on locally graded barely transitive groups

Cent. Eur. J. Math. • 11(7) • 2013 • 1188-1196DOI: 10.2478/s11533-013-0240-x

Central European Journal of Mathematics

On locally graded barely transitive groups

Research Article

Cansu Betin1∗, Mahmut Kuzucuoglu2†

1 Department of Mathematics, Atılım University, Ankara, 06836, Turkey

2 Department of Mathematics, Middle East Technical University, Ankara, 06531, Turkey

Received 20 June 2012; accepted 2 September 2012

Abstract: We show that a barely transitive group is totally imprimitive if and only if it is locally graded. Moreover, we obtainthe description of a barely transitive group G for the case G has a cyclic subgroup 〈x〉 which intersects non-triviallywith all subgroups and for the case a point stabilizer H of G has a subgroup H1 of finite index in H satisfying theidentity χ(H1) = 1, where χ is a multi-linear commutator of weight w.

MSC: 20E25

Keywords: Locally graded groups • Locally finite groups • Quasi-finite groups • Splitting automorphism© Versita Sp. z o.o.

1. Introduction

A group G is said to have a barely transitive representation if G acts on an infinite set Ω transitively and faithfully, andevery orbit of every proper subgroup is finite. An abstract group is called a barely transitive group, if it has a barelytransitive representation. Equivalently, an infinite group G is barely transitive if and only if G has a core free subgroupH of infinite index in G and for every proper subgroup K of G, the index |K :K ∩H| is finite where H corresponds toa stabilizer of a point [16]. If G is a barely transitive group with a point stabilizer H, then every infinite subset of thetransversal set of H generates the group G. Therefore, every barely transitive group is countable.Recall that an infinite group G is called quasi-finite (or a Schmidt group) if all of its proper subgroups are finite. Onecan easily see that an infinite group G is barely transitive in its regular permutation representation if and only if Gis quasi-finite. For this reason, there is a close structural relation between barely transitive groups and quasi-finitegroups. In particular, the quasi-finite group Cp∞ is the only abelian barely transitive group in its regular permutationrepresentation. There are also non-abelian quasi-finite barely transitive groups; for example, the two generated infinitequasi-finite groups constructed by Ol’shanskii [20, Theorem 2] are barely transitive on their right regular representations.∗ E-mail: [email protected]† E-mail: [email protected]

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C. Betin, M. Kuzucuoglu

For a transitive permutation group (G,Ω), if Ω has no non-trivial G-block, then G is called primitive; otherwise it iscalled imprimitive. Following Neumann [18], if an ascending chain of proper blocks of an imprimitive group G does nothave a maximal block, then G is said to be totally imprimitive. By definition, a barely transitive group is a group ofpermutations acting transitively on an infinite set. Therefore it is natural to ask whether an arbitrary barely transitivegroup is totally imprimitive or not? One can see that Ol’shanskii groups of exponent p, for a fixed prime p, given in [21]are primitive barely transitive. So, examples of primitive barely transitive groups exist. These Ol’shanskii groups aresimple. Indeed, for every normal subgroup N of a transitive group G, every N-orbit is a G-block [8, Theorem 4.2].Therefore, any primitive barely transitive group is simple. It is known that a locally finite barely transitive group cannotbe primitive [16]. For the primitivity of an arbitrary barely transitive group, we have the followingTheorem 1.1.A barely transitive group is totally imprimitive if and only if it is locally graded.

Recall that a group G is called locally graded if every finitely generated subgroup has a proper subgroup of finite index.Clearly, every locally finite group is a locally graded group. In fact, as we noted in [6], a barely transitive group islocally graded if and only if it is infinitely generated. This will be used freely in sequel.Question.Let G be a locally graded barely transitive group. Is G locally finite? In particular, can we provide some conditionsunder which G is locally finite?In [6], we have shown that if a point stabilizer of a locally graded barely transitive group G is permutable or (locallynilpotent)-by-soluble, then G is locally finite. Also, it is easy to see that a locally graded barely transitive group witha locally finite point stabilizer or with a point stabilizer of finite exponent, is locally finite.In this paper, the results given in [6] are generalized via multilinear commutators.Definition.Let x1, x2, . . . be group variables. Multilinear commutators of weight 1 in the variables xi are the variables xi themselves.By induction, multilinear commutators of weight w > 1 in the variables xi are commutators of the form χ = [χ1, χ2],where χ1 and χ2 are multilinear commutators with disjoint sets of variables of weight w1 and w2 for w = w1 + w2.A group satisfies the multilinear commutator identity χ = 1 if and only if the commutator verbal subgroup χ(G) obtainedby substituting the group in place of the variables in the commutator χ , is trivial.Theorem 1.2.If G is a barely transitive group with a point stabilizer H, having a subgroup H1 of finite index in H satisfying theidentity χ(H1) = 1 where χ is a multi-linear commutator of weight w, then one of the following holds:(A) G is locally finite and satisfies a multi-linear commutator identity derived from χ of weight at most 2w. In particular,

if H1 is soluble of derived length d, then G is soluble of derived length ≤ d+ 1.(B) G is a non-locally finite group with a locally finite maximal normal subgroup M such that G/M is infinite simple.Moreover, either(Bi) G/M is quasi-finite, or(Bii) M has a characteristic subgroup C satisfying χ and M/C is a finite abelian group.

Another tool which provides local finiteness is having a splitting automorphism. Let p be a prime number. An auto-morphism φ of a group G is called splitting of order p if φp = 1 and ggφgφ2 · · ·gφp−1 = 1 for all g in G. In the nextproposition we give a sufficient condition for a locally graded barely transitive group to be locally finite.

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Proposition 1.3.Let G be a locally graded barely transitive group. If G has a splitting automorphism of order p where p is prime, thenG is locally nilpotent. In particular, G is locally finite.

A group with an infinite derived subgroup in which every proper subgroup has a finite derived subgroup is called aMiller–Moreno group or a group of Miller–Moreno type. Bruno and Phillips showed that a locally graded Miller–Moreno group is Chernikov, in particular locally finite [7, Theorem 2]. Ol’shanskii groups can be given as examplesof barely transitive Miller–Moreno type groups. But they are not locally finite. Here we show that barely transitiveMiller–Moreno type groups have similar properties like Ol’shanskii groups; namely, we proveProposition 1.4.Every barely transitive group of Miller–Moreno type is finitely generated and quasi-finite.

In the second part of the paper we consider simple barely transitive groups. It is well known that if a group G has alocal system of simple subgroups, then it is simple; see [13, Theorem 4.4] or [11, Lemma 3.1]. On the other hand, thereare non-simple groups which are unions of their proper non-abelian simple subgroups; see [10, Theorem C]. Let S be theclass of groups which are unions of their non-abelian simple subgroups. One may ask “Which S-groups are simple?”For barely transitive groups we answer this question in the affirmative. Namely,Proposition 1.5.Every barely transitive S-group is simple.

Observe that the above question makes sense if the group is a union of proper simple subgroups. In this case, our proofshows that G is a union of finite simple subgroups.Corollary 1.6.Let G be a locally graded barely transitive group with a point stabilizer H. If H is an S-group, then G is locally finite.

Proposition 1.7.Let H be a point stabilizer of a barely transitive group G. Then any minimal normal subgroup of H is finite. In particular,if G is simple, then H is finite or H has no minimal normal subgroups.

The question of Hartley, “Does there exist a torsion-free barely transitive group?” is still open. Since by [17, Proposi-tion 1] every proper normal subgroup of a barely transitive group is locally finite, if a torsion-free barely transitive groupexists, then it must be simple. Ol’shanskii has constructed torsion-free groups satisfying for any non-identity x and yin G, 〈x〉 ∩ 〈y〉 6= 1; see [22, Theorem 31.4, p. 338]. Such groups are not simple so they do not have barely transitiverepresentations. Simple torsion-free groups in which any two subgroups intersect non-trivially were constructed byObraztsov in [19], but those groups do not have barely transitive representations as every proper subgroup of such agroup has a non-trivial FC-center, which is impossible by [17, Corollary]. For a barely transitive group G with anelement x satisfying 〈x〉 ∩ 〈y〉 6= 1, for any non-identity y ∈ G, we have a complete description of G.Theorem 1.8.Let G be a barely transitive group. If there exists x ∈ G satisfying for any non-identity y in G, 〈x〉 ∩ 〈y〉 6= 1, then G isperiodic and either G is isomorphic to Cp∞ for some prime p or G is a finitely generated quasi-finite group with a trivialpoint stabilizer. In the latter case, Z (G) 6= 1 and G/Z (G) is simple, and π(G) (the set of prime divisors of the orders ofelements of G) is finite.

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Barely transitive groups given as in the second case of Theorem 1.8 exist by constructions of Ol’shanskii. Recall thathe proved in [22, Theorem 28.1] that for sufficiently large prime p, there exist infinite groups G(∞) all of whose propersubgroups are cyclic of prime order p.Theorem 1.9 (Ol’shanskii, [22, Theorem 31.8]).For any sufficiently large prime p and any integer k ≥ 0, there is an extension K of the group G(∞) of exponent pby a cyclic central subgroup D of order pk such that every proper subgroup of K is cyclic and either contains D or iscontained in D.

In particular, there exist barely transitive quasi-finite groups with non-trivial center and moreover, the intersection ofany two non-trivial subgroups is non-trivial.We note that [2, Theorem 4.1] has a flaw in its proof. It contradicts our Theorem 1.8. In fact, the groups in Theorem 1.9form a counterexample to the statement of [2, Theorem 4.1].In the sequel, for basic properties of barely transitive groups we freely apply the results of [16, 17].2. Proofs

Lemma 2.1.(i) A locally graded barely transitive group cannot be a primitive permutation group.(ii) There exists no 2-transitive barely transitive group.

Proof. (i) In a primitive permutation group, the stabilizer of any point is a maximal subgroup; see [9, Corollary 1.5A].But by [1, Lemma 2.1] in a locally graded barely transitive group, no point stabilizer is contained in a maximal subgroup.Hence a locally graded barely transitive group cannot be primitive.(ii) Assume that (G,Ω) is a 2-transitive barely transitive group. Take any element α in Ω. Then Gα is transitiveon Ω \ α; see [8, Theorem 3.13]. Then for any point β of Ω \ α, the Gα orbit of β is Ω \ α which contradicts thefiniteness of orbits of proper subgroups.The following proposition gives some information about the cardinality of the support of an element in a primitive barelytransitive group.Proposition 2.2.If G is a primitive barely transitive group, then for any non-trivial element g in G, |suppg| is infinite.

Proof. Let G be a primitive barely transitive group and 1 6= g ∈ G be such that |suppg| is finite. Since G is barelytransitive, ⋂α∈Ω Gα = 1. So, there exists α ∈ Ω such that g is not in Gα . As G is primitive, by [9, Corollary 1.5A],

Gα is maximal. Thus 〈Gα , g〉 = G. Let Ω1, . . . ,Ωn be the Gα orbits of Ω meeting non-trivially with suppg. Thensuppg = suppg−1 ⊆ Ω1 ∪ · · · ∪ Ωn. So, Ω1 ∪ · · · ∪ Ωn is invariant under 〈Gα , g〉 = G. Transitivity of G implies thatΩ = Ω1 ∪ · · · ∪Ωn. It follows that Ω is finite as each Ωi is finite. This is impossible. Hence, for any non-trivial elementg in G, |suppg| is infinite.Lemma 2.3.Let G be a barely transitive group which acts on Ω and has a chain of proper blocks ∆1 < ∆2 < . . . Then each properblock ∆i is finite. If this chain is infinite, then Ω = ⋃∞

i=1 ∆i and G = ⋃∞i=1 G∆i where the proper subgroup G∆i is the

set stabilizer corresponding to ∆i.

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Proof. Let α be a fixed element in ∆1. Then Gα ≤ G∆i for all ∆i in the given chain. Since ∆i 6= Ω and G is transitiveon Ω, G∆i G and |G∆i :Gα | < ∞ by bare transitivity. Note that G∆i acts transitively on ∆i. Therefore we have|∆i| = |G∆i :Gα | <∞.If this chain is infinite, then ⋃∞

i=1 ∆i is an infinite block for G. Since each proper block is finite, ⋃∞i=1 ∆i = Ω.Let g ∈ G and αg = β for some β in Ω. Then there exists k ∈ N such that α, β ∈ ∆k . Then β ∈ ∆k ∩ ∆kg and so∆k = ∆kg. Hence g ∈ G∆k. It follows that G = ⋃∞

i=1 G∆i.Proof of Theorem 1.1. Assume that G is a totally imprimitive barely transitive group. Then, by Lemma 2.3,G = ⋃∞

i=1 G∆i where ∆1 < ∆2 < . . . is a chain of blocks for G. Assume that G is finitely generated. Say G =〈x1, x2, . . . , xn〉 for some xi ∈ G. As G = ⋃∞

i=1 G∆i, each xi is contained in G∆ki for some natural number ki. Saykj = max k1, k2, . . . , kn. Then, G = 〈x1, x2, . . . , xn〉 ≤ G∆kj G which is impossible. Thus G is infinitely generatedand hence G is locally graded.Conversely, suppose that G is a locally graded barely transitive group. If G is locally finite, then G = ⋃∞

i=1 Ni whereNi E G and N1 N2 . . . Take α ∈ Ω. By [8, Theorem 4.2], αNi is a block for G. If αNi = αNj for all j ≥ i, thenΩ = αG = αNi which is not possible as every proper subgroup has finite orbits. Hence G has a strictly increasinginfinite chain of blocks for G, i.e. G is totally imprimitive.Assume that G is not locally finite. By Lemma 2.1 (i), G cannot be primitive. Suppose that G is almost primitive whichmeans that G has a finite chain of blocks, say ∆1 < ∆2 < . . . < ∆n−1 < ∆n = Ω, such that ∆j+1 is the minimal blockcontaining ∆j for all j in 1, 2, . . . , n − 1. By Lemma 2.3, each proper block ∆j is finite. Set ∆ = ∆n−1 and constructΣ1 = ∆g : g ∈ G. As ∆ is finite and Ω is infinite, Σ1 is an infinite set. Since G acts on Ω transitively, so does it on Σ1.The fact that every proper subgroup of G has finite orbits on Ω implies that every proper subgroup of G has finite orbitson Σ1. If G is simple, then ⋂

x∈G Gx∆ = 1 and so the action of G on Σ1 is faithful. It follows that (G,Σ1) is a locallygraded barely transitive group. Suppose that (G,Σ1) has a proper block, say Γ. Without loss of generality, we mayassume ∆ ∈ Γ, see [9, p. 12]. By Lemma 2.3, Γ is finite; say Γ = ∆,∆g1, . . . ,∆gk for some natural number k . Then theset ∆ ⋃ ⋃k

i=1 ∆gi is a proper block for (G,Ω) which contradicts the maximality of ∆. Hence (G,Σ1) is a primitive locallygraded barely transitive group, which is not possible. Hence (G,Ω) is a locally graded barely transitive group that isneither locally finite nor simple. So, G has a non-trivial maximal normal subgroup, say M. Let α ∈ ∆1. As MG, αMis a block for G that contains α . Since ∆j+1 is a minimal block containing ∆j , there exists i ≤ n− 1 such that ∆i = αM.Then, M ≤ ⋂x∈G Gx

∆n−1 G. Since M is a maximal normal subgroup, M = ⋂x∈G Gx

∆n−1. Set Σ2 = ∆n−1x : x ∈ G.Now, by [16, Lemma 2.5], (G/M,Σ2) is a barely transitive group with a point stabilizer HM/M. Moreover, G/M is locallygraded because if it were finitely generated, say G/M = 〈x1M, . . . , xtM〉, then G = 〈x1, . . . , xt ,M〉. As two propersubgroups of G generate a proper subgroup [6, Remark], we have G = 〈x1, . . . , xt〉. This contradicts our assumption thatG is locally graded.Lemma 2.4.Let G be a locally graded barely transitive group with a point stabilizer H and a maximal normal subgroup N. Thenthe FC-center of H is contained in N. In particular, H is not an FC-group.

Proof. Let G be a locally graded barely transitive group and N be a maximal normal subgroup of G. Then, by[17, Proposition 1], G is not locally finite. If H has no FC-element, then we are done. Assume that the FC-center of His non-trivial. Let M = g ∈ G : |H :CH (g)| < ∞. Then, by assumption, M 6= 1 and, by [17, Proposition 2], M is anormal subgroup of G. If M = G, we claim that G is a minimal-non-FC group. Indeed, by bare transitivity, G has nosubgroup of finite index, so G is not an FC-group. Let K be a proper subgroup of G and t in K . As M = G, we have|H :CH (t)| <∞ and so |H∩K : CH (t)∩K | <∞. It follows that |H∩K : CK (t)∩H| <∞. Hence |K :CK (t)| <∞ and Kis an FC-group. So, G is a locally graded minimal-non-FC group. By [25, Lemma 8.14], G is locally finite which is acontradiction. Thus, M 6= G. Then, 〈M,N〉 is a proper normal subgroup of G; see [6, Remark]. Hence M 5 N. So, theFC-center of H is contained in N. If H is an FC-group, then H = H ∩M 5 N. As N is locally finite, H is locally finite.Then, G is locally finite which is a contradiction.

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Corollary 2.5.Let G be a simple locally graded barely transitive group and let H be a point stabilizer. Then the FC-center of H istrivial.

Corollary 2.6.Let G be a simple barely transitive group and let H be a point stabilizer. Then either H is finite or the FC-center of His trivial.

Proof. If G is finitely generated, then the result follows from [4, Corollary 4.7]. If G is infinitely generated, then G islocally graded and the result follows from Lemma 2.4.Proof of Theorem 1.2. By [6, Lemma 2.2] we may assume that G is a barely transitive group with a point stabilizerH satisfying the multilinear commutator identity χ of weight w. If G has no maximal normal subgroups, then G is aunion of proper normal subgroups and hence by [17, Proposition 1], G is a locally finite group. Let N be a proper normalsubgroup of G. Then |N :N∩H| is finite. Then, by [15, Theorem 1], there exists a characteristic subgroup C of N suchthat χ(C ) = 1 and |N :C | is finite. Then G/C acts on the finite group N/C . Since G has no subgroup of finite index,N/C is in the center of G/C . Then N satisfies the multilinear commutator identity for a commutator of weight 2w. Thisis true for all normal subgroups of G hence G satisfies the multilinear commutator of weight 2w. In particular, if H isabelian then G is metabelian and if H is soluble of derived length d, then G is soluble of class ≤ d+ 1.So we may assume that G has a maximal normal subgroup M. Then G/M is a simple barely transitive group with a pointstabilizer HM/M. If HM/M is finite, then G/M is a quasi-finite group and we are in case (Bi). So we may assume thatHM/M is an infinite group satisfying χ . Now |M :M ∩H| <∞, then again by [15, Theorem 1], M has a characteristicsubgroup C of finite index satisfying the same multilinear commutator identity χ . Then as above, M/C is central in G/Cand hence abelian.The groups constructed by Hartley form an example to (A), the groups constructed by Ol’shanskii form an example to (Bi).For the (Bii) part of the theorem we do not have examples but private communication with Ol’shanskii indicates that itshould be possible to construct extensions of Ol’shanskii groups of type G(∞) in Ol’shanskii’s notation.Proof of Proposition 1.3. Let φ be a splitting automorphism of G of order p. Take a finitely generated subgroupK of G and set K1 = 〈K,Kφ, . . . , Kφp−1〉. Then, K1 is a φ-invariant finitely generated subgroup of G. Since G isinfinitely generated, K1 is proper in G, and so, K1 is residually finite. Take any a in K1. Then, there exists NaEK1such that a /∈ Na and |K1/Na| is finite. Set Na1 = Na ∩ Nφ

a ∩ · · · ∩ Nφp−1a . Then, Na1 is a φ-invariant normalsubgroup of K1 of finite index. Hence, K1/Na1 is a finite group which admits a splitting automorphism of order p. Then

K1/Na1 is nilpotent of nilpotency class depending on the number of generators of K1/Na1 (consequently, it dependson the number of generators of K1) and p; see [12, Theorems 2 & 3] and [14, 6.4.2, 7.2.1]. Set Ψ: K1 → Cra∈K1K1/Na1with Ψ(k) = (kNa1 )a∈K1 . The map Ψ is a homomorphism with kernel ⋂a∈K1 Na1 = 1. Hence K1 can be embeddedin Cra∈K1K1/Na1 which is nilpotent. As K ≤ K1, K is nilpotent and so G is locally nilpotent. An infinite locally nilpotentgroup cannot be simple; see [23, 12.5.2]. So, G has a non-trivial proper normal subgroup, say M1. Then G is the unionof proper normal subgroups Mi, i.e., G = ⋃∞

i=1 Mi. By [17, Proposition 1], G is locally finite.Proof of Proposition 1.4. Let G be a barely transitive group of Miller–Moreno type. Then, by definition, G′ isinfinite and every proper subgroup of G has a finite derived subgroup. Since a group with a finite derived subgroup isan FC-group by [25, Theorem 1.1] or [23, 14.5.11], and G has no subgroup of finite index, we have G is an FC-groupif and only if G is abelian. But then G′ = 1. Hence G is a minimal non-FC-group. Let H be a point stabilizer of G.|Z (G) :H∩Z (G)| <∞ implies |Z (G)| is finite.Assume that G is locally finite. Then G = ⋃∞

i=1 Ni where Ni is a proper normal subgroup of G. Then each N ′i is finite.So, N ′i ≤ Z (G) for all i. Then, G′ = ⋃∞i=1 N ′i ≤ Z (G) and so |G′| ≤ |Z (G)| <∞ which is a contradiction. Hence G cannotbe locally finite. In [5] it is proven that a non-perfect barely transitive group is locally finite. So G is perfect. Then by[3, Theorem 1], either G is two generated and G/Z (G) is simple or G/Z (G) is an infinite non-abelian quasi-finite group.

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Assume that G is a two generated group such that G/Z (G) is simple. We need to say that G is quasi-finite. As G/Z (G)is simple, Z (G) is the maximal normal subgroup of G. Then, by [16, Lemma 2.5], G/Z (G) is a barely transitive groupwith a point stabilizer HZ (G)/Z (G). Then, by Corollary 2.6, HZ (G)/Z (G) is either finite or has trivial FC-center. SinceH is an FC-group, HZ (G)/Z (G) is an FC-group too. So, the first case, namely HZ (G)/Z (G) is finite, remains. Thus,|H| = |H : 1| = |H :H∩Z (G)| = |HZ (G) :Z (G)| is finite. Now let K be any proper subgroup of G. Then, K ∩ H and|K : K ∩H| are finite and so K is finite as required.Consider the last case in which G/Z (G) is quasi-finite. Since |Z (G)| is finite we have G is quasi-finite. If G is infinitelygenerated, then each finitely generated subgroup of G is finite, which is impossible as shown above.Proof of Proposition 1.5. Assume that G is a barely transitive S-group that is not simple. Then G has a non-trivial normal subgroup N and G = ⋃

S∈Σ S where Σ is the set of non-abelian simple subgroups of G. Take any S ∈ Σ.As G is not simple, S is a proper subgroup and so residually finite. A residually finite group is simple only if it is finite.So, S is finite and hence G is periodic.If N is finite, then N ≤ Z (G) as G has no subgroup of finite index [16, Lemma 2.2]. Take any x in N. Then there existsS in Σ such that x ∈ S. So, N ∩ S is a non-trivial normal subgroup of S. As S is simple, S = S ∩ N but this is notpossible as S is non-abelian.So, N is infinite. Assume that N is a proper subgroup of G. Then N is residually finite. In particular, N is locallygraded. As every normal subgroup of an S-group is an S-group [10], N is a periodic locally graded S-group. So, N issimple, see [10, Theorem A]. But this is not possible as N is an infinite residually finite group. Thus N = G.Proof of Corollary 1.6. Assume that G is a locally graded barely transitive group with a point stabilizer H.Suppose that H is an S-group. Then H = ⋃

S∈Σ S where Σ is the set of non-abelian simple subgroups of H. As everyelement in Σ is a proper subgroup of G, each S ∈ Σ is a residually finite simple group. Therefore each S ∈ Σ isfinite. Hence, H is periodic. As every periodic, locally graded S-group is simple [10, Theorem A], H is simple. But H isresidually finite. Hence, H is finite and so, G is locally finite.Recall that there exists no simple locally finite barely transitive group by [11]. Therefore, by Proposition 1.5 andCorollary 1.6, we deduce that if G is a locally graded barely transitive S-group, then a point stabilizer of G cannot bean S-group.Proof of Proposition 1.7. Let K be a minimal normal subgroup of H. Take an element x ∈ K . Then by residualfiniteness of H, there exists a normal subgroup Nx of H such that |H/Nx | < ∞ and x /∈ Nx . As K is minimal normaland x /∈ K ∩Nx , we have K ∩Nx = 1. Then,

K ∼= K/(K ∩Nx ) ∼= KNx/Nx ≤ H/Nx .

Therefore K is finite. We have a homomorphism φ : H → Aut(K ) with Kerφ = CH (K ). Then |H/CH (K )| ≤ |Aut(K )| <∞.In other words, K ≤ FC (H). Assume that G is simple. If H is not finite, then by Corollary 2.6, K ≤ FC (H) = 1.Proof of Theorem 1.8. Since for every nontrivial element y ∈ G, 〈x〉∩〈y〉 6= 1, we obtain that either G is periodicor torsion-free.Assume that G is periodic and x has order n. If G is infinitely generated, then any finitely generated subgroupGl = 〈g1, . . . gl, x〉 is a proper subgroup of G. Hence Gl is residually finite, and hence there exists a normal subgroup Mlsuch that Gl/Ml is finite and 〈x〉 ∩Ml = 1. By assumption, the equality 〈x〉 ∩Ml = 1 implies that Ml = 1 and henceG is locally finite. Then by [17], G is a union of an increasing sequence of proper normal subgroups Ni for i = 1, 2, . . .Therefore, there is a natural number k such that x is an element of Ni for all i ≥ k . Since Ni is residually finite, by theabove observation we have Ni is finite for all i ≥ k . As any finite normal subgroup of G is central [16, Lemma 2.2], G isabelian and isomorphic to Cp∞ for some prime p [16, Lemma 2.3].Then we may assume that G is finitely generated. The subgroup 〈x〉 is a finite cyclic subgroup of order n. If G hasan element y of order p, where p is a prime, then 〈x〉 ∩ 〈y〉 6= 1 implies that p |n. Let p1, p2, . . . , pl be the primes

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dividing n. Then only finitely many primes p1, p2, . . . , pl divide the order of elements of G. Let |〈xm〉| = p for some primep ∈ p1, p2 . . . pl and g be an element of order p in G. Then 〈g〉 ∩ 〈xm〉 6= 1 implies that 〈g〉 = 〈xm〉. In particular,〈xm〉 is the only subgroup of order p in G. Therefore, 〈xm〉 is a finite normal subgroup of G and hence, it is in the centerof G. Let H be a point stabilizer and let h ∈ H be an element of order p. Then h ∈ 〈xm〉 ≤ Z (G), which is not possibleas ⋂

x∈G Hx = 1. So, H = 1 and G is a barely transitive group with right regular representation. Since G is not locallyfinite, G′ = G [5]. By Grün’s Lemma, Z (G/Z (G)) = 1. If N/Z (G) is a normal subgroup of G/Z (G), then N is normal in Gand, moreover, as H = 1, N ≤ Z (G) and so G/Z (G) is simple and G is a finitely generated quasi-finite group.Now assume that G is a torsion-free group. Set Cx = ⋃∞i=1 CG(xi). Then Cx is a subgroup of G. Indeed, if a, b ∈ Cx ,then there exist natural numbers m and n such that a ∈ CG(xn) and b ∈ CG(xm). Then ab and a−1 are in CG(xnm) ≤ Cx .For any g1, g2 ∈ G, we have 〈x〉 ∩ 〈gi〉 6= 1. Then, xni ∈ 〈gi〉 for some natural number ni. So, 1 6= xn1n2 ∈ 〈g1〉 ∩ 〈g2〉.That is, intersection of any two non-trivial subgroups is non-trivial.Let g be a non-trivial element of G. Then 〈x〉 ∩ 〈g〉 6= 1 implies gn = xm for some m,n. Then g commutes with xm.Hence g ∈ CG(xm) and Cx = G. In fact, one can show with the same argument that Cy = G for all y ∈ G. Since G issimple, for any m ∈ N, the group CG(xm) is a proper subgroup. Then for any n ∈ N, the union of the infinite ascendingchain CG(x) ≤ CG(x2n) ≤ . . . ≤ CG(xin) ≤ . . . contains Cx = G. It follows that every finitely generated subgroup iscontained in CG(xin) for some i ∈ N. Hence G is locally graded.Now, G is a locally graded torsion-free barely transitive group. So every finitely generated subgroup is a residually finitegroup and the intersection of any two proper subgroups is nontrivial. Then, by [24, Theorem], every finitely generatedsubgroup is finite, but this is impossible as G is torsion-free.

Acknowledgements

The authors thank the referees for recommending various improvements concerning exposition of the article.

References

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