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Publicly Available Published by De Gruyter September 14, 2016

A group of generalized finitary automorphisms of an abelian group

Ulderico Dardano EMAIL logo and Silvana Rinauro
From the journal Journal of Group Theory

Abstract

We study the group IAut(A) generated by the inertial automorphisms of an abelian group A, that is, automorphisms γ with the property that each subgroup H of A has finite index in the subgroup generated by H and Hγ. Clearly, IAut(A) contains the group FAut(A) of finitary automorphisms of A, which is known to be locally finite. In a previous paper, we showed that IAut(A) is (locally finite)-by-abelian. In this paper, we show that IAut(A) is also metabelian-by-(locally finite). More precisely, IAut(A) has a normal subgroup Γ such that IAut(A)/Γ is locally finite and the derived subgroup Γ is an abelian periodic subgroup all of whose subgroups are normal in Γ. In the case when A is periodic, IAut(A) turns out to be abelian-by-(locally finite) indeed, while in the general case it is not even (locally nilpotent)-by-(locally finite). Moreover, we provide further details about the structure of IAut(A).

1 Introduction

A subgroup H of a group G is called inert if it is commensurable with its conjugates Hg:=g-1Hg for each gG (see [1]), that is, HHg has finite index in both H and Hg. Groups whose subgroups are all inert have been considered in [2] under the name TIN-groups. The class of TIN-groups contains Tarski monsters and FC-groups (see [19]), that is groups in which each element has finitely many conjugates, so it is a highly complex class. However there are no infinite locally-finite simple TIN-groups, as shown in [2]. This result has recently been extended to simple locally graded groups in [15]. On the other hand, soluble TIN-groups have been studied in [19] under some finiteness conditions.

According to [13], if H is a subgroup of an abelian group (A,+) and γ is endomorphism of A, then H is called γ-inert if H has finite index in H+Hγ. The interest in the concept of γ-inert subgroups comes from the study of the dynamical properties of a given endomorphism (see [14] and the bibliography therein). A subgroup H is called fully inert if it is γ-inert for every endomorphism γ of A. Fully inert subgroups of divisible abelian groups are investigated in [13]. Then from [11] we recall a definition.

Definition

An endomorphism γ of an abelian group (A,+) is called inertial endomorphism if (H+Hγ)/H is finite for each subgroup H. An inertial endomorphism which is bijective is called inertial automorphism.

The set of inertial endomorphisms of an abelian group A is a subring of the ring of all endomorphisms of A. We featured this subring in [9].

Inertial automorphisms of an abelian group may be used as a tool in the study of soluble groups whose many subgroups are inert. Recall that in [10] we have studied finitely generated groups in which subnormal subgroups are inert. Also recall that, on the other hand, in [12] it has been shown that if in a (non-abelian) group G all finitely generated subgroups H are strongly inert, that is the subgroups H and Hg both have finite index in their join H,Hg for each gG, then the group G is (locally finite)-by-abelian. Note that strongly inert subgroups are inert.

Recall that, when G is any group, an automorphism of G is called finitary if it acts as the identity map on a subgroup of finite index in G. Clearly finitary automorphisms of an abelian group are inertial. In [3] it has been shown that the group FAut(G) of all finitary automorphisms of a group G is both (locally finite)-by-abelian and abelian-by-(locally finite). When G is abelian, then FAut(G) is even locally finite ([21]).

In this paper we study the group IAut(A)generated by all inertial automorphisms of an abelian group A. Clearly IAut(A) contains both FAut(A) and the subgroup PAut(A) of automorphisms of A leaving each subgroup invariant. In [11] we have shown that IAut(A) is locally (center-by-finite), hence (locally finite)-by-abelian. Here we consider the question of when IAut(A) is abelian-by-(locally finite). From the main results of this paper, it follows:

Corollary A

If A is a periodic abelian group, then IAut(A) is center-by-(locally finite).

Example B

The group IAut((p)) is not (locally nilpotent)-by-(locally finite), provided p is an odd prime.

Recall that a group G is said to be metabelian if its derived subgroup G is abelian.

Corollary C

If A is an abelian group, then IAut(A) is metabelian-by-(locally finite).

Outline of the paper.

Since some of our statements and proofs are rather technical, in Section 2 we fix notation in a rather accurate way, recall some definitions and results, and introduce concepts that will be used in the paper.

In Section 3 we highlight the role played by stability groups with respect to finitary and inertial automorphisms of A since in our investigations we shall look for abelian normal subgroups Σ of IAut(A) such that the automorphisms induced by IAut(A) via conjugation on Σ are inertial. We also describe the group FAut(A) in the case when A is of a type which is relevant to our purposes (Proposition 3.6) and in the case when A splits on its torsion subgroup (Proposition 3.8).

In Section 4 by Theorem 4.2 we give a rather complete description of IAut(A) when A is periodic. It follows Corollary A. In Example 4.4 we see that it may happen that there is no abelian normal subgroup Λ of IAut(A) such that IAut(A)=ΛFAut(A).

In Section 5 by Theorem 5.1 we treat the case in which A is non-periodic. It follows Corollary C. In Example 5.2 we see that |IAut((p)(p))|=2.

In the final Section 6 we describe the group IAut(A) in detail in cases when A splits on its torsion subgroup, see Theorem 6.1 and Theorem 6.2. Thus we have Example B. Then in Example 6.3 we consider a group A for which the group IAut(A) is very large.

2 Preliminaries

Notation.

First of all, note that, since we are going to apply methods and results from [11] where we used the fact that inertial endomorphisms form a ring, in this paper we will regard abelian groups as right modules over their endomorphism ring and reserve the letter A for abelian groups, which are additively written.

It shall be remarked that in the previous paper [8] we gave a slightly more restrictive definition of inertial automorphism (and we used different notation). This is not a major problem since in the present paper we just treat the problem in greater generality, see Fact 2.12 below.

For undefined terminology, notation and basic facts we refer to [17] or [18]. The letter denotes the set of all prime numbers. If n, then π(n) denotes the set of prime divisors of n. If π, then Aπ:=Ap|pπ is called the π-component of A and π is the set of primes not in π. Moreover, T(A) and D(A), respectively, denote the torsion subgroup and the maximum divisible subgroup of the abelian group A.

If nA=0 for some 0n, we say that A is bounded. If p is a prime, by exponent m=exp(A) of a p-group A we mean the smallest m such that pmA=0, if A is bounded. Otherwise we say that A is unbounded and write exp(A)=. Furthermore, (π) is the additive group of rational numbers whose denominator n is such that π(n)π and (p):=(p)/, while (n) is the ring of integers modulo n. Also, r0(A) (resp. rp(A)) denotes the torsion-free rank (resp. the p-rank) of A, i.e., the cardinality of a maximal independent subset of A of elements with infinite order (resp. with p-power order). The rank of A is r0(A)+supprp(A).

Let ΓAut(A) and XA. As usual, if aγX for each aX and γΓ, we say that X is Γ-invariant and denote by γ|X the restriction of γ to X. If we have A=A1A2, where for i=1,2 the subgroup Ai is γ-invariant and γ|Ai=γiAut(Ai), we write γ=γ1γ2with respect toA=A1A2.

In the sequel, commutators are calculated in the holomorph group AAut(A). Moreover, if φ is an endomorphism of the additive abelian group A and aA, we use the notation [a,φ]:=aφ-a=a(φ-1).

Now we recall some known facts and introduce some notions which will be used in the rest of the paper.

Power automorphisms of an abelian group.

Recall that an automorphism leaving every subgroup invariant is called a power automorphism. Clearly, the set PAut(A) of power automorphisms of an abelian group A is a subgroup of Aut(A).

If A is an abelian p-group and α=i=0αipi (with 0αi<p) is an invertible p-adic integer, then, by a slight abuse of notation, one defines the power automorphism α of A by setting aα:=(i=0k-1αipi)a, for any aA of order pk. This gives an action on A of the group 𝒰p of units of the ring of p-adic integers, whose image is known to be the whole PAut(A). If exp(A)=, this action is faithful and, if exp(A)=:e<, its kernel is {α𝒰pα1 mod pe}. Thus we recall the following fundamental facts (see [18, Section 13.4.3 and point 9 of Exercises 13.4]).

Fact 2.1

Let A be an abelian group.

  1. If A is a p-group and exp(A)=, then PAut(A) is isomorphic to 𝒰p.

  2. If A is a p-group and e:=exp(A)<, then PAut(A) is isomorphic to the group of units of the ring (pe).

  3. If A is any periodic abelian group, then PAut(A) is isomorphic to the cartesian product of all the groups PAut(Ap) where Ap is the p-component of A and p ranges in .

  4. If A is non-periodic, then PAut(A)={±1}, i.e. PAut(A) consists only of the identity map and the map x-x.

  5. In any case PAut(A) is a central subgroup of Aut(A).

Invertible multiplications of an abelian group.

According to [11], let us consider a generalization of power automorphisms.

Definition 2.2

An automorphism γ is called an invertible multiplication of A if one of the following holds:

  1. A is periodic and γ is a power automorphism of A,

  2. A is non-periodic and

    1. there exist coprime integers m,n such that (na)γ=ma, for each aA.

Note that if (*) holds, even in the case that A is periodic, we necessarily have mnA=A and Aπ(mn)=0. Moreover, if A is periodic, then clearly (*) implies (a). In any case, by abuse of notation, we will write γ=m/n and say that γ acts on A as multiplication by m/n. Furthermore, if A is a p-group, α𝒰p and aγ=aα as defined above, we will write γ=α and say that γ acts on A as multiplication by α.

We warn the reader that we are using the word “multiplication” in a way different from [17] and that also in [8] the terminology is somewhat different. Moreover, sometimes we will omit writing the word “invertible”.

Proposition 2.3

The invertible multiplications of an abelian group A form a central subgroup of Aut(A).

Proof.

If A is periodic, the statement follows from Fact 2.1. Let then A be non-periodic. It is clear that the set of invertible multiplications is a subgroup of Aut(A). Let then γ,δAut(A) where (na)γ=ma, for each aA, with m,n coprime. Then for each xA we have

n(xδγ-xγδ)=n(xδ)γ-(nxγ)δ=m(xδ)-(mx)δ=0.

It follows

xδγ=xγδ,

since Aπ(nm)=0. Thus δγ=γδ. ∎

Inertial automorphisms of an abelian group.

In a lemma, we recall now some non-elementary facts that will be used in the sequel. They follow from [11, Propositions 2.2, 2.3, 3.3 and Theorem A].

Lemma 2.4

Let γ be an automorphism of an abelian group A.

  1. If A is torsion-free and γ is inertial, then γ is an invertible multiplication.

  2. If r0(A)=, then γ is inertial if and only if there are a subgroup A0 of finite index in A and an integer m such that γ acts as invertible multiplication by m on A0.

  3. If 0<r0(A)<, then γ is inertial if and only if

    1. there is a torsion-free γ-invariant subgroup V, which is finitely generated as a γ-submodule, and a rational number m/n (with m and n coprime integers) such that γ acts on V as the invertible multiplication by m/n,

    2. γ is inertial on the periodic group A/V,

    3. Aπ is bounded and A/Aπ is π -divisible, where π:=π(mn).

  4. If A is periodic, then γ inertial if and only if γ is inertial on each p-component Ap of A and acts as invertible multiplication on all but finitely many of them.

  5. If A is a p-group, then γ is inertial if and only if either γ acts as an invertible multiplication on a subgroup A0 of finite index in A or

    1. D:=D(A)0 has finite rank and A/D is infinite and bounded,

    2. there is a γ-invariant subgroup A1 of finite index in A such that γ acts as an invertible multiplication (by possibly different p-adics) on both A1/D and D.

Notice that if V is as in (2) above, then there are finitely many elements aiA such that, in the holomorph group of A, we have that V=a1,,anΓ is isomorphic to the direct sum of r0(A) copies of (π).

Corollary 2.5

Invertible multiplications of an abelian group A with r0(A)< are inertial.

Proof.

If A is periodic, the statement follows from Fact 2.1. Let then γ=m/n on A and V=Fγ where F0 is a free subgroup of A such that A/F is periodic. Then V/F is contained in the π-component of A/F. Hence V is torsion-free, as Aπ(mn)=0. Moreover γ=m/n on A/V, thus γ acts as a power automorphism on A/V. Thus, we may apply Lemma 2.4 (3) and deduce that γ is inertial. ∎

Now we state some facts that follow from [11, Theorem B and Corollary B].

Fact 2.6

Let A be an abelian group. Then:

  1. the subgroup IAut(A) is the set of the products γ1γ2-1 where γ1 and γ2 are both inertial automorphisms,

  2. if r0(A)<, then IAut(A) is the set of inertial automorphisms,

  3. the group IAut(A)/FAut(A) is abelian.

The group Q(A) of automorphisms of a non-periodic abelian group A.

Now we define a group Q(A) if inertial automorphisms of A of a particular type.

Lemma 2.7

Let A be a non-periodic abelian group and let p be a prime such that A/Ap is p-divisible and one of the following holds:

  1. Ap is finite,

  2. r0(A) is finite and Ap is bounded.

Then there is a unique subgroup C such that A=ApC and the automorphism γ(p):=1p (with respect to this decomposition) is an inertial automorphism of A.

Proof.

The existence of C follows from the fact that Ap is bounded (see [17]). For the uniqueness, if C1 is in the same conditions as C, we have that the group C/CC1 is trivial, as it is both bounded and p-divisible. Since C is a fully invariant subgroup of A, each γ(p) is inertial by Lemma 2.4. ∎

Definition 2.8

If π*(A) is the set of primes p as in Lemma 2.7 above, then we define Q(A):=γ(p)pπ*(A)×{±1}, where 1 and -1 denote the maps idA and -idA resp.

Proposition 2.9

If A is a non-periodic abelian group, then Q(A) is isomorphic to the multiplicative group of units of Q(π*(A)) and is a central subgroup of IAut(A).

Proof.

Apply Lemma 2.7, Proposition 2.3 and note that the rule -1-idA and pγ(p) (for each p) defines the wished isomorphism. ∎

The group IAut1(A) and finitary automorphisms of an abelian group A.

To answer the question in the Introduction, we will reduce a to a somewhat smaller group of automorphisms.

Definition 2.10

Let IAut1(A) be the set of inertial automorphisms of the abelian group A that act as the identity map on A/T(A).

Clearly IAut1(A) is a normal subgroup of IAut1(A). When A is non-periodic, we have

IAut1(A)PAut(A)=1.

When A is periodic, it holds

IAut1(A)=IAut(A).

Note that in the sequel we will apply the basic fact (1) below without reference.

Proposition 2.11

Let A be an abelian group. Then:

  1. an automorphism γ of A is finitary if and only if the subgroup A(γ-1) is finite,

  2. FAut(A)IAut1(A),

  3. FAut(A)=IAut1(A), provided r0(A)=.

Proof.

To prove statement (1), consider the endomorphism γ-1 of A and that A(γ-1)ker(γ-1). To prove (2) note that if γFAut(A), then by (1) we have A(γ-1)T(A). So that γ acts trivially on A/T(A). Finally, if r0(A)=, then each γIAut(G) acts as multiplication by m on a subgroup with finite index in A by Lemma 2.4 (2). By considering the endomorphism γ-m we have that A(γ-m)A/ker(φ-m) is finite. Therefore γ acts as the multiplication by m on A/T(A). If γIAut1(A), then m=1 and γFAut(A). ∎

Almost-power automorphisms of an abelian group.

Recall that the group of the so-called almost-power automorphisms of A, that is, automorphisms γ such that every subgroup of A contains a γ-invariant subgroup of finite index was introduced in [16], where generalized soluble groups in which subnormal subgroups are normal-by-finite (or core-finite, according to the terminology of [4] and [6]) are studied.

Clearly an almost-power automorphism γ has the following property:

  1. H and Hγ are commensurable for each HA,

which is stronger than the property of being inertial as defined in the Introduction.

Actually, in [8] we called inertial an automorphism γ with (), while the definition of inertial that we are using in the present paper is different (and is the same as in [11]). However, there is no risk of misunderstandings, since applying results from [8] we have the following.

Fact 2.12

Let A be an abelian group. Then:

  1. if A is periodic, then all inertial automorphisms are almost-power,

  2. if A is non-periodic, then the group of almost-power automorphism of A is IAut1(A)×{±1},

  3. if r0(A)<, then all inertial automorphisms γ have the property (),

  4. if r0(A)=, then an automorphism γ of A with the property () is almost-power.

3 Finitary automorphisms and stability groups

We begin this section by stating some basic facts that perhaps are already known (see [5]).

If ΓAut(A) and X is a subgroup of A, then we denote by StΓ(A,X) the stability group in Γ of the series AX0, that is, the set of γΓ such that X[A,γ]:=A(γ-1) and [X,γ]=0. When Γ=Aut(A), then we will write St(A,X):=StAut(G)(A,X).

If X is Γ-invariant, then each γΓ acts via conjugation on the abelian Γ-invariant subgroup Σ:=St(A,X) of Γ, according to the rule σγ-1σγ=:σγ for each σΣ. Similarly, γ acts on the additive group Hom(A/X,X) of homomorphisms A/XX by a corresponding formula, i.e. φγ|A/X-1φγ|X, where φHom(A/X,X) and γ|A/X denotes the group isomorphism induced by γ on A/X.

Thus, by extending the above actions (and the natural action of on any abelian group) both St(A,X) and Hom(A/X,X) are equipped with a structure of right Γ-module, where Γ denotes the group ring constructed in the usual way (see [18]).

Fact 3.1

Let A be an abelian group, ΓAut(A) and X a Γ-invariant subgroup of A. For each σStΓ(A,X), let σ be the well-defined homomorphism

a¯A/Xaσ-aX.

Then the map :σSt(A,X)σHom(A/X,X) is an isomorphism of Γ-modules, so that for each γAut(A) we have

(σγ)=γ|A/X-1(σ)γ|X.

Proof.

Let σ, τSt(A,X), and a¯A/X. On the one hand

a¯(σ+τ)=(aσ-a)+(aτ-a).

On the other hand we have (aσ-a)τ-(aσ-a)=0, thus

a¯(στ)=aστ-a=aτ+aσ-2a.

So is a group isomorphism, the inverse being the map

φHom(A/X,X)νφι+idSt(A,X),

where ν is the canonical homomorphism AA/X, ι is the embedding of X in A and id is the identity of A. Finally, for each γAut(A) and a¯A/X, we have a¯(σγ)=aσγ-a=aγ-1(σ-1)γ=a¯(γ|A/X-1(σ)γ|X) and is an isomorphism of Γ-modules. ∎

By this argument we have two technical lemmas. For the first one see [7].

Lemma 3.2

Let A be an abelian group, σ,γAut(A) and m1,m2Z. If σ stabilizes a series 0A1A, where γ=m1 on A1 and γ-1=m2 on A/A1, then σγ=σm1m2.

Proof.

Applying Fact 3.1 with X:=A1 and Γ=γ, for any a¯A/X, we have

a¯(σγ)=a¯γ|A/X-1(σ)γ|X=(m2a¯)(σ)γ|X=m1((m2a¯)σ)=a¯(m1m2(σ)).

Since is an isomorphism from a multiplicative to an additive group, it follows that σγ=σm1m2. ∎

Our next lemma deals with the case when A splits over X and will be used several times. In such a situation, once we have fixed a direct decomposition A=XK, we have an embedding Aut(K)Aut(A) given by γ1γ.

Lemma 3.3

Let A=XK, ΓAut(A), let X be a Γ-invariant subgroup of A and let ζ:A/XK the natural isomorphism. Furthermore, let Σ:=StΓ(A,X), Γ1:={γ|X1γΓ} and Γ2:={1ζ-1γ|A/XζγΓ}.

  1. If Γ1Γ, then Γ=CΓ(X)Γ1 and CΓ(A/X)=ΣΓ1.

  2. If Γ2Γ, then Γ=CΓ(A/X)Γ2 and CΓ(X)=ΣΓ2.

  3. If Γ1Γ2Γ, then Γ=Σ(Γ1×Γ2).

Proof.

To prove the first statement, for each γΓ, let γ¯:=γ|X1Γ1. Then δ:=γγ¯-1CΓ(X), hence Γ=CΓ(X)Γ1. Moreover CΓ(X) is normal in Γ, as X is Γ-invariant, and CΓ(X)Γ1=1, as it centralizes both X and K. If γCΓ(A/X), then δΣ and hence (1) is proved.

Statement (2) can be proved similarly and then (3) follows directly. ∎

Proposition 3.4

Let A be an abelian group and T:=T(A).

  1. If r0(A)<, then the automorphisms induced by FAut(A) via conjugation on St(A,T) are finitary.

  2. If r0(A)= and the quotient A/T is free abelian, then there is γFAut(A) which induces via conjugation on St(A,T) a non-finitary automorphism, provided FAut(T)1.

Proof.

(1) Denote A¯=A/T and fix γFAut(A). Then γ|A/T=1 by Proposition 2.11. So by Fact 3.1 with X:=T, for each σSt(A,T) we have

[σ,γ]=(σ-1σγ)=-σ+σγ=-σ+σγ|T=σ(γ|T-1)=:φσ.

Thus, we have to check that the set {φσσSt(A,T)} is finite. For each σ, we have that im(φσ)im(γ-1) has finite order, say n. On the other hand, ker(φσ)nA¯ and A¯/nA¯ is finite since r0(A¯)<.

(2) If A=TK, where K is free abelian on the infinite -basis {ai}, take γ0FAut(T){1}. Let tT such that tγ0t and γ:=γ01. For each i define σiSt(A,T) by the rule ai(σi-1):=t and aj(σi-1):=0 if ji. Then there are infinitely many [σi,γ], as aiker([σi,γ])aj for each ij. ∎

Example 3.5

If A=(p)(p), then St(A,T(A))FAut(A)=1.

Proof.

By Fact 3.1 we have that St(A,T(A))Hom((p),(p)) is infinite, while by using Proposition 2.11 (1) it is easy to check that FAut(A)=1. (see also Example 5.3). ∎

Despite the above example, we will see that for some relevant characteristic subgroups X of A, we have St(A,X)FAut(A), provided that one of the following holds:

  1. A/X is bounded and X has finite rank, as in Propositions 3.6 and 4.1 (2),

  2. A/X has finite rank and X is bounded, as in Theorem 6.1,

  3. A/X is finitely generated and X is periodic, as in Theorem 6.2.

Proposition 3.6

Let A be an abelian p-group such that D:=D(A) has finite rank and A/D is bounded. Then Σ:=St(A,D) is a bounded abelian p-group and there is a subgroup ΦFAut(A/D) such that

FAut(A)=ΣΦ,

where the automorphisms induced by Φ via conjugation on Σ are finitary and this action is faithful.

Proof.

First note that if σΣ, then [A,σ]=A(σ-1) is finite, since it is both of finite rank and bounded. Hence, we have σFAut(A). Consider a decomposition A=DB and apply Lemma 3.3, with X=D and Γ=FAut(A)=CΓ(X). Put Φ:=Γ2. Then FAut(A)=ΣΦ, as claimed.

Let γΦ. We have to show that set {[σ,γ]σΣ} is finite. Since γ|D=1, as in Proposition 3.4, we have

[σ,γ]=(σ-1σγ)=-σ+σγ=-σ+γ|A/D-1σ=(γ|A/D-1-1)σ=:φσ.

Thus, we have to count how many homomorphisms φσ there are. On the one hand, ker(φσ) contains ker(γ|A/D-1-1) which has finite index in A/D. On the other hand, the image of each φσ is contained in the finite subgroup D[pm], where pm is a bound for A/D. Therefore, there are only finitely many φσ, once γ is fixed.

Let us check that the action is faithful. Let 1γΦ and let bB with maximal order and bbγ. Then we have B=bB0 for some B0B and we can write bγ=nb+b0 with n,b0B0. If bnb, then there is σΣ such that B0(σ-1)=0 and b(σ-1)=d where dD has the same order as b. Thus, by Fact 3.1, bγ(σγ-1)=bγ(γ-1(σ-1))=d, while bγ(σ-1)=nd. Therefore, we have σγσ. Similarly, if b=nb, then there is σΣ such that b(σ-1)=0 and b0(σ-1)=d1 of order p. Then we have bγ(σγ-1)=0, while bγ(σ-1)=d1 and again σγσ. ∎

We now see that, in Proposition 3.6 the picture may be rather complicated. Recall that the FC-center of a group is the set of elements with finitely many conjugates or – equivalently – whose centralizer has finite index.

Example 3.7

There is an abelian p-group A such that D:=D(A) has finite rank, A/D is bounded and Σ:=St(A,D) is not contained in the FC-center of FAut(A).

Proof.

Write A=DB0, where D(p) and B0=ibiB is infinite and homogeneous. Fix σΣ such that b1(σ-1)=d, where d is an element of D of order p, and σ-1=0 on Dj1bj. For each i consider γiFAut(A) switching bib1 and acting trivially on D(j{1,i}bj). Then

σγi=γi-1(σ-1)+1.

Hence, biσγi=d+bi and bjσγi=bj for each ji. ∎

We now use a similar argument in the case when X=T(A).

Proposition 3.8

Let A be an abelian group with r0(A)< such that A/T is finitely generated (resp. T:=T(A) is bounded). Then Σ:=St(A,T) is a periodic (resp. bounded) abelian group and there is a subgroup Φ1FAut(T) such that

FAut(A)=ΣΦ1,

where Φ1 induces via conjugation on Σ finitary automorphisms.

In the case when A/T0 is finitely generated, this action by conjugation is faithful.

Proof.

In any case, we can write A=TK, where r:=r0(K)<. Recall that ΣHom(A/T,T). Note that ΣFAut(A). In fact, if σΣ, then we have that σ-1Hom(A/T,T) and A(σ-1) is an abelian group which is both finitely generated and periodic (resp. finite rank and bounded). Hence, A(σ-1) is finite that is σFAut(A).

Clearly Φ1:={φ1φFAut(T)}FAut(T) and Φ1FAut(A). Then, by Lemma 3.3 (1) we have that FAut(A)=ΣΦ1. By Proposition 3.4, Φ1 induces via conjugation on Σ finitary automorphisms.

If A/T is finitely generated, then ΣHom(A/T,T) is a periodic abelian group which is naturally isomorphic to the direct sum of r copies of T as a right Aut(A)-module. Therefore, the action of Φ1 on Σ is faithful. ∎

Example 3.9

In the notation of Proposition 3.8, if A=12(2), then Σ is not self-centralizing in FAut(A), that is, the action that Φ1 induces via conjugation on Σ is not faithful.

Proof.

We have FAut(A)/Σ𝒰12 and Σ3. ∎

4 The group IAut(A) when A is periodic

To give a detailed description of IAut(A) when A is a p-group, let us introduce some terminology.

Definition

If A is a p-group, by the essential exponente=eexp(A) of A we mean the smallest e such that peA is finite, or e= if A is unbounded. In the former case, this is equivalent to saying that A=A0A1A2, where A0 is finite, exp(A1)<eexp(A0) and A2 is the sum of infinitely many cyclic groups of order pe. In [8] we called critical a p-group of type A=BD with B infinite but bounded and D0 divisible with finite rank (see Lemma 2.4 (5)).

Critical groups will be a tool to describe IAut(A) when A is periodic.

Proposition 4.1

Let A be an abelian p-group and D:=D(A).

  1. If A is non-critical, then

    IAut(A)=PAut(A)FAut(A),

    where PAut(A)FAut(A) is either trivial or cyclic of order pm-e, according as A is unbounded or m:=exp(A)< and e:=eexp(A).

  2. If A=DB is critical, let

    Δ:={1nnp},Φ:={1φ0φ0FAut(B)},Ψ:={1γ0γ0IAut(B)}.

    Then

    IAut(A)=PAut(A)×(FAutAΔ).

    Moreover, FAutAΔ=CIAut(A)(D)=ΣΨ, where FAut(A)=ΣΦ and the following hold:

    1. Σ:=St(A,D) is an infinite abelian p-group, exp(Σ)=exp(B)=:m< and eexp(Σ)=eexp(B)=:e,

    2. Ψ=ΦΔIAut(B), where [Φ,Δ]=1 and Ψ induces via conjugation on Σ inertial automorphisms and this action is faithful,

    3. ΔPAut(B)𝒰((pm)), each δn:=1nΔ acts via conjugation on Σ as multiplication by n and FAut(A)Δ has order pm-e,

    4. ΦFAut(B) and Φ induces via conjugation on Σ finitary automorphisms.

Proof.

Let γΓ:=IAut(A).

(1) If A is non-critical, then, according to Lemma 2.4 (5), there exist a p-adic α and a subgroup A0 of finite index in A such that γ|A0=α. Thus, γ-1α acts on A0 as the identity map, that is, γ-1αFAut(A). Hence,

IAut(A)=PAut(A)FAut(A).

Further, if the multiplication by the p-adic number β=iβipi (0βi<p) is in PAut(A)FAut(A), then it is trivial on a subgroup B of finite index in A. Therefore, if exp(A)=, then exp(B)= and β=1. Otherwise, exp(B)e and then β1+i=e+1m-1βipimodpm (see Fact 2.1 and the definition of action given before). Thus, there are at most pm-e choices for such a β. On the other hand, the invertible multiplication on A by a p-adic number β1modpe is a finitary automorphisms since it acts trivially on A[pe].

(2) Let A=DB be critical. By Lemma 2.4 (5) there exists an invertible p-adic α such that γ|D=α. Thus, we get γ1:=γα-1CΓ(D). Clearly, we have PAut(A)CΓ(D)=1, so that IAut(A)=PAut(A)×CΓ(D).

Again by Lemma 2.4 (5), γ1 acts by multiplication by an integer n on a subgroup of finite index in A[pm] where A[pm]B. Therefore, if δn:=1nΔ with respect to A=DB, we have γ1δn-1FAut(A). Hence, CΓ(D)=FAut(A)Δ.

It is routine to verify that (i) holds, since Σ:=St(A,D)Hom(B,D). By Proposition 3.6, (iv) holds as well. By Lemma 3.3 (with X:=D, K:=B and so Γ2=Ψ), we have CΓ(D)=ΣΨ as stated in (2). Then, applying part (1) of the statement to B, we have [Φ,Δ]=1 and Ψ=ΦΔ=ΔΦ as stated in (ii). Moreover, FAut(A)Δ has order pm-e.

By Lemma 3.2, we have that Δ acts on Σ as in (iii). Thus, the whole of Ψ=ΔΦ acts via conjugation on Σ inducing inertial automorphisms and (ii) holds.

It remains to show that Ψ acts faithfully on Σ. Let φδnCΨ(Σ) with φΦ and δn:=1nΔ. On one hand, δn acts via conjugation on Σ as the multiplication by n by (iii). On the other hand, δn is finitary on Σ by (iv). Since eexp(Σ)=eexp(B) by (i), then multiplication by n is finitary on B. Thus, we have δnCΦ(Σ)=1 by Proposition 3.6. ∎

In next statement we regard FAut(Aπ) as naturally embedded in FAut(A).

Theorem 4.2

Let A be a periodic abelian group. Then there is a subgroup Δ of IAut(A) which is a direct product of finite abelian groups and such that

IAut(A)=PAut(A)FAut(A)Δ,

where Δ is trivial if A contains no non-trivial divisible subgroups.

Moreover, there are a set π of primes and subgroups Σ, Ψ of IAut(A) such that Σ is an abelian π-group with bounded primary components and

FAut(A)Δ=FAut(Aπ)×(ΣΨ),

where the automorphisms induced by Ψ via conjugation on Σ are inertial and this action is faithful.

Proof.

From Lemma 2.4 (4) we know that the group IAut(A) may be identified with PAut(A)DrpIAut(Ap). Apply Proposition 4.1 to each Ap. Let π be the set of primes p for which Ap is not critical. If pπ, we have

IAut(Ap)=PAut(Ap)FAut(Ap).

Otherwise, for each pπ, there are subgroups Δp,Σp,Ψp corresponding to Δ,Σ,Ψ in Proposition 4.1 such that IAut(Ap)=PAut(Ap)FAut(Ap)Δp and FAut(Ap)Δp=ΣpΨp. Now it is routine to verify that the statement follows by setting Δ:=DrpπΔp, Σ:=DrpπΣp, Ψ:=DrpπΨp, and recalling that FAut(A)=DrpFAut(Ap) as A=pAp. ∎

Proof of Corollary A.

In the notation of Theorem 4.2, we have that Δ is periodic abelian, PAut(A) is central and FAut(A) is locally finite as recalled in Section 2 and the Introduction, respectively. ∎

Lemma 4.3

If B0 is a subgroup of finite index in a bounded abelian group B, then there are subgroups B1 and B2 such that B2 is finite, B1B0 and B=B1B2.

Proof.

Clearly there is a finite subgroup F such that B=B0+F. Since B0 is separable and B0F is finite, then there is a finite subgroup B3B0F such that B0=B1B3 for some B1B0. Fix B1 and B2:=B3+F. On the one hand

B1+B2=B1+B3+F=B0+F=B.

On the other hand, by Dedekind’s law,

B1B2=B1(B3+F)
=B1(B0(B3+F))
=B1(B3+(B0F))
=B1B3=0.

Example 4.4

If A is a critical p-group (with p2) and ΛIAut(A) is such that CΓ(D)=FAut(A)Λ, then Λ is neither finite nor locally nilpotent.

Proof.

We use the same notation as in Proposition 4.1. Let n be a primitive root of 1modpm and consider δ:=1nΔ with respect to A=DB. Since ΔCΓ(D)=FAut(A)Λ, we can assume that φFAut(A) and λΛ. Hence, δ=λ=n on some subgroup B0 of finite index in B. By Lemma 4.3, B=B1B2 with B1B0 and B2 finite. Put A1:=D+B1 and note that λ|A1=1n with respect to A1=DB1.

It is sufficient to show that λΓ1 is infinite and not locally nilpotent, where Γ1 is the group of (inertial) automorphisms of A of type γ11 with respect to A=A1B2, with γ1IAut(A1). Thus, we may assume A1=A and Γ:=Γ1. Then multiplication by n is in Λ and Λ=ΔΓ.

We claim that ΔΓ=ΣΔ. In fact, by Proposition 4.1 we have that the group Δ𝒰(pm) acts faithfully by multiplication on the infinite abelian p-group Σ of exponent m and then Σ=[Σ,Δ] and ΔΓ=ΣΔ, as claimed. Thus, ΔΓ is not locally nilpotent, since the action of Δ on Σ is fixed-point-free. ∎

Remark that, in Theorem 4.2, when we consider the action of the above Ψ on the p-component Σp of the group Σ, we are concerned with subgroups of IAut(Σp)=PAut(Σp)FAut(Σp), where Σp is a bounded abelian p-group and PAut(Σp) is finite abelian.

5 The group IAut(A) when A is non-periodic

Let us state now our main results in the non-periodic case. Recall that metabelian groups G in which each subgroup of G is normal in G are called KI-groups and have been studied in a series of papers (see [20] and the references therein).

Theorem 5.1

Let A be a non-periodic abelian group. Then there is a central subgroup Q(A) of IAut(A), which is isomorphic to a multiplicative group of rational numbers, such that

IAut(A)=IAut1(A)×Q(A).

Moreover, there is a normal subgroup Γ of IAut1(A) such that:

  1. IAut1(A)/Γ is locally finite,

  2. Γ is a KI-group with periodic derived subgroup.

In particular, we have that if A is torsion-free, then IAut(A)=Q(A) is abelian, as IAut1(A)=1. Further, we will prove that in the statement of Theorem 5.1 one may take Γ to be the subgroup of IAut1(A) consisting of inertial automorphisms acting by multiplication on T(A). Unfortunately this subgroup need not be locally nilpotent, as in Example B.

Proof.

Let γ(p) and Q:=Q(A) as in Lemmata 2.7, 2.9 and Definition 2.8.

We first consider the case when r0(A)=. Let γIAut(A). By Fact 2.6 (1), we have γ=γ1γ2-1 with γ1,γ2 inertial. Further, by Lemma 2.4 (2), there is a subgroup A0 with finite index in A such that we have γ|A0=m/n=p1s1ptst (m,n coprime, pi prime, si). Also, IAut1(A)=FAut(A) and γ=m/n on A/T as well. If m/n=1, then γFAut(A). If m/n=-1, put γ0:=-1Q. Otherwise, since γ is invertible, we have that mA0=A0=nA0. Then, for each piπ:=π(mn), the pi-component of A is finite and A/T is pi-divisible. Consider then

γ0:=γ(p1)s1γ(pt)stQ.

In both cases, γγ0-1=1 on A0/(A0)π hence γγ0-1FAut(A). Thus,

IAut(A)=IAut1(A)×Q(A).

Moreover, (i) and (ii) are true with Γ=1, since IAut1(A)=FAut(A) is locally finite.

Let then r0(A)< and γIAut(A). By Fact 2.6 (2), γ is inertial. By Lemma 2.4 (3), we have that γ=m/n=p1s1ptst (m,n coprime, pi prime, si) on A/T. We also have that, for each piπ:=π(mn), the group A/T is pi-divisible and Api is bounded. Consider

γ0:=γ(p1)s1γ(pt)stQ.

Clearly γ0=m/n on A/T. Thus, γγ0-1 acts trivially on A/T and

IAut(A)=IAut1(A)×Q(A),

as stated.

Let Γ be the preimage of PAut(T) under the canonical homomorphism IAut1(A)IAut(T). Now statement (i) holds, since IAut1(A)/Γ is locally finite by Theorem 4.2. To check (ii), consider that the derived subgroup Γ of Γ stabilizes the series 0TA and therefore is abelian. Moreover, by Fact 2.6 (3), the subgroup Γ consists of finitary automorphisms. Thus, Γ is torsion and (ii) holds by Lemma 3.2. ∎

Proof of Corollary C.

Apply Theorem 5.1 and note that Q(A)Γ is metabelian where IAut(A)/Q(A)Γ is isomorphic to a quotient of IAut1(A)/Γ. ∎

When A/T is not finitely generated, it may happen that A has very few inertial automorphisms.

Proposition 5.2

Let A be a π-divisible non-periodic abelian group, where π is a set of primes. If T:=T(A) is a π-group, then IAut1(A)=1.

Proof.

If r0(A)=, then IAut1(A)=FAut(A). Moreover, if γFAut(A), then A(γ-1) is a finite π-group. Then A/ker(γ-1) is such. Hence, A=ker(γ-1) and FAut(A)=1.

If r0(A)<, by Lemma 2.4 (3) we have γ=1 on some free abelian subgroup VA such that A/V is periodic. Moreover, the π-component B/V of A/V is divisible. Then, by Lemma 2.4, parts (4) and (5), we have that γ is a multiplication on B/V. Furthermore, the group B/(V+T) is π-divisible and has non-trivial p-component for each pπ, since (V+T)/TV is free abelian. Thus, from γ=1 on B/T it follows that γ=1 on γ=1. Hence, γ stabilizes the series 0VB. However Hom(B/V,V)=0. Then γ=1 on B. Therefore, γ-1 induces a homomorphism A/BT which is necessarily 0 since A/B is a π-group. Thus, γ=1 on the whole group A. ∎

From Proposition 5.2 and Lemma 2.4 (3) we have

Example 5.3

If A=(p)(p), then IAut(A)={±1}.

6 The group IAut(A) when A splits on T(A)

The next two theorems consider cases in which A splits on its torsion subgroup T.

Theorem 6.1

Let A be an abelian group and T:=T(A). If r0(A)< and T is bounded, then Σ:=St(A,T) is a bounded abelian group and there is a subgroup Γ1 of IAut1(A) such that Γ1IAut(T) and

IAut1(A)=ΣΓ1,

where Γ1 induces via conjugation on Σ inertial automorphisms.

Proof.

We can write A=TK, where r:=r0(K)<. Note that the group ΣHom(A/T,T) is a periodic abelian group which is bounded as T.

Clearly Γ1:={γ1γIAut(T)}IAut(T). If γIAut(T), then γ1 (with respect to TK) is inertial by Lemma 2.4 (5), and so Γ1IAut1(A). Thus, we may apply Lemma 3.3 with Γ:=IAut1(A). We obtain IAut1(A)=ΣΓ1, as claimed.

By Proposition 4.1, we have

IAut(T)=FAut(T)PAut(T).

Hence, we have Γ1=Φ1Δ1, where Φ1:={φ1φFAut(T)}FAut(T) acts by conjugation on Σ by means of finitary automorphisms, by Proposition 3.8, and Δ1:={δ1δPAut(T)}PAut(T) acts via conjugation on Σ by means of multiplications, by Lemma 3.2. Therefore, the whole Γ1 induces by conjugation on Σ inertial automorphisms. ∎

We notice that the action of Γ1 on Σ in Theorem 6.1 need not be faithful, as already seen in Proposition 3.8.

Theorem 6.2

Let A be a non-periodic abelian group and T:=T(A). If A/T is finitely generated, then Σ:=St(A,T) is a periodic abelian group and there is a subgroup Γ1 of IAut1(A) such that Γ1IAut(T) and

IAut1(A)=ΣΓ1

where Γ1 induces via conjugation on Σ inertial automorphisms and this action is faithful.

If in addition T is unbounded, then IAut1(A) is not nilpotent-by-(locally finite). Further, if A2 is unbounded, then IAut1(A) is not even (locally nilpotent)-by-(locally finite).

Proof.

As in the proof of Theorem 6.1, we can write A=TK, where K is finitely generated. The group ΣHom(A/T,T) is a periodic abelian group which is isomorphic to the direct sum rT of r:=r0(A)>0 copies of T as a right Aut(A)-module.

Clearly Γ1:={γ1γIAut(T)}IAut(T). If γIAut(T), then γ1 (with respect to TK) is inertial by Lemma 2.4 (5). Hence, Γ1IAut1(A). Thus, we may apply Lemma 3.3 with Γ:=IAut1(A), and we obtain

IAut1(A)=ΣΓ1.

Let us investigate now the action of Γ1 via conjugation on Σ. Assume first that T is a p-group. Let γIAut(T). By Proposition 4.1, we have γ=γ0φ, where φFAut(T) and either γ0PAut(T) or T is a critical p-group and γ0 induces multiplications on both D(T) and T/D(T). Recall that Σ is Aut(A)-isomorphic to rT. In the former case, that is if γ0PAut(T), then γ01 acts via conjugation on Σ as a power automorphism (that is a multiplication). In the latter case, Σ is critical as well and γ01 induces invertible multiplications on both D(Σ) and Σ/D(Σ). Thus, γ01 acts via conjugation on Σ as an inertial automorphism of Σ, by Lemma 2.4 (5). In both cases, by Proposition 3.8, φ acts via conjugation on Σ as a finitary automorphism. Hence, γ1 acts via conjugation on Σ as an inertial automorphism.

In the general case, when T is any periodic group and γIAut(T), then γ1 (with respect to TK) acts via conjugation as an inertial automorphism on all primary components Σp of Σ, by what we have seen above and the fact that ΣpHom(A/T,Ap). Similarly, since γ1 acts as a multiplication on all but finitely many primary components Ap of A, it acts the same way on all but finitely many Σp. Thus, γ1 is inertial on Σ by Lemma 2.4 (4).

It is clear that the action via conjugation of Γ1 on Σ is faithful as the standard action of Γ1 on T is such.

To prove the last part of the statement, note that in the case when T is unbounded, there exists a non-periodic multiplication α of T. Note that the automorphism μ:=α1 (with respect to TK) belongs to Γ1. If, by way of contradiction, Σ,μ is nilpotent-by-(locally finite), then there is s{0} such that Σ,μs is nilpotent, so there is n such that [Σ,nμs]=0, and hence 0=Σ(μs-1)n=Σ(αs-1)n. This is a contradiction, since Σ is unbounded as T is.

Finally, if the group A2 is unbounded, then Σ2 is unbounded as well. Let α be a non-periodic multiplication of A2. Then μ:=α11 with respect to the group A=A2A2K acts as non-periodic multiplication (by α) of Σ2 acting fixed-point-free on a primary component. Thus, μ (and any non-trivial power of μ as well) does not belong to the locally nilpotent radical R of IAut1(A). Therefore, IAut1(A)/R is not locally finite. ∎

Proof of Example B.

It follows from the last part of Theorem 6.2. ∎

Finally, we note that, despite the above propositions, in the general case the group IAut1(A) may be large.

Example 6.3

There exists an abelian group A with r0(A)=1 and Ap(p) for each prime p such that IAut(A)=IAut1(A)×{±1}, IAut1(A)=ΣFAut(A), where Σ:=StIAut(A)(A,T(A))FAut(A) with the property that Σp(p) and IAut1(A)/FAut(A)Σ/T(Σ) is a divisible torsion-free abelian group with cardinality 20.

Moreover, any element of IAut1(A) induces a finitary automorphism on T(A).

Proof.

As in [11, Proposition A], we consider the group G:=BC, where B:=pbp, C:=pcp, and bp, cp have order p, p2, respectively, and p ranges over all primes. Consider the (aperiodic) element v:=(bp+pcp)pG and V:=v. We have that for each prime p there is an element d(p)G such that pd(p)=v-bp. Let A:=V+d(p)|p and T:=T(A). Then

A/T1/pp,

since A/T has torsion free rank 1 and v+T has p-height 1 for each p. Thus, T=T(B)p(p) and the p-component of A/V is generated by d(p)+V and has order p2, since pd(p)=v-bp.

Then Σp(p) and ΣFAut(A)=T(Σ), hence ΣFAut(A). Moreover, A=d(p)+V, where V=v is infinite cyclic and Ap=bp has order p. Also, Aut(A/T)={±1} and IAut(A)=IAut1(A)×{±1}.

We claim that if γIAut1(A) induces on T a finitary automorphism, then γΣFAut(A). In fact, Tγ is finite, so it is a π-component of A for some finite π. Thus, γγ0-1Σ, where γ0:=γ|Aπ1 with respect to A=AπK and clearly γ0FAut(A).

Finally, we prove the last part of the statement, from which it follows that IAut1(A)=ΣFAut(A). To this end, let γIAut1(A) and φ:=γ-1. Since AφT, there exists a non-zero integer n such that (nv)φ=0. We prove that TφAπ(n), which is finite. For any prime p, on the one hand, nd(p) is a p-element modulo nvkerφ, hence (nd(p))φAp, that implies

(pnd(p))φ=p(nd(p))φ=0.

On the other hand,

(pnd(p))φ=n(v-bp)φ=-n(bp)φ.

Hence, if p(n)φ, then Apφ=0. ∎


Dedicated to Martin L. Newell



Communicated by Andrea Lucchini


Acknowledgements

We thank the referee for her/his useful comments.

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Received: 2016-3-8
Revised: 2016-8-22
Published Online: 2016-9-14
Published in Print: 2017-3-1

© 2017 by De Gruyter

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