On Certain Characterizations of Barely Transitive Groups
AHMETARIKAN(*) - NADIRTRABELSI(**)
ABSTRACT- In the present work we study perfect barely transitive groups in which a point stabilizer has non-trivial Hirsch-Plotkin radical and give characterizations of these groups. We also prove that a perfect barely transitive group has two non-trivial proper subgroups which intersect trivially.
1. Introduction.
Let G be a permutation group on an infinite setV. If G acts transitively on Vand every orbitof every proper subgroup ofGis finite, thenGis called a barely transitive group. This class of groups was considered for the first time by Hartley in connection with the Heineken-Mohamed groups. It is well- known thatGcan be represented as a barely transitive group if and only ifG has a subgroup Hof infinite index such thatCoreGH T
g2GHg h1iand jM:M\Hjis finite for every proper subgroupMofG. The subgroupHis called a pointstabilizer inG.
Barely transitive groups are mostly considered in the locally finite case and it is well-known that these groups are locally nilpotentp-groups for a primep. In recent years some work has appeared on the non-locally finite case. Almost all such results can be found in the survey article [12].
In [2] locally graded simple non-periodic barely transitive groups are considered and it is proved that all point stabilizers have trivial Hirsch- Plotkin radical (that is, the maximal locally nilpotent normal subgroup is trivial). In the present paper we give certain characterizations of a perfect
(*) Indirizzo dell'A.: Gazi UÈniversitesi, Gazi EgÏitim FakuÈltesi, Matematik EgÏitimi Anabilim Dali
.
, 06500 Teknikokullar, Ankara, Turkey.E-mail: arikan@gazi.edu.tr
(**) Indirizzo dell'A.: Department of Mathematics Faculty of Sciences University Ferhat Abbas, Setif 19000, Algeria.
E-mail: nadir_trabelsi@yahoo.fr
simple barely transitive group with a point stabilizer whose Hirsch-Plokin radical is non-trivial.
LetXbe a class of groups. If every proper subgroup of a groupGis an X-group butGitself is not, thenGis called a minimal non-Xgroup.
Our main results are as follows:
THEOREM1.1. Let G be a simple barely transitive group with a point stabilizer H. If the Hirsch-Plotkin radical F(H)of H is non-trivial, then G is a periodic finitely generated minimal non-(locally finite) group. If in addition H is an FC-group, then G is a two-generated minimal non FC- group.
COROLLARY 1.2. Let G be a perfect barely transitive group with a point stabilizerH. IfH is soluble, thenGis a finitely generatedperiodic group with every proper subgroup locally finite. If in additionHis anFC- group, thenGis a two-generatedminimal non-FC group.
COROLLARY 1.3. Let G be a perfect barely transitive group with a point stabilizerH. IfHis hypercentral, then eitherGis locally finite orG is a finitely generatedperiodic group with every proper subgroup locally finite. If in addition H is an FC-group, then G is a two-generated minimal non-FC group.
2. Certain Characterizations.
Throughout the paper we use the term ``infinitely generated'' in the sense that the group is not finitely generated.
The following lemma is well-known. We just here take care to ensure that we have a proper perfect subgroup in the infinitely generated case.
LEMMA2.1. If G is a simple infinitely generatedgroup, then G con- tains a non-trivial proper perfect subgroup.
PROOF. We argue as in 1.B.3 Proposition of [9]. SinceGis non-abelian, there exist a;b2G such that c:[a;b]61. By assumption hcGi G.
Hence G has a finitely generated subgroup Y such that ha;bi hcYi.
Put Z: hc;Yi, thus we have c2 hcZi0 and hence hcZi hcZi0. Since
hcZi Z6G, we have thathcZiis a proper perfectsubgroup ofG. p
Now we give a characterization of perfect periodic non-locally finite barely transitive groups. We shall also see that the non-existence of perfect subgroups in the group plays a crucial role in determining the finitely generated structure. For example the non-existence of involutions in the group gives finite generation.
THEOREM2.2. Let G be a non-(locally finite) barely transitive group.
Then G has a finitely generatedsubgroup F such that hFGi G anda simple barely transitive homomorphic image G. Furthermore, if G is periodic and infinitely generated, then G is infinitely generatedand generatedby involutions, andis a union of a chain of infinite non-simple perfect subgroups generatedby involutions.
PROOF. Assume that G has no finitely generated subgroup F such thathFGi G. As in the first paragraph of proof of the Theorem of [2], G is countable and hence we can find a chain of finitely generated subgroupsFiof Gfori2Zsuch thatG S1
i1hFiGi. Since every proper normal subgroups ofGis locally finite, we see thatGis locally finite. But this is a contradiction. So Ghas a finitely generated subgroup F such that hFGi G.
LetHbe a pointstabilizer ofG. As every normal subgroup ofGdoes not containF,Ghas a maximal normal subgroupNandG:G=Nis a simple barely transitive group with a point stabilizerH.
Now assume thatGis periodic and infinitely generated. We first assume thatGis ap-group. IfGis not finitely generated, thenGcontains a proper non-trivial perfect subgroupPby Lemma 2.1. PutXCoreP(H\P), then P=X is a finite p-group by hypothesis. But since P is perfect, we have PXH. By arguing similarly, we obtain thatPHgfor everyg2Gas everyHgis also a pointstabilizer inG. Then
P\
g2G
Hg h1i:
But this is a contradiction. HenceGis finitely generated and thusGhas elementsy1;. . .;ys such thatG hy1;. . .;ysiN. Put F hy1;. . .;ysiand ECoreF(H\F). ThenF=E(F\N) is finite, sinceF is notequal toG.
SinceG=Nhas no proper subgroup of finite index, we haveFE(F\N).
Then GE(F\N)NHN. Butthen jG:Hj jN:N\Hj is finite, a contradiction.
SoGis notap-group. Now we have thatGis also infinitely generated by the above argument. Let Tbe the subgroup ofGgenerated by all 2-ele- ments inG. NowTis normal inGand hence eitherGTorT h1i. First assume thatT h1iand letPbe a proper non-trivial perfect subgroup of G. ThenPcontains a normal subgroupXHsuch thatP=Xis finite by hypothesis. Since G is periodic and has no involution, jP=Xj is odd and hence, by the Feit-Thompson theorem,PXH. Arguing as in the third paragraph we haveP h1i, a contradiction. Consequently,GT, i.e.Gis generated by 2-elements.
Letcbe an involution inG; thenhcGi Gand hence [c;G]G. SoGis generated by involutions, namely the conjugates ofcandGhas elements y1;. . .;yr such thatc[c;y1]. . .[c;yr]. Thus G has a finitely generated subgroup Y1 containing c such that hc;y1;. . .;yri hcY1i hc;Y1i 6G.
Now we have c2 hcY1i0 and hence Q: hcY1i hcY1i0. So Q is a proper perfectsubgroup ofGgenerated by involutions which are conjugates ofc.
We also have that for every finitely generated subgroup Z of G, Z is contained in a finitely generated subgroup U such that hcUi is perfect.
Hence sincehcGi G, there is a chain of finitely generated subgroupsYi ofGsuch thathcYiiis a perfect subgroup generated by involutions and
G [
i2Z
hcYii:
SinceGis notlocally finite andNis locally finite,Gis non-locally finite and hence only finitely many hcYii can be finite. Also there is a positive integer m such that for every n>m, H\ hcYni 6 h1i and hcYni 6H.
HencehcYniis non-simple for everyn>m. So the proof is complete. p This characterization shows that the structure of infinitely gener- ated barely transitive groups is very complicated. But in the rest of the paper we shall see that in many cases the tendency is to be finitely generated.
COROLLARY2.3. LetGbe a perfect periodic barely transitive group. If Gcontains no involution, thenGis finitely generated.
PROOF. Since G is periodic we see that no section ofG contains an involution. But ifGis infinitely generated, with the notation of Theorem 2.2, Gis generated by involutions, and we have a contradiction. p
3. Proof of Theorem 1.1.
PROOF OF THETHEOREM1.1. - By the Theorem in [2], we have thatG mustbe periodic and Gis not locally finite by [8]. Assume that Gis not finitely generated. Since G is countable (see [2]), we can write G fxi:i2Zg. DefineYn fx1;. . .;xng,Xn hH;Yni,RnCoreXnH as in [2] and [6]. As in the proof of Lemma 2.1 of [2],His notcontained in maximal subgroup ofGand henceXn 6Gfor every positive integern. We haveG S1
i1Yi,Rn/Xn andjXn=Rnjis finite.
We firstshow thatF(Rn)6 h1ifor alln1. For the contrary assume that F(Rn) h1i for some n2Z. Put PF(H), then PRn=Rn'P is finite. HenceHhas a non-trivial finite normal subgroup. This implies that jH:CH(P)jand thusjH:CH(g)jare both finite for everyg2P. This im- plies thatFCG(H)6 h1i, sinceHis inertinG(see [4] for the definitions).
Now by 2.2 of [4], we haveFCG(H)G. By the proof of Proposition 2 of [12],Gis now a minimal non-FCgroup. Itis well-known thatGis either a locally nilpotentp-group orGis quasi-finite by Theorem 1 of [3] and [10]. If Gis locally nilpotentp-group, then it is not simple, a contradiction. IfGis quasi-finite, then G is two-generated by [14] p. 97. But this is again a contradiction. Now we have thatF(Rn)6 h1iandPis infinite. Clearly we also have thatF(Xn)6 h1i. LetEbe a finitely generated subgroup ofG, then arguing as in the final paragraph of the proof of Lemma 3 of [6], we have E is nilpotent-by-finite, in particular Yn is nilpotent-by-finite for every positive integern. Now by [4],H=Pis anFC-group and thenH=Tis locally finite, whereT=PZ(G=P). ThusTis (locally nilpotent)-by-abe- lian. ClearlyH\Ynhas finite index inYnand thus it is finitely generated.
Since (H\Yn)T=TH\Yn=H\Yn\T is finite, H\Yn=T\Yn is fi- nite. PutTnT\Yn, t henjYn :Tnj jYn:H\YnkH\Yn:Tnjis finite.
This implies that Tn is finitely generated and thus Tn is nilpotent-by- abelian. AsYn is nilpotent-by-finite, it has a normal nilpotent subgroup Kn such that Yn=Kn is finite. Now Kn is a finitely generated nilpotent subgroup ofYnand clearly itis finite. ThusYnis finite for every positive integern. SinceGis the union ofYifori0,Gis locally finite. But this is a contradiction.
Now we may assume thatGis finitely generated. If His notanFC- group, then by [4] Theorem 1.4,H=Pis anFC-group. HenceH0=H0\Pis locally finite by [14] Theorem 4.32. Now by [4] Theorem 1.4 (ii),Pis ap- group for some primepand hence it is locally finite. This implies thatH0is locally finite. Since H is locally finite-by abelian, it is locally finite. As
jK:K\Hjfinite for every proper subgroupKofG,Kis locally finite, i. e.
every proper subgroup ofGis locally finite. Consequently,Gis a minimal non-(locally finite) group.
IfHis anFC-group, then as aboveGis a minimal non-FCgroup and hence Gis two-generated by [3] Theorem 1 and above similar arguments. p PROOF OFCOROLLARY1.2. - IfGis locally finite, then by Theorem 1 of [5] G is a p-group and Theorem 2 of [5] (or [1]) G cannotbe perfect, a contradiction. We also have thatGhas a maximal proper normal subgroup N such that GG=N is a simple barely transitive group with a point stabilizer H by Theorem 2.2. Since H is non-trivial soluble, we have F(H)6 h1iand hence by Theorem 1.1,Gis finitely generated and periodic with all proper subgroups locally finite. LetKbe a proper subgroup ofG andEbe a finitely generated subgroup ofK. SinceEN6G, we have that E=N\Eis finite and hence N\Eis finitely generated. As N is locally finite,N\Eis finite and thusEis finite. This means thatKis locally finite, i.e. every proper subgroup ofGis locally finite. The other properties follow
as in the proof of Theorem 1.2. p
The proof of Corollary 1.3 is also similar. Butitis notknown yet whether a locally nilpotent barely transitivep-group with all proper sub- groups hypercentral is not perfect.
COROLLARY 3.1. Let G be a simple barely transitive group with a point stabilizerH. IfCG(H)6 h1i, thenGis a periodic finitely generated minimal non-(locally finite) group.
PROOF. Let16a2CG(H), thenW:Hhai 6G. SinceGis simple, we have thatWis a pointstabilizer inGwitha2F(W). Hence the result fol-
lows by Theorem 1.1. p
4. Intersections of subgroups.
In this section we consider trivial intersections of proper subgroups of barely transitive groups. The classification in [15] is very useful for barely transitive groups, since every proper subgroup is residually finite.
THEOREM4.1. Let G be a perfect barely transitive group, then G has two proper non-trivial subgroups which intersect trivially.
PROOF. Assume that the assertion is false. First suppose that G is periodic. Then since every proper subgroup ofGis residually finite, by [15]
every proper subgroup ofGis isomorphic either to a finite group of (gen- eralized) quaternions or to a finite cyclicp-subgroup. Then clearly every proper subgroup ofGis finite.
IfGcontains a quaternion group, then since a quaternion group is a 2- group and proper subgroups ofGintersect non-trivially,Gis a 2-group. It is well-known that every infinite 2-group contains an infinite abelian sub- group, a contradiction. SoGcannot contain a quaternion group. SinceGis perfect,Gmust be two-generated. In this caseGhas a subgroupEof order pfor a primepwhich is contained in every proper subgroup ofG. ClearlyE is a normal subgroup of G, so for every proper subgroup H of G, CoreGH6 h1i. But this is a contradiction.
Now assume thatGis non-periodic. By hypothesisGmustbe torsion- free. LetKbe a proper subgroup ofG, thenKis residually finite by [2] p. 3 and hence K is isomorphic to a proper subgroup of (Q;) the additive group of rational numbers by [15]. So every proper subgroup of G is abelian. This implies that G is two-generated, say G ha;bi. Then
K: hai \ hbi 6 h1iby hypothesis. HenceKis a proper normal subgroup
ofG. But this gives a contradiction, sinceGis simple-see [13], p. 85. p Acknowledgments. Both authors are grateful to the referee for the careful reading and many useful comments.
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Manoscritto pervenuto in redazione il 6 maggio 2009.
