Inertial Automorphisms of an abelian Group

Inertial Automorphisms of an Abelian Group

ULDERICODARDANO(*) - SILVANARINAURO(**)

ABSTRACT- An automorphismsgof a group is inertial ifX\X^ghas finite index in bothXandX^gfor any subgroupX. We study inertial automorphisms of abelian groups and give characterization of them. In particular, if the group is periodic they have property thatX^hgi=Xhgiis bounded. We also study finitely generated groups of inertial automorphisms.

1. Introduction

Recently there has been some interest in groups whose subgroups are inert, that is commensurable to their conjugates (see [1], [7]). Here two subgroups X and Y of a group are told commensurable iff the index of X\Y in bothXandY is finite. Commensurability is an equivalence re- lation.

The study of the above groups yields to the consideration of inertial automorphisms, that is automorphisms mapping subgroups to commen- surable subgroups. In this paper we give characterization of these auto- morphisms of an abelian groupA. In Theorem 1 we show that whenAis torsion they reduce to so-calledalmost power automorphisms, that is au- tomorphisms with property (ap) below, which have been studied in [3] by Franciosi-de Giovanni-Newell. Further, we show that even in the non- periodic case almost power automorphisms coincide with automorphisms with the property (bp), which have been studied in [2] by Casolo.

(*) Indirizzo dell'A.: Dipartimento di Matematica e Applicazioni ``R.Caccioppo- li'', UniversitaÁ di Napoli ``Federico II'', Via Cintia - Monte S. Angelo, I-80126 Napoli,

E-mail: dardano@unina.it

(**) Indirizzo dell'A.: Dipartimento di Matematica e Informatica, UniversitaÁ della Basilicata, Via dell'Ateneo Lucano 10 - Contrada Macchia Romana, I-85100 Potenza, Italy.

E-mail: silvana.rinauro@unibas.it

More precisely, we consider following properties for an automorphismg of an abelian groupA. IfG is a group acting on a groupGandXG, as usual we denoteXG :ˆ \g2GX^gandX^G :ˆ hX^g jg2Gi.

(ap) 8XA jX=X_hgij51 (gis almost power);

(bp) 8XA jX^hgi=Xj51;

(in) 8XA jX=X\X^gj jX^g=X\X^gj51(gis inertial);

(bin)9k2N 8XA jX=X\X^gj jX^g=X\X^gj k.

By Theorem 1 we see that in the periodic case they correspond to the factthatgacts on a finite index subgroup (or modulo a finite subgroup) either as a power (elementary case) or pseudo-power automorphism (see Definition 1). Furthermore the above properties are equivalent to each of the following:

(d) there existsd2Nsuch thatjX=X_hgij dfor allXA.

(d) there exists d2Nsuch thatjX^hgi=Xj d for allXA.

Thus such ag is inertial in a ``bounded'' way. Moreover, all the relevant indeces are bounded each in terms of any other. We give also several ex- amples to show that many given bounds are best-possible.

Theorem 2 identifies inertial automorphisms of torsion-free abelian groups, while Theorem 3 treats the mixed case. It follows a characteriza- tion of ``bounded'' inertial automorphisms.

In the final section we apply the above results to finitely generated groups of inertial automorphisms.

Notation and Terminology

We use multiplicative notation for abelian group operation, except in Section 3. Most of our notation is standard and refers to [4] and [6]. In particular D(A), T(A), exp (A), Aut(A) denote the maximum divisible subgroup, the torsion subgroup, the exponent, the automorphism group of the abelian groupA, resp. Also, ifn2Z,A[n] is the subgroup of the ele- ments such thatxⁿˆ1 andp(n) denote the set of prime divisors ofn. Ifpa setof primes we denote byA_pthep-componentofAwhile byQ_pwe denote the group of rationals whose denominators can be written by a product of primes inponly.

Recall that a power automorphism is an automorphism mapping each subgroups into itself. Power automorphisms of an abelian non-periodic group are just the identity or the inversion map. Futher, power auto- morphisms of an abelianp-group are universal, that is have formx7!x^a, whereais an invertiblep-adic number.

If g2Aut(G) and CBA are such that g induces a power auto- morphism corresponding to the p-adic a on the p-group B=C we write g_jB=Cˆa. Clearly,B^{g a}C.

2. Periodic case

Let us start recalling that when dealing with torsion groups we can reduce top-groups.

PROPOSITION1. Let GˆDriGibe the direct product ofcoprime torsion groups andg2Aut(A). Thengis either(ap),(bp),(in)or(bin)iff it acts this way on finitely many G_i's and as a power on all the others. p Letus pointoutan easy factwhich involves some inertial automorphism that we can callelementary.

PROPOSITION2. Let A be an abelian group andg2Aut(A). Then:

i) Ifgis power on a finite index subgroup B, thengis power on a factor A=C withjCj jA=Bj.

ii) Ifgis power on a factor A=C with C finite, thengis power on a subgroup B withjA=Bj jCj.

PROOF. IfAis periodic, we can assumeAis ap-group and consider the p-adica such thatgˆa onBor on A=C, accordingly we are in the hy- potheses of statement (i) or (ii). For the non-periodic case in the same way let aˆ 1. Now denote by K and I the kernel and the image of the homomorphism g a, respectively. Then gˆa on BA iff BK and gˆaonA=Cif and only ifIC. ClearlyA=K'_gI. p Let us give a definition which will simplify statements and whose idea is in [2] and [3] already. It deals with non-elementary almost power auto- morphism, as we see in Proposition 3.

DEFINITION1. Let p be a prime, r;e2N and eˆp^e. An abelian p- group A is told(r;e)-bounded iff the largest divisible subgroup DˆD(A)of A has rankr and A=D has exponent dividing e.

Ifg2Aut(A)we say thatga pseudo-power automorphism ofA ifit is non-power and AˆDE where:

i) D is divisible with finite rank r, ii) E is infinite but has finite exponent e,

iii) gacts as a power automorphism on both D and E.

PROPOSITION3. Let p be a prime, r;e2Nand eˆp^e.

(i) Ifgis a pseudo-power automorphism ofan abelian(r;e)-bounded p-group, then for each XA we have

jX^hgi=Xj e^randjX=X_hgij e^r:

(ii) Ifp>2, there exists a pseudo-power automorphismgofan abelian (r;e)-bounded p-group A with a subgroup XA such that

jX^hgi=Xj ˆ jX=X_hgij ˆe^r:

PROOF. i) In the same notation as in Definition 1, ifX:ˆDX\XE, X:ˆ(D\X)(X\E) we have that X and X are g-invariantwhile the groupsX=X'X=X'(A\XB)=(A\X)'(AX\B)=(X\B) have rank rand exponente.

ii) Let A₁:ˆDV, where D is a PruÈfer p-group and Vˆ hvi has order p^e. Define g by g_jVˆ1 and g_jDˆa61 (mod p). Let X₁:ˆ hvdi, where d2D has order p^e. Write yˆv^md^mˆvⁿd^an2X₁\X^g₁, with 0m;n5p^e. Then nm and nma (mod p^e). Hence m(a 1)0 (mod p^e), so thatv^mˆ1 andX1\X₁^gˆ1.

LetnowAbe the direct product ofrcopies ofA₁by an infinite abelian group Y of exponent eand define gon each copy of A₁ as above and by g_jY ˆ1. Thengis pseudo-power onAand the statement holds whenXis the subgroup generated by thersubgroups corresponding toX1. p Notice that, even tough it is plain thatAsplits onDˆD(A), the fact thatgacts as a power on bothDandA=D(with finite rank and exponent, resp.) is not enough to guarantee thatgis either power or pseudo-power on A. To see this consider a groupAˆD hbiwhereDis a PruÈferp-group andbhas orderpwith 16ˆg_jDˆa1 (modp) andb^g ˆab, wherea2D has orderp. Clearlyg acts as a power on both Dand A=D. However, if AˆDEwithEˆE^g, thengacts trivially onE'g A=Dand therefore on the socleha;biofA, a contradiction.

On the other hand, if (i), (ii) and (iii) in Definition 1 hold with onlyA=D in the place ofE, thengis neverthless an almost power automorphism and there is a finiteCA and a subgroup Bsuch that g is either power or pseudo-power on bothA=C(see Theorem 2.11 in [2]) andB. The order ofC and the indexA=Bare restricted, as in the following statement.

PROPOSITION4. Let p be a prime, r;e2N, eˆp^e, s2Z anda a p- adic. IfA is an abelian(r;e)-bounded p-group andg2Aut(A)is such that g_jDˆa,g_A=Dˆs with D:ˆD(A), then

i) g acts as a power or pseudo-power on A=C where CˆD[pⁿ], with pⁿˆexp (ker_A=D(a s))and in particular

jCj ˆp^nre^r;

ii) gacts as a power or pseudo-power on a subgroup B, with jA=Bj e^r:

The above bounds are best-possible, as there exist abelian(r;e)-boun- ded p-groups A andg2AutA acting as power on DˆD(A)and A=D such that

iii) ifg acts as a power or pseudo-power on a factor A=C, then C contains D[pⁿ], whereg_jDˆaand pⁿˆexp (ker_A=D(a g)), hence

jCj p^nr;

iv) ifg acts as a power or a pseudo-power on a subgroup BA, then

jA=Bj e^r:

PROOF. i) LetKandEbe the kernels of the endomorphismsa sand g s of A, resp. Clearly ne and, even in case nˆe, we may assume K\DˆE\DˆD[pⁿ], as we can substitute s by s‡p^e. Obviously DA^{g s}, and since gˆs on no nontrivial images of D, we have that A^{g s}ˆD. As A=E'A^{g s}, we get AˆED. Thus g is pseudo-power on A=D[pⁿ].

ii) Let AˆDE and let A1ˆA[e]ˆD[e]E. Then gˆs on A₁=D[e]'_gA=D, so that gˆsonE₁A₁with index at mostjD[e]j ˆe^r. The statement holds withBˆD(E\E₁), asjA=Bj ˆ jE:E\E₁j e^r. iii) Let AˆA1 ArE, where for 0ir, AiˆDi heii, withDia PruÈferp-group,eiof ordereandEan infinite elementary abelian p-group. Letd_ibe an elementof ordereinD_ifor eachiandg2AutAsuch thatgˆaonDˆD₁ D_r,gˆsonEande^g_iˆe^s_id_i.

Letnow C be a g-invariantsubgroup of A such that A=Cˆ DC=CV=C and gˆs on V=C. For 1ir, there exist v_i2V and zi2Dsuch thateiˆvizi. Observe thatv^e_i 2C, asA=Dhas exponente, and so z^e_i 2C. Now, as gˆs on V=C, we get d_iz^s_i ^a2C and so d_i^{pe n}2C.

ThereforeCD[pⁿ].

iv) Let A1:ˆPV, where P is a PruÈfer p-group and Vˆ hvi has ordereˆp^e. Definegbygˆ1‡ponPandv^gˆvd, wheredis an element ofPof orderp^e. LetA⁰be the direct product ofrcopiesA_iofA1and define gonA⁰in the natural way componentwise. We show that:

() 8z2A⁰n1 z^gˆz) hzi \D6ˆ1:

In fact, ifrˆ1 we have that ifzˆvⁿd^mis such that the order ofd^mis not bigger than that ofvⁿ. Thennˆpⁿn1andmˆp^mm1, withp62p(n1m1) andm n. As z^gˆz, we getd^n‡pmˆ1, which impliesp^edividesnhence vⁿ ˆd^mˆ1 andzˆ1, a contradiction.

In the general case, zˆa₁. . .a_r, with a_iˆa^g_i 2A_i 8i, which implies thata_iˆ1 or a nontrivial powera^s_iⁱofa_iis inD, by the above. Choosing the biggestsiwe getthatz^siis inDn1. Thus () is proved.

LetnowAˆA⁰EwhereEis an infinite elementary abelianp-group and extendgbygˆ1 onE. Thengis power on bothDˆD(A) andA=D.

LetBbe a subgroup of finite index ofAsuch thatgacts as a pseudo- power on B. Then BˆDZ and gˆ1 on Z. By () we have that Z\A⁰ˆ1. Letzˆa⁰h2Zwitha⁰2A⁰andh2E. Thenz^p2A⁰\Zˆ1.

Then a⁰2D has order p. Hence z2DE and so BDE. Itfollows

jA=Bj e^r. p

Now we pointouta property of pseudo-power automorphisms corre- sponding to that in Proposition 2.

PROPOSITION 5. Let p be a prime, r;e2N, eˆp^e, A be an abelian (r;e)-bounded p-group andg2Aut(A), then

i) If g is pseudo-power on a finite index subgroup B, then g is pseudo-power on a factor A=C with bothjCj jA=Bje^randjCj jA=Bj². ii) If g is pseudo-power on a factor A=C with C finite, then g is pseudo-power on a subgroup B withjA=Bj jCje^r.

Consideration of the group in the proof of Proposition 4.ivshows that in Proposition 5.ii jA=Bj cannotbe bounded in terms of jCj only (take C:ˆD[p]).

PROOF. i) Let BˆDE where g is power on bothDand E. Also, AˆDE1 for someE1E. LetE2:ˆE^hgi₁ andC2:ˆD\E2ˆC₂^g. By applying Proposition 2 to the the groupE2=C2, we know thatgis power on a factorE2=Csuch thatjC=C2j jA=Bjand thereforegis pseudo-power on A=C.

On one hand, C2'E2=E has rank atmostr and exp (E2=E)ˆ exp (E1=E) divides e hence jC2j e^r. On the other hand, by applying Proposition 2 to the groupA=Ewe have thatgis power on a factorA=C3

withjC₃=Ej jA=Bj. HenceE₂E₁C₃, so thatjC₂j ˆ jE₂=E₁j jC₃=Ej jA=Bj.

ii) LetA=CˆD⁰=CE=C, where D⁰=C is divisible, Ehas finite ex- ponentand g is power on both D⁰=C and E=C. Observe that D⁰ˆDC.

Hence AˆDE and jD⁰=Dj jCj. Moreover g acts as a power on A=D⁰'_gE=C, then by Proposition 2,gacts as a power automorphisms on a subgroup A₁=D of index atmostjCj in A=D. The statement holds by

Proposition 4.ii. p

Let's recall now a key result, basically due to D. Robinson [7].

LEMMA1. Let a2A an abelian group andgan inertial automorphism.

Then

i) gacts as a power automorphism on any divisible torsion sub- group ofA;

ii) ifa is periodic, then a^hgiis finite.

LEMMA 2. Let A be an abelian p-group, g2Aut(G) and m2N. If jX=(X\X^g)j p^m for all finite XA, thenjX^hgi=Xj p^2m for all finite XA.

In Theorem 1 we will see that a similar statement holds even if the word

``finite'' is omitted.

PROOF. Fix X with order p^e. We first prove the statement when Xˆ haiis cyclic. Ifeˆ1, that isX^hgiˆ hai^hgi is an elementary abelianp- group, consider the epimorphismg2Z_p[x]7!a^g(g)2 hai^hgiofZ_p-modules.

By Lemma 1,hai^hgihas finite orderpⁿand so there isf 2Zp[x] with degree nsuch that

Z_p[x]

(f) 'ghai^hgi:

Ifn2m‡1 the statement is true. Assume false by contradiction. Then Xˆ ha;a^g2;. . .;a^g2mi has order p^m‡1. It follows that there is b2X\ X^gn f1g. Thus there are polynomialsg;h2Z_p[x] with degreem such that bˆa^g(g2) ˆa^gh(g2). Thereforef(x) dividesg(x²) xh(x²) which is not the constant 0 and has degree2m‡1, a contradiction.

If e>1, B:ˆ ha^{pe 1}i^hgi has order atmostp^(2m‡1), by case eˆ1, and hai^hgi=Bhas order atmostp^{(e 1)(2m‡1)}, by induction. The claim holds.

In the general case let Xˆ ha1i hariwith a_i of order p^ei and e₁‡ ‡e_rˆe. By the above, for each iwe have jha_ii^hgij p^(2m‡1)ei and

jX^hgi=Xj p^2me. p

Let's show that for periodic abelian groups all properties (ap), (bp), (in), (bin) are equivalent. Recall that (ap) and (bp)-automorphisms of abelian groups have been independently characterized in [3] and [2], respectively.

THEOREM1. Let A be a periodic abelian group andg2Aut(A). Then the following are equivalent to the fact thatgis inertial:

i) X=(X\X^g)is finite for all XA;

a) there exists a2Nsuch thatjX=(X\X^g)j a for all XA;

b) there is subgroup BA with finite index b, such thatgis power on all but finitely many primary components of B and pseudo-power on the other ones;

c) there is a subgroup CA with finite order c, such that g is power on all but finitely many primary components of A=C and pseudo- power on the other ones;

d) there exists d2Nsuch thatjX^hgi=Xj d for all XA;

d) there exists d2Nsuch thatjX=X_hgij d for all XA.

Moreover, there are functions (depending on A) bounding each of a, b, c, d, d in terms ofany other.

PROOF. Ifpis a prime greater that one among the abovea;b;c;d;d, then the corresponding statement tells thatgacts as a power automorphism on thep-componentofA. Thus, according to Proposition 1, we assume that Ais ap-group. LetDˆD(A) andrandethe rank ofDand the exponent of A=D, respectively (where this cardinals are possible infinite).

First of all recall that:

(c),(b) by Propositions 2 and 5. On the other side, ifAis periodic, (b) is equivalentto (ap) (by Theorem 2.8 of [3] with Gˆ hgi) and (c) is equivalentto (bp) (see Theorem 2.11 of [2]).

(i))(ap) Assume first Ahas exponentp. We show thatgis periodic (and so an almost power automorphism). By Lemma 1, the orbits induced by g are finite. Assume, by contradiction, they have unbounded orders.

Let's show that for all finite XA such that X\X^gˆ1 there exists X₁>X with the same property. Note thatX^hgi is finite by Lemma 1 and

choosea2AnX^hgiwhose orbit under the action ofghas length larger than the order ofK:ˆ haiX^hgi. Thusa^g62K otherwisehai^hgiK contradicting the choice ofa. LetnowX1:ˆ haiX. We will show thatX1\X₁^gˆ1. Let y2X₁\X₁^g. Hence9n;s2Z,9x;x₀2Xsuch thatyˆaⁿxˆa^sgx^g₀. Then a^sg2 haiX^hgiˆK while a^g62K. Thereforea^sˆ1 and itfollowsaⁿ2X^hgi while a^g62X^hgi. Therefore aⁿˆ1 as well. Itfollows yˆxˆx^g₀2X\ X^g ˆ1, as claimed. Now we define by induction a sequence of finite sub- groupsX_isuch thatX_i\X_i^gˆ1 andX_i5X_i‡1. ThusXv:ˆS

iX_iis infinite andX_v\X_v^g ˆ1, a contradiction. Thusgacts as an (ap)-automorphism on the socleA[p] ofA.

Let X be any subgroup of A. To show that X=X_hgi is finite, we may suppose X_hgiˆ1. Thus, sinceg acts as a power automorphism onD(see Lemma 1), we haveD\Xˆ1 andXis reduced. On the other hand, since we have seen thatgis almostpower onA[p] we getthatX[p] is finite. It follows thatXis finite.

(a))(d) Apply Lemma 2 to the groupA=X_hgi and getthat jX^hgi=Xj a²:

(d))(c) Trivially (d) implies (a). Therefore g is (ap) and, ifA is re- duced, by the above there is a power automorphismaofAsuch thatA^{g a} has finite order. In order to find a bound forcdepending ond, letus show that:

ifA is reduced, then jA^{g a}j d^e2‡3e2 or d, according to the fact that exp (A)ˆp^e51orˆ 1, respectively.

Letfirstexp (A)51. By contradiction, fix the smalleste2Nfor which the statement is false, where clearlye>0. We have thatgis power on no factorA=CwithjCj c(d;e):ˆd^e2‡3e2 . Then fix the largestn2Nsuch that

9AnA : jAnj ˆp^en and jA^hgi_n =Anj pⁿ

where the above holds fornˆ0, at least. DenoteCn:ˆA^hgin . We have p^(e‡1)n jC_nj jA_njdp^end and pⁿd:

The subgroup A^#ˆ fa2Aja^{pe 1}2Cng is such that exp (A^#=Cn)5p^e. Thus, by the choice ofe, there existsCsuch thatCnCA^#,gacts as a power onA^#=Cand

jCj ˆ jCnj jC=Cnj p^endc(d;e 1)d^e‡1c(d;e 1)ˆc(d;e) Letus show now thatgacts as a power on A=C, that is for alla2Awe have

a^g 2Chai. Ifa2A^#, then the thing is true by the above. Otherwiseahas order p^e even modulo Cn, by definition of A^#. In particular hai \Cnˆ hai \Anˆ1 andAn‡1:ˆAnhai ˆAn hai has orderp^n‡1. If a^g62Chai, thena^g62C_nhaiso that

pⁿ jCn=Anj ˆ jCnhai=Anhaij5jA^hgi_n‡1=An‡1j contradicting the choice ofn.

Letnow exp (A)ˆ 1andA be residually finite. We see that there is an Xsuch that

X\A^{g a}ˆ1 and A^{g a} X^hgi

whencejA^{g a}j jX^hgi=Xj d. To do this, proceed by induction and assume that for a maximal subgroup C1 of C:ˆA^{g a} there exist X1 such that X1\Cˆ1 andC1X^hgi₁ . Leta^{g a}2CnC1. AsAis residually finite andgis (ap) onA, by the abovegˆais power on a subgroupBˆB^hgiwith finite index inAand such thatB\CX₁hai ˆ1. AsAas infinite exponent, the same holds forBand we can pick and elementb2Bwith the same order asa. We have then a^{g a}ˆ(ab)^{g a}, thus CˆC1ha^{g a}i X^hgi where X:ˆX1habi.

Further ifc2X\C, t hen cˆx1a^mb^m for somex12X1, m2N, so t hat b^m2B\CX1hai ˆ1. By the choice ofb, we havea^mˆ1 as well. Itfollows cˆx₁2X₁\Cˆ1, as desired. The case A is residually finite is now complete.

In the caseA is reduced, butperhaps notresidually finite, letBbe a basic subgroup ofAandB1ˆB^hgi. ThenB1=Bis finite, so thatB1 is residually finite andA=B1is divisible. By the above there exists a subgroupCB1with ordercd^e2‡3e2 ord(according to exp (A)ˆp^e51orˆ 1), such thatgˆa onB₁=C. Then the kernel ofg aonA=CcontainsB₁=C, so that the image is divisible. On the other handA=Cis reduced, so thatgˆaonA=C.

In the general case, let AˆDR, where R is reduced. Then C1ˆR^hgi\Dhas order atmostd. On one hand, by the above, the reduced groupR^hgi=C₁has a subgroupC=C₁of order atmostthe abovecsuch thatg is power onR^hgi=C. On the other hand g is power on the divisible group DC=Cas well, so thatgis either power or pseudo-power onA=C. Clearly, according to exp (A)ˆp^e51orˆ 1,Chas order

cd^e2‡3e‡22 or d²:

(b))(d). Let X be any subgroup of A. Then by Proposition 3, j(X\B)=(X\B)_hgij e^r. Thus (d) holds with

jX=X^hgij be^r:

(d))(a) This is trivial. p

In the notation of Theorem 1, d is bounded by be^r and clearly d is bounded byce^r. Let us show that these bounds are best possible.

PROPOSITION6. For each prime p6ˆ2, r;e2N, eˆp^e, there exists an abelian(r;e)-bounded p-group such that:

i) there is an inertial automorphismgacting as a pseudo-power on a finite index subgroup B and a subgroup X of A such that

jX=X_hgij ˆ jA=Bje^r:

ii) there is an inertial automorphismgacting as a pseudo-power on a factor A=C where C is finite and a subgroup X of A such that

jX^hgi=Xj ˆ jCje^r:

PROOF. Let AˆDECC0 be a p-group where D is divisible with rankr,Eis the product of infinitely many copies of a cyclic group with ordereˆp^e,C0ˆ hc0iandCelementary abelian with orderspandp^e, resp.

Define, for the moment,g_DEC byg_jDˆa,g_jEˆ1,g_jCˆg₁whereaap-adic 61 (mod p), g₁ is an irreducibile automorphism of C and a such that Cˆ hai^hg1i. By Proposition 3 there isX0DEsuch thatjX0=X0hgij ˆe^rand jX₀^hgi=X0j ˆe^r.

i) Define g on the whole Aby g_jC0ˆ1. Clearlyg is pseudo-power on B:ˆDEC₀. LetX:ˆX₀M whereM is a maximal subgroup of C₀C con- taining neither C₀ norC. We haveX_hgiˆX_0hgiM_hgi, so thatjX=X_hgij ˆ jA=Bje^r.

ii) Definegon the wholeAbyc^g₀ˆc₀a. Clearly gis pseudo-power on A=C. IfX:ˆX₀C₀we havejX^hgi=Xj ˆ jCje^r. p

3. Torsion-free case

Recall (ap) and (bp)-automorphism of abelian torsion-free groups are just the identity or the inversion map (see [3] and [2] or just later). The picture is different for (in)-automorphisms. In this section we use additive notation for the abelian group operation inA.

Notice that if 06ˆn;m2Z, for each a2A, a torsion-free abelian group, there is at most oneb2Asuch that

() nbˆma:

If such a uniquebexists, we writebˆm

na. Thus whenm;nare coprime, () defines an automorphisma7!b if and only ifAˆmAˆnA. We say that this isthe rational power automorphismm

n.

The following result describes the structure of the groupIAut(A) of all inertial automorphisms of Ain terms ofA, in the caseA is torsion-free.

Denotep(A) the set of primespsuch thatpAˆA.

THEOREM2. Let(A;‡)be an abelian torsion-free group.

If g2 IAut(G), then g:x7!m

nx with m;n coprime integers and p(mn)p(A):

IfIAut(A)6ˆPAut(A)ˆ f1g, then A has finite rank r and ha1;. . .;asi^hgi'Qp. . .Qp (stimes)

where pˆp(mn), whenever the a_i's are Z-independent elements ofA.

Further, ifp(mn)p(A)thenm

n 2 IAut(A)so that IAut(A)' f‡1; 1g Dr

p2p(A)hpi:

PROOF. For anya2A, sincejhai:hai \ ha^gij ˆm2N, there exists an m2Z such that maˆna^g. As A is torsion-free, m;n can be choosen coprime. Letus show thatm

n does notdepend on a. Let a₁2A. If ha1i\ha^gi 6ˆ f0g, thena1ˆh

kafor someh;k2Z. Therefore a^g₁ˆh

ka^gˆh k

m naˆm

na₁:

Similarly, ifha1i \ ha^gi ˆ f0g, there will existm1;m2;n1;n22Zsuch that m

na‡m₁

n1 a₁ˆa^g‡a^g₁ˆ(a‡a₁)^gˆm₂

n2(a‡a₁)ˆm₂ n2 a‡m₂

n2a₁: Itfollows m

n ˆm2

n₂ ˆm1

n₁ and g7!m

n is a monomorphismIAut(A)!Q, the multiplicative group of the rationals.

Observe now that for each XA, by applying the monomorphism x7!nx we have X=X\X^g'nX=nX\mX. The latter has exponent at

most mand is finite if X has finite rank. Conversely, if the rank of the wholeAis infinite, we get a contradiction by takingXto be free of maximal rank.

Finally, if the rank ofAisr, thenAis isomorphic to a subgroupBof Q. . .Q(rtimes) containingZ. . .Z(rtimes). SinceBisp-divisible, itis clear thatBcontainsQp. . .Q_p(rtimes), wherepˆp(A). p Recall that in Theorem 1 we have seen that X=(X\X^g)51for each XAimpliesgis inertial indeed, providedAis periodic. However, this is no longer true if A is non-periodic, a counterexample being the map g:x7!1

2xin the product of infinitely many copies ofQ(see Theorem 2).

4. Mixed case

We start by observing that, by Theorem 2, if A is non-periodic with torsion subgroupTandgis inertial then there exist coprimem;n2Zsuch that

g_jSˆg_jA=Tˆm n

for each torsion-freeg-invariantsectionSofA. Definep(g):ˆp(mn).

Now we prove a generalization of Lemma 1. Recall that a p-group has finite exponent iff it has no nontrivial divisible quotients.(see Cor.

35.4 in [4]).

LEMMA 3. Let gan inertial automorphism ofan abelian group A. If a2A, then the torsion subgroup T ofVˆ hai^hgiis finite. Thus there is e2N such that V^e'Q_p(g)has finite index in V, provided a is aperiodic.

PROOF. We can assumeAˆV. Letgˆm

n onA=T. Define inductively A0:ˆ haiandAi‡1:ˆAiA^g_iA^g_i ¹. ThenAˆ S

i5vAiwhere, for eachi,Ai‡1=Ai

is generated by its elementsa^giAianda^{g i}Aiwith order dividingmandn, respectively. ThusT'TA0=A0andp(T)p(A=A0) consists of divisors of mn. In particularp(T)p(mn) is finite.

Suppose now A splits on T, that isAˆTBfor some B. Letthen B₁ˆBB^gB^{g 1}and note thatB₁=Bis finite. ThenB₁ˆT₁B, whereT₁is finite subgroup ofT. Thus by Lemma 1,KˆT₁^hgi is finite as well. More- over (BK)^gˆB^gKB1KˆBK and similarly (BK)^{g 1}BK. ThusBKis

g-invariant. ThereforeT=K'A=BK is finite, by Lemma 1 again. ThusT is finite, in this case, which applies ifThas finite exponent.

If, by contradiction,Thas infinite exponent there is a factorT=Rwhich is a PruÈferp-group. SinceR^hgi=Ris finite, by Theorem 1, thenT=R^hgiis a PruÈfer group andA=R^hgisplits on its torsion subgroupT=R^hgiwhich turns out to be finite, by the above, a contradiction.

Finally, ifeis the order ofT, we haveA^eˆ ha^ei^hgiis torsion-free and has

finite index inAby Lemma 1. p

We come now to a characterization of inertial automorphisms of an abelian group. Before we state some elementary facts as a Lemma.

LEMMA4. Let B a subgroup ofan abelian group A andpa set ofprimes.

i) IfA=B is ap'-group, then A isp-divisible iff B isp-divisible.

ii) IfA iftorsion-free, A=B periodic and B isp-divisible then A=B

is ap'-group and A isp-divisible. p

THEOREM 3. Let g be an automorphism ofan abelian non-periodic group A. Thengis inertial ifand only ifone ofthe following holds:

(a)there ag-invariant finite index subgroup B of A on whichgacts as a power.

(b) there are finitely many elements a_isuch that:

1. the subgroup V ˆ ha1;. . .;ari^hgiis torsion-free andginduces on it a rational power automorphismm

n where m, n are coprime integers;

2. the factor group A=V is torsion andginduces an almost power automorphism on it;

3. thep(mn)-component ofthe torsion subgroup T ofA has finite exponent.

Further, A^{ng m}is periodic. It has finite exponent iff () there exists k2Nsuch thatjX=(X\X^g)j k;8XA.

In case (b), V 'Qp. . .Qp (r times), where pˆp(mn), and g is almost power iff V is free abelian.

Aboutcase (a) see also Corollary 2 above. Condition (2) is characterized in Theorem 1 while (3) is always true ifgis almostpower, aspˆ ;in this case.

PROOF. LetTbe the torsion subgroup ofA. Suppose thatgis inertial.

By Theorem 1,gacts as an almost power automorphism on each periodic g-section ofA. In particular,gis almostpower onTand rational powerm onA=T (withm;ncoprime integers). n

If m

n ˆ 1, then g is an almost power automorphism ofA. To see this, letSbe a maximal independentsubsetof elements of infinite order of A. Then Yˆ hSi ˆDr_s2Shsiis free abelian and A=Y is periodic. Let s2S. By Lemma 3, the torsion subgroup R of hsi^hgi is finite and so hsi^hgi=hsi is finite too, since g is power on hsi^hgi=R. Hence hsi has a subgroup of finite index Ys which is g-invariantand so the subgroup ZˆDr_s2SYsis free abelian andg-invariant,A=Zis periodic andgacts as a power automorphism on Z. If Z has finit e rank we are in case (b) of Theorem 2.10 of [3] (withGˆ hgi) andgis almostpower onAand (b) in the statement holds. On the other hand, ifZhas infinite rank, then (a) holds. This can be shown by the same argument as in the proof of Theorem 2.10 of [3], keeping in mind that g act s as an almost power automorphism already on each periodic factor ofA. The claim is proved.

Ifm

n 6ˆ 1, by Theorem 2 a maximal independentsubsetfs1;. . .;srg of elements of infinite order ofAhas finite cardinality. Thus by Lemma 3 there ise2Nsuch thatV :ˆ hs^e₁;. . .;s^e_ri^hgiis torsion-free of rankr. Thus (1) and (2) hold.

Let us show that the p(mn)-componentof T has finite exponent.

Assume, by contradiction, thatThas an imageT=Xisomorphic to a PruÈfer p-group, withp2p(mn). By Theorem 1,X^hgi=Xis finite. Thus, without loss of generality, we may assume that X^hgiˆ1 and AˆVT. Also we can take 16ˆv2V and assumeV ˆ hvi^hgi has rank 1 and isp-divisible. Then AˆVT where Tˆ hc_njn2N₀;c₀ˆ1;c_n‡1^p ˆc_niis a PruÈfer p-group.

Consider then the subgroupL₁ofAgenerated by all elementsv^pic_i, with i2N (recall thatV is torsion-free and p-divisible, thus v^{p i} is uniquely identified). ClearlyL1=hviis the ``diagonal subgroup'' of thep-component of A=hvi, which is the direct product Vp=hvi Thvi=hvi of two PruÈferp- groups. Also letL₂=hvibe thep⁰-componentof the wholeA=hviand finally LˆL₁L₂. This is easily verified to be torsion-free, that isL\Tˆ1. On one hand,L'LT=TˆVT=T'VandLisp-divisible. On the other hand, A=L'Tis ap-group, so thatL=(L\L^g) is ap-group, which is finite asgis inertial. It follows LˆL^g. Thereforehvi ˆL\Vˆ(L\V)^gˆ hvi^g, con- tradicting the fact thatm

n 6ˆ 1.

Finallysuppose that()holdsand assume, by contradiction, that there existsa2Asuch thatb:ˆa^{ng m}has order greater thatk. We can assume a has infinite order (otherwise substitute withavwhere 16ˆv2V). Let X:ˆ hai. Thusb2XX^gand thereforejXX^g=X^gj>k, a contradiction.

Conversely, note that if (a) holds then g is (ap). Thus suppose that conditions (1), (2), (3) hold. Notice that gˆm

n on A=T, that is A^{ng m} is periodic.

We have to show that for eachXAitholdsXX^g(where bywe denote commensurability). We immediately notice that this is true ifX is periodic. In fact, in such a case XVX^gV and X=(X\X^g)' XV=(XV\X^gV) is finite, that isXX^g. Thusgacts as an inertial auto- morphism onT.

Letus show thatwhen A^{ng m}has finite exponent ethen jX=(X\X^g)j k:ˆdjmj^re^r

wheredis the bound given by Theorem 1.d. In factfor eacha2Awe have a^eng ˆa^em. So that X^engˆX^emX\X^g. AlsoS_hgiX\X^g, where Sis the torsion subgroup of X and jS=S_hgij d. Now, on one hand jX=X^emSj (ejmj)^r, asX=Sis torsion-free with rank at mostr. On the other handX^emS=X^emS_hgiis isomorphic to a factor ofS=S_hgi which has order at mostd. ThusjX=X^emS_hgij d(ejmj)^r, as claimed.

Let us treat the general case, whereTis possibly unbounded. We show thatXX^g. We claim thatwe may reduce to the case X is torsion-free, as ifSis the torsion subgroup ofX, thenS=S_hgiis finite, by the above. So, we may factor out byS_hgi(as conditions (1), (2), (3) are inherited by factors by torsionhgi-subgroups). Then we may assume thatShas finite ordersand X^sis torsion-free. SinceXX^s,from now on assume X is torsion-free.

We claim thatwe may factor out by anyg-invariant subgroup E with finite exponent e. In fact, ifuˆ jXX^gE=XEjis finite, then for anyx2X, (x^g)^u2XE. On the other hand x^{ng m}2T by the above. Thus x^{u(ng m)}2XE\TˆE, asXis torsion-free. Thereforex^{eu(ng m)}ˆ1 that is x^eungˆx^eum. Hence (X^eun)^gˆX^eum. Itfollows X^g (X^eun)^g ˆX^eumX, as wished.

By takingEˆT_p,we may assume that T is ap⁰-group. Moreover there are only finitely many primespfor which the action ofgon thep-compo- nentTpofTis notpower (Theorem 1). However, for each of these primes p1;. . .;plthere is ag-invariantsubgroup of finite exponentEpiofTpisuchg acts as a power automorphism on Tpi=Epi. So again we can take RˆE_pl E_pl and, by the above, assume that g acts as a power auto- morphism on T.

Proceed now by induction onr, the torsion-free rank ofA. Letrˆ1.

LetK=Xbe thep-componentofA=X. ThenKis is torsion-free, asTis ap⁰- group. By applying Lemma 4.(ii) to the groupA=T and to the subgroup VT=T, we getA=Tisp-divisible. Moreover, sinceA=KTis ap⁰-group by Lemma 4.(ii) we getK'KT=Tisp-divisible. Letb2K\V. Then for any prime p2p there exist k2K and v2V such that k^pˆbˆv^p. Hence (k ¹v)^pˆ1 so thatkˆv2K\V, asTis ap⁰-group. Itfollows thatK\V isp-divisible and so has finite index in V, asV has rank 1.

ThenVK=Kis finite. AsVKis inertized byg, so isK. ThereforeKK^g\Tis finite and soXX^g\Tis finite, too. NowXX^g=(XX^g\XT)'XX^gT=XT is finite, and (XX^g\XT)=X'XX^g\Tis finite as well. HenceXX^g.

Letnowr>1 and letr1be the rank ofX. Ifr15r, thenX^hgiThas rank r1, andXX^g, by induction onr. IfXhas rankr,Xcan be written asX1X2, where bothX₁andX₂have rank5r. Thusginertizes bothX₁andX₂, then XX^g.

The last part of the statement follows from Theorem 2.10 of [3] and

Theorem 2. p

COROLLARY1. Let Abe abelian group with torsion subgroup T and g2Aut(A). Then the following are equivalent:

i) gis (ap), that is almost power, ii) gis (bp),

iii) gis inertial on A and power on A=T.

Consideration ofV in the statement of previuos theorem is necessary, even whenAsplits on its torsion subgroup as we see in next Proposition.

PROPOSITION 7. There is an abelian group A with a non-inertial automorphism g which is inertial on both T and A=T, where T is the torsion subgroup ofA.

PROOF. Consider AˆDT with D'Q₆, T'Z₂¹, g_jTˆ1 and g_jDˆ3. Then there isV ˆV^gDwithV 'Q₃. Hencegis notinertial on A=V, which is periodic and divisible, as itis notpower on it. p Observe that there exist almost power automorphisms that are not boundedly inertial.

COROLLARY2. There exists an inertial automorphism ofan abelian groupAsuch thatjX=(X\X^g)jis unbounded whenXA.

PROOF. Letn,many coprime integers, withm

n 6ˆ1, letpˆp(nm) and consider the groupAˆVT, where V is isomorphic toQ_p andT is a periodicp⁰-group with infinite exponent. Letg2Aut(A) such thatg_jV ˆm and g_jTˆ1. By Theorem 3, g is (in) and even (ap) when m n

n ˆ 1. Also A^{ng m}T^{ng m}ˆT^{n m} has infinite exponent. Thus jX=(X\X^g)j is

unbounded. p

5. Groups of inertial automorphisms

Properties corresponding to (ap) and (bp) can be stated for a groupGof automorphisms of an abelian group, instead than for a singularg2Aut(A), as in [3] and [2], resp. So we define the following properties forG.

(AP) 8XA jX=X_Gj51;

(BP) 8XA jX^G=Xj51;

(CP) 8XA9YˆY^G Asuch thatXandY are commensurable.

Obviously both (AP) and (BP) imply (CP). From results in [3] (resp. [2]) itfollows thatgis (ap) (resp. (bp)) iffhgiis (AP) (resp. (bp)). We will see in Proposition 9 that these three properties coincide ifGis finitely generated.

On the other hand they are in fact different, as next Proposition 8 shows.

Observe also that any of (AP), (BP) or (CP) for G implies that G IAut(A), where the latter is the subgroup of Aut(A) filled by all inertial automorphism.

PROPOSITION8. There exist infinite p-groups A andGAutA which are:

(a) AP, not BP;

(b) BP, not AP;

(c) CP, neither AP nor BP.

PROOF. (a) LetBˆDr_i2Nhb_ii, where anyb_ihas orderpand letCˆ hci a group of orderp. LetAˆBCand consider, for anyi2N, the auto- morphism g_i of Afixing any elementof Band such that c^giˆb_ic. Then Gˆ hg_iji2Ni AutAis power on the subgroupBof finite index inA, but is notBP, asC^GˆA.

(b) Let AˆBC as in (a) and for any i2N consider the auto- morphismd_iofAfixing any elementofCand anyb_j, ifj6ˆiand such that

b^d_iⁱˆbic. The subgroup Dˆ hdiji2Ni of AutA is power on the group A=C, butis notAP, asBDˆ1.

(c) LetAˆA⁰A⁰⁰, withA⁰ˆB⁰C⁰andG as in (a) (withB⁰ andC⁰ instead ofBandC, respectively) andA⁰⁰ˆB⁰⁰C⁰⁰andDas in (b) (withB⁰⁰ andC⁰⁰instead ofBandC, respectively). The groupPˆGDAutAis power onB⁰A⁰⁰=C⁰⁰and hence isCP, butis neither AP nor BP. p PROPOSITION9. Letg₁;. . .;g_n automorphisms ofan abelian group A.

Then

i) ifg₁;. . .;g_nare(ap), thenhg₁;. . .;g_ni AutA is(AP).

ii) properties (AP), (BP) and (CP) are equivalent for finitely generated groupsGˆ hg₁;. . .;g_ni.

PROOF. (i) In the caseAis periodic, we may assumeAis ap-group and atleastone ofg_iis notelementary. ThenAˆDR, whereDˆD(A) has finite rank andRhas finite exponent. For 1inthere exists E_iR with finite index on whichg_iis power. So that eachg_iis power onE:ˆ \_iE_i andB:ˆDEhas finite index inA. By Theorem 2.8 of [3],Gis (AP) onA, as wished. For the general case, note that for 1inthere existsV_iA which is free abelian such thatg_iis power onViand (ap) on the torsion group A=Vi. So thatGis power onV :ˆ \iViand (AP) on the torsion groupA=V, by the above. Thus by Theorem 2.10 of [3],G is (AP) onA.

(ii) IfGis (CP), then its elements are (ap) and (bp), by Theorem 1 and Corollary 1. By the aboveGis (AP). It remains to show that if allg_iare (bp), thenGis (BP). Again, in the periodic case we may supposeAis ap-group and atleastone ofg_iis not elementary, so thatDˆD(A) has finite rank and exp (A=D)ˆ:e51. Then for 1inthere existsK_iˆK^g_iⁱAsuch that K_i=Dis finite and g_iis power onA=K_i. ThenK:ˆK₁. . .K_n isG-in- variantandK=Dis finite. ThusK[e] is finite and, by Theorem 2.11 of [2],G is (BP) onA. In the caseAis notperiodic, recall thatby Theorem 3 for 1inthere existsViAsuch thatA=Viis torsion andg_iis power on V_i. ThenV:ˆ \_iV_iisG-invariant. By the aboveG is (BP) on the periodic quotientA=V and by Theorem 2.12 of [2],G is (BP) on the wholeA. p THEOREM 4. Let A an abelian non-periodic group A and G ˆ hg₁;. . .;g_si AutA. Then eachg_iis inertial iff one of the following holds:

(a) there a finite index subgroup B of A on which each g_i acts as a power automorphism.

(b) there are finitely many elements a_i2A andg2G such that:

1. the subgroup V ˆ ha1;. . .;ari^hgi is torsion-free and each g_i in- duces on it a rational-power automorphismm_i

ni (m_i, n_icoprime);

2. the factor group A=V is torsion and each g_i acts as an almost- power automorphism on it;

3. thep(m1n1 msns)-component ofthe torsion subgroup T ofA has finite exponent.

PROOF. For eachg2 IAut(A) denotep(g):ˆp(mn)whereg_jA=T ˆm n. If allg_i's are (ap), thenG is (AP) by Proposition 9. Thus the statement holds by Theorem 2.10 of [3].

Otherwise, by Theorem 3, A has finite torsion-free rank r and for 1isthere is a torsion-free subgroupV_iwhich is the closure underhg_ii of r linearly independent elements such that g_iˆm_i

ni onV_i (with m_i,n_i coprime integers) and g_i is (ap) on the periodic factor group A=V_i. Put p_iˆp(m_in_i) andpˆp(m₁n₁ m_rn_r). Then by Theorem 3 thep_i-compo- nentofThas finite exponent and hence (3) in the statement holds.

Actually, there is g2G such that p(g)ˆp(g₁)[ [p(g_s)ˆp. This can be seen by induction ons, where it is enough to treat the casesˆ2.

Letjˆ1‡maxfi2Njpⁱdividesm₁m₂n₁n₂; pprime numberg and put gˆg₁g^t₂. Then gˆm₁m^t₂

n₁n^t₂ on A=T and the choice of t ensures p(g)ˆ p(g₁)[p(g₂).

PickZ-independentelementsb₁;. . .;b_r ofV₁\. . .\V_s. By Lemma 3 there existsa_i2 hb_iisuch thatha_ii^hgiis torsion-free with rank 1. Thus we obtain a torsion-free subgroupV ofAwith shape

V :ˆ ha1;. . .;ari^hgi:

Letus show thatVisg_i-invariantfor eachi. Letabe any elementofVand eˆexp (T_p). By Theorem 2,V=V_iis ap-group andVisp-divisible. Letb2V such thatb^eˆa. There is ap-numberusuch thatb^u 2V_i. SinceV_iˆV_i^giand g_iˆm_i

n_i on A=T we have b^{(nigi mi)u}2T\V_iˆ1. Itfollows b^{nigi mi}2T_p. Thereforea^{nigi mi}ˆb^{e(nigi mi)}ˆ1. ThusV^giˆV^nigiˆV^miˆV, as claimed.

Thus (1) holds, while (2) is trivial asVhas maximal torsion-free rank.

Conversely, suppose that conditions (1), (2), (3) hold. By Theorem 3, to show that g_i is inertial, we only have to prove that g_i is inertial on the periodic groupA=V_i, whereV_iˆ ha1;. . .;ari^hgii. Letpˆp(m1n1 msns) and letB_p⁰=V_ibep⁰-componentofA=V_i. ThenB_p⁰=V_i'_gi B_p⁰V=V, which is

thep⁰-componentofA=V and sog_iis inertial on it. Letp2pand letBp=Vi

bep-componentofA=Vi. SinceA=TV is ap⁰-group,Bpis contained inTV.

Sog_iis power on a subgroup of finite index ofBp=Bp\V'g_iBpV=Vwhich has finite exponent. Moreover g_i is power onB_p\V=V_i, which has finite rank. So by Theorem 2.12 of [2],g_iis (bp) and hence inertial onA=V_i. p

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Manoscritto pervenuto in redazione il 12 luglio 2011.