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Generalized nilpotency in uncountable groups

Generalized nilpotency in uncountable groups 1IntroductionA group G is said to have finite rank if there is a positive integer r such that every finitely generated subgroup of G can be generated by r elements; if such an integer does not exist, the group G is said to have infinite rank. It turns out that the structure of a (soluble) group of infinite rank is strongly influenced by that of its proper subgroups of infinite rank (see for instance the survey [4], where an almost complete reference list on the subject can be found, and the recent papers [6, 7] together with their reference lists). This kind of problem is actually a very special case of a more general one concerning the influence of “large” proper substructures on the whole structure. At first glance, it may appear that “infinite rank” is a not so natural concept of “largeness” in general, but, as far as groups are concerned, it is, because it is mainly defined in group theoretical terms. On the other hand, what seems to be in general a more natural concept of “largeness”, i.e. cardinality, does not appear to be strictly connected to the group theoretical structure itself, a priori. In 2016, F. de Giovanni and the second author [9] began the investigation of the influence of proper subgroups of large cardinality on the whole group. This investigation and subsequent ones (see [5], [3], [14], [15], [11], etc.) shows that, although in a mild form (compared with that of the rank), the influence is still there. The aim of this paper is to give a further contribution to the topic in connection with classes of groups that are close to being nilpotent.Let 𝔛{\mathfrak{X}} be a class of groups. The kind of problem we deal with in the first part of the paper is essentially the following one. Let G be a group of uncountable cardinality m{\mathfrak{m}} and assume that all proper subgroups of cardinality m{\mathfrak{m}} are X{\mathfrak{X}}-groups. Is it true that G is an X{\mathfrak{X}}-group as well? Of course, if we do not impose any additional condition on G we can have some problems. In fact, it has been shown by Shelah (without even appealing to the continuum hypothesis) that there are groups of cardinality ℵ1{\aleph_{1}} whose proper subgroups have cardinality strictly smaller than ℵ1{\aleph_{1}} (see [23]). Luckily, it turns out that these kind of monsters (named Jónsson groups of cardinality 𝔪{\mathfrak{m}}) admit simple homomorphic images of cardinality 𝔪{\mathfrak{m}} (see for instance [9]) and so we can get rid of them through the imposition of some generalized solubility condition.Among others results, we will show here that the above question answer positively whenever 𝔛{\mathfrak{X}} is one the following group classes (we refer to the next section for more precise definitions and statements): the class of (bounded) Engel groups (see Theorem 2.11 and Theorem 2.12), the class 𝒩1{\mathcal{N}_{1}} of groups whose subgroups are subnormal (see Corollary 2.6), the class 𝒩1f{\mathcal{N}_{1}^{f}} of groups whose subgroups are f-subnormal (see Corollary 2.7) and the class of Gruenberg groups (see Theorem 2.9). Notice here that many of these statements are actually corollaries of a more general result (see Theorem 2.4).Special attention is given to the class of hypercentral groups (see Section 3), for which we preliminary need to develop a theory describing how the hypercentral length of a group is connected to those of its proper subgroups. We employ such a theory to show that a (generalized soluble) uncountable group of cardinality 𝔪{\mathfrak{m}} whose proper subgroups of cardinality 𝔪{\mathfrak{m}} are hypercentral is itself hypercentral provided that the supremum of the hypercentral lengths of the proper subgroups of cardinality 𝔪{\mathfrak{m}} is smaller than the cofinality of the cardinal number 𝔪{\mathfrak{m}}.In Section 4, we see how to use the Generalized Continuum Hypothesis (GCH) to remove the assumption we made on the cofinality of the cardinality of the groups involved in some of the above-described statements. Notice that in this paper we assume the full strength of the Axiom of Choice.In the final part of the paper, we deal with groups whose large subgroups are subnormal and we generalize results of [3] showing that there is no need of any additional hypothesis on the cardinal number besides that of being uncountable.Most of our notation is standard and can be found in [20]. We use lower-case Gothic letters for cardinal numbers, while lower-case Greek letters for ordinal numbers.2Engel groups and groups with many subnormal subgroupsWe start the section with the following result, which deals with groups whose proper large subgroups are finite-by-soluble, i.e. admitting a finite normal subgroup with soluble factor group. This lemma (which is interesting in its own) will be employed in Corollary 2.7 but we put it here in order to not break up the discussion of the consequences of our first main theorem (Theorem 2.4). Recall that a group G is said to be locally graded if every non-trivial finitely generated subgroup of G contains a proper non-trivial subgroup of finite index. In the following, if 𝔪{\mathfrak{m}} is any cardinal number, we denote by cf⁡(𝔪){\operatorname{cf}(\mathfrak{m})} the cofinality of 𝔪{\mathfrak{m}}.Lemma 2.1.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a locally graded group of cardinality m{\mathfrak{m}} which has no simple homomorphic images of cardinality m{\mathfrak{m}}. If all proper subgroups of cardinality m{\mathfrak{m}} of G are finite-by-soluble, then G is (locally soluble)-by-finite.Proof.Suppose G is not (locally soluble)-by-finite, so G is perfect, admits no proper subgroups of finite index, and has a countable subgroup X that is not (locally soluble)-by-finite (see [10, Theorem 3.2]). Notice also that every proper normal subgroup of G is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G (see for instance [9, Corollary 2.6]) and so it is finite-by-soluble.Assume first that G has no proper normal subgroup of cardinality 𝔪{\mathfrak{m}}. Then G is the join of its proper normal subgroups (recall that there are no simple homomorphic images of cardinality 𝔪{\mathfrak{m}}), each of which is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, while the join of any finite number of proper normal subgroups is still a proper normal subgroup. Thus, each element x of X is contained in a proper normal subgroup Nx{N_{x}} of G. Since cf⁡(𝔪)>|X|{\operatorname{cf}(\mathfrak{m})>|X|}, we get that X is contained in the proper normal subgroup N=〈Nx:x∈X〉{N=\langle N_{x}:x\in X\rangle}. As we remarked above, N is finite-by-soluble, so X is finite-by-soluble and this gives a contradiction.Suppose now that G admits a proper normal subgroup N of cardinality 𝔪{\mathfrak{m}}. Clearly, G=X⁢N{G=XN} and it follows from the main result of [19] that G/N{G/N} is still locally graded, so G/N{G/N} cannot be finitely generated. Since X≃G/N{X\simeq G/N} is countable, we can find a chainN<X1<X2<⋯<Xn<Xn+1<⋯N<X_{1}<X_{2}<\cdots<X_{n}<X_{n+1}<\cdots\vspace*{-0.5mm}of proper subgroups of G such that G=⋃iXi{G=\bigcup_{i}X_{i}}. For each i, let Si{S_{i}} be the smallest normal subgroup of Xi{X_{i}} such that Xi/Si{X_{i}/S_{i}} is soluble. Obviously, Si≤Si+1{S_{i}\leq S_{i+1}} for every i, and S=⋃iSi{S=\bigcup_{i}S_{i}} is a countable normal subgroup of G such that G/S{G/S} is locally soluble, because every finitely generated subgroup of G is contained in some Xj{X_{j}}. Moreover, S is itself finite-by-soluble and so it contains a finite G-invariant subgroup T such that S/T{S/T} is soluble. However, G admits no proper subgroups of finite index, so T is abelian and hence G is locally soluble, the final contradiction. ∎An interesting (and direct) consequence of the above lemma is that all non-soluble finite pieces of the proper large subgroups vanish if we ask for G to admit no non-abelian simple homomorphic images.Corollary 2.2.Let m{\mathfrak{m}} be an uncountable cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a locally graded group of cardinality m{\mathfrak{m}} which has no non-abelian simple homomorphic images. If all proper subgroups of cardinality m{\mathfrak{m}} of G are finite-by-soluble, then G is locally soluble and, in particular, all proper subgroups of cardinality m{\mathfrak{m}} are soluble.In the above corollary we are not able to deduce that G itself is soluble. The problem here is that the structure of insoluble, locally soluble groups whose proper subgroups are soluble is not known; although such groups must be countable, they may occur as homomorphic images of uncountable groups. On the other hand, observe that if we also have a bound on both the finite piece and the derived length of the soluble residual of all proper large subgroups, then G is soluble by [11, Theorem D].Now, we define the group class for which we prove our first main theorem: a group G is an 𝒩′{\mathcal{N}^{\prime}}-group if and only if either |K:H|<∞{|K:H|<\infty} or HK<K{H^{K}<K} for all subgroups H≤K{H\leq K} of G. Of course, all nilpotent groups and all finite groups are 𝒩′{\mathcal{N}^{\prime}}-groups but the consideration of the locally dihedral 2-group shows that not every locally nilpotent group is an 𝒩′{\mathcal{N}^{\prime}}-group. Therefore, the class 𝒩′{\mathcal{N}^{\prime}} is not a local class, although, by its very definition, is closed with respect to forming subgroups and homomorphic images. On the other hand, our next lemma shows that this class is at least countably recognizable, i.e. a group is an 𝒩′{\mathcal{N}^{\prime}}-group if and only if all its countable subgroups are 𝒩′{\mathcal{N}^{\prime}}-groups.Lemma 2.3.The class N′{\mathcal{N}^{\prime}} is countably recognizable.Proof.Let G be a group whose countable subgroups are 𝒩′{\mathcal{N}^{\prime}}-groups and suppose by contradiction that G is not an 𝒩′{\mathcal{N}^{\prime}}-group. Then there is a proper subgroup X of G such that |G:X|=∞{|G:X|=\infty} and XG=G{X^{G}=G}. Let U1{U_{1}} be any countable subgroup of X and let 𝒯{\mathcal{T}} be a countably infinite subset of a given transversal to X in G. Put V1=〈U1,𝒯〉{V_{1}=\langle U_{1},\mathcal{T}\rangle}. Suppose we have defined countable subgroups Un{U_{n}} of X and Vn{V_{n}} of G such that |Vn:Un|=∞{|V_{n}:U_{n}|=\infty}; of course, by assumptions, UnVn<Vn{U_{n}^{V_{n}}<V_{n}}. Since XG=G{X^{G}=G}, for each v∈Vn∖UnVn{v\in V_{n}\setminus U_{n}^{V_{n}}}, there are finitely generated subgroups En⁢(v){E_{n}(v)} of X and Fn⁢(v){F_{n}(v)} of G such that v lies in En⁢(v)Fn⁢(v){E_{n}(v)^{F_{n}(v)}}. Put Un+1=〈Un,En(v):v∈Vn∖UnVn〉{U_{n+1}=\langle U_{n},E_{n}(v):v\in V_{n}\setminus U_{n}^{V_{n}}\rangle} and Vn+1=〈Un+1,Fn⁢(v)〉{V_{n+1}=\langle U_{n+1},F_{n}(v)\rangle}. Again, Un+1{U_{n+1}} is a subgroup of infinite index in Vn+1{V_{n+1}} because 𝒯{\mathcal{T}} lies in Vn+1{V_{n+1}}.Now, let U be the union of the subgroups Ui{U_{i}} and V the union of the subgroups Vj{V_{j}}; clearly, |V:U|=∞{|V:U|=\infty} since 𝒯⊆V{\mathcal{T}\subseteq V} and U≤X{U\leq X}. If v is any element of V, then it belongs to a certain Vn{V_{n}}, so it is either an element of UnVn{U_{n}^{V_{n}}} or it belongs to En⁢(v)Fn⁢(v){E_{n}(v)^{F_{n}(v)}}, which means that v lies in Un+1Vn+1{U_{n+1}^{V_{n+1}}}. In any case, v lies in UV{U^{V}} and so UV=V{U^{V}=V}, a contradiction.∎Our first main result shows that the group class 𝒩′{\mathcal{N}^{\prime}} is also recognizable by the behaviour of the proper large subgroups of a group. In what follows, if H and K are subgroups of a group, we denote by HK{H_{K}} the largest subgroup of H that is normalized by K; clearly, this coincides with the intersection of all conjugates of H by elements of K. Proposition 1.B.3 of [16] shows that hypothesis (2) in the following statement is satisfied in particular whenever the group is (locally soluble)-by-finite.Theorem 2.4.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are 𝒩′{\mathcal{N}^{\prime}}-groups,(1)every chief factor of G is finite or abelian.Then G is an N′{\mathcal{N}^{\prime}}-group.Proof.Let X be any countable subgroup of G; we show that X is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}}, so Lemma 2.3 completes the proof.Suppose first that G has no proper normal subgroup of cardinality 𝔪{\mathfrak{m}}. In this case, G is the join of its proper normal subgroups, each of which is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, while the join of any finite number of proper normal subgroups is still a proper normal subgroup of G. Thus, each element x of X is contained in a proper normal subgroup Nx{N_{x}} of G. The hypothesis on the cofinality yields that X is contained in the proper normal subgroup N=〈Nx:x∈X〉{N=\langle N_{x}:x\in X\rangle}. Now, G/N{G/N} is of cardinality 𝔪{\mathfrak{m}} and it cannot be a Jónsson group (see for instance [9, Corollary 2.6]), so it admits a proper subgroup M/N{M/N} of cardinality 𝔪{\mathfrak{m}}. Clearly, M is the required subgroup containing X.Suppose G admits a proper normal subgroup N of cardinality 𝔪{\mathfrak{m}}. Of course, we may assume that G=X⁢N{G=XN}. Let H=X∩N{H=X\cap N}. Since H is countable, then HN=H⁢[N,H]{H^{N}=H[N,H]} is a proper normal subgroup of N that is normalized by X. Thus X⁢HN<G{XH^{N}<G}, so HN{H^{N}} is of cardinality strictly smaller than 𝔪{\mathfrak{m}} and hence we may assume H≤HN={1}{H\leq H^{N}=\{1\}}. This assumption implies at once the possibility of a stronger one. In fact, now G=X⋉N{G=X\ltimes N} and if H is any proper G-invariant subgroup of cardinality 𝔪{\mathfrak{m}} of N, the subgroup XH is a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G and we are done. Therefore we may assume that N contains no proper G-invariant subgroups of cardinality 𝔪{\mathfrak{m}}.Let Y be a proper normal subgroup of X; in particular, L=Y⁢N{L=YN} is an 𝒩′{\mathcal{N}^{\prime}}-group and hence YL=Y⋉(YL∩N){Y^{L}=Y\ltimes(Y^{L}\cap N)} is a proper subgroup of YN. SinceL/YL∩N=YL/YL∩N×N/YL∩N,L/Y^{L}\cap N=Y^{L}/Y^{L}\cap N\times N/Y^{L}\cap N,it follows that [Y,N]{[Y,N]} lies in YL∩N{Y^{L}\cap N}, so [Y,N]{[Y,N]} is a proper subgroup of N. Moreover, [Y,N]{[Y,N]} is normalized by N and X, so it is a G-invariant subgroup of N and hence |[Y,N]|<𝔪{|[Y,N]|<\mathfrak{m}}. Let M be the join of all proper normal subgroups of X.If M=X{M=X}, then X is actually the join of countably many proper normal subgroups. Using the fact that the cofinality of 𝔪{\mathfrak{m}} is strictly larger than ω, we have that [X,N]{[X,N]} is of cardinality strictly smaller than 𝔪{\mathfrak{m}} and so we may assume G=X×N{G=X\times N}. Again by [9, Corollary 2.6], the subgroup N is not a Jónsson group, so it admits a proper subgroup M of cardinality 𝔪{\mathfrak{m}}. Of course, XM is the required subgroup, and we are done.Suppose now M<X{M<X}. Since [M,N]{[M,N]} is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, we can factor it out and assume that [M,N]={1}{[M,N]=\{1\}}. But then M is a normal subgroup of G and so we may even assume M={1}{M=\{1\}}. Obviously, this is equivalent to require that X is finite. If N′<N{N^{\prime}<N}, then N contains a proper subgroup E of countable index (see for instance [9, Lemma 2.3]) and X⁢EX{XE_{X}} is a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G, and we are done again. Thus N=N′{N=N^{\prime}} and in particular Z⁢(N)=Z2⁢(N){Z(N)=Z_{2}(N)} by a well-known result. Thus |Z⁢(N)|<𝔪{|Z(N)|<\mathfrak{m}} and we may assume that Z⁢(N)={1}{Z(N)=\{1\}}.Let K be any proper G-invariant subgroup of N; of course, |K|<𝔪{|K|<\mathfrak{m}}. Let g∈K{g\in K}; notice that |N:CN(g)|<𝔪{|N:C_{N}(g)|<\mathfrak{m}}. If CN⁢(g)≠N{C_{N}(g)\neq N}, then X⁢(CN⁢(g))X{X\big{(}C_{N}(g)\big{)}_{X}} is a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G and we are done. Thus CN⁢(g)=N{C_{N}(g)=N} for all g∈K{g\in K}, so K≤Z⁢(N)=1{K\leq Z(N)=1}. Thus N admits no non-trivial proper G-invariant subgroups, so it must be abelian and we have a contradiction.∎The first corollary we obtain is a somewhat stronger version of [11, Theorem A]. Notice that we will generalize this result in Theorem 3.7.Corollary 2.5.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are nilpotent,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is nilpotent.Proof.It follows from [11, Theorem B] that G is locally nilpotent, so we may apply Theorem 2.4 and get that G is an 𝒩′{\mathcal{N}^{\prime}}-group. If X is any countable subgroup of G, then |G:X|{|G:X|} is infinite and so XG<G{X^{G}<G}. Now, either XG{X^{G}} is of cardinality 𝔪{\mathfrak{m}} or G/XG{G/X^{G}} contains a proper subgroup M/XG{M/X^{G}} of cardinality 𝔪{\mathfrak{m}} (recall that G admits no Jónsson homomorphic images). In both cases, X is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}} and so it is nilpotent. It follows that the whole group G is nilpotent because the class of nilpotent groups is well known to be countably recognizable. ∎Recall that a group G is said to be an 𝒩1{\mathcal{N}_{1}}-group if all its subgroups are subnormal. It turns out that the class of 𝒩1{\mathcal{N}_{1}}-groups is countably recognizable (see [10, Theorem 2.7]) and that a group is an 𝒩1{\mathcal{N}_{1}}-group if and only if all its countable subgroups are subnormal (see also [13]). The following corollary can be now proved in a fashion similar to the previous one.Corollary 2.6.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are 𝒩1{\mathcal{N}_{1}}-groups,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is an N1{\mathcal{N}_{1}}-group.Recall that a subgroup H of a group G is f-subnormal in G if there is a finite chain of subgroupsH=H0≤H1≤…≤Hn=GH=H_{0}\leq H_{1}\leq\dots\leq H_{n}=Gsuch that for i=1,2,…,n{i=1,2,\dots,n} either the index |Hi:Hi-1|{|H_{i}:H_{i-1}|} is finite or Hi-1{H_{i-1}} is normal in Hi{H_{i}}. Let 𝒩1f{\mathcal{N}_{1}^{f}} be the class of all groups with only f-subnormal subgroups. It was proved in [2] that any 𝒩1f{\mathcal{N}_{1}^{f}}-group is a finite-by-soluble 𝒩′{\mathcal{N}^{\prime}}-group. On the other hand, in [13], we proved that the class 𝒩1f{\mathcal{N}_{1}^{f}} is countably recognizable. Our next corollary is therefore again a consequence of Theorem 2.4 (the proof being similar to that of Corollary 2.5, but with the replacement of [11, Theorem B] by Lemma 2.1).Corollary 2.7.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are 𝒩1f{\mathcal{N}_{1}^{f}}-groups,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is an N1f{\mathcal{N}_{1}^{f}}-group.Of course, similar corollaries hold for many other countably recognizable group classes contained in 𝒩1f{\mathcal{N}_{1}^{f}} (see for instance [13]), but, here, we just point out that the class of finite-by-nilpotent groups (i.e. groups admitting a finite normal subgroup with nilpotent factor group) is among these. Indeed, finite-by-nilpotent groups form a countably recognizable group class by [10, Theorem 3.6] and every subgroup H of a finite-by-nilpotent group G is obviously f-subnormal in G, so in particular this class is contained in 𝒩′{\mathcal{N}^{\prime}}. The following corollary should be seen in relation to [9, Theorem 4.6], which shows that a similar result holds for the class of nilpotent-by-finite groups.Corollary 2.8.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are finite-by-nilpotent,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is finite-by-nilpotent.The consideration of the locally dihedral 2-group shows that the assumption on the cofinality in the above results cannot be completely removed. However, we will see in Section 4 that using the Generalized Continuum Hypothesis we are able to prove all results for an arbitrary uncountable cardinal number.Engel groups are strictly related to Gruenberg groups (see after the proof of Theorem 2.9 for more details), that’s why we analyze the case of Gruenberg groups first. Recall that a group is called a Gruenberg group if it is generated by its ascendant abelian subgroups. Although the class of Gruenberg groups is not countably recognizable (see for instance [10]), our next result shows that it can be “uncountably” recognized.Theorem 2.9.Let G be a group of uncountable cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are Gruenberg groups,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is a Gruenberg group.Proof.It follows from [11, Theorem B] that G is locally nilpotent. Let N be any proper normal subgroup of G. If N is of cardinality 𝔪{\mathfrak{m}}, then all its cyclic subgroups are ascendant in G; if N is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, this is still the case since G/N{G/N} contains a proper subgroup of cardinality 𝔪{\mathfrak{m}}, which is a Gruenberg group. Let M be the subgroup generated by all proper normal subgroups of G. If G=M{G=M}, we are done. If M≠G{M\neq G}, then G/M{G/M} is cyclic of prime order; take g∈G{g\in G} with G=〈g〉⁢M{G=\langle g\rangle M}. We only need to show that 〈g〉{\langle g\rangle} is ascendant in G, so assume that this is not the case. Of course, G/G′{G/G^{\prime}} is cyclic of prime power order and actually G=〈g〉⁢G′{G=\langle g\rangle G^{\prime}}.Let V be any proper G-invariant subgroup of G′{G^{\prime}} such that G′/V{G^{\prime}/V} does not contain any non-trivial proper G-invariant subgroup. Since G is locally nilpotent, its chief factors are central and so G′/V≤Z⁢(G/V){G^{\prime}/V\leq Z(G/V)}. Thus G/V{G/V} is abelian and we have reached a contradiction. In particular, G′{G^{\prime}} admits no proper subgroup of finite index.Let 〈h〉=G′∩〈g〉{\langle h\rangle=G^{\prime}\cap\langle g\rangle}. If h≠1{h\neq 1}, then we can find a chief factor H/K{H/K} of G such that H=〈h〉⁢K≤G′{H=\langle h\rangle K\leq G^{\prime}}; if h=1{h=1}, we let H={1}{H=\{1\}}. In both cases, H<G′{H<G^{\prime}}. If G′{G^{\prime}} contains a proper G-invariant subgroup W of cardinality 𝔪{\mathfrak{m}} with W≥H{W\geq H}, then, since G′/W{G^{\prime}/W} admits no maximal (proper) G-invariant subgroups, we can define an ascending chainW≤W1<W2<…<Wα<Wα+1<⋯W\leq W_{1}<W_{2}<\dots<W_{\alpha}<W_{\alpha+1}<\cdotsof proper G-invariant subgroups of G′{G^{\prime}} whose union is G′{G^{\prime}}. Of course, 〈g〉{\langle g\rangle} is ascendant in every 〈g〉⁢Wα{\langle g\rangle W_{\alpha}} and so obviously 〈g〉{\langle g\rangle} is ascendant in 〈g〉⁢G′=G{\langle g\rangle G^{\prime}=G}, a contradiction.Now, suppose that G′{G^{\prime}} admits no proper G-invariant subgroup of cardinality 𝔪{\mathfrak{m}} containing H and let U be any proper G-invariant subgroup of G′{G^{\prime}} containing H. If G′/U{G^{\prime}/U} is abelian, we may find a proper subgroup L/U{L/U} of G′/U{G^{\prime}/U} such that |G′/L|=ℵ0{|G^{\prime}/L|=\aleph_{0}}. Then LG=∩x∈〈g〉Lx{L_{G}=\cap_{x\in\langle g\rangle}L^{x}} is G-invariant and |G′:LG|=ℵ0{|G^{\prime}:L_{G}|=\aleph_{0}}, a contradiction. Thus G′/U{G^{\prime}/U} is non-abelian and so there is a proper G-invariant subgroup V of G such that U<V<G′{U<V<G^{\prime}} and V/U{V/U} is not contained in Z⁢(G′/U){Z(G^{\prime}/U)}. Let G¯=G/U{\overline{G}=G/U} and notice that G¯′/CG¯′⁢(v⁢U){\overline{G}^{\prime}/C_{\overline{G}^{\prime}}(vU)} is of cardinality strictly smaller than 𝔪{\mathfrak{m}} for some v⁢U∈V/U{vU\in V/U} that is non-central in G¯′{\overline{G}^{\prime}}. Put C/U=CG¯′⁢(v⁢U){C/U=C_{\overline{G}^{\prime}}(vU)} and D=C〈g〉{D=C_{\langle g\rangle}}. Now, D is 〈g〉{\langle g\rangle}-invariant, contains U and |G′:D|<𝔪{|G^{\prime}:D|<\mathfrak{m}}. It follows that 〈g〉{\langle g\rangle} is ascendant in 〈g〉⁢D{\langle g\rangle D} and so a fortiori in 〈g〉⁢U{\langle g\rangle U}.Since G′/H{G^{\prime}/H} admits no G-invariant (proper) maximal subgroups, it is clearly possible to define an ascending chainH≤H1<H2<…<Hα<Hα+1<⋯H\leq H_{1}<H_{2}<\dots<H_{\alpha}<H_{\alpha+1}<\cdotsof proper G-invariant subgroups of G′{G^{\prime}} whose union is G′{G^{\prime}}. By what we have already proved, it follows that 〈g〉{\langle g\rangle} is ascendant in every 〈g〉⁢Hα{\langle g\rangle H_{\alpha}} and so obviously 〈g〉{\langle g\rangle} is ascendant in 〈g〉⁢G′=G{\langle g\rangle G^{\prime}=G}, the final contradiction.∎Observe that it is well known (and easy to see) that a countable and locally nilpotent group is a Gruenberg group. The consideration of the locally dihedral group C2⋉C3∞{C_{2}\ltimes C_{3^{\infty}}} (here C2{C_{2}} is a cyclic group of order 2 acting as the inversion on an infinite locally cyclic 3-group C3∞{C_{3^{\infty}}}) shows that either the uncountability of the cardinal or the locally nilpotent assumption cannot be omitted.Recall that a group G is called an Engel group if it coincides with the set of all its left (right) Engel elements, i.e. the set of all elements x of G such that, for all g∈G{g\in G}, we have [g,nx]=1{[g,_{n}x]=1} ([x,ng]=1{[x,_{n}g]=1}) for a suitable positive integer n=n⁢(g,x){n=n(g,x)}. Clearly, any locally nilpotent group is an Engel group and a well-known result of Zorn shows that any finite Engel group is nilpotent. Also notice that for soluble groups the properties of being Gruenberg, locally nilpotent or Engel coincide (see [21, 12.3.3]), so Theorem 2.9 can be rephrased as follows: if G is a soluble group of cardinality 𝔪{\mathfrak{m}} (cf⁡(𝔪)>ω{\operatorname{cf}(\mathfrak{m})>\omega}) whose proper subgroups of cardinality 𝔪{\mathfrak{m}} are Engel groups, then G is an Engel group. The solubility assumption can be weakened through the concept of strong local graduation. Recall that a group G is strongly locally graded if every section of G is locally graded (actually we will only need that every homomorphic image of G is locally graded). Strongly locally graded groups form a proper subclass of the class of all locally graded groups as shown by the consideration of any free non-abelian group, but this class is still large enough to contain for instance all locally (soluble-by-finite) groups.Before proving our main result concerning Engel groups, we need the following preparatory lemma on uncountable modules over Principal Ideal Domains.Lemma 2.10.Let m{\mathfrak{m}} be an uncountable cardinal number and let R be a Principal Ideal Domain of cardinality strictly smaller than m{\mathfrak{m}}. If M is an R-module of cardinality m{\mathfrak{m}}, then M contains a submodule of cardinality m{\mathfrak{m}} which is the direct sum of (infinitely many) non-trivial submodules of strictly smaller cardinality.Proof.Let E=ER⁢(M){E=E_{R}(M)} be the injective hull of M. Since R is a Principal Ideal Domain, it is well known that E is a direct sum of indecomposable injective R-modules and every indecomposable injective R-module is (the injective hull of) the residue field at a prime. It follows that every indecomposable injective R-module is of the same cardinality of R, so E is the direct sum of an infinite family {E}E∈ℰ{\{E\}_{E\in\mathcal{E}}} of non-trivial R-submodules of cardinality strictly smaller than 𝔪{\mathfrak{m}}. On the other hand, M is an essential R-submodule of E, so M∩Ei≠{0}{M\cap E_{i}\neq\{0\}}. The R-submodule ⊕E∈ℰ(M∩E){\bigoplus_{E\in\mathcal{E}}(M\cap E)} of M is the required one.∎Theorem 2.11.Let G be a group of uncountable cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are non-perfect Engel groups,(1)G is strongly locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is an Engel group.Proof.Suppose that G is not an Engel group, so there are elements x and y of G such that F=〈x,[y,x]〉{F=\langle x,[y,x]\rangle} is not an Engel group. Let N be any proper normal subgroup of G. If N is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, then G/N{G/N} contains a proper subgroup of cardinality 𝔪{\mathfrak{m}}, which is an Engel group; if |N|=𝔪{|N|=\mathfrak{m}}, then it is an Engel group by hypothesis. Suppose first that all proper normal subgroups of G are of cardinality strictly smaller than 𝔪{\mathfrak{m}}. As we have already seen many times, in such circumstances, F is contained in a proper normal subgroup of G and so it is an Engel group, a contradiction. Therefore, we may assume that G contains a proper normal subgroup M of cardinality 𝔪{\mathfrak{m}}; obviously, G=F⁢M{G=FM}. Since G/M{G/M} is locally graded and finitely generated, we may assume G/M{G/M} is even finite. Of course, all proper subgroups of G/M{G/M} are Engel groups, so it follows from [1, Theorem 2.1] that G/M{G/M} is either an Engel group or a minimal non-nilpotent group. In the former case, G/M{G/M} is nilpotent; in the latter case, it is soluble (see for instance [21, 9.1.9]). In both cases G′<G{G^{\prime}<G} and in particular G=F⁢G′=〈x〉⁢G′{G=FG^{\prime}=\langle x\rangle G^{\prime}}, which means that any nilpotent homomorphic image of G is cyclic.Let K be any proper normal subgroup of G such that G/K{G/K} is soluble: we prove that |K|=𝔪{|K|=\mathfrak{m}}. Suppose this is not the case and notice that the solubility of G/K{G/K} makes it possible to replace K in such a way that there is a normal abelian subgroup H/K{H/K} of cardinality 𝔪{\mathfrak{m}}. Moreover, an easy combination of [21, 12.3.3] and [11, Theorem B] shows that G/K{G/K} is locally nilpotent. In order to derive the contradiction that F is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}}, we factor out the normal subgroup F⁢K∩H{FK\cap H} and consequently assume that K=F∩H=1{K=F\cap H=1}; in particular, F≃G/H{F\simeq G/H} is cyclic since now G is locally nilpotent. By Lemma 2.10 the subgroup H contains a G-invariantsubgroup A of cardinality 𝔪{\mathfrak{m}} that is the direct product of G-invariant subgroups Ai{A_{i}}, i∈I{i\in I}, of cardinality strictly smaller than 𝔪{\mathfrak{m}}. Since the cofinality of 𝔪{\mathfrak{m}} is a limit ordinal, it follows that we may writeI=I1∪I2,{I=I_{1}\cup I_{2}},where I1∩I2=∅{I_{1}\cap I_{2}=\emptyset} and the G-invariant subgroupsA(I1)=〈Ai:i∈I1〉 and A(I2)=〈Ai:i∈I2〉A(I_{1})=\langle A_{i}:i\in I_{1}\rangle\quad\text{and}\quad A(I_{2})=\langle A%_{i}:i\in I_{2}\rangle\vspace*{-0.5mm}have both cardinality 𝔪{\mathfrak{m}}. Clearly, F⁢A⁢(I1)<G{F\,A(I_{1})<G} and we get the required contradiction.Therefore K is of cardinality 𝔪{\mathfrak{m}}. In this case, all proper subgroups of G/K{G/K} are locally nilpotent, and so [8, Lemma 3.2] yields that either G/K{G/K} is finite and minimal non-nilpotent, or it is locally nilpotent. In the former case, G′′′≤K{G^{\prime\prime\prime}\leq K} by [22]; in the latter case, G/K=F⁢K/K{G/K=FK/K} is nilpotent and so abelian. In both cases, G′′′≤K{G^{\prime\prime\prime}\leq K}. The arbitrariness of K shows that |G′′′|=𝔪{|G^{\prime\prime\prime}|=\mathfrak{m}} and that G′′′{G^{\prime\prime\prime}} is the last term of the derived series of G. This is clearly in contradiction with the hypotheses of the statement.∎Let n be any positive integer. By an n-Engel group is meant a group G such that [x,ny]=1{[x,_{n}y]=1} for all x,y∈G{x,y\in G}; that is, every element is both left and rightn-Engel. It follows that the class of n-Engel groups is the variety determined by the law [x,ny]=1{[x,_{n}y]=1}. Of course, any nilpotent group of class n is an n-Engel group but n-Engel groups need not be nilpotent. The final theorem of this section shows that when we deal with n-Engel groups, the additional assumptions in Theorem 2.11 can be dropped. In order to prove such a result, we need preliminary to observe that [11, Theorem A] does not really need any assumption on the cofinality whenever the cardinality 𝔪{\mathfrak{m}} is assumed to be uncountable and there is a bound on the nilpotency classes of the proper subgroups of cardinality 𝔪{\mathfrak{m}}: this can be seen from the proof of [11, Theorem A], but it is obtained also as a consequence of a more general result concerning hypercentrality in the next section (see Theorem 3.8).Theorem 2.12.Let k be a positive integer and let G be a group of uncountable cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are k-Engel groups,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is a k-Engel group.Proof.It follows from [17, Corollary 6] that every locally graded k-Engel group is locally nilpotent, so [11, Theorem B] shows that G is locally nilpotent. Suppose that G is not a k-Engel group, so there is a finitely generated subgroup F=〈x,y〉{F=\langle x,y\rangle} which is not contained in any proper subgroup of cardinality 𝔪{\mathfrak{m}} or in any proper normal subgroup N of G (since G/N{G/N} contains a subgroup L/N{L/N} such that L is of cardinality 𝔪{\mathfrak{m}}); of course, F is nilpotent, of class c, say.Now, if X is any normal subgroup of cardinality 𝔪{\mathfrak{m}} of G, then G=F⁢X{G=FX} and so G/X{G/X} is nilpotent of class at most c. It follows that S=γc+1⁢(G){S=\gamma_{c+1}(G)} lies in X. Moreover, if |γh⁢(G)|<𝔪{|\gamma_{h}(G)|<\mathfrak{m}} for some positive integer h, we can easily find a proper normal subgroup M of G such that both |M|{|M|} and |G/M|{|G/M|} are of cardinality 𝔪{\mathfrak{m}} (see for instance [9, Lemma 2.4]). Then FM is a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G and we obtain a contradiction. Therefore S is the smallest normal subgroup of cardinality 𝔪{\mathfrak{m}} of G and it is also the smallest term of the lower central series of G.A well-known theorem of Zelmanov asserts that any torsion-free nilpotent k-Engel group is nilpotent of class bounded by an integer depending only on k, so a torsion-free locally nilpotent k-Engel group is nilpotent. Let T be the periodic part of G. If |T|<𝔪{|T|<\mathfrak{m}}, then all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G/T{G/T} are nilpotent and so it follows from the remark just before the statement that G/T{G/T} is nilpotent, a contradiction since in this case |S|≤|T|<𝔪{|S|\leq|T|<\mathfrak{m}}. Thus |T|=𝔪{|T|=\mathfrak{m}} and S≤T{S\leq T}.Suppose first that G/S{G/S} is non-periodic and choose a prime q. Then G contains a normal subgroup Q of index q. Let θ be the word [x1,kx2]{[x_{1},_{k}x_{2}]} and notice now that the verbal subgroup θ⁢(Q){\theta(Q)} is trivial. An application of [18, Theorem 2.3.5] therefore yields that θ⁢(G){\theta(G)} is a q-group. The arbitrariness of the prime q shows that θ⁢(G)={1}{\theta(G)=\{1\}} and so that G is a k-Engel group, a contradiction.It is thus possible to assume that G is periodic; in particular, F is finite. Since S is the smallest normal subgroup of cardinality 𝔪{\mathfrak{m}} of G, the finite group F∩S{F\cap S} is contained in a proper G-invariant subgroup of S (which is actually a term of a chief series of G) that has cardinality strictly smaller than 𝔪{\mathfrak{m}}. In order to show that F is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G, it is therefore safe to assume G=F⋉S{G=F\ltimes S}.Clearly, S coincides with the subgroup generated by all its proper G-invariant subgroups, so that, if S is non-abelian, it contains a proper G-invariant subgroup K (of cardinality strictly smaller than 𝔪{\mathfrak{m}}) which is not contained in Z⁢(S){Z(S)}. Let x∈K∖Z⁢(S){x\in K\setminus Z(S)} and notice that 1<|S:CS(x)|<𝔪{1<|S:C_{S}(x)|<\mathfrak{m}}. The intersection C=⋂g∈FCS⁢(x)g{C=\bigcap_{g\in F}C_{S}(x)^{g}} is now an F-invariant subgroup of cardinality 𝔪{\mathfrak{m}}, hence giving the contradiction F⁢C<G{FC<G}. Thus S is abelian and so it contains a proper subgroup U of countable index for which we can repeat the above trick (here, U takes the place of K), thus obtaining the final contradiction.∎3HypercentralityThe aim of this section is to describe uncountable groups whose proper “large” subgroups are hypercentral. In order to carry out such an analysis we first need to understand how the hypercentral length of the whole group is connected to those of the “small” subgroups of the group. This leads to Corollaries 3.5 and 3.6, results that naturally generalize the well-known fact that the class of (nilpotent) hypercentral groups is countably recognizable.Let G be a group. We say that an element g of G is a Černikov element if and only if for any sequence x0,x1,…,xn,xn+1,…{x_{0},x_{1},\dots,x_{n},x_{n+1},\dots} of elements of G there is a certain non-negative integer m such that[…⁢[[g,x0],x1],…,xm]=1.{[\dots[[g,x_{0}],x_{1}],\dots,x_{m}]=1}.This definition is motivated by the theorem of Černikov which shows that hypercentrality is countably recognizable (see for instance [20, Theorem 2.19]). Although in a slightly different (but equivalent) form, this definition has been introduced for the first time in [12], where it was showed that it is connected to some “ring-theoretic questions” in group theory.The following result shows that the set of all Černikov element is a subgroup and actually a well-known subgroup for that matter.Lemma 3.1.Let G be a group. An element g of G belongs to the hypercentre Z¯{\overline{Z}} of G if and only if g is a Černikov element.Proof.Suppose that g lies in Z¯{\overline{Z}} but it is not a Černikov element. Then there is a sequence x0,x1,…,xn,xn+1,…{x_{0},x_{1},\dots,x_{n},}{x_{n+1},\dots} such that […[g,x0],x1],…,xm]≠1{[\dots[g,x_{0}],x_{1}],\dots,x_{m}]\neq 1} for all non-negative integers m. Let λ0{\lambda_{0}} be the smallest ordinal number such that g lies in Zλ0⁢(G){Z_{\lambda_{0}}(G)}. Then λ0{\lambda_{0}} is not limit and [g,x0]{[g,x_{0}]} lies in a term Zλ1⁢(G){Z_{\lambda_{1}}(G)} of the upper central series of G with λ1<λ0{\lambda_{1}<\lambda_{0}}; of course, λ1{\lambda_{1}} can be chosen successor. Clearly, [[g,x0],x1]{[[g,x_{0}],x_{1}]} lies in Zλ2⁢(G){Z_{\lambda_{2}}(G)} with λ2<λ1{\lambda_{2}<\lambda_{1}} and so on. Since the ordinal numbers are well ordered we have a contradiction.Suppose g is a Černikov element outside Z¯{\overline{Z}}. In order to achieve a contradiction, we can put Z¯=1{\overline{Z}=1}. Since Z⁢(G)={1}{Z(G)=\{1\}}, we can find an element x0{x_{0}} such that [g,x0]≠1{[g,x_{0}]\neq 1}. Since [g,x0]{[g,x_{0}]} does not belong to Z⁢(G){Z(G)}, we can find an element x1∈G{x_{1}\in G} such that [[g,x0],x1]≠1{[[g,x_{0}],x_{1}]\neq 1}. Continuing in this way we find a sequence x0,x1,…,xn,xn+1,…{x_{0},x_{1},\dots,x_{n},x_{n+1},\dots} of elements of G for which there is no non-negative integer m such that[…⁢[[g,x0],x1],…,xm]=1.[\dots[[g,x_{0}],x_{1}],\dots,x_{m}]=1.\vspace*{-0.5mm}This contradicts the definition of Černikov element.∎Now we need to introduce the concept of bounded Černikov element. Let G be a group. We say that 1 is the only Černikov 0-element of G and we put 𝒞0={1}{\mathcal{C}_{0}=\{1\}}. Let λ be a non-zero ordinal number and assume that we have defined the set 𝒞μ{\mathcal{C}_{\mu}} of all Černikov μ-elements of G for all ordinal numbers μ<λ{\mu<\lambda}. If λ is limit, we define the set 𝒞λ{\mathcal{C}_{\lambda}} of all Černikov λ-elements of G simply as the union of all sets 𝒞μ{\mathcal{C}_{\mu}} with μ<λ{\mu<\lambda}. If λ is successor, we define 𝒞λ{\mathcal{C}_{\lambda}} as the set of all elements g∈G{g\in G} such that, for all x∈G{x\in G}, there is an ordinal μx<λ{\mu_{x}<\lambda} such that [g,x]∈𝒞μx{[g,x]\in\mathcal{C}_{\mu_{x}}}.It is clear that g is a Černikov 1-element if and only if g lies in Z⁢(G){Z(G)}. Notice that a Černikov μ-element is also a Černikov λ-element whenever μ<λ{\mu<\lambda}, and that, for all g∈G{g\in G}, the commutator [g,x]{[g,x]} is a Černikov λ-element for all x∈G{x\in G} if and only if g is a Černikov (λ+1){(\lambda+1)}-element. Our next statement shows that for any ordinal number λ, the set 𝒞λ{\mathcal{C}_{\lambda}} is a subgroup.Lemma 3.2.Let G be a group and let λ be an ordinal number. An element g of G belongs to Zλ⁢(G){Z_{\lambda}(G)} if and only if g is a Černikov λ-element.Proof.We work by induction on λ, being the assertion obvious for λ=0,1{\lambda=0,1}. Suppose first that λ>1{\lambda>1} is limit. In this case, g lies in Zλ⁢(G){Z_{\lambda}(G)} if and only if it lies in Zμ⁢(G){Z_{\mu}(G)} for some ordinal μ<λ{\mu<\lambda}. By induction, the last statement is equivalent to require that g is a Černikov μ-element for some μ<λ{\mu<\lambda}, and, by definition, this is equivalent to require that g is a Černikov λ-element.Suppose now that λ>1{\lambda>1} is successor. In this case, g lies in Zλ⁢(G){Z_{\lambda}(G)} if and only if, for all x∈G{x\in G}, we have [g,x]∈Zλ-1⁢(G){[g,x]\in Z_{\lambda-1}(G)}. By induction, the last statement is equivalent to require that, for all x∈G{x\in G}, [g,x]∈𝒞λ-1{[g,x]\in\mathcal{C}_{\lambda-1}}, and this is equivalent to require that g∈𝒞λ{g\in\mathcal{C}_{\lambda}}. ∎Combining Lemma 3.1 and Lemma 3.2, we see that an element is a Černikov element if and only if it is a Černikov λ-element for some ordinal λ. The above characterizations of (bounded) Černikov elements will be subsequently used without any further reference, due to the large number of times it should be cited.Lemma 3.3.Let G be a group and let g be an element of the hypercentre of G. Let κ be the smallest ordinal number such that g lies in Zκ⁢(G){Z_{\kappa}(G)}. Then, for any ordinal λ<κ{\lambda<\kappa} which is either 0 or successor, there is a sequence of elements x0,x1,…,xm{x_{0},x_{1},\dots,x_{m}} of G such that λ is the smallest ordinal number for which[…⁢[[g,x0],x1],…,xm]∈Zλ⁢(G).{[\dots[[g,x_{0}],x_{1}],\dots,x_{m}]\in Z_{\lambda}(G)}.Proof.We work by induction on κ, being the statement true if κ=0,1{\kappa=0,1}. Suppose κ>1{\kappa>1}. Now, observe that κ is successor and g is a Černikov κ-element that is not a Černikov (κ-1){(\kappa-1)}-element.Suppose first κ-1{\kappa-1} is successor. Then there exists x∈G{x\in G} such that [g,x]{[g,x]} is a Černikov (κ-1){(\kappa-1)}-element that is not a Černikov (κ-2){(\kappa-2)}-element. Applying induction hypothesis completes the proof in this case.Assume now that κ-1{\kappa-1} is limit. Then, for any ordinal μ<κ-1{\mu<\kappa-1}, there are xμ∈G{x_{\mu}\in G} and a successor ordinal μ<κμ<κ-1{\mu<\kappa_{\mu}<\kappa-1} such that [g,xμ]{[g,x_{\mu}]} is a Černikov κμ{\kappa_{\mu}}-element that is not a Černikov (κμ-1){(\kappa_{\mu}-1)}-element. If we fix λ<κ{\lambda<\kappa}, which is either 0 or successor, then λ<κ-1{\lambda<\kappa-1}. Using the induction hypothesis on κλ>λ{\kappa_{\lambda}>\lambda} and [g,xλ]{[g,x_{\lambda}]}, we are able to complete the proof.∎Lemma 3.4.Let λ be an infinite ordinal number of cardinality m{\mathfrak{m}} and let G be a group. Then any subset X of cardinality at most m{\mathfrak{m}} of G is contained in a subgroup Y of cardinality at most m{\mathfrak{m}} of G such that Zκ⁢(G)∩Y=Zκ⁢(Y){Z_{\kappa}(G)\cap Y=Z_{\kappa}(Y)} for all κ≤λ{\kappa\leq\lambda}.Proof.We define the subgroup Y by recursion, but first we need a procedure to associate with every element g of G a set 𝒰g{\mathcal{U}_{g}} of elements of Gencoding the information concerning which term, if any, of the upper central series of G, it belongs. Thus, let 1≠g∈G{1\neq g\in G}. If g does not belong to the hypercentre Z¯{\overline{Z}} of G, we can find a sequence x0,x1,…,xn,xn+1,…{x_{0},x_{1},\dots,x_{n},x_{n+1},\dots} of elements of G such that […[g,x0],x1],…,xm]≠1{[\dots[g,x_{0}],x_{1}],\dots,x_{m}]\neq 1} for all non-negative integers m. In this case we put 𝒰g={x0,x1,…,xn,xn+1,…}{\mathcal{U}_{g}=\{x_{0},x_{1},\dots,x_{n},x_{n+1},\dots\}}.Suppose g lies in Z¯{\overline{Z}} and let μ be the smallest ordinal number such that g lies in Zμ⁢(G){Z_{\mu}(G)}. For any successor ordinal number κ with κ≤λ{\kappa\leq\lambda} and κ<μ{\kappa<\mu}, choose a sequence of elements x0κ,x1κ,…,xmκκ{x_{0}^{\kappa},x_{1}^{\kappa},\dots,x_{m_{\kappa}}^{\kappa}} such that[…⁢[[g,x0κ],x1κ],…,xmκκ]∈Zκ⁢(G)∖Zκ-1⁢(G){[\ldots[[g,x_{0}^{\kappa}],x_{1}^{\kappa}],\ldots,x_{m_{\kappa}}^{\kappa}]\inZ%_{\kappa}(G)\setminus Z_{\kappa-1}(G)}(this is possible by Lemma 3.3). Let 𝒰gκ={x0κ+,x1κ+,…,xmκ+κ+}{\mathcal{U}_{g}^{\kappa}=\{x_{0}^{\kappa^{+}},x_{1}^{\kappa^{+}},\dots,x_{m_{%\kappa^{+}}}^{\kappa^{+}}\}} and 𝒰g=⋃κ𝒰gκ{\mathcal{U}_{g}=\bigcup_{\kappa}\mathcal{U}_{g}^{\kappa}} (put 𝒰g0=∅{\mathcal{U}_{g}^{0}=\emptyset}).It is easy to see that in all cases 𝒰g{\mathcal{U}_{g}} is of cardinality at most 𝔪{\mathfrak{m}}. Now, let Y0=〈X〉{Y_{0}=\langle X\rangle} and suppose we have defined a subgroup Yn{Y_{n}} of G of cardinality at most 𝔪{\mathfrak{m}}. Then we putYn+1=〈Yn,𝒰g:g∈Yn〉.Y_{n+1}=\langle Y_{n},\mathcal{U}_{g}:g\in Y_{n}\rangle.Finally, put Y=⋃iYi{Y=\bigcup_{i}Y_{i}}; we show that Y is the required subgroup. Suppose not and let α be the smallest ordinal number such that α≤λ{\alpha\leq\lambda} and Zα⁢(G)∩Y≠Zα⁢(Y){Z_{\alpha}(G)\cap Y\neq Z_{\alpha}(Y)}. Of course, α≠0{\alpha\neq 0} is successor, so Zα-1⁢(G)∩Y=Zα-1⁢(Y){Z_{\alpha-1}(G)\cap Y=Z_{\alpha-1}(Y)}. Now, if x is any element of Zα⁢(G)∩Y{Z_{\alpha}(G)\cap Y}, it follows that [x,y]∈Zα-1⁢(G)∩Y≤Zα-1⁢(Y){[x,y]\in Z_{\alpha-1}(G)\cap Y\leq Z_{\alpha-1}(Y)} for all y∈Y{y\in Y}, and so Zα⁢(G)∩Y<Zα⁢(Y){Z_{\alpha}(G)\cap Y<Z_{\alpha}(Y)}. Let u∈Zα⁢(Y)∖(Zα⁢(G)∩Y){u\in Z_{\alpha}(Y)\setminus\big{(}Z_{\alpha}(G)\cap Y\big{)}} and let n be such that u∈Yn{u\in Y_{n}}. If u∉Z¯{u\not\in\overline{Z}}, then Yn+1{Y_{n+1}} contains 𝒰u{\mathcal{U}_{u}} and so u does not belong even to the hypercentre of Y (see Lemma 3.1), a contradiction. Thus u lies in Z¯{\overline{Z}}; let β be the smallest ordinal such that u lies in Zβ⁢(G){Z_{\beta}(G)}. If β≤α{\beta\leq\alpha}, we get u∈Zβ⁢(G)≤Zα⁢(G){u\in Z_{\beta}(G)\leq Z_{\alpha}(G)} and so u lies in Zα⁢(G)∩Y{Z_{\alpha}(G)\cap Y}, a contradiction. Thus β>α{\beta>\alpha}. In this case, for each successor ordinal γ≤α{\gamma\leq\alpha}, we have[⋱⁢[[g,x0γ],x1γ],…,xmγγ]∈(Zγ⁢(G)∖Zγ-1⁢(G))∩Y=Zγ⁢(Y)∖Zγ-1⁢(Y)[\ddots[[g,x_{0}^{\gamma}],x_{1}^{\gamma}],\ldots,x_{m_{\gamma}}^{\gamma}]\in%\big{(}Z_{\gamma}(G)\setminus Z_{\gamma-1}(G)\big{)}\cap Y=Z_{\gamma}(Y)%\setminus Z_{\gamma-1}(Y)and 𝒰uγ⊆Y{\mathcal{U}_{u}^{\gamma}\subseteq Y}. It follows that u is not a Černikov α-element of Y, so it belongs to Z¯⁢(Y)∖Zα⁢(Y){\overline{Z}(Y)\setminus Z_{\alpha}(Y)}, the final contradiction.∎Corollary 3.5.Let λ be an infinite ordinal number of cardinality m{\mathfrak{m}} and let G be a group. If Zλ⁢(X)=X{Z_{\lambda}(X)=X} for any subgroup X of cardinality at most m{\mathfrak{m}}, then Zλ⁢(G)=G{Z_{\lambda}(G)=G}.Proof.Suppose the statement is false. Of course, G is hypercentral since the class of all hypercentral groups is countably recognizable. Now, since Zλ⁢(G)<G{Z_{\lambda}(G)<G}, we have that Zλ⁢(G)<Zλ+1⁢(G){Z_{\lambda}(G)<Z_{\lambda+1}(G)}; choose x in Zλ+1⁢(G)∖Zλ⁢(G){Z_{\lambda+1}(G)\setminus Z_{\lambda}(G)}. By Lemma 3.4, the element x is contained in a subgroup Y of cardinality at most 𝔪{\mathfrak{m}} such that Zκ⁢(G)∩Y=Zκ⁢(Y){Z_{\kappa}(G)\cap Y=Z_{\kappa}(Y)} for all κ≤λ+1{\kappa\leq\lambda+1}. It follows that Zλ⁢(Y)<Zλ+1⁢(Y){Z_{\lambda}(Y)<Z_{\lambda+1}(Y)}, a contradiction.∎Corollary 3.6.Let λ be an infinite ordinal number of cardinality m{\mathfrak{m}} and let G be a group. Suppose that for each subgroup X of cardinality at most m{\mathfrak{m}} there is an ordinal μ=μ⁢(X)<λ{\mu=\mu(X)<\lambda} such that Zμ⁢(X)=X{Z_{\mu}(X)=X}. Then Zκ⁢(G)=G{Z_{\kappa}(G)=G} for some ordinal κ<λ{\kappa<\lambda}.Proof.Suppose the statement is false and let G be a counterexample with smallest possible λ; since the class of nilpotent groups is countably recognizable, we have λ>ω{\lambda>\omega}. Moreover, by Corollary 3.5, we can also assume that λ is limit. Then there is a strictly increasing sequence of successor ordinal numbers {λi}i∈I{\{\lambda_{i}\}_{i\in I}} converging to λ (from the bottom). The minimality of λ shows that for each i∈I{i\in I} there is a subgroup Xi{X_{i}} of cardinality at most 𝔪{\mathfrak{m}} such that Zλi⁢(Xi)<Xi{Z_{\lambda_{i}}(X_{i})<X_{i}}. Clearly, X=⋃i∈IXi{X=\bigcup_{i\in I}X_{i}} is a subgroup of cardinality at most 𝔪{\mathfrak{m}} and hence there is μ=μ⁢(X)<λ{\mu=\mu(X)<\lambda} such that Zμ⁢(X)=X{Z_{\mu}(X)=X}. On the other hand, there is j∈I{j\in I} such that λj>μ{\lambda_{j}>\mu}, so we reach the contradiction sinceZλj⁢(Xj)<Xj=Zμ⁢(Xj)≤Zλj⁢(Xj).Z_{\lambda_{j}}(X_{j})<X_{j}=Z_{\mu}(X_{j})\leq Z_{\lambda_{j}}(X_{j}).The statement is proved.∎We are now ready to deal with uncountable groups whose proper large subgroups are hypercentral. Here we are not able to obtain a perfect analogous of the theorems in the previous section without the additional requirement that the hypercentral lengths of the proper larger subgroups are not too big. In any case, the following theorems generalize [11, Theorem A] and Corollary 2.5.Theorem 3.7.Let m{\mathfrak{m}} be an uncountable cardinal number and let λ be an ordinal number such that𝔫=|λ|<cf⁡(𝔪).{\mathfrak{n}=|\lambda|<\operatorname{cf}(\mathfrak{m})}.Let G be a group of cardinality m{\mathfrak{m}} such that:(1)Zλ⁢(X)=X{Z_{\lambda}(X)=X} for all proper subgroups X of cardinality 𝔪{\mathfrak{m}} of G, (1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then Zλ⁢(G)=G{Z_{\lambda}(G)=G}.Proof.Suppose that the statement is false and let G be a counterexample with smallest possible λ. It follows from [11, Theorem B] that G is locally nilpotent.Suppose first that λ=c{\lambda=c} is finite and assume by contradiction that G is not nilpotent of class at most c. Then there is a finitely generated subgroup X which is not nilpotent of class at most c. If all proper normal subgroups of G have cardinality strictly smaller than 𝔪{\mathfrak{m}}, we have that G is the join of all its proper normal subgroups. In this case, X is easily seen to be contained in a proper normal subgroup of G and consequently to be nilpotent of class at most c, a contradiction.Therefore G admits a proper normal subgroup M of cardinality 𝔪{\mathfrak{m}}; clearly, G=X⁢M{G=XM}. We now try to find a contradiction by showing that X is contained in some proper subgroup of cardinality 𝔪{\mathfrak{m}} of G; to this end, we often replace G with factor groups. Since M is nilpotent, we can easily choose M such that M′{M^{\prime}} is of cardinality strictly smaller that 𝔪{\mathfrak{m}}. Of course, we can even assume M′={1}{M^{\prime}=\{1\}} and so that M is actually abelian. Now, we may further assume that X∩M={1}{X\cap M=\{1\}}. Let N be any proper normal subgroup of X. Clearly, NM is nilpotent, so there is a non-negative integer d such that Zd+1⁢(M⁢N)/Zd⁢(M⁢N){Z_{d+1}(MN)/Z_{d}(MN)} is of cardinality 𝔪{\mathfrak{m}}. It follows that M∩Zd+1⁢(M⁢N)/M∩Zd⁢(M⁢N){M\cap Z_{d+1}(MN)/M\cap Z_{d}(MN)} has cardinality 𝔪{\mathfrak{m}} and so we may assume M∩Zd⁢(M⁢N)={1}{M\cap Z_{d}(MN)=\{1\}} and M=M∩Zd+1⁢(M⁢N){M=M\cap Z_{d+1}(MN)}. Now, N is centralized by M.If X is the join of all its proper normal subgroups, then it is the join of finitely many such subgroups (recall that X is finitely generated). Thus we may safely assume that [M,X]={1}{[M,X]=\{1\}}. Since M contains a proper subgroup L of cardinality 𝔪{\mathfrak{m}}, it then follows that LX is a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G, a contradiction.Thus X is simple and so cyclic of prime order. Now, Lemma 2.10 shows that M contains a proper G-invariant subgroup L of cardinality 𝔪{\mathfrak{m}}. Clearly, X⁢L<G{XL<G}, a contradiction showing that λ is infinite.Now, by Corollary 3.5 there is a subgroup X of cardinality at most 𝔫{\mathfrak{n}} such that Zλ⁢(X)<X{Z_{\lambda}(X)<X}. If all proper normal subgroups of G are of cardinality strictly smaller than 𝔪{\mathfrak{m}}, we easily get that G is the join of all its proper normal subgroups. Thus, for each x∈X{x\in X} there is a proper normal subgroup Nx{N_{x}} of G such that x lies in Nx{N_{x}}. Since 𝔫<cf⁡(𝔪){\mathfrak{n}<\operatorname{cf}(\mathfrak{m})} we have that N=〈Nx:x∈X〉{N=\langle N_{x}:x\in X\rangle} is a normal subgroup of cardinality strictly smaller than 𝔪{\mathfrak{m}}, so X is contained in the proper normal subgroup N and hence in a proper subgroup of cardinality 𝔪{\mathfrak{m}}. This contradiction shows that G contains a proper normal subgroup M of cardinality 𝔪{\mathfrak{m}}.Now, Zλ⁢(M)=M{Z_{\lambda}(M)=M} and |λ|<cf⁡(|M|){|\lambda|<\operatorname{cf}(|M|)}, so there is an ordinal number μ<λ{\mu<\lambda} such that Zμ+1⁢(M)/Zμ⁢(M){Z_{\mu+1}(M)/Z_{\mu}(M)} is of cardinality 𝔪{\mathfrak{m}}; of course, we can choose μ smallest possible with respect to such a condition. It follows that |L′|<𝔪{|L^{\prime}|<\mathfrak{m}}, where L=Zμ+1⁢(M){L=Z_{\mu+1}(M)}. We know that G=X⁢L{G=XL} and in order to show that X is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}} we can certainly assume L′={1}{L^{\prime}=\{1\}} and subsequently that X∩L={1}{X\cap L=\{1\}}; in particular, G=X⋉L{G=X\ltimes L} and L contains no proper G-invariant subgroups of cardinality 𝔪{\mathfrak{m}}.Let N be any proper normal subgroup of X. By hypothesis, Zλ⁢(N⁢L)=N⁢L{Z_{\lambda}(NL)=NL} and, since |λ|<cf⁡(𝔪){|\lambda|<\operatorname{cf}(\mathfrak{m})}, there is an ordinal number μ<λ{\mu<\lambda} such that |Zμ+1⁢(N⁢L)/Zμ⁢(N⁢L)|=𝔪{|Z_{\mu+1}(NL)/Z_{\mu}(NL)|=\mathfrak{m}}. It follows that L∩Zμ+1⁢(N⁢L)=L{L\cap Z_{\mu+1}(NL)=L} and |Zμ⁢(N⁢L)|<𝔪{|Z_{\mu}(NL)|<\mathfrak{m}}. Thus, if needed, we can assume that Zμ⁢(N⁢L)={1}{Z_{\mu}(NL)=\{1\}} and so that N is a normal subgroup of G.If X is the join of all its proper normal subgroups, then it is the join of at most 𝔫{\mathfrak{n}} such proper normal subgroups. Thus we may safely assume that [L,X]={1}{[L,X]=\{1\}}. Since L contains a proper subgroup U of cardinality 𝔪{\mathfrak{m}}, we get a contradiction. Therefore X is simple and so cyclic of prime order. Now, Lemma 2.10 shows that L contains a proper G-invariant subgroup V of cardinality 𝔪{\mathfrak{m}}, the last contradiction.∎Theorem 3.8.Let m{\mathfrak{m}} be an uncountable cardinal number and let λ be an ordinal number such that𝔫=|λ|<cf⁡(𝔪).{\mathfrak{n}=|\lambda|<\operatorname{cf}(\mathfrak{m})}.\vspace*{-0.5mm}Let G be a group of cardinality m{\mathfrak{m}} such that:(1)for each proper subgroup X of cardinality 𝔪{\mathfrak{m}} of G there is an ordinal μ=μ⁢(X)<λ{\mu=\mu(X)<\lambda} such that Zμ⁢(X)=X{Z_{\mu}(X)=X}, (1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then Zκ⁢(G)=G{Z_{\kappa}(G)=G} for some ordinal κ<λ{\kappa<\lambda}.Proof.Theorem 3.7 yields that λ is infinite and Zλ⁢(G)=G{Z_{\lambda}(G)=G}. The hypothesis |λ|<cf⁡(𝔪){|\lambda|<\operatorname{cf}(\mathfrak{m})} shows that there is an ordinal μ<λ{\mu<\lambda} such that Zμ+1⁢(G)/Zμ⁢(G){Z_{\mu+1}(G)/Z_{\mu}(G)} is of cardinality 𝔪{\mathfrak{m}} and so there is a subgroup L of G such that both |L|{|L|} and |G:L|{|G:L|} have cardinality 𝔪{\mathfrak{m}}. Now, any subgroup X of cardinality at most 𝔫{\mathfrak{n}} is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}} and so Zκ⁢(X)=X{Z_{\kappa}(X)=X} for some ordinal κ<λ{\kappa<\lambda}. The statement follows from Corollary 3.6.∎Finally, we see that in the torsion-free case no restriction on the hypercentral lengths is required.Theorem 3.9.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a torsion-free group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are hypercentral,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is hypercentral.Proof.It follows from [11, Theorem B] that G is locally nilpotent. Assume by contradiction that G is not hypercentral and let X be any countable subgroup of G that is not hypercentral.Suppose first that G has no proper normal subgroup of cardinality 𝔪{\mathfrak{m}}. In this case, G is the join of its proper normal subgroups, each of which is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, while the join of any finite number of proper normal subgroups is still a proper normal subgroup of G. Thus, each element x of X is contained in a proper normal subgroup Nx{N_{x}} of G. The hypothesis on the cofinality yields that X is contained in the proper normal subgroup N=〈Nx:x∈X〉{N=\langle N_{x}:x\in X\rangle}. Now, G/N{G/N} is of cardinality 𝔪{\mathfrak{m}} and it cannot be a Jónsson group (see for instance [9, Corollary 2.6]), so it admits a proper subgroup M/N{M/N} of cardinality 𝔪{\mathfrak{m}}. Clearly, M is hypercentral and so X is such, a contradiction.Assume that G contains a proper normal subgroup N of cardinality 𝔪{\mathfrak{m}}; of course, G=X⁢N{G=XN}. Clearly, all proper subgroups of G/N{G/N} are hypercentral. If G/N{G/N} is minimal non-hypercentral, then it is also periodic (see [8]). It follows that G is the isolator of a hypercentral subgroup and so it is itself hypercentral (see [18, 2.3.9]). On the other hand, if G/N{G/N} is hypercentral, then G′{G^{\prime}} is a proper subgroup of G. In this case, there is a proper normal subgroup L of G such that G/L{G/L} is periodic and so again G is hypercentral by [18, 2.3.9].∎Again the example of the locally dihedral 2-groups shows that we cannot extend the above results to countable cardinalities.4Removing the cofinality assumptionThe aim of this section is to show that under a very powerful assumption (namely, the Generalized Continuum Hypothesis) we are able to remove the hypothesis on the cofinality of the (uncountable) cardinalities of the groups involved in some of our previous statements. Thus all cardinalities in these statements will be subject to the sole condition of being uncountable.We start with a very useful lemma, which holds under GCH, and then we describe how to employ such a result to achieve our goal.Lemma 4.1.Assume that the Generalized Continuum Hypothesis holds and let G be a group of uncountable cardinality m{\mathfrak{m}}. Let X be a subgroup of Aut⁡(G){\operatorname{Aut}(G)} containing Inn⁡(G){\operatorname{Inn}(G)} and such that |X:Inn(G)|≤ℵ0{|X:\operatorname{Inn}(G)|\leq\aleph_{0}}. If(1)there is no maximal X-invariant subgroup N of G such that G/N{G/N} is of cardinality 𝔪{\mathfrak{m}} and(1)all proper X-invariant subgroups of G have cardinality strictly smaller than 𝔪{\mathfrak{m}}, then every countable subgroup of G is contained in a proper X-invariant subgroup of G.Proof.Since all proper X-invariant subgroups of G are of cardinality strictly smaller than 𝔪{\mathfrak{m}}, an easy application of Zorn’s lemma yields that G coincides with the subgroup generated by all its proper X-invariant subgroups. It follows from the fact that |X:Inn(G)|≤ℵ0{|X:\operatorname{Inn}(G)|\leq\aleph_{0}} and [9, Lemma 2.4] that G cannot be abelian, so in particular CG⁢(X)≤Z⁢(G)<G{C_{G}(X)\leq Z(G)<G} and G contains proper X-invariant subgroups that are not contained in Z⁢(G){Z(G)}. Let N be any such subgroup and put |N|=𝔫<𝔪{|N|=\mathfrak{n}<\mathfrak{m}}. Clearly, G/CG⁢(N){G/C_{G}(N)} embeds into Aut⁡(N){\operatorname{Aut}(N)} and so its cardinality is at most 𝔫𝔫=2𝔫{\mathfrak{n}^{\mathfrak{n}}=2^{\mathfrak{n}}}. Moreover, CG⁢(N){C_{G}(N)} is obviously X-invariant and, since Z⁢(G)≱N{Z(G)\not\geq N}, we have CG⁢(N)<G{C_{G}(N)<G}. Thus CG⁢(N){C_{G}(N)} is of cardinality strictly smaller than 𝔪{\mathfrak{m}} and hence 2𝔫=|G/CG⁢(N)|=𝔪{2^{\mathfrak{n}}=|G/C_{G}(N)|=\mathfrak{m}}. Now, it easily follows from the Generalized Continuum Hypothesis that every proper non-central X-invariant subgroup of G is of cardinality 𝔫{\mathfrak{n}}. Furthermore, by hypothesis, the factor group G/Z⁢(G){G/Z(G)} (which is of cardinality 𝔪{\mathfrak{m}}) contains non-trivial proper X-invariant subgroups, and hence all proper X-invariant subgroups of G are of cardinality at most 𝔫{\mathfrak{n}}.Let H be any countable subgroup of G. Since G is generated by its proper X-invariant subgroups, for any x∈X{x\in X} we can find a proper X-invariant subgroup Nx{N_{x}} of G containing x. Of course, the subgroup 〈Nx:x∈X〉{\langle N_{x}:x\in X\rangle} is an X-invariant subgroup of cardinality 𝔫⋅ℵ0=𝔫<𝔪{\mathfrak{n}\cdot\aleph_{0}=\mathfrak{n}<\mathfrak{m}} and we are done.∎We now described how to employ the above lemma in order to remove the restriction on the cofinality in the results of Section 2. First of all, an application of Lemma 4.1 with X=Inn⁡(G){X=\operatorname{Inn}(G)} takes care of the situation described in the second paragraph of the proof of Lemma 2.1, which is the only paragraph where the hypothesis on the cofinality has been used. Thus the hypothesis on the cofinality can be removed from Lemma 2.1.In a similar way we can deal with the second paragraph of the proof of Theorem 2.4. From paragraph 4 on, things change. An application of Lemma 2.1 shows that N contains a maximal G-invariant subgroup L such that N/L{N/L} is of cardinality 𝔪{\mathfrak{m}}. Of course, in this case we can assume L={1}{L=\{1\}}. Now, all proper normal subgroups Y of X are centralized by N (since [Y,N]={1}{[Y,N]=\{1\}}) and so the join M of all proper normal subgroups of X is centralized by N. If M=X{M=X}, then G=X×N{G=X\times N} and we can argue as in the last part of paragraph 5. The proof now goes along the same lines of the original one from paragraph 6 on (but notice that one could actually simplify some tiny bit of it). This takes care of Theorem 2.4.The proofs of Corollaries 2.5, 2.6, 2.7 and 2.8 need no change besides application of the new statements of previous results.5Embedding propertiesThe aim of this short section is to provide an easy proof of the main results contained in [3, Section 2]. We will show that the hypothesis on the regularity of the cardinal that was required in [3] is actually a non-essential one.Lemma 5.1.Let m{\mathfrak{m}} be an uncountable cardinal number. Let G=E⁢A{G=EA} be a group that is the product of a normal abelian subgroup A of cardinality m{\mathfrak{m}} and a finitely generated subgroup E. If there is a positive integer m such that [A,mE]{[A,_{m}E]} is of cardinality strictly smaller than m{\mathfrak{m}}, then Z⁢(G){Z(G)} is of cardinality m{\mathfrak{m}}.Proof.We may assume that m is such that B=[A,mE]{B=[A,_{m}E]} is of cardinality 𝔪{\mathfrak{m}} and |[B,E]|<𝔪{|[B,E]|<\mathfrak{m}}. LetE=〈x1,…,xn〉{E=\langle x_{1},\dots,x_{n}\rangle}and, for each i=1,…,n{i=1,\dots,n}, consider the mapsφi:b∈B↦[b,xi]∈[B,E].\varphi_{i}:b\in B\mapsto[b,x_{i}]\in[B,E].Clearly, B/Ker⁡(φi){B/\operatorname{Ker}(\varphi_{i})} is of cardinality strictly smaller than 𝔪{\mathfrak{m}} and so U=⋂i=1nKer⁡(φi){U=\bigcap_{i=1}^{n}\operatorname{Ker}(\varphi_{i})} is a subgroup of Z⁢(G){Z(G)} of cardinality 𝔪{\mathfrak{m}}. The statement is proved.∎Notice that the condition on E in the above lemma is certainly satisfied whenever E is subnormal in G. We employ the above result in the next lemma, which is actually the first step in proving the more general Corollary 5.4. In the following, [20, Theorem 7.42] is referred to as “Roseblade’s theorem”.Lemma 5.2.Let k be a positive integer and let G be an uncountable group of cardinality m{\mathfrak{m}} in which all subgroups of cardinality m{\mathfrak{m}} are subnormal of defect at most k. If G contains an abelian normal subgroup of cardinality m{\mathfrak{m}}, then all subgroups of G are subnormal of defect at most k.Proof.Let A be a normal abelian subgroup of G of cardinality 𝔪{\mathfrak{m}}. Let c=f⁢(k){c=f(k)}, where f is the Roseblade’s function described in [20, p. 71, Part 2]. Let g be any element of G and put B=γc+1⁢(〈g〉⁢A){B=\gamma_{c+1}(\langle g\rangle A)}; obviously, B≤A{B\leq A}. If |B|=𝔪{|B|=\mathfrak{m}}, then B/B∩〈g〉{B/B\cap\langle g\rangle} contains a proper 〈g〉{\langle g\rangle}-invariant subgroup C/B∩〈g〉{C/B\cap\langle g\rangle} of cardinality 𝔪{\mathfrak{m}} (see Lemma 2.10). Since all subgroups of cardinality 𝔪{\mathfrak{m}} are subnormal, it follows that [〈g〉⁢C,B]=[〈g〉,B]{[\langle g\rangle C,B]=[\langle g\rangle,B]} is a proper 〈g〉{\langle g\rangle}-invariant subgroup of B. Clearly, there is a proper subgroup D of B such that |B:D|<𝔪{|B:D|<\mathfrak{m}} and D≥[〈g〉,B]{D\geq[\langle g\rangle,B]}. However, all proper subgroups of 〈g〉⁢A/D{\langle g\rangle A/D} are subnormal of defect at most k and so B=γc+1⁢(〈g〉⁢A)≤D{B=\gamma_{c+1}(\langle g\rangle A)\leq D}, a contradiction. Thus B is of cardinality strictly smaller than 𝔪{\mathfrak{m}}. It follows now from Lemma 5.1 that A contains a subgroup Z of cardinality 𝔪{\mathfrak{m}} which is centralized by g. Clearly, Z contains a subgroup U that is the direct product of two subgroups U1{U_{1}} and U2{U_{2}} of cardinality 𝔪{\mathfrak{m}} and such that U∩〈g〉={1}{U\cap\langle g\rangle=\{1\}}. Thus 〈g〉=〈g〉⁢U1∩〈g〉⁢U2{\langle g\rangle=\langle g\rangle U_{1}\cap\langle g\rangle U_{2}} is a subnormal subgroup of defect at most k. The arbitrariness of g in G yields that G is a Baer group.Let E be any finitely generated subgroup of G. Then E is subnormal in G and so Lemma 5.1 shows that A contains a subgroup V of cardinality 𝔪{\mathfrak{m}} which is centralized by E. The argument in the last part of the above paragraph shows that E is the intersection of two subgroups of cardinality 𝔪{\mathfrak{m}} and so that E is subnormal of defect at most k. Lemma 7.41 (ii) of [20] completes the proof.∎Theorem 5.3.Let k be a positive integer and let G be an uncountable group of cardinality m{\mathfrak{m}} in which all subgroups of cardinality m{\mathfrak{m}} are subnormal of defect at most k. If G contains no normal subgroups that are Jónsson groups of cardinality m{\mathfrak{m}}, then all subgroups of G are subnormal of defect at most k; in particular, G is nilpotent.Proof.Let c=f⁢(k){c=f(k)}, where f is the Roseblade’s function described in [20, p. 71, Part 2], and put L=γc+1⁢(G){L=\gamma_{c+1}(G)}. If L is of cardinality 𝔪{\mathfrak{m}}, then it contains a proper subgroup H of cardinality 𝔪{\mathfrak{m}}. Then H is subnormal in L and hence L/HL{L/H^{L}} is a non-trivial factor group whose subgroups are subnormal of length bounded by k. Then L/HL{L/H^{L}} is nilpotent and in particular L′<L{L^{\prime}<L}. If |L′|=𝔪{|L^{\prime}|=\mathfrak{m}}, then L=γc+1⁢(G)≤L′{L=\gamma_{c+1}(G)\leq L^{\prime}}, a contradiction. Thus L/L′{L/L^{\prime}} is of cardinality 𝔪{\mathfrak{m}} and G/L′{G/L^{\prime}} contains a normal abelian subgroup of cardinality 𝔪{\mathfrak{m}}. It follows from Lemma 5.2 that all subgroups of G/L′{G/L^{\prime}} are subnormal of defect at most k and hence G/L′{G/L^{\prime}} is nilpotent of class at most c, a contradiction.∎Corollary 5.4.Let k be a positive integer and let G be an uncountable group of cardinality m{\mathfrak{m}} in which all subgroups of cardinality m{\mathfrak{m}} are subnormal of defect at most k. If G contains an abelian subgroup of cardinality m{\mathfrak{m}}, then all subgroups of G are subnormal of defect at most k; in particular, G is nilpotent.Proof.Let A be an abelian subgroup of G of cardinality 𝔪{\mathfrak{m}}. It follows that A is subnormal in G of defect m, say. If m=1{m=1}, then G is nilpotent by Lemma 5.2 and Roseblade’s theorem. If m>1{m>1}, then A is subnormal in AG{A^{G}} of defect strictly smaller than m and so by induction AG{A^{G}} is nilpotent. However, all subgroups of G/AG{G/A^{G}} are subnormal of defect at most k and so G/AG{G/A^{G}} is nilpotent as well. It follows from [9, Corollary 2.6] that G admits no normal subgroups that are Jónsson groups of cardinality 𝔪{\mathfrak{m}}. Therefore, an application of Theorem 5.3 (and Roseblade’s theorem) completes the proof.∎ http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Forum Mathematicum de Gruyter

Generalized nilpotency in uncountable groups

Ferrara, Maria; Trombetti, Marco
Forum Mathematicum , Volume 34 (3): 15 – May 1, 2022
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de Gruyter
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© 2022 Walter de Gruyter GmbH, Berlin/Boston
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0933-7741
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1435-5337
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10.1515/forum-2021-0137
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Abstract

1IntroductionA group G is said to have finite rank if there is a positive integer r such that every finitely generated subgroup of G can be generated by r elements; if such an integer does not exist, the group G is said to have infinite rank. It turns out that the structure of a (soluble) group of infinite rank is strongly influenced by that of its proper subgroups of infinite rank (see for instance the survey [4], where an almost complete reference list on the subject can be found, and the recent papers [6, 7] together with their reference lists). This kind of problem is actually a very special case of a more general one concerning the influence of “large” proper substructures on the whole structure. At first glance, it may appear that “infinite rank” is a not so natural concept of “largeness” in general, but, as far as groups are concerned, it is, because it is mainly defined in group theoretical terms. On the other hand, what seems to be in general a more natural concept of “largeness”, i.e. cardinality, does not appear to be strictly connected to the group theoretical structure itself, a priori. In 2016, F. de Giovanni and the second author [9] began the investigation of the influence of proper subgroups of large cardinality on the whole group. This investigation and subsequent ones (see [5], [3], [14], [15], [11], etc.) shows that, although in a mild form (compared with that of the rank), the influence is still there. The aim of this paper is to give a further contribution to the topic in connection with classes of groups that are close to being nilpotent.Let 𝔛{\mathfrak{X}} be a class of groups. The kind of problem we deal with in the first part of the paper is essentially the following one. Let G be a group of uncountable cardinality m{\mathfrak{m}} and assume that all proper subgroups of cardinality m{\mathfrak{m}} are X{\mathfrak{X}}-groups. Is it true that G is an X{\mathfrak{X}}-group as well? Of course, if we do not impose any additional condition on G we can have some problems. In fact, it has been shown by Shelah (without even appealing to the continuum hypothesis) that there are groups of cardinality ℵ1{\aleph_{1}} whose proper subgroups have cardinality strictly smaller than ℵ1{\aleph_{1}} (see [23]). Luckily, it turns out that these kind of monsters (named Jónsson groups of cardinality 𝔪{\mathfrak{m}}) admit simple homomorphic images of cardinality 𝔪{\mathfrak{m}} (see for instance [9]) and so we can get rid of them through the imposition of some generalized solubility condition.Among others results, we will show here that the above question answer positively whenever 𝔛{\mathfrak{X}} is one the following group classes (we refer to the next section for more precise definitions and statements): the class of (bounded) Engel groups (see Theorem 2.11 and Theorem 2.12), the class 𝒩1{\mathcal{N}_{1}} of groups whose subgroups are subnormal (see Corollary 2.6), the class 𝒩1f{\mathcal{N}_{1}^{f}} of groups whose subgroups are f-subnormal (see Corollary 2.7) and the class of Gruenberg groups (see Theorem 2.9). Notice here that many of these statements are actually corollaries of a more general result (see Theorem 2.4).Special attention is given to the class of hypercentral groups (see Section 3), for which we preliminary need to develop a theory describing how the hypercentral length of a group is connected to those of its proper subgroups. We employ such a theory to show that a (generalized soluble) uncountable group of cardinality 𝔪{\mathfrak{m}} whose proper subgroups of cardinality 𝔪{\mathfrak{m}} are hypercentral is itself hypercentral provided that the supremum of the hypercentral lengths of the proper subgroups of cardinality 𝔪{\mathfrak{m}} is smaller than the cofinality of the cardinal number 𝔪{\mathfrak{m}}.In Section 4, we see how to use the Generalized Continuum Hypothesis (GCH) to remove the assumption we made on the cofinality of the cardinality of the groups involved in some of the above-described statements. Notice that in this paper we assume the full strength of the Axiom of Choice.In the final part of the paper, we deal with groups whose large subgroups are subnormal and we generalize results of [3] showing that there is no need of any additional hypothesis on the cardinal number besides that of being uncountable.Most of our notation is standard and can be found in [20]. We use lower-case Gothic letters for cardinal numbers, while lower-case Greek letters for ordinal numbers.2Engel groups and groups with many subnormal subgroupsWe start the section with the following result, which deals with groups whose proper large subgroups are finite-by-soluble, i.e. admitting a finite normal subgroup with soluble factor group. This lemma (which is interesting in its own) will be employed in Corollary 2.7 but we put it here in order to not break up the discussion of the consequences of our first main theorem (Theorem 2.4). Recall that a group G is said to be locally graded if every non-trivial finitely generated subgroup of G contains a proper non-trivial subgroup of finite index. In the following, if 𝔪{\mathfrak{m}} is any cardinal number, we denote by cf⁡(𝔪){\operatorname{cf}(\mathfrak{m})} the cofinality of 𝔪{\mathfrak{m}}.Lemma 2.1.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a locally graded group of cardinality m{\mathfrak{m}} which has no simple homomorphic images of cardinality m{\mathfrak{m}}. If all proper subgroups of cardinality m{\mathfrak{m}} of G are finite-by-soluble, then G is (locally soluble)-by-finite.Proof.Suppose G is not (locally soluble)-by-finite, so G is perfect, admits no proper subgroups of finite index, and has a countable subgroup X that is not (locally soluble)-by-finite (see [10, Theorem 3.2]). Notice also that every proper normal subgroup of G is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G (see for instance [9, Corollary 2.6]) and so it is finite-by-soluble.Assume first that G has no proper normal subgroup of cardinality 𝔪{\mathfrak{m}}. Then G is the join of its proper normal subgroups (recall that there are no simple homomorphic images of cardinality 𝔪{\mathfrak{m}}), each of which is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, while the join of any finite number of proper normal subgroups is still a proper normal subgroup. Thus, each element x of X is contained in a proper normal subgroup Nx{N_{x}} of G. Since cf⁡(𝔪)>|X|{\operatorname{cf}(\mathfrak{m})>|X|}, we get that X is contained in the proper normal subgroup N=〈Nx:x∈X〉{N=\langle N_{x}:x\in X\rangle}. As we remarked above, N is finite-by-soluble, so X is finite-by-soluble and this gives a contradiction.Suppose now that G admits a proper normal subgroup N of cardinality 𝔪{\mathfrak{m}}. Clearly, G=X⁢N{G=XN} and it follows from the main result of [19] that G/N{G/N} is still locally graded, so G/N{G/N} cannot be finitely generated. Since X≃G/N{X\simeq G/N} is countable, we can find a chainN<X1<X2<⋯<Xn<Xn+1<⋯N<X_{1}<X_{2}<\cdots<X_{n}<X_{n+1}<\cdots\vspace*{-0.5mm}of proper subgroups of G such that G=⋃iXi{G=\bigcup_{i}X_{i}}. For each i, let Si{S_{i}} be the smallest normal subgroup of Xi{X_{i}} such that Xi/Si{X_{i}/S_{i}} is soluble. Obviously, Si≤Si+1{S_{i}\leq S_{i+1}} for every i, and S=⋃iSi{S=\bigcup_{i}S_{i}} is a countable normal subgroup of G such that G/S{G/S} is locally soluble, because every finitely generated subgroup of G is contained in some Xj{X_{j}}. Moreover, S is itself finite-by-soluble and so it contains a finite G-invariant subgroup T such that S/T{S/T} is soluble. However, G admits no proper subgroups of finite index, so T is abelian and hence G is locally soluble, the final contradiction. ∎An interesting (and direct) consequence of the above lemma is that all non-soluble finite pieces of the proper large subgroups vanish if we ask for G to admit no non-abelian simple homomorphic images.Corollary 2.2.Let m{\mathfrak{m}} be an uncountable cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a locally graded group of cardinality m{\mathfrak{m}} which has no non-abelian simple homomorphic images. If all proper subgroups of cardinality m{\mathfrak{m}} of G are finite-by-soluble, then G is locally soluble and, in particular, all proper subgroups of cardinality m{\mathfrak{m}} are soluble.In the above corollary we are not able to deduce that G itself is soluble. The problem here is that the structure of insoluble, locally soluble groups whose proper subgroups are soluble is not known; although such groups must be countable, they may occur as homomorphic images of uncountable groups. On the other hand, observe that if we also have a bound on both the finite piece and the derived length of the soluble residual of all proper large subgroups, then G is soluble by [11, Theorem D].Now, we define the group class for which we prove our first main theorem: a group G is an 𝒩′{\mathcal{N}^{\prime}}-group if and only if either |K:H|<∞{|K:H|<\infty} or HK<K{H^{K}<K} for all subgroups H≤K{H\leq K} of G. Of course, all nilpotent groups and all finite groups are 𝒩′{\mathcal{N}^{\prime}}-groups but the consideration of the locally dihedral 2-group shows that not every locally nilpotent group is an 𝒩′{\mathcal{N}^{\prime}}-group. Therefore, the class 𝒩′{\mathcal{N}^{\prime}} is not a local class, although, by its very definition, is closed with respect to forming subgroups and homomorphic images. On the other hand, our next lemma shows that this class is at least countably recognizable, i.e. a group is an 𝒩′{\mathcal{N}^{\prime}}-group if and only if all its countable subgroups are 𝒩′{\mathcal{N}^{\prime}}-groups.Lemma 2.3.The class N′{\mathcal{N}^{\prime}} is countably recognizable.Proof.Let G be a group whose countable subgroups are 𝒩′{\mathcal{N}^{\prime}}-groups and suppose by contradiction that G is not an 𝒩′{\mathcal{N}^{\prime}}-group. Then there is a proper subgroup X of G such that |G:X|=∞{|G:X|=\infty} and XG=G{X^{G}=G}. Let U1{U_{1}} be any countable subgroup of X and let 𝒯{\mathcal{T}} be a countably infinite subset of a given transversal to X in G. Put V1=〈U1,𝒯〉{V_{1}=\langle U_{1},\mathcal{T}\rangle}. Suppose we have defined countable subgroups Un{U_{n}} of X and Vn{V_{n}} of G such that |Vn:Un|=∞{|V_{n}:U_{n}|=\infty}; of course, by assumptions, UnVn<Vn{U_{n}^{V_{n}}<V_{n}}. Since XG=G{X^{G}=G}, for each v∈Vn∖UnVn{v\in V_{n}\setminus U_{n}^{V_{n}}}, there are finitely generated subgroups En⁢(v){E_{n}(v)} of X and Fn⁢(v){F_{n}(v)} of G such that v lies in En⁢(v)Fn⁢(v){E_{n}(v)^{F_{n}(v)}}. Put Un+1=〈Un,En(v):v∈Vn∖UnVn〉{U_{n+1}=\langle U_{n},E_{n}(v):v\in V_{n}\setminus U_{n}^{V_{n}}\rangle} and Vn+1=〈Un+1,Fn⁢(v)〉{V_{n+1}=\langle U_{n+1},F_{n}(v)\rangle}. Again, Un+1{U_{n+1}} is a subgroup of infinite index in Vn+1{V_{n+1}} because 𝒯{\mathcal{T}} lies in Vn+1{V_{n+1}}.Now, let U be the union of the subgroups Ui{U_{i}} and V the union of the subgroups Vj{V_{j}}; clearly, |V:U|=∞{|V:U|=\infty} since 𝒯⊆V{\mathcal{T}\subseteq V} and U≤X{U\leq X}. If v is any element of V, then it belongs to a certain Vn{V_{n}}, so it is either an element of UnVn{U_{n}^{V_{n}}} or it belongs to En⁢(v)Fn⁢(v){E_{n}(v)^{F_{n}(v)}}, which means that v lies in Un+1Vn+1{U_{n+1}^{V_{n+1}}}. In any case, v lies in UV{U^{V}} and so UV=V{U^{V}=V}, a contradiction.∎Our first main result shows that the group class 𝒩′{\mathcal{N}^{\prime}} is also recognizable by the behaviour of the proper large subgroups of a group. In what follows, if H and K are subgroups of a group, we denote by HK{H_{K}} the largest subgroup of H that is normalized by K; clearly, this coincides with the intersection of all conjugates of H by elements of K. Proposition 1.B.3 of [16] shows that hypothesis (2) in the following statement is satisfied in particular whenever the group is (locally soluble)-by-finite.Theorem 2.4.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are 𝒩′{\mathcal{N}^{\prime}}-groups,(1)every chief factor of G is finite or abelian.Then G is an N′{\mathcal{N}^{\prime}}-group.Proof.Let X be any countable subgroup of G; we show that X is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}}, so Lemma 2.3 completes the proof.Suppose first that G has no proper normal subgroup of cardinality 𝔪{\mathfrak{m}}. In this case, G is the join of its proper normal subgroups, each of which is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, while the join of any finite number of proper normal subgroups is still a proper normal subgroup of G. Thus, each element x of X is contained in a proper normal subgroup Nx{N_{x}} of G. The hypothesis on the cofinality yields that X is contained in the proper normal subgroup N=〈Nx:x∈X〉{N=\langle N_{x}:x\in X\rangle}. Now, G/N{G/N} is of cardinality 𝔪{\mathfrak{m}} and it cannot be a Jónsson group (see for instance [9, Corollary 2.6]), so it admits a proper subgroup M/N{M/N} of cardinality 𝔪{\mathfrak{m}}. Clearly, M is the required subgroup containing X.Suppose G admits a proper normal subgroup N of cardinality 𝔪{\mathfrak{m}}. Of course, we may assume that G=X⁢N{G=XN}. Let H=X∩N{H=X\cap N}. Since H is countable, then HN=H⁢[N,H]{H^{N}=H[N,H]} is a proper normal subgroup of N that is normalized by X. Thus X⁢HN<G{XH^{N}<G}, so HN{H^{N}} is of cardinality strictly smaller than 𝔪{\mathfrak{m}} and hence we may assume H≤HN={1}{H\leq H^{N}=\{1\}}. This assumption implies at once the possibility of a stronger one. In fact, now G=X⋉N{G=X\ltimes N} and if H is any proper G-invariant subgroup of cardinality 𝔪{\mathfrak{m}} of N, the subgroup XH is a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G and we are done. Therefore we may assume that N contains no proper G-invariant subgroups of cardinality 𝔪{\mathfrak{m}}.Let Y be a proper normal subgroup of X; in particular, L=Y⁢N{L=YN} is an 𝒩′{\mathcal{N}^{\prime}}-group and hence YL=Y⋉(YL∩N){Y^{L}=Y\ltimes(Y^{L}\cap N)} is a proper subgroup of YN. SinceL/YL∩N=YL/YL∩N×N/YL∩N,L/Y^{L}\cap N=Y^{L}/Y^{L}\cap N\times N/Y^{L}\cap N,it follows that [Y,N]{[Y,N]} lies in YL∩N{Y^{L}\cap N}, so [Y,N]{[Y,N]} is a proper subgroup of N. Moreover, [Y,N]{[Y,N]} is normalized by N and X, so it is a G-invariant subgroup of N and hence |[Y,N]|<𝔪{|[Y,N]|<\mathfrak{m}}. Let M be the join of all proper normal subgroups of X.If M=X{M=X}, then X is actually the join of countably many proper normal subgroups. Using the fact that the cofinality of 𝔪{\mathfrak{m}} is strictly larger than ω, we have that [X,N]{[X,N]} is of cardinality strictly smaller than 𝔪{\mathfrak{m}} and so we may assume G=X×N{G=X\times N}. Again by [9, Corollary 2.6], the subgroup N is not a Jónsson group, so it admits a proper subgroup M of cardinality 𝔪{\mathfrak{m}}. Of course, XM is the required subgroup, and we are done.Suppose now M<X{M<X}. Since [M,N]{[M,N]} is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, we can factor it out and assume that [M,N]={1}{[M,N]=\{1\}}. But then M is a normal subgroup of G and so we may even assume M={1}{M=\{1\}}. Obviously, this is equivalent to require that X is finite. If N′<N{N^{\prime}<N}, then N contains a proper subgroup E of countable index (see for instance [9, Lemma 2.3]) and X⁢EX{XE_{X}} is a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G, and we are done again. Thus N=N′{N=N^{\prime}} and in particular Z⁢(N)=Z2⁢(N){Z(N)=Z_{2}(N)} by a well-known result. Thus |Z⁢(N)|<𝔪{|Z(N)|<\mathfrak{m}} and we may assume that Z⁢(N)={1}{Z(N)=\{1\}}.Let K be any proper G-invariant subgroup of N; of course, |K|<𝔪{|K|<\mathfrak{m}}. Let g∈K{g\in K}; notice that |N:CN(g)|<𝔪{|N:C_{N}(g)|<\mathfrak{m}}. If CN⁢(g)≠N{C_{N}(g)\neq N}, then X⁢(CN⁢(g))X{X\big{(}C_{N}(g)\big{)}_{X}} is a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G and we are done. Thus CN⁢(g)=N{C_{N}(g)=N} for all g∈K{g\in K}, so K≤Z⁢(N)=1{K\leq Z(N)=1}. Thus N admits no non-trivial proper G-invariant subgroups, so it must be abelian and we have a contradiction.∎The first corollary we obtain is a somewhat stronger version of [11, Theorem A]. Notice that we will generalize this result in Theorem 3.7.Corollary 2.5.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are nilpotent,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is nilpotent.Proof.It follows from [11, Theorem B] that G is locally nilpotent, so we may apply Theorem 2.4 and get that G is an 𝒩′{\mathcal{N}^{\prime}}-group. If X is any countable subgroup of G, then |G:X|{|G:X|} is infinite and so XG<G{X^{G}<G}. Now, either XG{X^{G}} is of cardinality 𝔪{\mathfrak{m}} or G/XG{G/X^{G}} contains a proper subgroup M/XG{M/X^{G}} of cardinality 𝔪{\mathfrak{m}} (recall that G admits no Jónsson homomorphic images). In both cases, X is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}} and so it is nilpotent. It follows that the whole group G is nilpotent because the class of nilpotent groups is well known to be countably recognizable. ∎Recall that a group G is said to be an 𝒩1{\mathcal{N}_{1}}-group if all its subgroups are subnormal. It turns out that the class of 𝒩1{\mathcal{N}_{1}}-groups is countably recognizable (see [10, Theorem 2.7]) and that a group is an 𝒩1{\mathcal{N}_{1}}-group if and only if all its countable subgroups are subnormal (see also [13]). The following corollary can be now proved in a fashion similar to the previous one.Corollary 2.6.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are 𝒩1{\mathcal{N}_{1}}-groups,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is an N1{\mathcal{N}_{1}}-group.Recall that a subgroup H of a group G is f-subnormal in G if there is a finite chain of subgroupsH=H0≤H1≤…≤Hn=GH=H_{0}\leq H_{1}\leq\dots\leq H_{n}=Gsuch that for i=1,2,…,n{i=1,2,\dots,n} either the index |Hi:Hi-1|{|H_{i}:H_{i-1}|} is finite or Hi-1{H_{i-1}} is normal in Hi{H_{i}}. Let 𝒩1f{\mathcal{N}_{1}^{f}} be the class of all groups with only f-subnormal subgroups. It was proved in [2] that any 𝒩1f{\mathcal{N}_{1}^{f}}-group is a finite-by-soluble 𝒩′{\mathcal{N}^{\prime}}-group. On the other hand, in [13], we proved that the class 𝒩1f{\mathcal{N}_{1}^{f}} is countably recognizable. Our next corollary is therefore again a consequence of Theorem 2.4 (the proof being similar to that of Corollary 2.5, but with the replacement of [11, Theorem B] by Lemma 2.1).Corollary 2.7.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are 𝒩1f{\mathcal{N}_{1}^{f}}-groups,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is an N1f{\mathcal{N}_{1}^{f}}-group.Of course, similar corollaries hold for many other countably recognizable group classes contained in 𝒩1f{\mathcal{N}_{1}^{f}} (see for instance [13]), but, here, we just point out that the class of finite-by-nilpotent groups (i.e. groups admitting a finite normal subgroup with nilpotent factor group) is among these. Indeed, finite-by-nilpotent groups form a countably recognizable group class by [10, Theorem 3.6] and every subgroup H of a finite-by-nilpotent group G is obviously f-subnormal in G, so in particular this class is contained in 𝒩′{\mathcal{N}^{\prime}}. The following corollary should be seen in relation to [9, Theorem 4.6], which shows that a similar result holds for the class of nilpotent-by-finite groups.Corollary 2.8.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are finite-by-nilpotent,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is finite-by-nilpotent.The consideration of the locally dihedral 2-group shows that the assumption on the cofinality in the above results cannot be completely removed. However, we will see in Section 4 that using the Generalized Continuum Hypothesis we are able to prove all results for an arbitrary uncountable cardinal number.Engel groups are strictly related to Gruenberg groups (see after the proof of Theorem 2.9 for more details), that’s why we analyze the case of Gruenberg groups first. Recall that a group is called a Gruenberg group if it is generated by its ascendant abelian subgroups. Although the class of Gruenberg groups is not countably recognizable (see for instance [10]), our next result shows that it can be “uncountably” recognized.Theorem 2.9.Let G be a group of uncountable cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are Gruenberg groups,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is a Gruenberg group.Proof.It follows from [11, Theorem B] that G is locally nilpotent. Let N be any proper normal subgroup of G. If N is of cardinality 𝔪{\mathfrak{m}}, then all its cyclic subgroups are ascendant in G; if N is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, this is still the case since G/N{G/N} contains a proper subgroup of cardinality 𝔪{\mathfrak{m}}, which is a Gruenberg group. Let M be the subgroup generated by all proper normal subgroups of G. If G=M{G=M}, we are done. If M≠G{M\neq G}, then G/M{G/M} is cyclic of prime order; take g∈G{g\in G} with G=〈g〉⁢M{G=\langle g\rangle M}. We only need to show that 〈g〉{\langle g\rangle} is ascendant in G, so assume that this is not the case. Of course, G/G′{G/G^{\prime}} is cyclic of prime power order and actually G=〈g〉⁢G′{G=\langle g\rangle G^{\prime}}.Let V be any proper G-invariant subgroup of G′{G^{\prime}} such that G′/V{G^{\prime}/V} does not contain any non-trivial proper G-invariant subgroup. Since G is locally nilpotent, its chief factors are central and so G′/V≤Z⁢(G/V){G^{\prime}/V\leq Z(G/V)}. Thus G/V{G/V} is abelian and we have reached a contradiction. In particular, G′{G^{\prime}} admits no proper subgroup of finite index.Let 〈h〉=G′∩〈g〉{\langle h\rangle=G^{\prime}\cap\langle g\rangle}. If h≠1{h\neq 1}, then we can find a chief factor H/K{H/K} of G such that H=〈h〉⁢K≤G′{H=\langle h\rangle K\leq G^{\prime}}; if h=1{h=1}, we let H={1}{H=\{1\}}. In both cases, H<G′{H<G^{\prime}}. If G′{G^{\prime}} contains a proper G-invariant subgroup W of cardinality 𝔪{\mathfrak{m}} with W≥H{W\geq H}, then, since G′/W{G^{\prime}/W} admits no maximal (proper) G-invariant subgroups, we can define an ascending chainW≤W1<W2<…<Wα<Wα+1<⋯W\leq W_{1}<W_{2}<\dots<W_{\alpha}<W_{\alpha+1}<\cdotsof proper G-invariant subgroups of G′{G^{\prime}} whose union is G′{G^{\prime}}. Of course, 〈g〉{\langle g\rangle} is ascendant in every 〈g〉⁢Wα{\langle g\rangle W_{\alpha}} and so obviously 〈g〉{\langle g\rangle} is ascendant in 〈g〉⁢G′=G{\langle g\rangle G^{\prime}=G}, a contradiction.Now, suppose that G′{G^{\prime}} admits no proper G-invariant subgroup of cardinality 𝔪{\mathfrak{m}} containing H and let U be any proper G-invariant subgroup of G′{G^{\prime}} containing H. If G′/U{G^{\prime}/U} is abelian, we may find a proper subgroup L/U{L/U} of G′/U{G^{\prime}/U} such that |G′/L|=ℵ0{|G^{\prime}/L|=\aleph_{0}}. Then LG=∩x∈〈g〉Lx{L_{G}=\cap_{x\in\langle g\rangle}L^{x}} is G-invariant and |G′:LG|=ℵ0{|G^{\prime}:L_{G}|=\aleph_{0}}, a contradiction. Thus G′/U{G^{\prime}/U} is non-abelian and so there is a proper G-invariant subgroup V of G such that U<V<G′{U<V<G^{\prime}} and V/U{V/U} is not contained in Z⁢(G′/U){Z(G^{\prime}/U)}. Let G¯=G/U{\overline{G}=G/U} and notice that G¯′/CG¯′⁢(v⁢U){\overline{G}^{\prime}/C_{\overline{G}^{\prime}}(vU)} is of cardinality strictly smaller than 𝔪{\mathfrak{m}} for some v⁢U∈V/U{vU\in V/U} that is non-central in G¯′{\overline{G}^{\prime}}. Put C/U=CG¯′⁢(v⁢U){C/U=C_{\overline{G}^{\prime}}(vU)} and D=C〈g〉{D=C_{\langle g\rangle}}. Now, D is 〈g〉{\langle g\rangle}-invariant, contains U and |G′:D|<𝔪{|G^{\prime}:D|<\mathfrak{m}}. It follows that 〈g〉{\langle g\rangle} is ascendant in 〈g〉⁢D{\langle g\rangle D} and so a fortiori in 〈g〉⁢U{\langle g\rangle U}.Since G′/H{G^{\prime}/H} admits no G-invariant (proper) maximal subgroups, it is clearly possible to define an ascending chainH≤H1<H2<…<Hα<Hα+1<⋯H\leq H_{1}<H_{2}<\dots<H_{\alpha}<H_{\alpha+1}<\cdotsof proper G-invariant subgroups of G′{G^{\prime}} whose union is G′{G^{\prime}}. By what we have already proved, it follows that 〈g〉{\langle g\rangle} is ascendant in every 〈g〉⁢Hα{\langle g\rangle H_{\alpha}} and so obviously 〈g〉{\langle g\rangle} is ascendant in 〈g〉⁢G′=G{\langle g\rangle G^{\prime}=G}, the final contradiction.∎Observe that it is well known (and easy to see) that a countable and locally nilpotent group is a Gruenberg group. The consideration of the locally dihedral group C2⋉C3∞{C_{2}\ltimes C_{3^{\infty}}} (here C2{C_{2}} is a cyclic group of order 2 acting as the inversion on an infinite locally cyclic 3-group C3∞{C_{3^{\infty}}}) shows that either the uncountability of the cardinal or the locally nilpotent assumption cannot be omitted.Recall that a group G is called an Engel group if it coincides with the set of all its left (right) Engel elements, i.e. the set of all elements x of G such that, for all g∈G{g\in G}, we have [g,nx]=1{[g,_{n}x]=1} ([x,ng]=1{[x,_{n}g]=1}) for a suitable positive integer n=n⁢(g,x){n=n(g,x)}. Clearly, any locally nilpotent group is an Engel group and a well-known result of Zorn shows that any finite Engel group is nilpotent. Also notice that for soluble groups the properties of being Gruenberg, locally nilpotent or Engel coincide (see [21, 12.3.3]), so Theorem 2.9 can be rephrased as follows: if G is a soluble group of cardinality 𝔪{\mathfrak{m}} (cf⁡(𝔪)>ω{\operatorname{cf}(\mathfrak{m})>\omega}) whose proper subgroups of cardinality 𝔪{\mathfrak{m}} are Engel groups, then G is an Engel group. The solubility assumption can be weakened through the concept of strong local graduation. Recall that a group G is strongly locally graded if every section of G is locally graded (actually we will only need that every homomorphic image of G is locally graded). Strongly locally graded groups form a proper subclass of the class of all locally graded groups as shown by the consideration of any free non-abelian group, but this class is still large enough to contain for instance all locally (soluble-by-finite) groups.Before proving our main result concerning Engel groups, we need the following preparatory lemma on uncountable modules over Principal Ideal Domains.Lemma 2.10.Let m{\mathfrak{m}} be an uncountable cardinal number and let R be a Principal Ideal Domain of cardinality strictly smaller than m{\mathfrak{m}}. If M is an R-module of cardinality m{\mathfrak{m}}, then M contains a submodule of cardinality m{\mathfrak{m}} which is the direct sum of (infinitely many) non-trivial submodules of strictly smaller cardinality.Proof.Let E=ER⁢(M){E=E_{R}(M)} be the injective hull of M. Since R is a Principal Ideal Domain, it is well known that E is a direct sum of indecomposable injective R-modules and every indecomposable injective R-module is (the injective hull of) the residue field at a prime. It follows that every indecomposable injective R-module is of the same cardinality of R, so E is the direct sum of an infinite family {E}E∈ℰ{\{E\}_{E\in\mathcal{E}}} of non-trivial R-submodules of cardinality strictly smaller than 𝔪{\mathfrak{m}}. On the other hand, M is an essential R-submodule of E, so M∩Ei≠{0}{M\cap E_{i}\neq\{0\}}. The R-submodule ⊕E∈ℰ(M∩E){\bigoplus_{E\in\mathcal{E}}(M\cap E)} of M is the required one.∎Theorem 2.11.Let G be a group of uncountable cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are non-perfect Engel groups,(1)G is strongly locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is an Engel group.Proof.Suppose that G is not an Engel group, so there are elements x and y of G such that F=〈x,[y,x]〉{F=\langle x,[y,x]\rangle} is not an Engel group. Let N be any proper normal subgroup of G. If N is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, then G/N{G/N} contains a proper subgroup of cardinality 𝔪{\mathfrak{m}}, which is an Engel group; if |N|=𝔪{|N|=\mathfrak{m}}, then it is an Engel group by hypothesis. Suppose first that all proper normal subgroups of G are of cardinality strictly smaller than 𝔪{\mathfrak{m}}. As we have already seen many times, in such circumstances, F is contained in a proper normal subgroup of G and so it is an Engel group, a contradiction. Therefore, we may assume that G contains a proper normal subgroup M of cardinality 𝔪{\mathfrak{m}}; obviously, G=F⁢M{G=FM}. Since G/M{G/M} is locally graded and finitely generated, we may assume G/M{G/M} is even finite. Of course, all proper subgroups of G/M{G/M} are Engel groups, so it follows from [1, Theorem 2.1] that G/M{G/M} is either an Engel group or a minimal non-nilpotent group. In the former case, G/M{G/M} is nilpotent; in the latter case, it is soluble (see for instance [21, 9.1.9]). In both cases G′<G{G^{\prime}<G} and in particular G=F⁢G′=〈x〉⁢G′{G=FG^{\prime}=\langle x\rangle G^{\prime}}, which means that any nilpotent homomorphic image of G is cyclic.Let K be any proper normal subgroup of G such that G/K{G/K} is soluble: we prove that |K|=𝔪{|K|=\mathfrak{m}}. Suppose this is not the case and notice that the solubility of G/K{G/K} makes it possible to replace K in such a way that there is a normal abelian subgroup H/K{H/K} of cardinality 𝔪{\mathfrak{m}}. Moreover, an easy combination of [21, 12.3.3] and [11, Theorem B] shows that G/K{G/K} is locally nilpotent. In order to derive the contradiction that F is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}}, we factor out the normal subgroup F⁢K∩H{FK\cap H} and consequently assume that K=F∩H=1{K=F\cap H=1}; in particular, F≃G/H{F\simeq G/H} is cyclic since now G is locally nilpotent. By Lemma 2.10 the subgroup H contains a G-invariantsubgroup A of cardinality 𝔪{\mathfrak{m}} that is the direct product of G-invariant subgroups Ai{A_{i}}, i∈I{i\in I}, of cardinality strictly smaller than 𝔪{\mathfrak{m}}. Since the cofinality of 𝔪{\mathfrak{m}} is a limit ordinal, it follows that we may writeI=I1∪I2,{I=I_{1}\cup I_{2}},where I1∩I2=∅{I_{1}\cap I_{2}=\emptyset} and the G-invariant subgroupsA(I1)=〈Ai:i∈I1〉 and A(I2)=〈Ai:i∈I2〉A(I_{1})=\langle A_{i}:i\in I_{1}\rangle\quad\text{and}\quad A(I_{2})=\langle A%_{i}:i\in I_{2}\rangle\vspace*{-0.5mm}have both cardinality 𝔪{\mathfrak{m}}. Clearly, F⁢A⁢(I1)<G{F\,A(I_{1})<G} and we get the required contradiction.Therefore K is of cardinality 𝔪{\mathfrak{m}}. In this case, all proper subgroups of G/K{G/K} are locally nilpotent, and so [8, Lemma 3.2] yields that either G/K{G/K} is finite and minimal non-nilpotent, or it is locally nilpotent. In the former case, G′′′≤K{G^{\prime\prime\prime}\leq K} by [22]; in the latter case, G/K=F⁢K/K{G/K=FK/K} is nilpotent and so abelian. In both cases, G′′′≤K{G^{\prime\prime\prime}\leq K}. The arbitrariness of K shows that |G′′′|=𝔪{|G^{\prime\prime\prime}|=\mathfrak{m}} and that G′′′{G^{\prime\prime\prime}} is the last term of the derived series of G. This is clearly in contradiction with the hypotheses of the statement.∎Let n be any positive integer. By an n-Engel group is meant a group G such that [x,ny]=1{[x,_{n}y]=1} for all x,y∈G{x,y\in G}; that is, every element is both left and rightn-Engel. It follows that the class of n-Engel groups is the variety determined by the law [x,ny]=1{[x,_{n}y]=1}. Of course, any nilpotent group of class n is an n-Engel group but n-Engel groups need not be nilpotent. The final theorem of this section shows that when we deal with n-Engel groups, the additional assumptions in Theorem 2.11 can be dropped. In order to prove such a result, we need preliminary to observe that [11, Theorem A] does not really need any assumption on the cofinality whenever the cardinality 𝔪{\mathfrak{m}} is assumed to be uncountable and there is a bound on the nilpotency classes of the proper subgroups of cardinality 𝔪{\mathfrak{m}}: this can be seen from the proof of [11, Theorem A], but it is obtained also as a consequence of a more general result concerning hypercentrality in the next section (see Theorem 3.8).Theorem 2.12.Let k be a positive integer and let G be a group of uncountable cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are k-Engel groups,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is a k-Engel group.Proof.It follows from [17, Corollary 6] that every locally graded k-Engel group is locally nilpotent, so [11, Theorem B] shows that G is locally nilpotent. Suppose that G is not a k-Engel group, so there is a finitely generated subgroup F=〈x,y〉{F=\langle x,y\rangle} which is not contained in any proper subgroup of cardinality 𝔪{\mathfrak{m}} or in any proper normal subgroup N of G (since G/N{G/N} contains a subgroup L/N{L/N} such that L is of cardinality 𝔪{\mathfrak{m}}); of course, F is nilpotent, of class c, say.Now, if X is any normal subgroup of cardinality 𝔪{\mathfrak{m}} of G, then G=F⁢X{G=FX} and so G/X{G/X} is nilpotent of class at most c. It follows that S=γc+1⁢(G){S=\gamma_{c+1}(G)} lies in X. Moreover, if |γh⁢(G)|<𝔪{|\gamma_{h}(G)|<\mathfrak{m}} for some positive integer h, we can easily find a proper normal subgroup M of G such that both |M|{|M|} and |G/M|{|G/M|} are of cardinality 𝔪{\mathfrak{m}} (see for instance [9, Lemma 2.4]). Then FM is a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G and we obtain a contradiction. Therefore S is the smallest normal subgroup of cardinality 𝔪{\mathfrak{m}} of G and it is also the smallest term of the lower central series of G.A well-known theorem of Zelmanov asserts that any torsion-free nilpotent k-Engel group is nilpotent of class bounded by an integer depending only on k, so a torsion-free locally nilpotent k-Engel group is nilpotent. Let T be the periodic part of G. If |T|<𝔪{|T|<\mathfrak{m}}, then all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G/T{G/T} are nilpotent and so it follows from the remark just before the statement that G/T{G/T} is nilpotent, a contradiction since in this case |S|≤|T|<𝔪{|S|\leq|T|<\mathfrak{m}}. Thus |T|=𝔪{|T|=\mathfrak{m}} and S≤T{S\leq T}.Suppose first that G/S{G/S} is non-periodic and choose a prime q. Then G contains a normal subgroup Q of index q. Let θ be the word [x1,kx2]{[x_{1},_{k}x_{2}]} and notice now that the verbal subgroup θ⁢(Q){\theta(Q)} is trivial. An application of [18, Theorem 2.3.5] therefore yields that θ⁢(G){\theta(G)} is a q-group. The arbitrariness of the prime q shows that θ⁢(G)={1}{\theta(G)=\{1\}} and so that G is a k-Engel group, a contradiction.It is thus possible to assume that G is periodic; in particular, F is finite. Since S is the smallest normal subgroup of cardinality 𝔪{\mathfrak{m}} of G, the finite group F∩S{F\cap S} is contained in a proper G-invariant subgroup of S (which is actually a term of a chief series of G) that has cardinality strictly smaller than 𝔪{\mathfrak{m}}. In order to show that F is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G, it is therefore safe to assume G=F⋉S{G=F\ltimes S}.Clearly, S coincides with the subgroup generated by all its proper G-invariant subgroups, so that, if S is non-abelian, it contains a proper G-invariant subgroup K (of cardinality strictly smaller than 𝔪{\mathfrak{m}}) which is not contained in Z⁢(S){Z(S)}. Let x∈K∖Z⁢(S){x\in K\setminus Z(S)} and notice that 1<|S:CS(x)|<𝔪{1<|S:C_{S}(x)|<\mathfrak{m}}. The intersection C=⋂g∈FCS⁢(x)g{C=\bigcap_{g\in F}C_{S}(x)^{g}} is now an F-invariant subgroup of cardinality 𝔪{\mathfrak{m}}, hence giving the contradiction F⁢C<G{FC<G}. Thus S is abelian and so it contains a proper subgroup U of countable index for which we can repeat the above trick (here, U takes the place of K), thus obtaining the final contradiction.∎3HypercentralityThe aim of this section is to describe uncountable groups whose proper “large” subgroups are hypercentral. In order to carry out such an analysis we first need to understand how the hypercentral length of the whole group is connected to those of the “small” subgroups of the group. This leads to Corollaries 3.5 and 3.6, results that naturally generalize the well-known fact that the class of (nilpotent) hypercentral groups is countably recognizable.Let G be a group. We say that an element g of G is a Černikov element if and only if for any sequence x0,x1,…,xn,xn+1,…{x_{0},x_{1},\dots,x_{n},x_{n+1},\dots} of elements of G there is a certain non-negative integer m such that[…⁢[[g,x0],x1],…,xm]=1.{[\dots[[g,x_{0}],x_{1}],\dots,x_{m}]=1}.This definition is motivated by the theorem of Černikov which shows that hypercentrality is countably recognizable (see for instance [20, Theorem 2.19]). Although in a slightly different (but equivalent) form, this definition has been introduced for the first time in [12], where it was showed that it is connected to some “ring-theoretic questions” in group theory.The following result shows that the set of all Černikov element is a subgroup and actually a well-known subgroup for that matter.Lemma 3.1.Let G be a group. An element g of G belongs to the hypercentre Z¯{\overline{Z}} of G if and only if g is a Černikov element.Proof.Suppose that g lies in Z¯{\overline{Z}} but it is not a Černikov element. Then there is a sequence x0,x1,…,xn,xn+1,…{x_{0},x_{1},\dots,x_{n},}{x_{n+1},\dots} such that […[g,x0],x1],…,xm]≠1{[\dots[g,x_{0}],x_{1}],\dots,x_{m}]\neq 1} for all non-negative integers m. Let λ0{\lambda_{0}} be the smallest ordinal number such that g lies in Zλ0⁢(G){Z_{\lambda_{0}}(G)}. Then λ0{\lambda_{0}} is not limit and [g,x0]{[g,x_{0}]} lies in a term Zλ1⁢(G){Z_{\lambda_{1}}(G)} of the upper central series of G with λ1<λ0{\lambda_{1}<\lambda_{0}}; of course, λ1{\lambda_{1}} can be chosen successor. Clearly, [[g,x0],x1]{[[g,x_{0}],x_{1}]} lies in Zλ2⁢(G){Z_{\lambda_{2}}(G)} with λ2<λ1{\lambda_{2}<\lambda_{1}} and so on. Since the ordinal numbers are well ordered we have a contradiction.Suppose g is a Černikov element outside Z¯{\overline{Z}}. In order to achieve a contradiction, we can put Z¯=1{\overline{Z}=1}. Since Z⁢(G)={1}{Z(G)=\{1\}}, we can find an element x0{x_{0}} such that [g,x0]≠1{[g,x_{0}]\neq 1}. Since [g,x0]{[g,x_{0}]} does not belong to Z⁢(G){Z(G)}, we can find an element x1∈G{x_{1}\in G} such that [[g,x0],x1]≠1{[[g,x_{0}],x_{1}]\neq 1}. Continuing in this way we find a sequence x0,x1,…,xn,xn+1,…{x_{0},x_{1},\dots,x_{n},x_{n+1},\dots} of elements of G for which there is no non-negative integer m such that[…⁢[[g,x0],x1],…,xm]=1.[\dots[[g,x_{0}],x_{1}],\dots,x_{m}]=1.\vspace*{-0.5mm}This contradicts the definition of Černikov element.∎Now we need to introduce the concept of bounded Černikov element. Let G be a group. We say that 1 is the only Černikov 0-element of G and we put 𝒞0={1}{\mathcal{C}_{0}=\{1\}}. Let λ be a non-zero ordinal number and assume that we have defined the set 𝒞μ{\mathcal{C}_{\mu}} of all Černikov μ-elements of G for all ordinal numbers μ<λ{\mu<\lambda}. If λ is limit, we define the set 𝒞λ{\mathcal{C}_{\lambda}} of all Černikov λ-elements of G simply as the union of all sets 𝒞μ{\mathcal{C}_{\mu}} with μ<λ{\mu<\lambda}. If λ is successor, we define 𝒞λ{\mathcal{C}_{\lambda}} as the set of all elements g∈G{g\in G} such that, for all x∈G{x\in G}, there is an ordinal μx<λ{\mu_{x}<\lambda} such that [g,x]∈𝒞μx{[g,x]\in\mathcal{C}_{\mu_{x}}}.It is clear that g is a Černikov 1-element if and only if g lies in Z⁢(G){Z(G)}. Notice that a Černikov μ-element is also a Černikov λ-element whenever μ<λ{\mu<\lambda}, and that, for all g∈G{g\in G}, the commutator [g,x]{[g,x]} is a Černikov λ-element for all x∈G{x\in G} if and only if g is a Černikov (λ+1){(\lambda+1)}-element. Our next statement shows that for any ordinal number λ, the set 𝒞λ{\mathcal{C}_{\lambda}} is a subgroup.Lemma 3.2.Let G be a group and let λ be an ordinal number. An element g of G belongs to Zλ⁢(G){Z_{\lambda}(G)} if and only if g is a Černikov λ-element.Proof.We work by induction on λ, being the assertion obvious for λ=0,1{\lambda=0,1}. Suppose first that λ>1{\lambda>1} is limit. In this case, g lies in Zλ⁢(G){Z_{\lambda}(G)} if and only if it lies in Zμ⁢(G){Z_{\mu}(G)} for some ordinal μ<λ{\mu<\lambda}. By induction, the last statement is equivalent to require that g is a Černikov μ-element for some μ<λ{\mu<\lambda}, and, by definition, this is equivalent to require that g is a Černikov λ-element.Suppose now that λ>1{\lambda>1} is successor. In this case, g lies in Zλ⁢(G){Z_{\lambda}(G)} if and only if, for all x∈G{x\in G}, we have [g,x]∈Zλ-1⁢(G){[g,x]\in Z_{\lambda-1}(G)}. By induction, the last statement is equivalent to require that, for all x∈G{x\in G}, [g,x]∈𝒞λ-1{[g,x]\in\mathcal{C}_{\lambda-1}}, and this is equivalent to require that g∈𝒞λ{g\in\mathcal{C}_{\lambda}}. ∎Combining Lemma 3.1 and Lemma 3.2, we see that an element is a Černikov element if and only if it is a Černikov λ-element for some ordinal λ. The above characterizations of (bounded) Černikov elements will be subsequently used without any further reference, due to the large number of times it should be cited.Lemma 3.3.Let G be a group and let g be an element of the hypercentre of G. Let κ be the smallest ordinal number such that g lies in Zκ⁢(G){Z_{\kappa}(G)}. Then, for any ordinal λ<κ{\lambda<\kappa} which is either 0 or successor, there is a sequence of elements x0,x1,…,xm{x_{0},x_{1},\dots,x_{m}} of G such that λ is the smallest ordinal number for which[…⁢[[g,x0],x1],…,xm]∈Zλ⁢(G).{[\dots[[g,x_{0}],x_{1}],\dots,x_{m}]\in Z_{\lambda}(G)}.Proof.We work by induction on κ, being the statement true if κ=0,1{\kappa=0,1}. Suppose κ>1{\kappa>1}. Now, observe that κ is successor and g is a Černikov κ-element that is not a Černikov (κ-1){(\kappa-1)}-element.Suppose first κ-1{\kappa-1} is successor. Then there exists x∈G{x\in G} such that [g,x]{[g,x]} is a Černikov (κ-1){(\kappa-1)}-element that is not a Černikov (κ-2){(\kappa-2)}-element. Applying induction hypothesis completes the proof in this case.Assume now that κ-1{\kappa-1} is limit. Then, for any ordinal μ<κ-1{\mu<\kappa-1}, there are xμ∈G{x_{\mu}\in G} and a successor ordinal μ<κμ<κ-1{\mu<\kappa_{\mu}<\kappa-1} such that [g,xμ]{[g,x_{\mu}]} is a Černikov κμ{\kappa_{\mu}}-element that is not a Černikov (κμ-1){(\kappa_{\mu}-1)}-element. If we fix λ<κ{\lambda<\kappa}, which is either 0 or successor, then λ<κ-1{\lambda<\kappa-1}. Using the induction hypothesis on κλ>λ{\kappa_{\lambda}>\lambda} and [g,xλ]{[g,x_{\lambda}]}, we are able to complete the proof.∎Lemma 3.4.Let λ be an infinite ordinal number of cardinality m{\mathfrak{m}} and let G be a group. Then any subset X of cardinality at most m{\mathfrak{m}} of G is contained in a subgroup Y of cardinality at most m{\mathfrak{m}} of G such that Zκ⁢(G)∩Y=Zκ⁢(Y){Z_{\kappa}(G)\cap Y=Z_{\kappa}(Y)} for all κ≤λ{\kappa\leq\lambda}.Proof.We define the subgroup Y by recursion, but first we need a procedure to associate with every element g of G a set 𝒰g{\mathcal{U}_{g}} of elements of Gencoding the information concerning which term, if any, of the upper central series of G, it belongs. Thus, let 1≠g∈G{1\neq g\in G}. If g does not belong to the hypercentre Z¯{\overline{Z}} of G, we can find a sequence x0,x1,…,xn,xn+1,…{x_{0},x_{1},\dots,x_{n},x_{n+1},\dots} of elements of G such that […[g,x0],x1],…,xm]≠1{[\dots[g,x_{0}],x_{1}],\dots,x_{m}]\neq 1} for all non-negative integers m. In this case we put 𝒰g={x0,x1,…,xn,xn+1,…}{\mathcal{U}_{g}=\{x_{0},x_{1},\dots,x_{n},x_{n+1},\dots\}}.Suppose g lies in Z¯{\overline{Z}} and let μ be the smallest ordinal number such that g lies in Zμ⁢(G){Z_{\mu}(G)}. For any successor ordinal number κ with κ≤λ{\kappa\leq\lambda} and κ<μ{\kappa<\mu}, choose a sequence of elements x0κ,x1κ,…,xmκκ{x_{0}^{\kappa},x_{1}^{\kappa},\dots,x_{m_{\kappa}}^{\kappa}} such that[…⁢[[g,x0κ],x1κ],…,xmκκ]∈Zκ⁢(G)∖Zκ-1⁢(G){[\ldots[[g,x_{0}^{\kappa}],x_{1}^{\kappa}],\ldots,x_{m_{\kappa}}^{\kappa}]\inZ%_{\kappa}(G)\setminus Z_{\kappa-1}(G)}(this is possible by Lemma 3.3). Let 𝒰gκ={x0κ+,x1κ+,…,xmκ+κ+}{\mathcal{U}_{g}^{\kappa}=\{x_{0}^{\kappa^{+}},x_{1}^{\kappa^{+}},\dots,x_{m_{%\kappa^{+}}}^{\kappa^{+}}\}} and 𝒰g=⋃κ𝒰gκ{\mathcal{U}_{g}=\bigcup_{\kappa}\mathcal{U}_{g}^{\kappa}} (put 𝒰g0=∅{\mathcal{U}_{g}^{0}=\emptyset}).It is easy to see that in all cases 𝒰g{\mathcal{U}_{g}} is of cardinality at most 𝔪{\mathfrak{m}}. Now, let Y0=〈X〉{Y_{0}=\langle X\rangle} and suppose we have defined a subgroup Yn{Y_{n}} of G of cardinality at most 𝔪{\mathfrak{m}}. Then we putYn+1=〈Yn,𝒰g:g∈Yn〉.Y_{n+1}=\langle Y_{n},\mathcal{U}_{g}:g\in Y_{n}\rangle.Finally, put Y=⋃iYi{Y=\bigcup_{i}Y_{i}}; we show that Y is the required subgroup. Suppose not and let α be the smallest ordinal number such that α≤λ{\alpha\leq\lambda} and Zα⁢(G)∩Y≠Zα⁢(Y){Z_{\alpha}(G)\cap Y\neq Z_{\alpha}(Y)}. Of course, α≠0{\alpha\neq 0} is successor, so Zα-1⁢(G)∩Y=Zα-1⁢(Y){Z_{\alpha-1}(G)\cap Y=Z_{\alpha-1}(Y)}. Now, if x is any element of Zα⁢(G)∩Y{Z_{\alpha}(G)\cap Y}, it follows that [x,y]∈Zα-1⁢(G)∩Y≤Zα-1⁢(Y){[x,y]\in Z_{\alpha-1}(G)\cap Y\leq Z_{\alpha-1}(Y)} for all y∈Y{y\in Y}, and so Zα⁢(G)∩Y<Zα⁢(Y){Z_{\alpha}(G)\cap Y<Z_{\alpha}(Y)}. Let u∈Zα⁢(Y)∖(Zα⁢(G)∩Y){u\in Z_{\alpha}(Y)\setminus\big{(}Z_{\alpha}(G)\cap Y\big{)}} and let n be such that u∈Yn{u\in Y_{n}}. If u∉Z¯{u\not\in\overline{Z}}, then Yn+1{Y_{n+1}} contains 𝒰u{\mathcal{U}_{u}} and so u does not belong even to the hypercentre of Y (see Lemma 3.1), a contradiction. Thus u lies in Z¯{\overline{Z}}; let β be the smallest ordinal such that u lies in Zβ⁢(G){Z_{\beta}(G)}. If β≤α{\beta\leq\alpha}, we get u∈Zβ⁢(G)≤Zα⁢(G){u\in Z_{\beta}(G)\leq Z_{\alpha}(G)} and so u lies in Zα⁢(G)∩Y{Z_{\alpha}(G)\cap Y}, a contradiction. Thus β>α{\beta>\alpha}. In this case, for each successor ordinal γ≤α{\gamma\leq\alpha}, we have[⋱⁢[[g,x0γ],x1γ],…,xmγγ]∈(Zγ⁢(G)∖Zγ-1⁢(G))∩Y=Zγ⁢(Y)∖Zγ-1⁢(Y)[\ddots[[g,x_{0}^{\gamma}],x_{1}^{\gamma}],\ldots,x_{m_{\gamma}}^{\gamma}]\in%\big{(}Z_{\gamma}(G)\setminus Z_{\gamma-1}(G)\big{)}\cap Y=Z_{\gamma}(Y)%\setminus Z_{\gamma-1}(Y)and 𝒰uγ⊆Y{\mathcal{U}_{u}^{\gamma}\subseteq Y}. It follows that u is not a Černikov α-element of Y, so it belongs to Z¯⁢(Y)∖Zα⁢(Y){\overline{Z}(Y)\setminus Z_{\alpha}(Y)}, the final contradiction.∎Corollary 3.5.Let λ be an infinite ordinal number of cardinality m{\mathfrak{m}} and let G be a group. If Zλ⁢(X)=X{Z_{\lambda}(X)=X} for any subgroup X of cardinality at most m{\mathfrak{m}}, then Zλ⁢(G)=G{Z_{\lambda}(G)=G}.Proof.Suppose the statement is false. Of course, G is hypercentral since the class of all hypercentral groups is countably recognizable. Now, since Zλ⁢(G)<G{Z_{\lambda}(G)<G}, we have that Zλ⁢(G)<Zλ+1⁢(G){Z_{\lambda}(G)<Z_{\lambda+1}(G)}; choose x in Zλ+1⁢(G)∖Zλ⁢(G){Z_{\lambda+1}(G)\setminus Z_{\lambda}(G)}. By Lemma 3.4, the element x is contained in a subgroup Y of cardinality at most 𝔪{\mathfrak{m}} such that Zκ⁢(G)∩Y=Zκ⁢(Y){Z_{\kappa}(G)\cap Y=Z_{\kappa}(Y)} for all κ≤λ+1{\kappa\leq\lambda+1}. It follows that Zλ⁢(Y)<Zλ+1⁢(Y){Z_{\lambda}(Y)<Z_{\lambda+1}(Y)}, a contradiction.∎Corollary 3.6.Let λ be an infinite ordinal number of cardinality m{\mathfrak{m}} and let G be a group. Suppose that for each subgroup X of cardinality at most m{\mathfrak{m}} there is an ordinal μ=μ⁢(X)<λ{\mu=\mu(X)<\lambda} such that Zμ⁢(X)=X{Z_{\mu}(X)=X}. Then Zκ⁢(G)=G{Z_{\kappa}(G)=G} for some ordinal κ<λ{\kappa<\lambda}.Proof.Suppose the statement is false and let G be a counterexample with smallest possible λ; since the class of nilpotent groups is countably recognizable, we have λ>ω{\lambda>\omega}. Moreover, by Corollary 3.5, we can also assume that λ is limit. Then there is a strictly increasing sequence of successor ordinal numbers {λi}i∈I{\{\lambda_{i}\}_{i\in I}} converging to λ (from the bottom). The minimality of λ shows that for each i∈I{i\in I} there is a subgroup Xi{X_{i}} of cardinality at most 𝔪{\mathfrak{m}} such that Zλi⁢(Xi)<Xi{Z_{\lambda_{i}}(X_{i})<X_{i}}. Clearly, X=⋃i∈IXi{X=\bigcup_{i\in I}X_{i}} is a subgroup of cardinality at most 𝔪{\mathfrak{m}} and hence there is μ=μ⁢(X)<λ{\mu=\mu(X)<\lambda} such that Zμ⁢(X)=X{Z_{\mu}(X)=X}. On the other hand, there is j∈I{j\in I} such that λj>μ{\lambda_{j}>\mu}, so we reach the contradiction sinceZλj⁢(Xj)<Xj=Zμ⁢(Xj)≤Zλj⁢(Xj).Z_{\lambda_{j}}(X_{j})<X_{j}=Z_{\mu}(X_{j})\leq Z_{\lambda_{j}}(X_{j}).The statement is proved.∎We are now ready to deal with uncountable groups whose proper large subgroups are hypercentral. Here we are not able to obtain a perfect analogous of the theorems in the previous section without the additional requirement that the hypercentral lengths of the proper larger subgroups are not too big. In any case, the following theorems generalize [11, Theorem A] and Corollary 2.5.Theorem 3.7.Let m{\mathfrak{m}} be an uncountable cardinal number and let λ be an ordinal number such that𝔫=|λ|<cf⁡(𝔪).{\mathfrak{n}=|\lambda|<\operatorname{cf}(\mathfrak{m})}.Let G be a group of cardinality m{\mathfrak{m}} such that:(1)Zλ⁢(X)=X{Z_{\lambda}(X)=X} for all proper subgroups X of cardinality 𝔪{\mathfrak{m}} of G, (1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then Zλ⁢(G)=G{Z_{\lambda}(G)=G}.Proof.Suppose that the statement is false and let G be a counterexample with smallest possible λ. It follows from [11, Theorem B] that G is locally nilpotent.Suppose first that λ=c{\lambda=c} is finite and assume by contradiction that G is not nilpotent of class at most c. Then there is a finitely generated subgroup X which is not nilpotent of class at most c. If all proper normal subgroups of G have cardinality strictly smaller than 𝔪{\mathfrak{m}}, we have that G is the join of all its proper normal subgroups. In this case, X is easily seen to be contained in a proper normal subgroup of G and consequently to be nilpotent of class at most c, a contradiction.Therefore G admits a proper normal subgroup M of cardinality 𝔪{\mathfrak{m}}; clearly, G=X⁢M{G=XM}. We now try to find a contradiction by showing that X is contained in some proper subgroup of cardinality 𝔪{\mathfrak{m}} of G; to this end, we often replace G with factor groups. Since M is nilpotent, we can easily choose M such that M′{M^{\prime}} is of cardinality strictly smaller that 𝔪{\mathfrak{m}}. Of course, we can even assume M′={1}{M^{\prime}=\{1\}} and so that M is actually abelian. Now, we may further assume that X∩M={1}{X\cap M=\{1\}}. Let N be any proper normal subgroup of X. Clearly, NM is nilpotent, so there is a non-negative integer d such that Zd+1⁢(M⁢N)/Zd⁢(M⁢N){Z_{d+1}(MN)/Z_{d}(MN)} is of cardinality 𝔪{\mathfrak{m}}. It follows that M∩Zd+1⁢(M⁢N)/M∩Zd⁢(M⁢N){M\cap Z_{d+1}(MN)/M\cap Z_{d}(MN)} has cardinality 𝔪{\mathfrak{m}} and so we may assume M∩Zd⁢(M⁢N)={1}{M\cap Z_{d}(MN)=\{1\}} and M=M∩Zd+1⁢(M⁢N){M=M\cap Z_{d+1}(MN)}. Now, N is centralized by M.If X is the join of all its proper normal subgroups, then it is the join of finitely many such subgroups (recall that X is finitely generated). Thus we may safely assume that [M,X]={1}{[M,X]=\{1\}}. Since M contains a proper subgroup L of cardinality 𝔪{\mathfrak{m}}, it then follows that LX is a proper subgroup of cardinality 𝔪{\mathfrak{m}} of G, a contradiction.Thus X is simple and so cyclic of prime order. Now, Lemma 2.10 shows that M contains a proper G-invariant subgroup L of cardinality 𝔪{\mathfrak{m}}. Clearly, X⁢L<G{XL<G}, a contradiction showing that λ is infinite.Now, by Corollary 3.5 there is a subgroup X of cardinality at most 𝔫{\mathfrak{n}} such that Zλ⁢(X)<X{Z_{\lambda}(X)<X}. If all proper normal subgroups of G are of cardinality strictly smaller than 𝔪{\mathfrak{m}}, we easily get that G is the join of all its proper normal subgroups. Thus, for each x∈X{x\in X} there is a proper normal subgroup Nx{N_{x}} of G such that x lies in Nx{N_{x}}. Since 𝔫<cf⁡(𝔪){\mathfrak{n}<\operatorname{cf}(\mathfrak{m})} we have that N=〈Nx:x∈X〉{N=\langle N_{x}:x\in X\rangle} is a normal subgroup of cardinality strictly smaller than 𝔪{\mathfrak{m}}, so X is contained in the proper normal subgroup N and hence in a proper subgroup of cardinality 𝔪{\mathfrak{m}}. This contradiction shows that G contains a proper normal subgroup M of cardinality 𝔪{\mathfrak{m}}.Now, Zλ⁢(M)=M{Z_{\lambda}(M)=M} and |λ|<cf⁡(|M|){|\lambda|<\operatorname{cf}(|M|)}, so there is an ordinal number μ<λ{\mu<\lambda} such that Zμ+1⁢(M)/Zμ⁢(M){Z_{\mu+1}(M)/Z_{\mu}(M)} is of cardinality 𝔪{\mathfrak{m}}; of course, we can choose μ smallest possible with respect to such a condition. It follows that |L′|<𝔪{|L^{\prime}|<\mathfrak{m}}, where L=Zμ+1⁢(M){L=Z_{\mu+1}(M)}. We know that G=X⁢L{G=XL} and in order to show that X is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}} we can certainly assume L′={1}{L^{\prime}=\{1\}} and subsequently that X∩L={1}{X\cap L=\{1\}}; in particular, G=X⋉L{G=X\ltimes L} and L contains no proper G-invariant subgroups of cardinality 𝔪{\mathfrak{m}}.Let N be any proper normal subgroup of X. By hypothesis, Zλ⁢(N⁢L)=N⁢L{Z_{\lambda}(NL)=NL} and, since |λ|<cf⁡(𝔪){|\lambda|<\operatorname{cf}(\mathfrak{m})}, there is an ordinal number μ<λ{\mu<\lambda} such that |Zμ+1⁢(N⁢L)/Zμ⁢(N⁢L)|=𝔪{|Z_{\mu+1}(NL)/Z_{\mu}(NL)|=\mathfrak{m}}. It follows that L∩Zμ+1⁢(N⁢L)=L{L\cap Z_{\mu+1}(NL)=L} and |Zμ⁢(N⁢L)|<𝔪{|Z_{\mu}(NL)|<\mathfrak{m}}. Thus, if needed, we can assume that Zμ⁢(N⁢L)={1}{Z_{\mu}(NL)=\{1\}} and so that N is a normal subgroup of G.If X is the join of all its proper normal subgroups, then it is the join of at most 𝔫{\mathfrak{n}} such proper normal subgroups. Thus we may safely assume that [L,X]={1}{[L,X]=\{1\}}. Since L contains a proper subgroup U of cardinality 𝔪{\mathfrak{m}}, we get a contradiction. Therefore X is simple and so cyclic of prime order. Now, Lemma 2.10 shows that L contains a proper G-invariant subgroup V of cardinality 𝔪{\mathfrak{m}}, the last contradiction.∎Theorem 3.8.Let m{\mathfrak{m}} be an uncountable cardinal number and let λ be an ordinal number such that𝔫=|λ|<cf⁡(𝔪).{\mathfrak{n}=|\lambda|<\operatorname{cf}(\mathfrak{m})}.\vspace*{-0.5mm}Let G be a group of cardinality m{\mathfrak{m}} such that:(1)for each proper subgroup X of cardinality 𝔪{\mathfrak{m}} of G there is an ordinal μ=μ⁢(X)<λ{\mu=\mu(X)<\lambda} such that Zμ⁢(X)=X{Z_{\mu}(X)=X}, (1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then Zκ⁢(G)=G{Z_{\kappa}(G)=G} for some ordinal κ<λ{\kappa<\lambda}.Proof.Theorem 3.7 yields that λ is infinite and Zλ⁢(G)=G{Z_{\lambda}(G)=G}. The hypothesis |λ|<cf⁡(𝔪){|\lambda|<\operatorname{cf}(\mathfrak{m})} shows that there is an ordinal μ<λ{\mu<\lambda} such that Zμ+1⁢(G)/Zμ⁢(G){Z_{\mu+1}(G)/Z_{\mu}(G)} is of cardinality 𝔪{\mathfrak{m}} and so there is a subgroup L of G such that both |L|{|L|} and |G:L|{|G:L|} have cardinality 𝔪{\mathfrak{m}}. Now, any subgroup X of cardinality at most 𝔫{\mathfrak{n}} is contained in a proper subgroup of cardinality 𝔪{\mathfrak{m}} and so Zκ⁢(X)=X{Z_{\kappa}(X)=X} for some ordinal κ<λ{\kappa<\lambda}. The statement follows from Corollary 3.6.∎Finally, we see that in the torsion-free case no restriction on the hypercentral lengths is required.Theorem 3.9.Let m{\mathfrak{m}} be a cardinal number such that cf⁡(m)>ω{\operatorname{cf}(\mathfrak{m})>\omega} and let G be a torsion-free group of cardinality m{\mathfrak{m}} such that:(1)all proper subgroups of cardinality 𝔪{\mathfrak{m}} of G are hypercentral,(1)G is locally graded,(1)G has no simple homomorphic image of cardinality 𝔪{\mathfrak{m}}. Then G is hypercentral.Proof.It follows from [11, Theorem B] that G is locally nilpotent. Assume by contradiction that G is not hypercentral and let X be any countable subgroup of G that is not hypercentral.Suppose first that G has no proper normal subgroup of cardinality 𝔪{\mathfrak{m}}. In this case, G is the join of its proper normal subgroups, each of which is of cardinality strictly smaller than 𝔪{\mathfrak{m}}, while the join of any finite number of proper normal subgroups is still a proper normal subgroup of G. Thus, each element x of X is contained in a proper normal subgroup Nx{N_{x}} of G. The hypothesis on the cofinality yields that X is contained in the proper normal subgroup N=〈Nx:x∈X〉{N=\langle N_{x}:x\in X\rangle}. Now, G/N{G/N} is of cardinality 𝔪{\mathfrak{m}} and it cannot be a Jónsson group (see for instance [9, Corollary 2.6]), so it admits a proper subgroup M/N{M/N} of cardinality 𝔪{\mathfrak{m}}. Clearly, M is hypercentral and so X is such, a contradiction.Assume that G contains a proper normal subgroup N of cardinality 𝔪{\mathfrak{m}}; of course, G=X⁢N{G=XN}. Clearly, all proper subgroups of G/N{G/N} are hypercentral. If G/N{G/N} is minimal non-hypercentral, then it is also periodic (see [8]). It follows that G is the isolator of a hypercentral subgroup and so it is itself hypercentral (see [18, 2.3.9]). On the other hand, if G/N{G/N} is hypercentral, then G′{G^{\prime}} is a proper subgroup of G. In this case, there is a proper normal subgroup L of G such that G/L{G/L} is periodic and so again G is hypercentral by [18, 2.3.9].∎Again the example of the locally dihedral 2-groups shows that we cannot extend the above results to countable cardinalities.4Removing the cofinality assumptionThe aim of this section is to show that under a very powerful assumption (namely, the Generalized Continuum Hypothesis) we are able to remove the hypothesis on the cofinality of the (uncountable) cardinalities of the groups involved in some of our previous statements. Thus all cardinalities in these statements will be subject to the sole condition of being uncountable.We start with a very useful lemma, which holds under GCH, and then we describe how to employ such a result to achieve our goal.Lemma 4.1.Assume that the Generalized Continuum Hypothesis holds and let G be a group of uncountable cardinality m{\mathfrak{m}}. Let X be a subgroup of Aut⁡(G){\operatorname{Aut}(G)} containing Inn⁡(G){\operatorname{Inn}(G)} and such that |X:Inn(G)|≤ℵ0{|X:\operatorname{Inn}(G)|\leq\aleph_{0}}. If(1)there is no maximal X-invariant subgroup N of G such that G/N{G/N} is of cardinality 𝔪{\mathfrak{m}} and(1)all proper X-invariant subgroups of G have cardinality strictly smaller than 𝔪{\mathfrak{m}}, then every countable subgroup of G is contained in a proper X-invariant subgroup of G.Proof.Since all proper X-invariant subgroups of G are of cardinality strictly smaller than 𝔪{\mathfrak{m}}, an easy application of Zorn’s lemma yields that G coincides with the subgroup generated by all its proper X-invariant subgroups. It follows from the fact that |X:Inn(G)|≤ℵ0{|X:\operatorname{Inn}(G)|\leq\aleph_{0}} and [9, Lemma 2.4] that G cannot be abelian, so in particular CG⁢(X)≤Z⁢(G)<G{C_{G}(X)\leq Z(G)<G} and G contains proper X-invariant subgroups that are not contained in Z⁢(G){Z(G)}. Let N be any such subgroup and put |N|=𝔫<𝔪{|N|=\mathfrak{n}<\mathfrak{m}}. Clearly, G/CG⁢(N){G/C_{G}(N)} embeds into Aut⁡(N){\operatorname{Aut}(N)} and so its cardinality is at most 𝔫𝔫=2𝔫{\mathfrak{n}^{\mathfrak{n}}=2^{\mathfrak{n}}}. Moreover, CG⁢(N){C_{G}(N)} is obviously X-invariant and, since Z⁢(G)≱N{Z(G)\not\geq N}, we have CG⁢(N)<G{C_{G}(N)<G}. Thus CG⁢(N){C_{G}(N)} is of cardinality strictly smaller than 𝔪{\mathfrak{m}} and hence 2𝔫=|G/CG⁢(N)|=𝔪{2^{\mathfrak{n}}=|G/C_{G}(N)|=\mathfrak{m}}. Now, it easily follows from the Generalized Continuum Hypothesis that every proper non-central X-invariant subgroup of G is of cardinality 𝔫{\mathfrak{n}}. Furthermore, by hypothesis, the factor group G/Z⁢(G){G/Z(G)} (which is of cardinality 𝔪{\mathfrak{m}}) contains non-trivial proper X-invariant subgroups, and hence all proper X-invariant subgroups of G are of cardinality at most 𝔫{\mathfrak{n}}.Let H be any countable subgroup of G. Since G is generated by its proper X-invariant subgroups, for any x∈X{x\in X} we can find a proper X-invariant subgroup Nx{N_{x}} of G containing x. Of course, the subgroup 〈Nx:x∈X〉{\langle N_{x}:x\in X\rangle} is an X-invariant subgroup of cardinality 𝔫⋅ℵ0=𝔫<𝔪{\mathfrak{n}\cdot\aleph_{0}=\mathfrak{n}<\mathfrak{m}} and we are done.∎We now described how to employ the above lemma in order to remove the restriction on the cofinality in the results of Section 2. First of all, an application of Lemma 4.1 with X=Inn⁡(G){X=\operatorname{Inn}(G)} takes care of the situation described in the second paragraph of the proof of Lemma 2.1, which is the only paragraph where the hypothesis on the cofinality has been used. Thus the hypothesis on the cofinality can be removed from Lemma 2.1.In a similar way we can deal with the second paragraph of the proof of Theorem 2.4. From paragraph 4 on, things change. An application of Lemma 2.1 shows that N contains a maximal G-invariant subgroup L such that N/L{N/L} is of cardinality 𝔪{\mathfrak{m}}. Of course, in this case we can assume L={1}{L=\{1\}}. Now, all proper normal subgroups Y of X are centralized by N (since [Y,N]={1}{[Y,N]=\{1\}}) and so the join M of all proper normal subgroups of X is centralized by N. If M=X{M=X}, then G=X×N{G=X\times N} and we can argue as in the last part of paragraph 5. The proof now goes along the same lines of the original one from paragraph 6 on (but notice that one could actually simplify some tiny bit of it). This takes care of Theorem 2.4.The proofs of Corollaries 2.5, 2.6, 2.7 and 2.8 need no change besides application of the new statements of previous results.5Embedding propertiesThe aim of this short section is to provide an easy proof of the main results contained in [3, Section 2]. We will show that the hypothesis on the regularity of the cardinal that was required in [3] is actually a non-essential one.Lemma 5.1.Let m{\mathfrak{m}} be an uncountable cardinal number. Let G=E⁢A{G=EA} be a group that is the product of a normal abelian subgroup A of cardinality m{\mathfrak{m}} and a finitely generated subgroup E. If there is a positive integer m such that [A,mE]{[A,_{m}E]} is of cardinality strictly smaller than m{\mathfrak{m}}, then Z⁢(G){Z(G)} is of cardinality m{\mathfrak{m}}.Proof.We may assume that m is such that B=[A,mE]{B=[A,_{m}E]} is of cardinality 𝔪{\mathfrak{m}} and |[B,E]|<𝔪{|[B,E]|<\mathfrak{m}}. LetE=〈x1,…,xn〉{E=\langle x_{1},\dots,x_{n}\rangle}and, for each i=1,…,n{i=1,\dots,n}, consider the mapsφi:b∈B↦[b,xi]∈[B,E].\varphi_{i}:b\in B\mapsto[b,x_{i}]\in[B,E].Clearly, B/Ker⁡(φi){B/\operatorname{Ker}(\varphi_{i})} is of cardinality strictly smaller than 𝔪{\mathfrak{m}} and so U=⋂i=1nKer⁡(φi){U=\bigcap_{i=1}^{n}\operatorname{Ker}(\varphi_{i})} is a subgroup of Z⁢(G){Z(G)} of cardinality 𝔪{\mathfrak{m}}. The statement is proved.∎Notice that the condition on E in the above lemma is certainly satisfied whenever E is subnormal in G. We employ the above result in the next lemma, which is actually the first step in proving the more general Corollary 5.4. In the following, [20, Theorem 7.42] is referred to as “Roseblade’s theorem”.Lemma 5.2.Let k be a positive integer and let G be an uncountable group of cardinality m{\mathfrak{m}} in which all subgroups of cardinality m{\mathfrak{m}} are subnormal of defect at most k. If G contains an abelian normal subgroup of cardinality m{\mathfrak{m}}, then all subgroups of G are subnormal of defect at most k.Proof.Let A be a normal abelian subgroup of G of cardinality 𝔪{\mathfrak{m}}. Let c=f⁢(k){c=f(k)}, where f is the Roseblade’s function described in [20, p. 71, Part 2]. Let g be any element of G and put B=γc+1⁢(〈g〉⁢A){B=\gamma_{c+1}(\langle g\rangle A)}; obviously, B≤A{B\leq A}. If |B|=𝔪{|B|=\mathfrak{m}}, then B/B∩〈g〉{B/B\cap\langle g\rangle} contains a proper 〈g〉{\langle g\rangle}-invariant subgroup C/B∩〈g〉{C/B\cap\langle g\rangle} of cardinality 𝔪{\mathfrak{m}} (see Lemma 2.10). Since all subgroups of cardinality 𝔪{\mathfrak{m}} are subnormal, it follows that [〈g〉⁢C,B]=[〈g〉,B]{[\langle g\rangle C,B]=[\langle g\rangle,B]} is a proper 〈g〉{\langle g\rangle}-invariant subgroup of B. Clearly, there is a proper subgroup D of B such that |B:D|<𝔪{|B:D|<\mathfrak{m}} and D≥[〈g〉,B]{D\geq[\langle g\rangle,B]}. However, all proper subgroups of 〈g〉⁢A/D{\langle g\rangle A/D} are subnormal of defect at most k and so B=γc+1⁢(〈g〉⁢A)≤D{B=\gamma_{c+1}(\langle g\rangle A)\leq D}, a contradiction. Thus B is of cardinality strictly smaller than 𝔪{\mathfrak{m}}. It follows now from Lemma 5.1 that A contains a subgroup Z of cardinality 𝔪{\mathfrak{m}} which is centralized by g. Clearly, Z contains a subgroup U that is the direct product of two subgroups U1{U_{1}} and U2{U_{2}} of cardinality 𝔪{\mathfrak{m}} and such that U∩〈g〉={1}{U\cap\langle g\rangle=\{1\}}. Thus 〈g〉=〈g〉⁢U1∩〈g〉⁢U2{\langle g\rangle=\langle g\rangle U_{1}\cap\langle g\rangle U_{2}} is a subnormal subgroup of defect at most k. The arbitrariness of g in G yields that G is a Baer group.Let E be any finitely generated subgroup of G. Then E is subnormal in G and so Lemma 5.1 shows that A contains a subgroup V of cardinality 𝔪{\mathfrak{m}} which is centralized by E. The argument in the last part of the above paragraph shows that E is the intersection of two subgroups of cardinality 𝔪{\mathfrak{m}} and so that E is subnormal of defect at most k. Lemma 7.41 (ii) of [20] completes the proof.∎Theorem 5.3.Let k be a positive integer and let G be an uncountable group of cardinality m{\mathfrak{m}} in which all subgroups of cardinality m{\mathfrak{m}} are subnormal of defect at most k. If G contains no normal subgroups that are Jónsson groups of cardinality m{\mathfrak{m}}, then all subgroups of G are subnormal of defect at most k; in particular, G is nilpotent.Proof.Let c=f⁢(k){c=f(k)}, where f is the Roseblade’s function described in [20, p. 71, Part 2], and put L=γc+1⁢(G){L=\gamma_{c+1}(G)}. If L is of cardinality 𝔪{\mathfrak{m}}, then it contains a proper subgroup H of cardinality 𝔪{\mathfrak{m}}. Then H is subnormal in L and hence L/HL{L/H^{L}} is a non-trivial factor group whose subgroups are subnormal of length bounded by k. Then L/HL{L/H^{L}} is nilpotent and in particular L′<L{L^{\prime}<L}. If |L′|=𝔪{|L^{\prime}|=\mathfrak{m}}, then L=γc+1⁢(G)≤L′{L=\gamma_{c+1}(G)\leq L^{\prime}}, a contradiction. Thus L/L′{L/L^{\prime}} is of cardinality 𝔪{\mathfrak{m}} and G/L′{G/L^{\prime}} contains a normal abelian subgroup of cardinality 𝔪{\mathfrak{m}}. It follows from Lemma 5.2 that all subgroups of G/L′{G/L^{\prime}} are subnormal of defect at most k and hence G/L′{G/L^{\prime}} is nilpotent of class at most c, a contradiction.∎Corollary 5.4.Let k be a positive integer and let G be an uncountable group of cardinality m{\mathfrak{m}} in which all subgroups of cardinality m{\mathfrak{m}} are subnormal of defect at most k. If G contains an abelian subgroup of cardinality m{\mathfrak{m}}, then all subgroups of G are subnormal of defect at most k; in particular, G is nilpotent.Proof.Let A be an abelian subgroup of G of cardinality 𝔪{\mathfrak{m}}. It follows that A is subnormal in G of defect m, say. If m=1{m=1}, then G is nilpotent by Lemma 5.2 and Roseblade’s theorem. If m>1{m>1}, then A is subnormal in AG{A^{G}} of defect strictly smaller than m and so by induction AG{A^{G}} is nilpotent. However, all subgroups of G/AG{G/A^{G}} are subnormal of defect at most k and so G/AG{G/A^{G}} is nilpotent as well. It follows from [9, Corollary 2.6] that G admits no normal subgroups that are Jónsson groups of cardinality 𝔪{\mathfrak{m}}. Therefore, an application of Theorem 5.3 (and Roseblade’s theorem) completes the proof.∎

Journal

Forum Mathematicum – de Gruyter

Published: May 1, 2022

Keywords: Uncountable group; Engel group; hypercentral group; 20F45; 20F19

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