M. De Falco a, F. de Giovanni a,∗, C. Musella a, Y.P. Sysak b
a Dipartimento di Matematica e Applicazioni, Università di Napoli Federico II,Complesso Universitario Monte S. Angelo, Via Cintia, I-80126 Napoli, Italyb Institute of Mathematics, Ukrainian National Academy of Sciences,vul. Tereshchenkivska 3, 01601 Kiev, Ukraine
a r t i c l e i n f o a b s t r a c t
Article history:Received 26 October 2012Available online 29 March 2014Communicated by Gernot Stroth
MSC:20F19
Keywords:Metahamiltonian groupGroup of infinite rank
A group is metahamiltonian if all its non-abelian subgroupsare normal. It is proved here that a (generalized) solublegroup of infinite rank is metahamiltonian if and only if allits subgroups of infinite rank are either abelian or normal.
It is well known that a group G has only normal subgroups (i.e. is a Dedekind group) ifand only if G is either abelian or the direct product of a quaternion group of order 8 anda periodic abelian group with no elements of order 4. The structure of groups for whichthe set of non-normal subgroups is small in some sense has been studied by many authors
✩ This work was partially supported by MIUR-PRIN 2009 (Teoria dei Gruppi e Applicazioni) – the firstthree authors are members of GNSAGA (INdAM); the last author is grateful for the hospitality from theUniversity of Napoli Federico II and for the financial support from the Istituto Nazionale di Alta Matematicaduring the preparation of this paper.* Corresponding author.
136 M. De Falco et al. / Journal of Algebra 407 (2014) 135–148
in several different situations. A group G is called metahamiltonian if every non-abeliansubgroup of G is normal. Metahamiltonian groups were introduced and studied in aseries of relevant papers by G.M. Romalis and N.F. Sesekin (see [11–13]); they proved inparticular that the commutator subgroup of any (generalized) soluble group with suchproperty is finite with prime-power order.
Recall also that a group G is said to have finite (Prüfer) rank r if every finitelygenerated subgroup of G can be generated by at most r elements, and r is the leastpositive integer with such property. A classical theorem of A.I. Mal’cev [9] states thata locally nilpotent group of infinite rank must contain an abelian subgroup of infiniterank. The investigation of the influence on a (generalized) soluble group of the behaviorof its subgroups of infinite rank has been developed in a series of recent papers (seefor instance [2–6]). In particular, M.J. Evans and Y. Kim [7] have proved that if G isa (generalized) soluble group in which all subgroups of infinite rank are normal, theneither G is a Dedekind group or it has finite rank.
The aim of this paper is to provide a further contribution to this topic, characterizingmetahamiltonian groups in terms of their subgroups of infinite rank. In fact, we willprove the following result.
Theorem. Let G be a locally (soluble-by-finite) group in which every non-abelian subgroupof infinite rank is normal. Then either G has finite rank or it is metahamiltonian.
Actually, the above theorem holds within the universe of strongly locally gradedgroups, a large class of generalized soluble groups that can be defined as follows.
Recall that a group G is locally graded if every finitely generated non-trivial subgroupof G contains a proper subgroup of finite index. Let D be the class of all periodic locallygraded groups, and let D be the closure of D by the operators P, P, R, L (for thedefinitions of these and other relevant operators on group classes we refer to the firstchapter of [10]). It is easy to prove that any D-group is locally graded, and that theclass D is closed with respect to forming subgroups. Moreover, N.S. Černikov [1] provedthat every D-group with finite rank contains a locally soluble subgroup of finite index.Obviously, all residually finite groups belong to D, and hence the consideration of anyfree non-abelian group shows that the class D is not closed with respect to homomorphicimages. We shall say that a group G is strongly locally graded if every section of G is aD-group. Thus strongly locally graded groups form a large subgroup and quotient closedclass of generalized soluble groups, containing in particular all locally (soluble-by-finite)groups.
Most of our notation is standard and can be found in [10].
2. Proofs
If G is any group of infinite rank, we shall denote by M(G) the intersection of allnormal subgroups of infinite rank of G. This characteristic subgroup will play a relevant
M. De Falco et al. / Journal of Algebra 407 (2014) 135–148 137
role in our considerations, and our first step is to prove that, in the situation of our maintheorem, the subgroup M(G) has finite rank.
Lemma 1. Let G be a locally (soluble-by-finite) group of infinite rank whose non-abeliansubgroups of infinite rank are normal. If the subgroup M(G) has infinite rank, then it isabelian. Moreover, either M(G) has prime exponent or it is divisible and its torsion parthas finite rank.
Proof. Clearly, M(G) has no proper G-invariant subgroups of infinite rank, so that allits proper subgroups of infinite rank are abelian, and hence M(G) itself must be abelian(see [4], Theorem A′). Suppose first that M(G) is periodic, so that it is the direct productof its primary components. Then M(G) is a p-group for some prime number p, and itssocle S has infinite rank, so that M(G) = S has exponent p.
Assume now that M(G) is non-periodic, so that the subgroup T consisting of allelements of finite order of M(G) has finite rank. Then the torsion-free group M(G)/Thas infinite rank, and hence the characteristic subgroup M(G)p of M(G) has infiniterank for each prime p. Therefore M(G)p = M(G) for all p, and so M(G) is divisible. �Lemma 2. Let G be a locally (soluble-by-finite) group of infinite rank whose non-abeliansubgroups of infinite rank are normal. If the subgroup M(G) has infinite rank, there existsan element z in G \M(G) such that every proper 〈zk〉-invariant subgroup of M(G) hasfinite rank and [M(G), zk] = M(G) for each positive integer k.
Proof. The subgroup M = M(G) is abelian by Lemma 1, so that M lies in its centralizerCG(M) and all subgroups of G properly containing CG(M) are non-abelian. Since M
has infinite rank, it follows that G/CG(M) is a Dedekind group; moreover, CG(M) �= G,because M has no proper G-invariant subgroups of infinite rank. Let Z/CG(M) be thecentre of G/CG(M), and consider any element z of Z \ CG(M). The centralizer
CM (z) = CM
(⟨z, CG(M)
⟩)
is a proper G-invariant subgroup of M , and hence it has finite rank. On the other hand,the G-invariant subgroup
[M, z] =[M,
⟨z, CG(M)
⟩]
is isomorphic to M/CM (z), so that it has infinite rank and then [M, z] = M .Assume for a contradiction that zn belongs to CG(M) for some integer n > 1, and
suppose first that M has prime exponent p. Then n = mpk, where m is coprime to p,and the subgroup 〈zm,M〉 is nilpotent (see for instance [10], Part 2, Lemma 6.34); inparticular, [M, zm] �= M and so zm belongs to CG(M) by the above argument. Appli-cation of Maschke’s theorem yields now that M is the direct product of infinitely manyfinite 〈z〉-invariant subgroups. In particular, M contains a proper 〈z〉-invariant subgroup
138 M. De Falco et al. / Journal of Algebra 407 (2014) 135–148
K such that 〈z〉 ∩M � K and the index |M : K| is finite. Then the product K〈z〉 is anon-abelian subgroup of infinite rank, and hence it is normal in G, so that also
K = K〈z〉 ∩M
is normal in G, a contradiction. It follows that M cannot have prime exponent, and soit is divisible by Lemma 1. As z induces an automorphism of finite order on M , we havethat M is the direct product of infinitely many 〈z〉-invariant divisible subgroups of finiterank. Then M contains a proper subgroup L such that 〈z〉 ∩M � L and M/L has finiterank, and the argument used above gives a contradiction also in this case. Therefore z hasinfinite order and 〈z〉 ∩ CG(M) = {1}. Let k be any positive integer. Then CM (zk) is aproper G-invariant subgroup of M , so that it has finite rank. Since [M, zk] is isomorphicto M/CM (zk), it follows that [M, zk] has infinite rank. But the subgroup [M, zk] isnormal in G, so that [M, zk] = M . Moreover, if H is any 〈zk〉-invariant subgroup of Mof infinite rank, the product H〈zk〉 is a normal subgroup of G, so that
H = H⟨zk
⟩∩M
is likewise normal in G and hence H = M . The statement is proved. �In order to show that M(G) has finite rank, we also need the following result on the
dimension of certain modules over the group algebra of an infinite cyclic group.
Lemma 3. Let F be a field, F 〈z〉 the group algebra of the infinite cyclic group 〈z〉 over F ,and let M be an F 〈z〉-module satisfying one of the following conditions:
(a) F has prime characteristic p and for each n � 1 every proper F 〈zpn〉-submodule ofM has finite dimension over F ;
(b) F has characteristic 0 and the F -envelope of every proper Z〈z〉-submodule of M hasfinite dimension over F .
Then M has finite dimension over F .
Proof. Assume for a contradiction that the M has infinite dimension over F . The groupalgebra F 〈z〉 is a principal ideal domain in which every ideal is generated by a polynomialin z over F , and all irreducible F 〈z〉-modules have finite dimension. Therefore M isgenerated by its proper F 〈z〉-submodules, and we may consider a non-zero element x0of M such that the submodule A = x0F 〈z〉 has minimal dimension over F , n say. Theelements
x0, x0z, . . . , x0zn
are linearly dependent over F , and so there exists a polynomial
M. De Falco et al. / Journal of Algebra 407 (2014) 135–148 139
f(t) = tn + a1tn−1 + · · · + an
over F such that x0f(z) = 0. The mapping
α : x �−→ xf(z)
is an F 〈z〉-endomorphism of M , and hence the kernel ker α and the image Im α = MF 〈z〉are F 〈z〉-submodules of M . As M/ker α and Im α are isomorphic, it follows that eitherker α = M or ker α has finite dimension over F and Mf(z) = M .
Assume first that ker α = M , i.e. xf(z) = 0 for each element x of M . In particular, ifx �= 0, the minimal choice of n yields that the elements
x, xz, . . . , xzn−1
are linearly independent over F and generate the irreducible submodule xF 〈z〉, so thatxF 〈z〉 has dimension n over F , and M can be decomposed as a direct sum of irre-ducible submodules. Therefore M has a proper submodule of infinite dimension over F ,contradicting the conditions of the statement.
Suppose now that ker α has finite dimension over F and Mf(z) = M . Then thereexists in M a sequence of elements
x1, x2, . . . , xm, xm+1, . . .
such that x0 = x1f(z) and xm = xm+1f(z) for all m � 1. Let m be any positive integer,and assume that there exist elements
contradicting the choices of the coefficients. This contradiction proves that the elements
x0, x0z, . . . , x0zn−1, x1, x1z, . . . , x1z
n−1, . . . , xm, xmz, . . . , xmzn−1, . . .
are linearly independent over F .
140 M. De Falco et al. / Journal of Algebra 407 (2014) 135–148
Suppose that F has prime characteristic p. Dividing the polynomial tpn −1 by f(t) inthe field F (t), we obtain that xm+1(zp
n − 1) belongs to xmF 〈z〉 for each m � 1, and so
xpm
(zp
n+1 − 1)
= xpm
(zp
n − 1)p ∈ xp(m−1)F 〈z〉.
Thus the F -envelope of the elements
x0, x0z, . . . , x0zn−1, xp, xpz, . . . , xpz
n−1, . . . , xpm, xpmz, . . . , xpmzn−1, . . .
is a proper F 〈zpn+1〉-submodule of M and it has infinite dimension over F , contrary tocondition (a).
Assume finally that F has characteristic 0. Then the elements
x0, x0z, . . . , x0zn−1, x1, x1z, . . . , x1z
n−1, . . . , xm, xmz, . . . , xmzn−1, . . .
generate a proper Z〈z〉-submodule of M , whose F -envelope has infinite dimension over F ,contradicting in this case condition (b). This last contradiction proves that M has finitedimension over F . �Lemma 4. Let G be a locally (soluble-by-finite) group of infinite rank whose non-abeliansubgroups of infinite rank are normal. Then the subgroup M(G) has finite rank.
Proof. Assume for a contradiction that the subgroup M = M(G) has infinite rank. ThenLemma 1 yields that either M is abelian of prime exponent p or M is divisible abelianand its torsion part has finite rank, and in the latter case we may suppose without lossof generality that M is torsion-free. Moreover, by Lemma 2 there exists an element z ofG such that every proper 〈zk〉-invariant subgroup of M has finite rank for each positiveinteger k. Let F denote either the field with p elements (if M has exponent p) or thefield of rational numbers (when M is torsion-free). Then M is an F 〈z〉-module satisfyingthe condition of Lemma 3, and hence it has finite dimension over F , a contradiction.Therefore M has finite rank. �Lemma 5. Let G be a group of infinite rank such that the subgroup M(G) has finite rank,and let N be a normal subgroup of G such that G/N has finite rank. Then N containsa proper G-invariant subgroup of infinite rank.
Proof. Clearly, N has infinite rank, and so it properly contains the subgroup M(G).Thus there exists a normal subgroup K of G of infinite rank which does not contain N .Moreover, K/K ∩N has finite rank, and hence K ∩N is a proper G-invariant subgroupof N of infinite rank. �Lemma 6. Let G be a locally (soluble-by-finite) group of infinite rank whose non-abeliansubgroups of infinite rank are normal, and let N be a non-abelian normal subgroup of Gcontained in G′. Then |G′ : N | � 2.
M. De Falco et al. / Journal of Algebra 407 (2014) 135–148 141
Proof. Since N is non-abelian, all subgroups of infinite rank of the group G/N arenormal, and so G/N is a Dedekind group (see [7], Theorem C). Therefore G′/N hasorder at most 2. �
As we mentioned in the introduction, the commutator subgroup of any (generalized)soluble metahamiltonian group is finite. Our next step is to prove that in our case thecommutator subgroup must have finite rank.
Lemma 7. Let G be a locally (soluble-by-finite) group whose non-abelian subgroups ofinfinite rank are normal. Then the commutator subgroup G′ of G has finite rank.
Proof. Assume for a contradiction that G′ has infinite rank. The subgroup M(G) hasfinite rank by Lemma 4, so that we may replace G by G/M(G) and suppose withoutloss of generality that M(G) = {1}. Let N be any normal subgroup of infinite rankof G. Then the factor group G/N is metahamiltonian; in particular the commutatorsubgroup G′N/N of G/N is finite, and so G′ ∩N has finite index in G′. It follows thatG is residually metahamiltonian and G′ is residually G-finite (i.e. it has a system ofG-invariant subgroups of finite index with trivial intersection). Application of Lemma 6yields that G′ contains a proper abelian G-invariant subgroup A such that the index|G′ : A| is finite; in particular, A has infinite rank.
Suppose first that A is periodic. Clearly, A cannot be decomposed into the directproduct of two G-invariant subgroups of infinite rank, and so it contains an infiniteG-invariant subgroup B of prime exponent. As G′ is residually G-finite, there exists inB a descending chain
B1 > B2 > · · · > Bn > · · ·
of G-invariant subgroups of finite index such that
⋂
n∈N
Bn = {1}.
Let K be any non-abelian subgroup of finite rank of G. Then B ∩ K is finite, and sothere exists a positive integer m such that Bm ∩K = {1}. For each positive integer n,the product BnK is a non-abelian subgroup of infinite rank, and hence it is normal in G,so that also
K =⋂
n�m
BnK
is normal in G. Therefore G is metahamiltonian, a contradiction as G′ is infinite.Assume now that A is not periodic, and let T be the subgroup consisting of all
elements of finite order of A. Then A/T is infinite and hence T has finite rank, so that
142 M. De Falco et al. / Journal of Algebra 407 (2014) 135–148
replacing G by G/T we may suppose that A is torsion-free (note here that with thisreplacement the condition that A is residually finite could be lost). Clearly, CG(A) �= G
and every subgroup of G properly containing CG(A) is non-abelian and has infinite rank,so that G/CG(A) is a Dedekind group. Put Z/CG(A) = Z(G/CG(A)), and consider anelement z of Z \ CG(A). The subgroups CA(z) and [A, z] are normal in G; moreover,A/CA(z) is isomorphic to [A, z], and so it is torsion-free. As CA(z) �= A, the index|A : CA(z)| is infinite, so that CA(z) must have finite rank, and hence the factor groupG/CA(z) is likewise a counterexample. Replacing now G by G/CA(z), we may assumethat A ∩ 〈z〉 = {1}, and we have again that CA(z) has finite rank. Let L be the set ofall G-invariant subgroups of infinite rank of A. Clearly, A∩N has infinite rank for eachnormal subgroup of infinite rank N of G, and so the subgroup
C =⋂
L∈L
L
coincides with the intersection of all normal subgroups of infinite rank of G. Thus C hasfinite rank by Lemma 4. Let L be any element of L. Then L is not contained in CA(z), sothat the product L〈z〉 is a non-abelian subgroup of infinite rank, and hence it is normalin G and G/L〈z〉 is a Dedekind group. Therefore (G′)2 is contained in L〈z〉 for all L ∈ L,and so
(G′)2 �
⋂
L∈L
L〈z〉 = C〈z〉.
In particular, A2 is contained in
C〈z〉 ∩A = C,
so that A2 has finite rank, which is impossible, as A is a torsion-free abelian group ofinfinite rank. This last contradiction completes the proof of the lemma. �Lemma 8. Let G be a group such that M(R/S) has finite rank for each section R/S of G,and let E be a finitely generated subgroup of G. If every subgroup of infinite rank of Gcontaining E is normal and there exists an abelian E-invariant subgroup of G of infiniterank, then E is normal in G.
Proof. Let A be an abelian subgroup of infinite rank of G such that AE = A. Obviouslywe may suppose that E has finite rank, and it is clearly enough to show that in thesubgroup EA there exists a collection of subgroups of infinite rank whose intersectioncoincides with E. Thus without loss of generality it can be assumed that G = EA. ThenE ∩ A is a normal subgroup of G, and replacing G by G/E ∩ A we may also supposethat E ∩ A = {1}. Let L be the set of all normal subgroups of infinite rank of G. Thenthe product EL is a normal subgroup of G for each element L of L, and hence
M. De Falco et al. / Journal of Algebra 407 (2014) 135–148 143
K =⋂
L∈L
EL
is likewise normal in G. The subgroup B = A ∩K is normal in G = EA, and
K = EA ∩K = EB.
Let X be any proper E-invariant subgroup of B, and assume that EX = K; then
B = EX ∩B = X,
and this contradiction shows that EX is a proper subgroup of K, so that X must havefinite rank. Application of Lemma 5 to the group K yields that B has finite rank. As[A,E] is contained in A∩K = B, it follows that [A,E] has finite rank. For each elementx of E, the factor group A/CA(x) is isomorphic to [A, x], and hence has finite rank. ButE is finitely generated, and so also A/CA(E) has finite rank. Thus the centralizer CA(E)has infinite rank, and hence it contains a subgroup V of infinite rank, which is residuallyfinite and such that E ∩ V = {1}. If W is the set of all subgroups of finite index of V ,and W is any element of W, the product EW is a normal subgroup of G, and so
E =⋂
W∈W
EW
is likewise normal in G. The lemma is proved. �Lemma 9. Let G be a locally (soluble-by-finite) group whose non-abelian subgroups ofinfinite rank are normal, and let E be a finitely generated non-abelian subgroup of G. IfG contains an E-invariant subgroup N of infinite rank which is nilpotent with class atmost 2, then E is normal in G.
Proof. It can obviously be assumed that E has finite rank and, as in the proof ofLemma 8, that G = EN , so that in particular N is normal in G. The commutatorsubgroup G′ of G has finite rank by Lemma 7, so that in particular N ′ has finite rankand N/N ′ has infinite rank. Clearly, every subgroup of infinite rank of G/N ′ containingEN ′/N ′ is normal, and so it follows from Lemma 8 that EN ′ is a normal subgroup ofG. For each element x of E, the mapping
θx : g �−→ [g, x]N ′
is a homomorphism from N into [N,E]N ′/N ′; if Kx is the kernel of θx, the factor groupN/Kx is isomorphic to a subgroup of [N,E]N ′/N ′, and so it is abelian of finite rank.Let {x1, . . . , xt} be a finite set of generators of E. Then
K =t⋂Kxi
i=1
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is an E-invariant subgroup of N such that
[K,E] � N ′ � Z(N);
moreover, N/K has finite rank, and hence K has infinite rank. It follows that for everyelement x of E the mapping
ϕx : z �−→ [z, x]
is a homomorphism from K into [K,E] with kernel CK(x), and so K/CK(x) has finiterank. As
CK(E) =t⋂
i=1CK(xi),
it follows that also K/CK(E) has finite rank. Therefore CK(E) is a nilpotent group ofinfinite rank, and hence it contains an abelian subgroup A of infinite rank, so that thesubgroup E is normal in G by Lemma 8. �
As a consequence, we have that the statement of our main theorem holds when thegroup contains large nilpotent normal subgroups.
Corollary 10. Let G be a locally (soluble-by-finite) group whose non-abelian subgroupsof infinite rank are normal. If G contains a normal subgroup of infinite rank which isnilpotent with class at most 2, then G is metahamiltonian.
Proof. It follows from Lemma 9 that every finitely generated non-abelian subgroup ofG is normal. As any non-abelian subgroup of G is the join of its finitely generatednon-abelian subgroups, the group G is metahamiltonian. �Lemma 11. Let the group G = AH be the product of an abelian minimal normal subgroupA of finite rank and an abelian subgroup H of infinite rank. Then there exist a subgroupK of H of infinite rank and a proper K-invariant subgroup B of A such that A/B isperiodic.
Proof. If the centralizer CH(A) has infinite rank, it is enough to choose a proper subgroupB of A such that A/B is periodic and put K = CH(A). Suppose now that CH(A) hasfinite rank. Since A is an infinite minimal normal subgroup of G, it is torsion-free divisibleand the group G cannot contain a nilpotent normal subgroup of finite index. Clearly,A∩H = {1} and so, replacing G by the factor group G/CH(A), it can be assumed withoutloss of generality that CH(A) = {1}. Then H is isomorphic to an abelian subgroup ofinfinite rank of the automorphism group of A. Consider A as a vector space over the fieldQ of rational numbers with basis {a1, . . . , an}, so that H can be identified with a group
M. De Falco et al. / Journal of Algebra 407 (2014) 135–148 145
of invertible n×n matrices over Q. Let π be the set of all prime divisors of denominatorsof entries of all such matrices. Then H is actually represented by invertible matrices overthe subring Qπ consisting of all rational numbers whose denominators are π-numbers.It follows that the set π must be infinite, because otherwise H is finitely generated. Letp be an element of π. It is easy to show that the set of all matrices in H whose entrieshave denominators prime to p is a proper subgroup K of H of infinite rank. Since K
is formed only by invertible matrices over the subring Q(p), consisting of all rationalnumbers whose denominators are coprime to p, it follows that the subgroup
B = Q(p)a1 + · · · + Q(p)an
is K-invariant, and of course the factor group A/B is periodic. �Lemma 12. Let G be a group whose non-abelian subgroups of infinite rank are normal.If G = AH, where A is an abelian normal subgroup of finite rank and H is an abeliansubgroup of infinite rank, then G is metahamiltonian.
Proof. Assume for a contradiction that G is not metahamiltonian. Clearly, the subgroupH can be chosen to be maximal abelian. As the commutator subgroup G′ of G is con-tained in A, the intersection G′ ∩H is a subgroup of the centre Z(G). Suppose first thatG′/G′ ∩H is finite. Then the centralizer CH(G′/G′ ∩H) has finite index in H, and sothe product
G′CH
(G′/G′ ∩H
)
is a normal subgroup of infinite rank which is nilpotent of class at most 2, contradictingthe statement of Corollary 10. Therefore G′/G′∩H is infinite. Let X be any G-invariantsubgroup of G such that
G′ ∩H < X � G′.
Then H is properly contained in XH, so that XH is not abelian and hence it is normalin G. Moreover, the factor group
G/XH = AH/XH A/X(A ∩H)
is abelian, and so G′ is contained in XH. It follows that
G′ = G′ ∩XH = X(G′ ∩H
)= X,
which proves that G′/G′ ∩H is a chief factor of G. On the other hand, the factor groupG/G′ ∩ H has infinite commutator subgroup, so that it is not metahamiltonian, andreplacing G by G/G′ ∩H it can be assumed that G′ is a minimal normal subgroup of G.
146 M. De Falco et al. / Journal of Algebra 407 (2014) 135–148
Clearly, G′ is also a minimal normal subgroup of the group G′H, and hence applicationof Lemma 11 to such group yields that there exists a subgroup K of infinite rank of Hand a proper K-invariant subgroup B of G′ such that G′/B is periodic. Observe herethat the subgroup B cannot be trivial, because the minimal normal subgroup G′ of Gis not periodic. If K centralizes G′, the product G′K is an abelian normal subgroup ofinfinite rank, and so G is metahamiltonian by Corollary 10. This contradiction shows thatCG′(K) = CG′(AK) is a proper G-invariant subgroup of G′, and hence CG′(K) = {1}.Thus BK is a non-abelian subgroup of infinite rank, and so it is normal in G. Then
B = BK ∩G′
is likewise normal in G, and hence B = G′. This last contradiction completes the proofof the lemma. �
The metabelian step of the main theorem is a special case of the following result.
Corollary 13. Let G be a group whose non-abelian subgroups of infinite rank are normal.If G contains a normal subgroup N of infinite rank such that N ′′ is finite, then G ismetahamiltonian.
Proof. The commutator subgroup N ′ of N has finite rank by Lemma 7. As the group N issoluble-by-finite, it contains an abelian subgroup H of infinite rank (see for instance [8]).Consider in N the normal subgroup of infinite rank W = N ′H. The factor group W/N ′′ ismetahamiltonian by Lemma 12, so that in particular it has finite commutator subgroup.It follows that W ′ is finite, so that the index |W : CW (W ′)| is finite and CW (W ′) hasinfinite rank; since CW (W ′) is a nilpotent normal subgroup of N with class at most 2, thegroup N is metahamiltonian by Corollary 10. In particular, N ′ is finite, and applying nowCorollary 10 to the centralizer CN (N ′), we obtain that G itself is metahamiltonian. �
We are now ready to complete the proof of the theorem.
Proof of Theorem. Assume for a contradiction that the statement is false. Then thecommutator subgroup G′ of G is an infinite non-abelian group by Corollary 13; more-over, G′ has finite rank by Lemma 7. Let N be any proper non-abelian G-invariantsubgroup of G′. Then all subgroups of infinite rank of G/N are normal, so that G/N
is a hamiltonian group (see [7], Theorem C) and hence G′/N has order 2; it follows inparticular that N cannot contain proper non-abelian G-invariant subgroups. Let U/N
be an abelian subgroup of index 2 in G/N , and consider an element g of G \ U . If X isany proper non-abelian U -invariant subgroup of N , the intersection X ∩ Xg is normalin G, and so it is abelian. On the other hand, U/X ∩Xg contains an abelian G-invariantsubgroup V/X ∩ Xg of finite index, and V is metabelian, contradicting Corollary 13.This contradiction shows that N has no proper non-abelian U -invariant subgroups, andso N = U ′ because U cannot be metabelian. Replacing now G by U , we may suppose
M. De Falco et al. / Journal of Algebra 407 (2014) 135–148 147
that all proper G-invariant subgroups of G′ are abelian. Let A be a maximal abelianG-invariant subgroup of G′, so that G′/A is a chief factor of G.
The commutator subgroup G′ contains a locally soluble subgroup S of finite index(see [1]), and there exists a positive integer k such that S(k) is hypercentral (see [10],Part 2, Lemma 10.39). It follows that G is hyperabelian-by-finite, and so it containsan abelian subgroup H of infinite rank. Then the metabelian subgroup AH cannot benormal in G by Corollary 13, so that AH is abelian and [A,H] = {1}; in particular, thecentralizer CG(A) has infinite rank. As all subgroups of G properly containing CG(A) arenon-abelian, G/CG(A) is a Dedekind group, and hence it contains an abelian subgroupW/CG(A) of index at most 2. Thus W ′ lies in CG(A), but it is not contained in A, since Wcannot be metabelian. Moreover, W ′ = G′ because all proper G-invariant subgroups ofG′ are abelian. Therefore A is contained in Z(G′H).
If the chief factor G′/A is finite, then AH is an abelian subgroup of finite index inG′H, and so G′H contains also an abelian G-invariant subgroup of finite index. ThusG′H is metahamiltonian by Corollary 10, and so G is likewise metahamiltonian by Corol-lary 13, a contradiction. Therefore G′/A must be infinite, and hence abelian, so that G′′
is contained in A. Assume that [G′, H] is abelian. Then
(G′H
)′ = G′′[G′, H]
is likewise abelian, so that (G′H)′ lies in A, because A is a maximal abelian G-invariantsubgroup of G′. It follows that the normal subgroup G′H is nilpotent of class 2, andhence G is metahamiltonian by Corollary 10. This contradiction proves that [G′, H] isnot abelian, so that G′′[G′, H] = G′ and (G′H)′ = G′. Then the factor group G′H/A ismetabelian, and so it is metahamiltonian by Corollary 13. This forces the commutatorsubgroup G′/A of G′H/A to be finite, and this last contradiction completes the proof ofthe theorem. �References
[1] N.S. Černikov, A theorem on groups of finite special rank, Ukrainian Math. J. 42 (1990) 855–861.[2] M. De Falco, F. de Giovanni, C. Musella, N. Trabelsi, Groups with restrictions on subgroups of
infinite rank, Rev. Mat. Iberoam. (2014), in press.[3] M. De Falco, F. de Giovanni, C. Musella, N. Trabelsi, Groups whose proper subgroups of infinite
rank have finite conjugacy classes, Bull. Aust. Math. Soc. 89 (2014) 41–48.[4] M.R. Dixon, M.J. Evans, H. Smith, Locally (soluble-by-finite) groups with all proper non-nilpotent
subgroups of finite rank, J. Pure Appl. Algebra 135 (1999) 33–43.[5] M.R. Dixon, M.J. Evans, H. Smith, Groups with all proper subgroups (finite rank)-by-nilpotent,
Arch. Math. (Basel) 72 (1999) 321–327.[6] M.R. Dixon, Z.Y. Karatas, Groups with all subgroups permutable or of finite rank, Cent. Eur. J.
Math. 10 (2012) 950–957.[7] M.J. Evans, Y. Kim, On groups in which every subgroup of infinite rank is subnormal of bounded
defect, Comm. Algebra 32 (2004) 2547–2557.[8] M.I. Kargapolov, On soluble groups of finite rank, Algebra Logika 1 (1962) 37–44.[9] A.I. Mal’cev, On certain classes of infinite soluble groups, Mat. Sb. 28 (1951) 567–588; Amer. Math.
Soc. Transl. Ser. 2, 1956, pp. 1–21.[10] D.J.S. Robinson, Finiteness Conditions and Generalized Soluble Groups, Springer, Berlin, 1972.
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