(a),h〉⁢〈a⁢ga〉 does not belong to 〈a⁢ga〉⁢〈gao⁡(a),h〉.

Let now an be an element of P having order (n+1)⋅o⁡(a) for n≥1. Then the subgroup Li=〈gano(a)i,h,ano(a)i〉 is not permutable in

for all i=1,…,n+1 as its quotient by 〈ano(a)i〉 is not such in Li-1/〈ano(a)i〉 for the above reasons. Thus we may construct non-permutability chains of arbitrarily large lengths, and this is absurd. ∎

Lemma 2.18.

Let G be a locally graded k-quasihamiltonian group for some positive integer k. Let N be a normal abelian subgroup of G such that the torsion part T of G is contained in N and G/N is locally cyclic and periodic. Suppose also that 〈z〉⁢T has a finite commutator subgroup for all aperiodic elements z of G. If the 0-rank of G is finite and ≥2, then G is finite-by-abelian.

Proof.

Let H¯=G/N, take an aperiodic element z of N, and put

with H¯ naturally acting onto T. It is easy to see that G¯ is locally isomorphic to a section of G, and hence it must be k-quasihamiltonian. Indeed, let h¯=g⁢N∈H¯, and take g1∈G∖T such that

which is possible by hypothesis and the fact that G/T is abelian; then the section 〈g,g1〉⁢T/〈go⁡(h¯)〉 is isomorphic to (〈h¯〉⋉T)×〈z〉, and the arbitrariness of h¯ proves our claim.

Then, by Proposition 2.16, the group G¯ must be finite-by-quasihamiltonian and hence finite-by-abelian. Thus there is a normal subgroup M of T such that H¯⋉T/M is abelian so that T/M is central in G/M, and we may assume T is central in G factoring out M.

Using induction on k, we may now assume G is a minimal counterexample with respect to k and k≥2. Thus there is a finitely generated subgroup F of G which is not (k-1)-quasihamiltonian and is such that G/F⁢T is periodic. We factor out the periodic part of F and assume therefore F being torsion-free and permutable in G; put M=T⁢F, and note that G/M is abelian of finite rank (in the usual sense of abelian groups).

Now let f∈F and g∈G be aperiodic elements such that [g,f]≠1, in particular fg=f⁢t for some 1≠t∈T. Clearly, |〈g〉:C〈g〉(f)|=o(t). The group 〈g〉/C〈g〉⁢(f)⋉(T×〈f〉) is isomorphic to a section of G since T⁢〈f〉∩〈g〉=1; otherwise, the subgroup 〈f,g〉 would be abelian. If g1 is another element of G such that some power of it is equal to g⁢m1 for some m1∈M, then 〈g〉/C〈g〉⁢(f) embeds in 〈g1〉/C〈g1〉⁢(f). This shows that the group

is locally k-quasihamiltonian and hence k-quasihamiltonian. Suppose G/CG⁢(f) is infinite. Then, by Proposition 2.16, the above group is finite-by-quasihamiltonian, and T contains elements of the same orders of those contained in G/CG⁢(f). But T is central, and hence this is a contradiction. Thus G/CG⁢(f) is finite, and f has finitely many conjugates in G.

It follows that each f in F has finitely many conjugates, and hence it is possible to find a finite subgroup of T such that T⁢〈f〉 is normal in G. Therefore, there exists a finite subgroup U of T such that F⁢U/U is central in G/U. We factor out this subgroup. If f is any element of G, then 〈f〉 is central in G is finite-by-quasihamiltonian. If it is finite-by-abelian, then G′ would be finite, so, by induction on the 0-rank, we may assume the 0-rank of G being 2.

In this case, suppose first that the commutator subgroup of G has elements of infinitely many pairwise coprime orders, and let p1,…,pn be n of these prime orders. Then there are two infinite cyclic subgroups, say 〈a〉 and 〈b〉, such that ab=a⁢t with o⁡(t)=p1⋅…⋅pn. Clearly, T⁢〈a〉∩〈b〉={1}, and the quotient

is a direct product of n non-quasihamiltonian groups, which means it contains a non-permutability chain of length n. The arbitrariness of n gives a contradiction.

The index |FG:F| is finite, but then G/FG is periodic and hence even finite-by-quasihamiltonian. Thus there is a finite normal subgroup N/FG of G/FG such that G/N is quasihamiltonian and periodic. The non-abelian primary components of G/N have finite exponent, which means they are center-by-finite since G/T⁢F has finite rank (in the usual sense of abelian groups) and hence finite commutator subgroup by Schur’s theorem. But they are also in a finite number by what we have shown above, and hence G′⁢N/N is finite. Therefore, G′ is finitely generated and even finite. ∎

Lemma 2.19.

Let G be a locally graded m-quasihamiltonian group for some positive integer m. If N is any normal abelian subgroup of G such that the torsion part T of G is contained in N and G/N is locally cyclic and periodic, then T≤F⁢C⁢(G).

Proof.

Suppose that T≰F⁢C⁢(G), and let x∈T∖F⁢C⁢(G). For all n≥1, there are aperiodic elements g1,…,gn∈G such that

The hypotheses show that there is g∈G with

and kn=1,kn-1|kn-2,…,k2|k1. Let i∈{1,…,n-1}. If 〈x〉〈gki〉⁢〈gki〉 permutes with 〈gki+1〉, then 〈x〉〈gki〉 would be normalized by 〈gki+1〉, which is not possible. The arbitrariness of n contradicts the assumption that G is m-quasihamiltonian. ∎

Proposition 2.20.

Let G be a locally graded k-quasihamiltonian group for some positive integer k. If the 0-rank of G is strictly greater than 1, then G is finite-by-quasihamiltonian.

Proof.

Let T be the periodic part of G, and let x be any element of infinite order. Proposition 2.16 shows that the subgroup T⁢〈x〉 is finite-by-quasihamiltonian so that T is finite-by-abelian (see [13, Theorem 2.4.11]), and hence we may even assume T abelian, replacing G by G/T′.

We now let k≥2, working by induction on k, and let E be a finitely generated subgroup of G which is not (k-1)-quasihamiltonian. If G/T has infinite 0-rank, then there is an element of infinite order modulo E so that E is normal in G by [2, Lemma 2.6]. It follows that G/E is abelian (see [13, Theorem 2.4.11]), which means G′ is finitely generated and hence even finite, being locally finite. It is therefore possible to consider only the case in which G/T has finite 0-rank.

Suppose now there is an aperiodic element g not commuting with infinitely many primary components of T. Let Tp be a p-primary component of T such that

Lemma 2.17, Proposition 2.16 and the fact that the 0-rank of G is ≥2 yield the existence of an infinite cyclic subgroup

moreover, it is even possible to choose Tp such that [H,〈g〉]∩Tp={1}. Factoring out Tp′ for the moment being, we obtain [h,g]=1. Now 〈g,ho⁡(tp)〉 does not permute with 〈tp⁢h〉, where tp is any element of Tp not commuting with g. We can repeat the process in the subgroup Tp′⁢〈g,ho⁡(tp)〉 getting a contradiction to the fact that G is 𝔄k. This shows that any aperiodic element of G does not commute with finitely many primary components of T.

Let Tp be any p-component of T for some prime p, and assume first that it has finite exponent. Let x∈G be an aperiodic element such that Lx=〈x〉⁢Tp has infinite commutator subgroup. Again, we factor out Tp′ and take y∈G∖T with Lx∩〈y〉={1}. Since Lx is finite-by-quasihamiltonian by Proposition 2.16, there is a finite normal subgroup Nx of Lx such that Lx/Nx is quasihamiltonian. Let C=C〈x〉⁢(Lx/Nx). Then, applying Lemma 2.15 to 〈y〉⁢Lx/〈C,Nx〉〈y,x〉, we reach a contradiction.

If Tp has infinite exponent, then Lemma 2.17 shows that Tp⁢〈z〉 has finite commutator subgroup for any aperiodic element z of G. Summing up, the commutator subgroup of T⁢〈z〉 is finite for any z∈G.

Now let F be any non-(k-1)-quasihamiltonian subgroup of G with maximal 0-rank. Factoring out the commutator subgroup of the normal subgroup FT (here we use Lemma 2.19 and what we have just proved) makes it possible to apply Lemma 2.18 to each subgroup L of G such that L/T⁢F is periodic and locally cyclic, and hence L is finite-by-abelian. Since G/T has finite rank, we can find a finite normal subgroup R of G such that, factoring it out, we get TF central in G.

If the 0-rank of G is strictly greater than 2, by induction on the 0-rank, it follows that G/〈f〉 is finite-by-abelian for every non-trivial element f of F. Thus G′ is finitely generated and so finite. We are leaving with the case 0-rank 2. In this case, suppose first that the commutator subgroup of G has elements of infinitely many pairwise coprime orders, and let p1,…,pn be n of these prime orders. Then there are two infinite cyclic subgroups, say 〈a〉 and 〈b〉, such that ab=a⁢t with o⁡(t)=p1⋅…⋅pn. Clearly, T⁢〈a〉∩〈b〉={1} and the quotient

is a direct product of n non-quasihamiltonian groups, which means it contains a non-permutability chain of length n. The arbitrariness of n gives a contradiction.

Therefore, G′ has finitely many non-trivial primary components. Now G/F is quasihamiltonian and periodic so that it contains only finitely many non-abelian primary components. These primary components must have finite exponents, and since G/T⁢F has finite rank (in the usual sense of abelian groups), these non-abelian primary components must even be center-by-finite (recall that T is central) and hence have a finite commutator subgroup by Schur’s theorem. Thus G/F has a finite commutator subgroup, which means that G′ is finitely generated and so finite. ∎

Theorem 2.21.

A locally graded group G is finite-by-quasihamiltonian if and only if it belongs to the class A∞.

Proof.

If G belongs to the class 𝔄∞, then Proposition 2.16 and Proposition 2.20 show that it is finite-by-quasihamiltonian.

Conversely, suppose G has a finite normal subgroup N such that G/N is quasihamiltonian. Clearly, there is some non-negative integer m (depending only on the order of N) such that if ⋯≤Hn+1≤Hn≤⋯≤H1≤H0 is a descending chain of subgroups of G with either Hi+1=Hi or Hi+1 not permutable in Hi, then Hj∩N=Hj+1∩N for all j≥m. Thus

for all j≥m. As Hj⁢N/N is quasihamiltonian, it follows that Hj+1 is permutable in Hj, and hence Hj=Hj+1 for all j≥m. This shows that Hm is quasihamiltonian, and hence G is (m+1)-quasihamiltonian. ∎