Groups with a small set of generators

@ Applied General Topology c Universidad Polit´ecnica de Valencia Volume 4, No. 2, 2003 pp. 327–350 Groups with a small set of generators Dikran Dikranjan, Umberto Marconi and Roberto Moresco∗ Dedicated to Professor S. Naimpally on the occasion of his 70th birthday. Abstract. Following [22] we study the class Sof all groups that admit a small set of generators. Here we adopt also another notion of smallness (P-small) introduced by Prodanov in the case of abelian groups. We push further some results obtained in [22] (by adding some new members of S) and partially resolve an open question posed in [22]. We show that in most cases the groups in Sadmit a P-small set of generators. 2000 AMS Classification: 20B30, 20F16, 20K45, 22C05, 54H11. Keywords: group, large set, small set, permutation group, linear group, compact group, profinite group. 1. Introduction. The question of measuring the size of a set of generators of a group is certainly a relevant one. In the case of topological group one puts topological restrictions on the set of generators to ensure smallness (see [19, 20] for the so called “suitable sets” – “small” sets of generators born in the Theory of Cohomology of infinite Galois groups in the work of Tate and Douady [8]). In the case of discrete groups the following notion of smallness was proved to be a very useful property in this respect in [22]. A subset Bof a group Gis large if G=F·B=B·Ffor some finite set Fof G. This property has been largely studied in the literature also under different names (big, discretely syndetic, relatively dense). A set S⊆Gis small if for every finite set Fthe sets S·Fand F·Shave a large complement in G[2, 3] (clearly, only infinite groups may have small sets). The role of small and large ∗The first author was partially supported by Research Grant of the Italian MURST in the framework of the project “Nuove prospettive nella teoria degli anelli, dei moduli e dei gruppi abeliani” 2000. The second and third author were supported by the Grant “Progetti di ricerca di Ateneo, 2001” of the University of Padova. 328 D. Dikranjan, U. Marconi, R. Moresco subsets of groups in number theory, compact representations of groups and dynamics can hardly be overestimated [7, 13, 14]. The question when an infinite group may have a small set of generators was addressed in [22]. Let us denote by Sthe class of groups with small set of generators. It was proved in [22] that Scontains all groups that have an infinite abelian normal subgroup (in particular, all groups with infinite center) as well as all solvable groups. Call a subset Sof an abelian group Gsmall in the sense of Prodanov (briefly, P-small) if there exist x1, . . . , xn. . . in Gsuch that the sets {S+xn}nare pairwise disjoint. It is easy to see that when S−Sis not large, then Sis P-small. This was the motivation for the introduction of P-small sets in [23]. It was noticed by Gusso [15] that P-small sets of the abelian groups are small. Their advantage is also that they are much easier to understand and construct. The main contributions of the present paper go in two directions. In §2.1 we define approriate versions of P-smallness in non-abelian groups, as well as other versions of smallness that turn out to be stronger than smallness or P-smallness in many cases. We show that in most of the cases the small generated groups from [22] have a set of generators satisfying also a much stronger property of smallness. On the other hand, we add to the list of groups in Sfound in [22] some new classes of non-abelian groups with a small set of generators proceding in two ways. In the case of permutation groups and linear groups our arguments are purely algebric. We offer also another approach to the problem that heavily leans on topology. This allows us to produce a wealth of small and P-small sets and to provide many examples of compact-like topological groups that belong to S. Finally, we partially answer a question from [22] by showing that Scontains all compact groups with eventual exception of the topologically finitely generated pro-pgroups and the profinite groups with trivial Frattini subgroup. 1.1. Notation and terminology. We denote by Nand Pthe sets of natural and prime numbers, respectively; by Zthe integers, by Qthe rationals, by R the reals, by Tthe unit circle group R/Z, by Z(p) the cyclic group of order p and by Zpthe p-adic integers (p∈P). The cardinality of continuum 2ωwill be denoted by c. If Sis a set, we denote by P(S) its power set. Let Gbe a group. We denote by 1 the neutral element of Gand by Z(G) the center of G. For a subset Sof Gwe denote by hSithe subgroup generated by S. The group Gis divisible if for every n∈Nand g∈Gthere exists x∈G with xn=g. The semidirect product of the groups Gand Kis denoted by GoK. Abelian groups are mostly written additively. In particular, 0 denotes the neutral element of an additively written abelian group Gand for A⊆Gand n∈Nwe let A(n)=A+· · · +A |{z } n when n > 0 and A(n)={0}for completeness. Topological groups are Hausdorff. A topological group Gis precompact if its completion is compact, pseudocompact if every continuous real-valued function on Gis bounded. For a topological group Gwe denote by c(G) the connected Groups with a small set of generators 329 component of the identity and by ψ(G) the pseudocharacter G(the minimum cardinality of a set Uof neighborhoods of 1 such that TU∈U U={1}). Unless explicitly stated, all groups are assumed to be infinite. If Xis a topological space and A⊆X, the closure of Ais denoted by A. For undefined symbols or notions see [7], [9], [12], or [17]. We denote by S(G) and SP(G) the collections of all small and P-small sets of a group Grespectively. 2. The various levels of smallness. Let us give the following more precise form of largeness. Definition 2.1. Let Gbe a group. A subset Bof Gis left large (resp. right large) if for some finite set Fthe union F·B(resp. B·F) of left (resp. right) translates of Bcovers G. The following trivial equalities are helpful when passing to complements: {g∈G:g·F6⊆ A}= (G\A)·F−1,{g∈G:g·F6⊆ G\A}=A·F−1(1) Clearly, they imply that G\Ais right large iff there exists a finite Fsuch that Acontains no left translate gF of F. Or, G\Ais not right large iff A contains left translates gF of every finite set F. Definition 2.2. A subset Sof a group Gis called: (a) left small if for every finite set Fthe sets S·Fand F·Shave a left large complement; right small is defined analogously. (b) weakly left small if for every finite set Fthe set S·Fhas a left large complement; weakly right small is defined analogously. (c) n-small, for a positive n∈N, if Sis small and the sets (S−1·S)n−1 and (S·S−1)n−1are not large. (d) microscopic if for every n∈Nthe sets (S−1·S)nand (S·S−1)nare small. A set is small iff it is left small and right small. Clearly the 1-small sets are precisely the small ones. The 2-small sets are those small sets Ssuch that S−1·Sand S·S−1are not large. For n > 1 we denote by Sn(G) the family of n-small sets of G. In additive notation, S∈Sn+1(G) for an abelian group G and n > 0 iff (S−S)(n)=S(n)−S(n)is not large, i.e., S(n)∈S2(G) (indeed, 2-small implies P-small which in turn implies small by [15]). The equalities (1) give the following useful criterion mentioned in [1] and [22] in the bilateral version of smallness: Lemma 2.3. A set Sis weakly left small (left small) iff for every finite set F there exists a finite Ksuch that the set S·Fcontains no right translate Kg (and the set F·Scontains no right translate Kg). 330 D. Dikranjan, U. Marconi, R. Moresco 2.1. Left and right smallness in the sense of Prodanov. Call a set Sof a group G: (a) right small in the sense of Prodanov (briefly right P-small) if there exist x1, . . . , xn. . . in Gsuch that the sets {S·xn}nare pairwise disjoint (or, equivalently, xn·x−1 m6∈ S−1·Sfor n6=m). (b) left small in the sense of Prodanov (briefly left P-small) if there exist x1, . . . , xn. . . in Gsuch that the sets {xn·S}nare pairwise disjoint (or, equivalently, x−1 m·xn6∈ S·S−1for n6=m). (c) strongly right P-small if there exist x1, . . . , xn. . . in Gsuch that the sets {Sf·xn}nare pairwise disjoint for every f∈G(or, equivalently, xn·x−1 m6∈ (S−1·S)ffor n6=m); strongly left P-small is defined analogously. (d) (strongly)P-small if it is (strongly) left and (strongly) right P-small. It is easy to see that Sis left P-small (strongly left P-small) if and only if S−1is right P-small (strongly right P-small). Here we give separately the smallnes conditions in terms of the difference sets S·S−1and S−1·S. Claim 2.4. Let Sbe a subset of a group G. Then: (al)Sis left P-small iff there exists an infinite set Xsuch that (S·S−1)∩(X·X−1) = {1}; (ar)Sis right P-small iff there exists an infinite set Xsuch that (S−1·S)∩(X−1·X) = {1}; (b) Sis strongly left P-small iff there exists an infinite set Xsuch that (∀f∈G)(S·S−1)f∩(X·X−1) = {1}. Consequently, each one of the properties weakly left (right) small, left (right) small, left (right) P-small, P-small, n-small, microscopic, is invariant under left and right translations. It is clear that strongly P-small implies P-small (by taking f= 1 in the definitions (c) and (d) above). In the non-abelian case, P-small need not imply small (cf. Example 2.17). The final part of the claim shows that microscopic implies small. Remark 2.5. (1) Assume that Sis right large. Then S·F=Gfor some finite F. Then for every infinite set Xone of the sets Sf (f∈F) contains infinitely many members X1of X. In particular, there exist x, y ∈Xwith x6=yand x, y ∈Sf. This gives xy−1∈X·X−1∩S·S−1. Thus X·X−1∩S·S−16={1}for every infinite X. (More precisely, for every infinite Xthere exists an infinite X1⊆Xsuch that X1·X−1 1⊆ S·S−1). Hence Sis not left P-small. Therefore left P-small sets cannot be right large (but can be left large, cf. Example 2.18). The same conclusion may be obtained by using (a) in the next Lemma 2.6. Groups with a small set of generators 331 (2) If S−1·Scontains a finite index subgroup H, then Sis not right Psmall. Indeed, if X⊆Gis an infinite set, then some coset aH will contain an infinite subset X1of X, hence X−1 1·X1⊆H⊆S−1·S, so Sis nor right P-small by (ar) of the preceding Claim. In the next lemma we give some connection between these notions of smallnes. Note that S−1·Sis symmetric, so all three versions of “large” coincide for S−1·S. Lemma 2.6. Let Sbe a subset of a group G. (a) If Sis left (right) P-small, then it is also weakly left (right) small. (b) If S−1·Sis not large then Sis right P-small, if S·S−1is not large, then Sis left P-small, (c) If Sis 2-small then it is P-small. (d) If Sis strongly left P-small, then it is left small. (e) If Sis strongly P-small, then it is small. (f) If Sis microscopic, then it is n-small for every n∈N. In particular, microscopic sets are 2-small. If Gis abelian, then the first implication can be reversed. Proof. (a) There exist x1, . . . , xn, . . . in Gsuch that x−1 m·xn6∈ S·S−1for n6=m. Let Fbe a finite subset of G. To see that S·Fhas a left large complement choose nsuch that |F|< n and take K={x−1 1, . . . , x−1 n}. Then S·Fcontains no right translates K·gof K. Indeed, assume that K·g⊆S·F for some g∈G. Then there exist xi6=xjin Ksuch that for some f∈Fone can find s, s1∈Swith x−1 ig=sf and x−1 jg=s1f. The second equation gives g−1·xj=f−1·s−1 1. Multiplying this by the first equation we get x−1 ixj=ss−1 1. Hence x−1 ixj∈S·S−1, that leads to contradiction. Analogously one proves that right P-small implies weakly right small. (b) Indeed, there exists x1, . . . , xn. . . in Gsuch that xn6∈ S−1·S· {x1, . . . , xn−1}, so that xn·x−1 m6∈ S−1·Sfor m < n. Since S−1·Sis symmetric, we have also xm·x−1 n6∈ S−1·S. Thus Sis right P-small. One can see analogously, that S is left P-small whenever S·S−1is not large. (c) Follows from (b). (d) To prove that Sis left small take any finite set F. We can assume without loss of generality that 1 ∈F. We have to find a finite set Ksuch that S·Fcontains no right translates K·gand F·Scontains no right translates K·g. The former property was already checked above in (a). To check the latter one note that by assumption there exist x1, . . . , xn, . . . in Gsuch that x−1 m·xn6∈ Sf·(S−1)ffor n6=mand for every f∈F. Take n > |F|and K={x−1 1, . . . , x−1 n}. Assume that K·g⊆F·Sfor some g∈G. Then there exist xi6=xjin Ksuch that for some f∈Fone can find t, t1∈Swith x−1 ig=ft and x−1 jg=ft1. The second equation gives g−1xj=t−1 1·f−1. 332 D. Dikranjan, U. Marconi, R. Moresco Multiplying this with the first equation we get x−1 ixj=ftt−1 1f−1. Hence x−1 ixj∈Sf·(S−1)f, that leads to contradiction. (e) Follows directly from (d). (f) It suffices to observe that Sis small (indeed S−1·Scontains a translated set of S). For the reverse implication in the Abelian case, it suffices to apply (a) and (b) to the set (S−S)(n). In the next diagram we give all the implications that are always valid between the various symmetric smallness properties we have introduced so far. microscopic -... -n-small -... -2-small  3 QQ Qs (∗) small P-small Q Q Qk   + (+) strongly P-small The arrow (+) becomes an equivalence for abelian groups, so that the diagram becomes linear with P-small=strongly P-small placed between 2-small and small. The inverse arrow of (∗) fails in two ways: a P-small set need not be small (as mentioned above) and S·S−1,S−1·Sneed not be large for a P-small set S(an example for the simultaneous failure of these implications is given in 2.17). The following question remains open even in the abelian case: Problem 2.7. Do the notions 2-small and strongly P-small coincide? Now we produce a series of examples of small sets on Zthat are not P-small. Example 2.8. (a) Let Prdenote the set of all products of at most rodd primes and let Ar=Pr∪ −Pr. It was proved in [1] that the set Pris small for every r∈N, hence Aris small as well. Brun proved in 1919 that P9+P9contains all sufficiently large even integers. This was improved by Rademacher in 1924 who brought rdown to 7 and Selberg improved it to r= 3 in 1950. According to Goldbach’s conjecture P1+P1contains all even integers nwith |n|>4. In these terms, the set Ar−Arcontains the subgroup of all even integers, hence it is surely large for r≥3 (actually, Aris not P-small by Remark 2.5) (2)). This shows that if Goldbach’s conjecture has positive answer, then the set A1of all odd primes (positive and negative) is small but not P-small. Obviously, to the same result leads the positive answer to the (unsolved) conjugate Goldbach’s conjecture (every even integer is a difference of two odd primes). (b) Let f(x) = ax2+bx +c, with a, b, c, ∈Z,a6= 0. Then Sf={f(n) : n∈Z}is small, but not P-small. Since these properties are invariant under translation, we can assume without loss of generality that c= 0. We shall see that Sf−Sfalways contains a non-zero subgroup H, so that it is not P-small by Remark 2.5. Indeed, if b6= 0, then for every n∈Zone has 2bn =f(n)−f(−n)∈Sf−Sf, so H= 2bZ⊆Sf−Sf. If b= 0, then 4an =f(n+ 1) −f(n−1) ∈Sf−Sf, so H= 4aZ⊆Sf−Sf. Furthermore Sf is small because limn→+∞(f(n+ 1) −f(n)) = +∞[1]. Groups with a small set of generators 333 2.2. Smallness vs topology, measure, asymptotical density and growth. It was proved in [2, Proposition 3.7] that every compact subset of a non-compact topological group is small. In fact, one can prove the following much stronger property. Lemma 2.9. A compact set Kof a non-compact topological group is always microscopic. Proof. Indeed, K−1is compact as well. Hence K−1·Kand K·K−1are compact as continuous images under multiplication of the compact spaces K−1×Kand K×K−1. For every n∈Nand for every finite set Fthe sets F·(K·K−1)n and F·(K−1·K)nare still compact, hence small by [2, Proposition 3.7].  We shall apply this lemma very often when Gis countable, so surely noncompact. Also we shall have often Ka converging sequence. This is why we make use of the following modified version of a notion introduced by Protasov and Zelenyuk [29] (the original version was for Gabelian and Hreplaced by G). Definition 2.10. A sequence a1, a2, . . . , an, . . . in an infinite group Gis a Tsequence if the subgroup Hgenerated by the set {a1, a2, . . . , an, . . .}admits a Hausdorff group topology such that an→1. Since the underlying set of every convergent sequence, along with its limit, is a compact set, we get Corollary 2.11. The underlying set of a T-sequence {an}∞ n=1 of an infinite group Gis a microscopic set. Recall, that a Banach measure on a group Gis a finitely additive invariant measure on Gsuch that every subset of Gis measurable and µ(G) = 1 (see [10, 11] for the existence of Banach measure on all abelian groups and the existence of groups that admit no Banach measure). Iv. Prodanov noted that Banach measures are a very convenient tool when dealing with large sets. Indeed, if µis a right invariant Banach measure on the group G, then obviously every right large set has a positive Banach measure, while every right P-small set has measure zero (hence a right P-small set is never right large in such a group). Therefore, an infinite group Gthat admits a right invariant Banach measure is never a finite union of right P-small sets (see [27] or [7, p. 29] for an elementary proof of this fact in the Abelian case that makes no recourse to Banach measures). It is proved in [1, Theorem 3.4] that every compact group Gadmits closed small sets Aof positive Haar measure arbitrarily close to 1 hence small sets need not be neither left nor right P-small. (In this case one can apply also Steinhouse-Weil theorem to conclude that A−1·Ahas non-empty interior, so A−1·Ais large.) Example 2.12. Here we give examples of small sets in Zwith various levels of smallness. 334 D. Dikranjan, U. Marconi, R. Moresco (1) Let f(x)∈Z[x] be polynomial of degree d > 0. Then the set Sf= {f(n) : n∈Z}is large if d= 1. It follows from the criterion given in [1] that Sfis small if d > 1. According to Example 2.8 Sf−Sfis large if d≤2. As Sf−Sf⊇Sg, where g(x) = f(x+1)−f(x) (so deg g=d−1), it is easy to see that in general (Sf−Sf)(m)= (Sf)(m)−(Sf)(m)is large if m≥2d−2. Hence, Sfis never microscopic. The above implications cannot be inverted for arbitrary polynomials f(for f(x) = x2d−xthe set Sf−Sfcontains all even integers, hence it is large), but seemingly this can be done for monomials f(x) = axd(e.g., Sfis 2-small iff d > 2, etc.). In this way one can construct n-small sets of Zthat are not n+ 1-small for every n∈N. (2) For every n > 1 let rnbe the rest of n2modulo 2k, where 2kis the greatest power of 2 with 2k≤n. Then S=±{n2−rn:n > 1}is microscopic by the above corollary (as n2−rnconverges to 0 in the 2-adic topology of Z) even if the function defining Sgrows slower than the polynomial function n7→ n2. Remark 2.13. It will be desireable to understand better the various kinds of small sets of Z. Here we propose two more points of view. (a) One can connect smallness or largeness of subsets of Zwith density (for an infinite set A⊆Zlet the density of Abe the limit limn|A∩[−n,n]| 2n, whenever it exists). Clearly, every large set has positive density (cf. [1, Proposition 1.13]). One can build examples of small sets of positive density ([1, Proposition 1.14]). On the other hand, by Schnilermann’s approach, for every set Awith positive density A(n)=Zfor nsufficiently large. Hence a set of positive density cannot be microscopic. We do not know whether such a set can be n-small for some n > 1. (b) Another approach to the smallness of a set A={an}of positive integers is to study the asymptotic behavior of the ratio an+1 an. There exist T-sequences {an}in Zsuch that limn an+1 an= 1 (cf. Example 2.12 (2), note that n2≥an=n2−rn≥n2−nby the choice rn< n, so limn an+1 an= 1 holds true). Hence there exists a microscopic set A={an}in Zsuch that the ratio an+1 anconverges to 1. It is known that when limn an+1 an=∞or the limit is a transcendental real number, then {an}is a T-sequence [29], while every algebraic number α≥1 admits a sequence {an}with limn an+1 an=αthat is not a T-sequence [29]. We do not know whether the condition limn an+1 an=α > 1 may imply that Ais microscopic, strongly small or P-small (it yields Ais small as an+1 −an→ ∞ when α > 1, so that [1, Proposition 1.2] applies). If yes, then we obtain new examples of small (microscopic) sets that are not the underlying set of a T-sequence. 2.3. Smallness of subgroups and transversals. Let us recall that a left transversal of a subgroup Hof a group Gis a subset Tof Gsuch that G=T·Hand (T−1·T)∩H={1}.(2) Groups with a small set of generators 335 Lemma 2.14. Let Hbe a subgroup of Gand let Tbe a left transversal of H. Then: (a) if His infinite, then Tis right P-small (or, equivalently, T−1is left P-small); (b) if Tis infinite, then His P-small; moreover, if His small, then His microscopic. (c) if (T−1·T)∩Hf={1}for every f∈G, then His microscopic if Tis infinite (and Tis strongly right P-small if His infinite). Proof. (a) If His infinite, then (2) implies that Tis right P-small since every infinite subset {xn}of Hwill witness right P-smallness. This yields T−1is left P-small. (b) The part (T−1·T)∩H={1}of (2), in view of H=H−1·H, witnesses His right P-small. If T1is a right transversal of H, then analogous argument proves that His left P-small. Since H−1=H, in both cases His P-small. Clearly, His not large when Tis infinite. Since His a subgroup, this implies that His microscopic, if His small. (c) For t6=t1in Tand every f∈Gthe cosets tHfand t1Hfare disjoint by hypothesis. Then His strongly left P-small by definition. Since His a subgroup, this implies that His strongly P-small, hence small (by Lemma 2.6). As Tis infinite, Hmust be microscopic, according to item (b).  Item (a) cannot be inverted (cf. Example 2.18). Corollary 2.15. Let Gbe a group and H, D be infinite subgroups of G. (a) if D∩H={1}, then both Hand Dare P-small. (b) if D∩Hf={1}, for every f∈Gthen both Hand Dare strongly P-small and microscopic. Corollary 2.16. Let Hbe a normal subgroup of a group G. (a) If Hhas infinite index, then His strongly P-small and microscopic. (b) If His infinite, then any transversal Tof His strongly P-small (hence, small). (c) If His infinite and has infinite index, then H∪Tis small for any transversal Tof H. Proof. (a) To see that His strongly left P-small fix a left transversal Tof H. Then it is also a right transversal. For t6=t1in Tthe cosets Ht =tH and Ht1=t1Hare disjoint. Moreover, Hf=Hfor every f∈F, so that Hftand Hft1remain disjoint for every f∈Fand t6=t1in T, i.e., (T−1·T)∩H={1} and (T·T−1)∩H={1}for every f∈G. Therefore, His strongly right P-small by Lemma 2.14. Since His a subgroup, this implies that His strongly P-small, hence small. Therefore, His also microscopic. (b) To see that Tis strongly left P-small let Fbe a finite set of G. Take any countably infinite subset {xn}of H. Then for every f∈Fthe sets Tfxnare pairwise disjoint. Indeed, if z∈Tfxn∩Tfxm, then zx−1 n=tfand zx−1 m=tf 1 342 D. Dikranjan, U. Marconi, R. Moresco for some integer n,Gis perfect if G=G0. The group Gis hyperabelian if G contains no perfect subgroups beyond the trivial one. Proposition 3.7. (a) If G/G0is infinite then Ghas a set of generators that is microscopic and strongly P-small; (b) All free groups belong to SP∩ M. (c) G∈ S2if Gis an infinite solvable group. (d) G∈ S2when G/G(n)is infinite for some n. (e) G∈ SPif Ghas an infinite abelian normal subgroup (in particular, if Ghas infinite center). Proof. (a) Follows from Lemma 3.3 and Theorem 3.6 (for abelian groups, microscopic implies strongly P-small). (b) Follows from (a) and Lemma 3.3 since F/F 0is infinite for a free group F. (c) Pick the smallest ksuch that the factor G(k)/G(k+1) is infinite (since G is infinite such a kmust exist). Then by (a) the (infinite) subgroup G(k)has a microscopic and strongly P-small set of generators (recall that microscopic implies 2-small). To see that this yields G∈ S2we need the following property. If Nis a normal subgroup of a group Gwith N∈ S2, then |G/N|<∞implies G∈ S2. Indeed, let Sbe a 2-small set of generators of N. Chose a finite transversal Fof N. Then the set S1=S∪Fis 2-small. Indeed, it suffices to see that if Sis 2-small, i.e., S∈S(N) and S−1·S,S·S−1are not large in Nthen also for any x∈Gthe union S1=S∪ {x}is 2-small in G. Now S1·S−1 1= (S·S−1)∪Sx−1∪xS−1∪ {1}. Now Sx−1∪xS−1∪ {1}is a small set, and S·S−1is not large (by Remark 2.22), therefore their union S1·S−1 1 cannot be large (as the difference between a large set and a small one is a large set [2, Theorem 1.4]). (d) The quotient G/G(n)is an infinite solvable group, so (c) and Lemma 3.3 apply. (e) Follows from (a) of Lemma 3.3 and Theorem 3.6.  Remark 3.8. (i) For the class S(b) and (e) were proved in [22, Proposition 6] and [22, Prop. 10] respectively. (ii) In the proof of (c) above, it is proved implicitly that the union of a 2small set and a finite set is still 2-small. We do not know if this property holds for microscopic sets as well (this holds true in abelian groups, but the above proof needs a non-abelian version of the property). (iii) A more careful analysis of the proof shows that every solvable group admits a skinny set of generators, i.e., a finite union of strongly left (or right) P-small sets, that are surely small. Note that there are small sets that are not skinny, since skinny sets have Banach measure zero, while small sets need not have this property. It follows from (e) that AoAut(A)∈ S when Ais an infinite abelian group (since Ais a normal subgroup of AoAut(A)). Groups with a small set of generators 343 3.2. Linear groups and permutation groups. According to Proposition 3.7 (e) the linear group GLn(K) has a small set of generators for every infinite field Kand n≥1 (its center is infinite). On the other hand, smaller linear groups (as the one of all upper (resp., lower) triangular matrices in GLn(K)) are solvable, hence again have small sets of generators (by Proposition 3.7). The following two instances cannot be obtained directly from that proposition. Theorem 3.9. The group G=SLn(A)has a small set of generators for every infinite euclidean domain Aand n > 1. Proof. Let T+(T−) denote the subgroup of all upper (resp., lower) triangular matrices in G=SLn(K) and let Ddenote the subgroup of all diagonal matrices in G. Then one can easily check that D,T−and T+, along with the finite set {π1, . . . , πs}of all matrices in G, having a single non-zero entry equal to ±1 on each row and column, generate the group G. Now the subgroups T−and T+ are solvable, hence small generated. The same holds for the abelian group D. Hence Lemma 3.1 applies to conclude G∈ S. Example 3.10. For some euclidean domains A(e.g., A=Z) the groups SLn(A) are actually finitely generated. This holds true when the additive group (A, +) is torsion-free and finitely generated (so isomorphic to Zmfor some m∈N). Indeed, for every i6=jand for every generator bof the additive group (A, +) consider the matrix αb ij =In+bEij, where Eij is the matrix with only one non-zero entry 1 placed at position ij. Then the matrices αb ij along with the matrices πigenerate SLn(A). For a short proof by induction note that starting with an arbitrary matrix ξ∈SLn(A) after appropriate permutation of the rows and columns (achieved by multiplication by various πi) and multiplication by matrices αb ij one can arrange to obtain from ξa matrix (aij) having the entry a11 6= 0 with minimal δ(a11), where δdenotes the Euclidean norm in A, all other entries on the first row or column are zero (as ξ∈SLn(A), the entry a11 is necessarily invertible). Now the inductive hypothesis applies to the minor obtained by removing the first row and the first column. Theorem 3.11. Let Xbe an infinite set, let S(X)be the group of all permutations of Xand let Sω(X)be the subgroup of S(X)of all permutations of finite support. Then S(X)∈ SPand Sω(X)has a set of generators that is microscopic and strongly P-small. Proof. First we prove that the group G=Sω(X) has a microscopic strongly Psmall set of generators. Obviously the set Tof all transpositions of Ggenerates G. We show that this set is microscopic and strongly P-small. Indeed, for any subset A⊆Xdenote by GAthe subgroup of all permutations with support contained in A. Since the set T=T−1is symmetric and invariant under conjugation, to check that Tis strongly P-small it suffices to check that it is right P-small. Indeed, S=T · T =T−1· T =T · T −1is not large, as every element of Sis either a 2-cycle, or a 3-cycle or a product of two disjoint 344 D. Dikranjan, U. Marconi, R. Moresco 2-cycles. Hence G6=S·GFfor any finite F⊆X. Now we can conclude with Remark 2.5 that Tis strongly left P-small and strongly right P-small, hence Tis small. Since an analogous argument shows that Snis not large for every n∈Nwe conclude that Tis also microscopic. The normal subgroup Sω(X) of S(X) belongs to SP, so Lemma 3.3 implies S(X)∈ SP. For a direct alternative proof of S(X)∈ S, one can apply Lemma 3.1 since Gis an infinite normal subgroup of S(X) of infinite index (any permutation of infinite order of Xwill witness it).  The famous theorem of Graham Higman, Bernhard Neumann and Hanna Neumann [18] says that every countable group is isomorphic to a subgroup of a 2-generated group. In this spirit we have: Theorem 3.12. Every group is a subgroup of a small generated group and a quotient of an M-generated group. Proof. Let Gbe an infinite group. It suffices to embed Gin the group S(G) of all permutations of the set G. For the second assertion it suffices to write Gas a quotient of a free group and apply Lemma 3.1  It is still an open question whether Scoincides with the class Gof all groups [22]. It follows from the above theorem that S=Gis equivalent to either of the following invariance properties of S: (i) stability under taking subgroups; (ii) stability under taking quotients. Another open question from [22] is whether every uncountable group Gadmits a small set of size |G|. If this is true for all groups of size ω1, then Shelah’s group G([25], let us recall that all proper subgroups of Gare countable) admits a small uncountable set S, then Swill generate G, so that Gwill be small generated. In the next section we study the question when G∈ S for groups Gthat admit a non-discrete group topology (close to being compact). 3.3. Some (compact-like) topological groups have small set of generators. It was shown in [22] that all non-metrizable compact groups belong to Sand it was asked whether the class Scontains all compact groups. Here we obtain further progress in this direction. Theorem 3.13. Scontains all compact groups that are not totally disconnected. Proof. Let Gbe a compact group that is not totally disconnected. Then c(G)6= {1}, hence it is an infinite normal subgroup of G. If it has infinite index, then G∈ S by Lemma 3.1. Hence it remains to consider the case when c(G) has finite index. In this case it suffices to prove c(G)∈ S, this will imply G∈ S. Hence it suffices to prove that every compact connected group belongs to S. Let us see first that every connected compact Lie group Gbelongs to S. Indeed, let Tbe a maximal torus of G. Then Ghas a finite number of Borel subgroups B1, . . . , Bncontaining Tand they generate G([21, §25.2, Corollary Groups with a small set of generators 345 B]). Since the subgroups Biare solvable, we can apply Lemma 3.1 and conclude G∈ S. We prove now that every compact connected group Gbelongs to S. If G is abelian then we apply Theorem 3.6. Otherwise, the quotient G/Z(G) is a product of simple connected Lie groups [26], [19, Th. 9.24]. In particular, some non-trivial quotient of Gis a compact connected Lie group, so that we can apply the first part of the argument and Lemma 3.3.  Following the above proof one can expect that the above theorem extends to all locally compact groups. Indeed, it suffices to prove that every connected locally compact group belongs to S. Since such groups are projectively Lie groups (i.e., inverse limits of Lie groups), it suffices to prove that every simple Lie group belongs to S. This makes us believe that this extension to the locally compact case is possible. As far as compact groups are concerned, the above theorem reduces the problem to totally disconnected compact groups, i.e., profinite groups. Let us recall, that profinite (pro-p) groups are inverse limits of finite (finite p-torsion) groups. A topological group Gis said to be topologically finitely generated if it has a dense finitely generated subgroup. We have no proof at hand for the following conjecture (see Remark 2.5 for a detailed comment). Conjecture 3.14. Every topologically finitely generated pro-pgroup belongs to Sfor every prime number p. The Frattini subgroup Φ(G) of a profinite group Gis the intersection of all maximal open subgroups of G. It is easy to see that Φ(G) is a closed normal subgroup of G. Conjecture 3.15. Every profinite group with trivial Frattini subgroup belongs to S. Since non-metrizable profinite groups already belong to S, to verify this conjecture one should consider only metrizable profinite groups Gwith Φ(G) = {1}. Then there will be countably many maximal open subgroups Mn. For every maximal open subgroup Mthe largest normal subgroup MGof Gcontained in Mis still open in G. Thus we obtain a countable family Nnof open normal subgroups, such that each one is an intersection of maximal open subgroups, and TnNn={1}. Since Gis compact, this implies that the open normal subgroups Un=Tn k=1 Nkform a local base at 1. Let us see now that a positive answer to these two conjecture would imply that every compact group is small generated. Theorem 3.16. If Scontains all topologically finitely generated pro-pgroups for every prime pand all profinite group with trivial Frattini subgroup, then all compact groups belong to S. Proof. Let Gbe a compact group. If Gis not totally disconnected Theorem 3.13 applies. Hence from now on we suppose that Gis totally disconnected, 346 D. Dikranjan, U. Marconi, R. Moresco i.e., profinite. Then its Frattini subgroup Φ(G) is pronilpotent ([24, Corollary 2.8.4]), hence isomorphic to the direct produt QpGpof its p-Sylow subgroups Gp([24, Proposition 2.3.8]). If the subgroup Φ(G) of Ghas infinite index, then consider the quotient G1=G/Φ(G) that has Φ(G1) = {1}(cf. [24, Proposition 2.8.2 (a)]), so G1∈ S by Conjecture 3.15 and G∈ S by Lemma 3.1. Therefore, from now on we assume that the subgroup Φ(G) has finite index, i.e., Φ(G) is open. Hence it suffices to prove that Φ(G)∈ S. If at least two of the groups Gp are infinite, then we are done by Lemma 3.1. If infinitely many groups Gpare non-zero then again Lemma 3.1 works. So it remains the case when precisely one of the groups Gpis infinite and there exists finitely many non-trivial groups Gq(with q6=p) and all of them are finite. In other words, Gpis open in G. Hence we can assume without loss of generality that Gis a pro-p-group. Then by [24, Lemma 2.8.7 (b)] the quotient group G/Φ(G) is an abelian group, hence G/Φ(G)∈ S in case it is infinite. Therefore, we can assume from now on that G/Φ(G) is finite, i.e., Φ(G) is open in G. Then Gis topologically finitely generated by [24, Proposition 2.8.10], so that Conjecture 3.14 applies now.  According to this theorem, if there exists a compact group without a small set of generators, then there exists also a profinite metrizable group Gwith this property, that is either a topologically finitely generated pro-pgroup, or has Φ(G) = {1}. Remark 3.17. Let us discuss here our grounds to believe that Conjecture 3.14 is true. When Gis a topologically finitely generated pro-pgroup, the topology of Gcoincides with its pro-finite topology (that has as typical neighborhoods of 1 all subgroups of finite index of G), since by Serre’s theorem [28, §4.3] every finite index subgroup of Gis open. Moreover, the Frattini series {Φn(G)}n∈N (defined inductively by Φ1(G) = Φ(G) and Φn+1(G) = Φ(Φn(G)) for n≥1) forms a fundamental system of neighborhoods of 1 in Gfor the pro-finite topology of G. Taking into account that Φ(G) = GpG0, where Gp={xp:x∈G} (cf. [24, Lemma 2.8.7 (c)]), this describes completely the topological group G in purely algebraic terms. It was proved by E. Zel0manov [30] (see also [24, Theorem 4.8.5c]) that the torsion finitely generated pro-pgroups are finite. So our conjecture holds true for torsion groups. On the other hand, it holds true also for all profinite groups of finite rank (a profinite group Gis said to be of finite rank dif every closed subgroup of Ghas at most dtopological generators, [28, Chap. 8]). Indeed, it is known that every profinite groups of finite rank Gadmits closed normal subgroups C≤Nsuch that Nhas finite index in G, Cis nilpotent and N/C is solvable. Now it is clear that G∈ S when Cis infinite (as then C∈ S as a solvable group so that Lemma 3.1 applies). When Cis finite, then Nand N/C are necessarily infinite, so that N/C ∈ S (by Proposition 3.7) and consequently G/C ∈ S. Now Lemma 3.1 applies again. Hence the conjecture remains to be proved only for the topologically finitely generated pro-pgroups of infinite rank. It should be noted that unlike the finite p-groups, the pro-pgroups may have a trivial center even in the case of very nice groups. For example, if K1 Groups with a small set of generators 347 denotes the subgroup 1 + pZppZp pZp1 + pZpof the linear group GL2(Zp), then the profinite group G=SL2(Zp)∩K1has trivial center. Nevertheless, one can prove explicitly that this group belongs to S. Indeed, one has to observe that the subgroups T+and T−of upper and lower triangular matrices, respectively, in Gand the finite set of all permutation matrices generate G. Then one notes also that both groups T+and T−are soluble, so that Lemma 3.1 applies. The topologically finitely generated free pro-pgroups belong to M⊆Sas they have infinite abelian quotients. A subset Aof a topological group is said to be (totally)bounded if for every non-empty open set Uof Gthere exists a finite set Fsuch that A⊆F·U,A is said to be σ-bounded if Ais a countable union of bounded subsets of G. Proposition 3.18. If Gis a σ-bounded topological group, then Gadmits a closed normal subgroup Nof infinite index with ψ(G/N)≤ω. If ψ(G)> ω, then Nmust be infinite and Gis small generated. Proof. If Gis discrete, then the conclusion is obvious. Otherwise, since the union of finitely many bounded sets is again a bounded set, there exists an increasing chain K1⊆K2⊆. . . ⊆Kn⊆. . . of bounded sets of Gsuch that G=S∞ n=1 Kn. We shall use in the sequel the following property of a bounded set K: for every neighborhood Vof 1 there exists a symmetric neighborhood Uof 1 such that Ux⊆Vfor every x∈K. Let U06=Gbe a symmetric open neighborhood of 1. Define inductively a decreasing chain U0⊇U1⊇. . . ⊇Un⊇. . . (1) of symmetric neighborhoods of 1 such that for each n (a) U2 n⊂Un−1and the inclusion is proper; (b) Ux n⊆Un−1for every x∈Kn. Note that a) yields Un⊆Un−1. Therefore the set N=TnUnis a closed subgroup of G. Let x∈G. Then there exists n∈Nsuch that x∈Kn. Since N=Tm≥nUm, one can easily see that Nx≤N. Therefore Nis a normal subgroup of G. Let us show now that the subgroup Nhas infinite index. Indeed, let f:G→G/N be the canonical quotient map. Since all inclusions in (a) are chosen to be proper, then also all inclusions f(Un)⊃f(Un+1) are proper as f(Un) = f(Un+1) yields Un⊆N·Un+1 ⊆U2 n+1, a contradiction. Therefore G/N is infinite and obviously ψ(G/N)≤ω. Finally, Nfinite would immediately imply ψ(G)≤ω. Consequently, Nis infinite and Gis small generated by Lemma 3.1 (a).  Remark 3.19. For an alternative argument in the case of a non-metrizable locally compact σ-compact group Gsee [22]. Note that every group with those properties satisfies also the hypothesis of our proposition as locally compact non-metrizable groups necessarily have uncountable pseudocharacter. The local compactness was exploited in [22] to get metrizability of the quotient G/N that in general has countable pseudocharacter, as the above proposition shows. 348 D. Dikranjan, U. Marconi, R. Moresco A topological group Gis totally minimal if every continous surjective homomorphism with domain Gis open. Theorem 3.20. A topological group G∈ S in each of the following eight cases: (a) Gis σ-bounded with ψ(G)> ω; (b) Gis σ-compact with ψ(G)> ω; (c) Gis SIN-group with ψ(G)> ω; (d) Gis precompact with ψ(G)> ω; (e) ([22, Theorem 14]) Gis compact and non-metrizable; (f) Gis pseudocompact and non-compact; (g) Gis totally minimal, precompact and non-metrizable; (h) Gis not totally disconnected and c(G)has infinite index (in particular, if it is not open in G). Proof. (a) By Proposition 3.18 we get an infinite normal subgroup Nof infinite index. Now Lemma 3.1 gives G∈ S. (b) Every compact set is bounded, thus (a) applies. (c) Let us recall that a SIN-group has (by definition) a base of invariant neighborhoods of 1 (so that SIN stands for small invariant neighborhoods). Now the proof of Proposition 3.18 can be carried out in this case to get a normal subgroup Nof infinite index with ψ(G/N)≤ω. Now the assumption ψ(G)> ω yields Nis infinite. (d) Since every precompact group is a SIN-group, we can apply (c). (e) Since every non-metrizable compact group has uncountable pseudocharacter, we can apply (d). (f) Every pseudocompact group is precompact ([5]) and every pseudocompact group of countable pseudocharacter is compact [4]. Therefore Gis precompact of uncountable pseudocharacter, so that (d) applies. (g) Indeed, if Gis compact then this follows from (e). Now assume that Gis not compact. Then the completion Kof Gis a compact non-metrizable group. Then Khas a closed normal subgroup Nof infinite index with ψ(K/N) = ω by Proposition 3.18. Since K/N is compact, we get χ(K/N) = ψ(K/N) = ω, so that K/N is metrizable. Therefore, Nis infinite as Kis not metrizable. Moreover, N∩Gis dense in Nby the total minimality of G([7, Theorem 4.3.3]), hence N∩Gis infinite. Moreover, N∩Ghas infinite index in G, since otherwise it would be an open subgroup of Gand consequently N=G∩Nwould be an open subgroup of K, a contradiction. Therefore, G∈ S by Lemma 3.1 (a). (h) c(G) is an infinite closed normal subgroup of G. Since it has infinite index in G, Lemma 3.1 (a) applies.  It follows from (h) that if there exists an infinite topological group failing to have a small set of generators then there exists also such a group Gthat is either connected or totally diconnected. In the former case the arc-component of Gis either trivial or has finite index (note that the the arc-component need not be closed even if Gis compact, hence this need not immediately imply, by connectendness of G, that Gis also arc-wise connected). Groups with a small set of generators 349 4. Questions. (i) Does Scontain all perfect groups? (ii) Does Scontain all hyperabelian groups of class ωwith finite factors (i.e., all groups Gsuch that T∞ n=1 G(n)={1}and all quotients G(n)/G(n+1) are finite)? (iii) Does every countable group have a small set of generators? (iv) For which commutative rings Kthe group SLn(K) has a small set of generators? Positive answers to (i) and (ii) yield that Scontains all groups. Indeed, if the derived series of Gstops after a finite number of steps, then G(n)is perfect for some n. If G/G(n)is infinite, then Lemma 3.3 and Proposition 3.7 apply to give G∈ S. Otherwise, G(n)is infinite and G∈ S iff G(n)∈ S (i.e., (i) applies). 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Received January 2002 Revised September 2002 Umberto Marconi, Roberto Moresco Dipartimento di Matematica Pura e Applicata, Universit`a di Padova, via Belzoni 7, 35131 Padova, Italy E-mail address:[email protected], [email protected] Dikran Dikranjan Dipartimento di Matematica e Informatica, Universit`a di Udine, via delle Scienze 206, 33100 Udine, Italy E-mail address:[email protected]