# Anticentral and Antisemicentral Elements of Infinite Soluble Groups

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## Abstract

An element *a* of a group G is said to be anticentral in *G* if *G* ^{′} = {[*g*, *a*] : *g* ∈ *G*} and antisemicentral in *G* if {[*g*,*a*] : *g* ∈ *G*} contains a normal subgroup of *G* of finite index in *G* ^{′}. We study such elements in various types of infinite soluble group.

## Keywords

Infinite soluble group Nilpotent group Locally nilpotent group Anticentral elements## Mathematics Subject Classification (2010)

20F16 20F18 20F19## 1 Introduction

The notion of an anticentral element of a group has been around for a decade, though I have been unaware of it until very recently. The earliest published work on it, as far as I know, is Ladisch [4]. Most of the past work concerns finite groups, but it turns out that some work of mine in a different context is relevant for extending this to certain types of infinite groups.

*a*of a group

*G*is anticentral if

*a*

*G*

^{′}=

*a*

^{ G }, equivalently if

*G*

^{′}= {[

*g*,

*a*] :

*g*∈

*G*}. This concept lies somehow between the notions of a fixed-point-free (fpf for short) automorphism and a splitting automorphism. Suppose

*G*=<

*a*>

*H*is the split extension of its normal subgroup

*H*by the cyclic group <

*a*>. Let

*α*∈Aut

*G*denote conjugacy on

*G*by

*a*. Define maps

*γ*=

*γ*(

*α*) and

*ψ*

_{ i }=

*ψ*

_{ i }(

*α*) of

*G*into itself for integers

*i*≥ 1 by

Generally, when working with a group *G* and its automorphism group we actually work in the holomorph of *G*. Thus, for example, in the above *g* ^{− 1} ⋅ *g* *α* actually is the commutator [*g*,*a*] in the usual sense but in the holomorph of *G*. Note that *γ* need not be a homomorphism. In spite of this, we still use the symbol ker*γ* for {*g* ∈ *G* : *g* *γ* = 1}. A similar remark applies to *ψ* _{ i }.

If *H* *ψ* _{ n } = {1}, we say *α* is a *n*-splitting of *H* and if *α* has order *n* with *H* *ψ* _{ n } = {1} we just say *α* is a splitting of *H*. Suppose *α* has order *n* on *H* and order *m* on *G* ^{′}, so *m* divides *n*. All the following are presumably well known.

### **Lemma 1**

*Suppose*H

*is finite and consider the following four statements.*

- a)
*α**acts**fpf*^{ l y }*on**G*^{′}*.* - b)
a

*is an anticentral element of*G*; equivalently**G*^{′}=*G**γ**.* - c)
*α**yields an*n*-splitting automorphism of**G*^{′}*;**equivalently**G*^{′}*ψ*_{ n }= {1}*.* - d)
*α**yields a splitting automorphism of**G*^{′}*;**equivalently**G*^{′}*ψ*_{ m }= {1}*.*

*Then*a)

*implies*b)

*and*d), b)

*implies*c)

*and*d)

*implies*c)

*. There are no further*

*implications in general between*a), b), c)

*and*d)

*.*

Here we consider how far these implications extend to infinite groups. (For locally finite groups, these and related questions have already been considered in [1] and [2]). The connections between b), c), and d) are easy to resolve, are quite general, and hence are uninteresting (but see Lemmas 2 and 3 below). However, the implications of a) are much more delicate, restricted and interesting. One cannot be too ambitious because of the following easy example. Let *H* be the free group of rank *n* ≥ 2 on *x* _{1},*x* _{2},…,*x* _{ n } and *α* the automorphism of *H* that permutes the *x* _{ i } cyclically. Then *α* acts fpf^{ l y } on *H* but is not anticentral in *G* =< *α* > *H*; for \(y=[{x_{2}^{2}}, x_{1}]\in G^{\prime } \backslash G\gamma \); also *y* *ψ* _{ n }≠ 1. In fact *y* *ψ* _{ r }≠ 1 for all *r* ≥ 1. (We only need the square if *n* = 2).

If we weaken our assumptions in a) to *α* being almost fpf (afpf for short), meaning that *C* _{ H }(*α*) is finite, in b) to *α* (or equivalently the element *a*) being antisemicentral, meaning that *G* *γ* contains a normal subgroup of *G* of finite index in *G* ^{′} and c) to *α* inducing an *n*-splitting on a normal subgroup of *G* of finite index in *G* ^{′}, then our conclusions become much more general. They are also more symmetric than the finite case above.

### **Theorem 1**

*Let*H

*be a group,*

*α*

*an automorphism of*H

*and*G

*the split extension of*H

*by*<

*α*>

*.*

- a)
*If*H*is locally nilpotent and if**α**is fpf and of order 2, then**α**is an anticentral element of*G*.* - b)
*For each**m*> 2*,**there exists a finitely generated torsion-free nilpotent**group*H*of class 2 and a fpf automorphism**α**of*H*of order*m*such that**α**is not anticentral in*G*.* - c)
*There exists an abelian-by-(finite cyclic),**polycyclic group*H*and a fpf automorphism**α**of*H*of order 2 such that**α**is not anticentral in*G*.* - d)
*If*H*is the wreath product of two infinite cyclic groups (so*H*is torsion-free,**finitely generated and metabelian), then*H*has a fpf automorphism**α**of order 2 such that**α**is not anticentral in*G*.* - e)
*If*H*is abelian, then**α**is anticentral in*G*.*

Part e) is very easy; we have only included it to point out that we cannot reduce nilpotent of class 2 in b) or metabelian in c) or d) just to abelian. In passing, we remark that in the notation above and of Theorem 1 in particular, the condition *H* = *H* *γ* is much stronger than the condition *G* ^{′} = *G* *γ*, even although always *G* ^{′}≤ *H* and *G* *γ* = *H* *γ*. For example, if *H* is finitely generated and soluble-by-finite (actually finitely generated hyper (abelian or finite) would do), if *α* has finite prime order *p* and if *H* = *H* *γ*, then *H* is a finite nilpotent *p* ^{′}-group, see [6, 10.53].

An FAR group is a soluble-by-finite group with finite Hirsch number that also satisfies the minimal condition on *p*-subgroups for every prime *p*. See [5], specially Chapter 5, for alternative definitions and basic properties of FAR groups. Note that all soluble-by-finite groups of finite rank and hence all minimax groups and all polycyclic-by-finite groups are FAR groups. The following two theorems are little more than reinterpretations of results in [11] and [10].

### **Theorem 2**

*Let*G

*be a nilpotent*

*FAR group,*n

*a positive integer and*a

*an element of*G

*of order dividing*n

*. Let*

*α*

*denote conjugation by*a

*. The following are equivalent.*

- a)
*α**induces a afpf automorphism on**G*^{′}*.* - b)
a

*is an antisemicentral element of*G*.* - c)
*There exists a normal subgroup*N*of*G*of finite index in**G*^{′}*such that**α**induces an*n*-splitting on*N*.*

### **Theorem 3**

*Let*G

*be an FAR group,*p

*a prime and*a

*an element of*G

*of order*p

*. Let*

*α*

*denote conjugation by*a

*. The following are equivalent.*

- a)
*α**induces a afpf automorphism on**G*^{′}*.* - b)
a

*is an antisemicentral element of*G*.* - c)
*There exists a normal subgroup*N*of*G*of finite index in**G*^{′}*such that**α**induces a*p*-splitting on*N*.*

We say that a subset of a group *G* is of finite index in *G* if it contains a subgroup of *G* of finite index. With *G* and *α* as in Theorem 2 and using the notation above, the corollary to the theorem of [11] says in particular that the following statements are equivalent: ker*γ* is finite; *G* *γ* is of finite index in *G*; ker*ψ* _{ n } is of finite index in *G*; *G* *ψ* _{ n } is finite. The first three of these applied to the action of *α* on *G* ^{′} and also the very simple Lemma 2 below yield Theorem 2. (Note that any FAR group of finite exponent is finite, so if we have *X* ≤ *Y* ≤ *Z*, where *Z* is an FAR group, *Y* is normal in *Z* and *X* has finite index in *Y*, then *X* contains a normal subgroup of *Z* of finite index in *Y*). In a similar way, Theorem 3 follows from [10, Theorem 1].

## 2 The Proofs

*G*be a group,

*a*an element of

*G*of finite order

*n*and denote conjugation by

*a*on

*G*by

*α*. With

*γ*and

*ψ*

_{ r }defined as above, we have

*G*

*γ*

*ψ*

_{ n }= {1}; that is,

*G*

*γ*⊆ ker

*ψ*

_{ n }. Clearly

*G*

*γ*⊆

*G*

^{′}. Suppose

*α*restricted to

*G*

^{′}has order

*m*. If

*g*∈

*G*, then

*a*

^{ m }∈

*C*

_{ G }(

*G*

^{′}), so

*g*

*γ*

*ψ*

_{ m }∈

*C*

_{ G }(

*G*

^{′}). Therefore,

*G*

*γ*

*ψ*

_{ m }⊆

*ζ*

_{1}(

*G*

^{′}), the centre of

*G*

^{′}. Also

*m*divides

*n*, say

*m*

*d*=

*n*. Then for

*x*∈

*G*

^{′}we have

*x*

*ψ*

_{ n }= (

*x*

*ψ*

_{ m })

^{ d }. We have proved the following.

### **Lemma 2**

*With the notation above* *G* *γ* *ψ* _{ n } = {1},*G* *γ* *ψ* _{ m } ⊆ *ζ* _{1}(*G* ^{′})*and*(*G* *γ* *ψ* _{ m })^{ d } = {1}*.*

Continuing the notation above, if *a* is anticentral in *G*, then *G* ^{′} = *G* *γ* and if *a* is only antisemicentral then, there is a normal subgroup *N* of *G* of finite index in *G* ^{′} with *N* ⊆ *G* *γ*. The above yields the following.

### **Lemma 3**

*If*a

*is anticentral in*G

*, then*

*G*

^{′}

*ψ*

_{ n }= {1}

*and*

*α*

*is an*n

*-splitting of*

*G*

^{′}

*.*

*If*a

*is antisemicentral in*G

*, then with*N

*as above*

*α*

*is an*n

*-splitting of*N

*and*

### **Lemma 4**

*With the notation above, if*G

*is finite and if*a), b), c)

*and*d)

*are as below, then*a)

*implies*b)

*and*d)

*,*b)

*implies*c)

*and*d)

*implies*c)

*.*

- a)
*α**is fpf on**G*^{′}*.* - b)
a

*is anticentral in*G*.* - c)
*G*^{′}*ψ*_{ n }= {1}*.* - d)
*G*^{′}*ψ*_{ m }= {1}*.*

### *Proof*

If *α* is fpf on *G* ^{′}, then from [3, 10.1.1] we have *G* ^{′} = *G* ^{′} *γ* ⊆ *G* *γ* ⊆ *G* ^{′}, so *G* ^{′} = *G* *γ* and hence a) implies b). Lemma 3 yields that b) implies c). Since *x* *ψ* _{ n } = (*x* *ψ* _{ m })^{ d } for all x in *G* ^{′}, so d) implies c). Finally, a) implies that < *a* > ∩*G* ^{′} =< 1 > and, via [3, 10.1.1], that *G* ^{′} = *G* ^{′} *γ*. Then Lemma 2 applied to < *a* > *G* ^{′}/ < *a* ^{ m } > yields that *G* ^{′} *ψ* _{ m } = {1}. Thus a) implies d). □

### *Proof Proof of Lemma 1*

By Lemma 4, we have that a) implies b) and d), b) implies c), and d) implies c). We have now only to construct suitable counter examples.

Let *H* =< *x* > have order 8 and let *α* be inversion on H. Let G be the split extension of H by < *α* > and *a* = *α*. Then *G* ^{′} = *G* *γ* = *H* *γ* =< *x* ^{2} >,*m* = *n* = 2,*x* ^{4} *α* = *x* ^{4}≠ 1 and *H* *ψ* _{2} = {1}. Thus, *a* = *α* satisfies b), c) and d) but not a).

Let *H* =< *x* > × < *y*,*z* >, where |*x*| = 8,|*y*| = 4,|*z*| = 2 and *y* ^{ z } = *y* ^{− 1}. Let *α* ∈Aut*H* invert < *x* > and centralize < *y*,*z* > and set *G* =< *α* > *H*. Then *G* ^{′} =< *x* ^{2},*y* ^{2} >,*m* = *n* = 2 and *G* ^{′} *ψ* _{2} = {1}. Thus, *a* = *α* satisfies c) and d). But *G* *γ* = *H* *γ* =< *x* ^{2} > ≠*G* ^{′}, so a does not satisfy b).

Let *H* =< *x* > × < *y* >, where x and y have order 8. Define the map *α* of H by (*x* ^{ i } *y* ^{ j })*α* = *x* ^{ i } *y* ^{ i+ 3j }. Then *x* *α* ^{2} = *x* *y* ^{4},*x* *α* ^{4} = *x* and *y* *α* ^{2} = *y*. Thus, *α* is an automorphism of H of order 4. Then *n* = 4,*G* has order 256,*G* ^{′} =< *y* >= *G* *γ* and *α* satisfies b) and hence c). However *G* ^{′} =< *y* > is centralized by *α* ^{2}, so *m* = 2 and *y* *ψ* _{2} = *y* ⋅ *y* *α* = *y* ^{4}≠ 1, so *α* does not satisfy d). That is neither b) nor c) implies d). □

### **Lemma 5**

*Let* H *be a locally nilpotent group with a fpf automorphism* *α* *of order 2. Then* H *is abelian and inverted by* *α*,*H* *ψ* _{2} = {1}*and* *α* *is a splitting automorphism of* H*.*

### *Proof*

If X is a finitely generated subgroup of H, then so is *Y* =< *X*,*X* *α* > and, assuming *X*≠ < 1 >,*α* induces an fpf automorphism on Y of order 2. Thus, we may assume that H is finitely generated (and nilpotent). If T denotes the torsion subgroup of H, then T is finite and *H*/*T* is torsion-free.

By [8, Theorem 1], the group H has an abelian *α*-invariant normal subgroup A inverted by *α* and of finite index in H. Suppose *T* =< 1 >. Then H is isomorphic to a unipotent (and hence connected by [7, 6.6]) matrix group over the integers, see [7, Point 3, p. 23]. Then the Zariski closure of A in H is H itself and hence H is abelian ([7, 5.11]). In particular *α* inverts H.

*T*≠ < 1 > and set

*Z*=

*C*

_{ T }(

*H*). Then Z is finite of odd order (since

*α*is fpf), non-trivial and normalized by

*α*. Also

*α*inverts Z. Further,

*α*induces an fpf automorphism on

*H*/

*Z*by [9, Lemma 13 c)]. By induction

*H*/

*Z*is abelian and inverted by

*α*. Let

*h*∈

*H*. Then

*h*

*α*=

*h*

^{− 1}

*z*for some

*z*∈

*Z*. Now

*z*

*α*=

*z*

^{− 1}and

*α*

^{2}= 1. Hence

*z*is central. Therefore

*z*

^{2}= 1. But

*Z*has odd order. Consequently,

*z*= 1,

*h*

*α*=

*h*

^{− 1}and

*α*inverts

*H*. The remaining parts of the lemma now follow.□

### *Proof Proof of Theorem 1*

Part a). Now H is abelian and inverted by *α* by Lemma 5 and *G* =< *α* > *H*. With *g* *γ* = [*g*,*α*] as usual, for the integer s and *h* ∈ *H* we have (*α* ^{ s } ⋅ *h*)*γ* = *h* *γ* = *h* ^{− 2} and so *G* *γ* = *H* *γ* = *H* ^{2}. Also *α* centralizes *H*/*H* ^{2}, so *G*/*H* ^{2} is abelian and *G* ^{′}≤ *H* ^{2} = *G* *γ* ⊆ *G* ^{′}. Thus *G* ^{′} = *G* *γ* and *α* is an anticentral element of G.

Part b). Suppose for some *m* > 2, we have constructed a suitable H and *α*. Let *d* > 1, let c be a primitive mdth root of unity in the complex numbers **C** and set *C* = **Z**[*c*] ≤**C**. Then c is a fpf automorphism of C of order md and additively C is free abelian of finite rank.

Set *K* = *H* × *C*, extend *α* and c to K by letting *α* centralize C and letting c centralize H and set *β* = *α* *c*. Then *β* is a fpf automorphism of K of order md. Since *α* is not anticentral in *G* =< *α* > *H*, there exists x in *G* ^{′} such that *x*≠[*h*,*α*] for any h in H. Set *L* =< *β* > *K*. Then *G* ^{′} = *H* ^{′}[*H*,*α*] = *H* ^{′}[*H*,*β*] ≤ *L* ^{′}. Suppose *x* = [*y*,*β*] for some y in L. Then *x* = [*k*,*β*] for some k in K. But *x* ∈ *H*, so *x* = [*h*,*α*] for some *h* ∈ *H*, a contradiction. Thus *L* ^{′}≠{[*y*,*β*] : *y* ∈ *L*} and hence we have extended our counter example for m to one for *m* *d*,*d* > 1. Consequently, we now have only to construct examples for every odd prime p and for 2^{2}. We do these together by giving an example for each *m* = *p* ^{ r } ≥ 3 and every prime p.

*m*=

*p*

^{ r }≥ 3 and set

*f*=

*ϕ*(

*m*) =

*m*(

*p*− 1)/

*p*, so

*f*≥ 2. Let

*ω*be a primitive mth root of unity in

**C**and set

*J*=

**Z**[

*ω*] = ⊕

_{0≤i<f }

**Z**

*ω*

^{ i }≤

**C**. Now

*v*=

*v*

*ω*, where \(v={\sum }_{0\leq i<f}v_{i}\omega ^{i}\) and the

*v*

_{ i }are integers. Then

*v*

_{ f− 1}= −(

*p*− 1)

*v*

_{ f− 1}, so

*p*

*v*

_{ f− 1}= − 1. But

*v*

_{ f− 1}is an integer. Therefore, no such

*v*exists.

Let *H* be the 3 by 3 lower unitriangular group Tr_{1}(3,*J*) over *J*. If *h* = (*h* _{ i j }) ∈ *H*, represent *h* by *h* = (*h* _{12},*h* _{23},*h* _{13}) since these three entries of *h* determine *h*. Let *α* = diag(*ω* ^{2},*ω*,1) ∈ *G* *L*(3,*J*). If *y* = (*u*,*v*,*w*) ∈ *H*, then *y* ^{ α } = (*u* *ω*,*v* *ω*,*w* *ω* ^{2}). In particular, conjugation on *H* by *α* is a fpf automorphism of *H* of order *m* and *G* =< *α*,*H* >≤ *G* *L*(3,*J*) is the split extension of *H* by < *α* >.

Let *x* = (0,0,1) ∈ *H* and suppose *x* = [*y*,*α*],*y* ∈ *H* as above. Then *y* *x* = *y* ^{ α }, so (*u*,*v*,1 + *w*) = (*u* *ω*,*v* *ω*,*w* *ω* ^{2}). Therefore, *u* = *v* = 0 and 1 + *w* = *w* *ω* ^{2}. Hence 1 + *w*(*ω* + 1) = *w*(*ω* + 1)*ω*. By the above, no such element *w*(*ω* + 1) exists. Thus, *x*≠[*y*,*α*] for any *y* in *H*, so *x*≠[*g*,*α*] for any *g* in *G*. But *x* = [(0,1,0),(1,0,0)] ∈ *H* ^{′}≤ *G* ^{′}. Therefore, *α* is not an anticentral element of *G*. The construction is complete.

Part c). Let *p* be an odd prime, *ω* a primitive *p* th root of 1 in the complex numbers **C** and *A* = **Z**[*ω*] = ⊕_{0≤i<p− 1} **Z** *ω* ^{ i } ≤**C**. Let *h* ∈Aut*A* denote multiplication by *ω* and let *H* =< *h* > *A* denote the split extension of *A* by < *h* >.

*α*of

*H*into itself by

*a*

*α*= −

*a*for any

*a*in

*A*and

*α*is a fpf automorphism of

*H*of order 2.

*G*denotes the split extension of

*H*by <

*α*>. Let

*γ*denote the map of

*G*into itself given by

*g*

*γ*= [

*g*,

*α*]. Now

*H*

^{′}= {[

*a*,

*h*] :

*a*∈

*A*} = (

*ω*− 1)

*A*=

*A*. Also

*h*

*γ*= 1

_{ A }(the multiplicative identity of

*A*, not the identity 1.0 of

*H*), so

*G*/

*A*is abelian. Thus

*G*

^{′}≤

*A*=

*H*

^{′}≤

*G*

^{′}and

*G*

^{′}=

*A*. Clearly,

*A*

*γ*= {− 2

*a*:

*a*∈

*A*} = 2

*A*and

*G*

*γ*=

*H*

*γ*. If 0 <

*i*<

*p*, then

*H*

*γ*⋅ 2

*A*/2

*A*= {2

*A*,1 +

*ω*+ ⋯ +

*ω*

^{ i− 1}+ 2

*A*: 0 <

*i*<

*p*}. Therefore, the cardinality of

*H*

*γ*⋅ 2

*A*/2

*A*is

*p*. However (

*G*

^{′}: 2

*A*) = (

*A*: 2

*A*) = 2

^{ p− 1}. Since

*p*is odd, so 2

^{ p− 1}>

*p*. Consequently,

*G*

^{′}≠

*G*

*γ*and hence

*α*is not anticentral in

*G*.

If we take *p* = 2 in the above construction, the group *G* and map *α* do exist with *A* infinite cyclic, but they do not have the required properties. However, there does exist a suitable example with (*H* : *A*) = 2, but with *A* of rank 3, as we now show.

Let *A* =< *b*,*c*,*d* > be a free abelian group of rank 3. Define *h* ∈Aut*A* of order 2 by *b* ^{ h } = *c*,*c* ^{ h } = *b* & *d* ^{ h } = *d* ^{− 1} and set *H* =< *h* > *A* (the split extension). Then define *α* : *H* → *H* by *a* *α* = *a* ^{− 1} and (*h* *a*)*α* = *h* *d* *a* ^{− 1} for all *a* in *A*. Again simple calculations show that *α* is a fpf automorphism of *H* of order 2.

*G*=<

*α*>

*H*denote the split extension of

*A*by <

*α*>. Then [

*a*,

*α*] =

*a*

^{− 2}and [

*h*

*a*,

*α*] =

*d*

*a*

^{− 2}for all

*a*in

*A*. Therefore

*b*,

*h*] =

*b*

^{− 1}

*c*∉ <

*d*>

*A*

^{2}. Therefore,

*G*

*γ*≠

*G*

^{′}and hence

*α*is not anticentral in

*G*.

Part d). Let *H* =< *x*,*Y* >, where *Y* is the free abelian group on the *y* _{ i } for *i* ∈**Z**, the integers, and *x* acts on *Y* via \({y_{i}^{x}}=y_{i + 1}\) for each *i* (so *H* is the wreath product of two infinite cyclic groups). Define automorphisms *𝜃* and *ϕ* of *H* by *x* *𝜃* = *x* *y* _{0},*y* *𝜃* = *y*,*x* *ϕ* = *x* and *y* *ϕ* = *y* ^{− 1}, this for all *y* in *Y*. Then set *α* = *𝜃* *ϕ* ∈Aut*H*. It is easy to check that *α* has order 2.

*y*

*α*=

*y*

^{− 1}, so

*α*is fpf on

*Y*. Let

*r*be a positive integer. Then

*y*∈

*Y*with (

*x*

^{ r }

*y*)

*α*=

*x*

^{ r }

*y*, then

*x*

^{ r }

*y*

_{0}

*y*

_{1}…

*y*

_{ r− 1}

*y*

^{− 1}=

*x*

^{ r }

*y*, so

*y*

_{0}

*y*

_{1}…

*y*

_{ r− 1}∈

*Y*

^{2}, which is false. If (

*x*

^{−r }

*y*)

*α*=

*x*

^{−r }

*y*, then we obtain

*y*

_{−r }…

*y*

_{− 2}

*y*

_{− 1}∈

*Y*

^{2}, which again is false. Consequently,

*α*is a fpf automorphism of

*H*.

*G*=<

*α*>

*H*. Now

*G*

^{′}contains \([y_{1},x]=y_{1}^{-1}y_{2}\in y_{1}y_{2}Y^{2}\). Clearly [

*y*,

*α*] =

*y*

^{− 2}∈

*Y*

^{2}. Also

*y*

_{1},

*x*]≠[

*g*,

*α*] for all

*g*∈

*G*,

*G*

^{′}≠

*G*

*γ*in our notation above and hence

*α*is not an anticentral element of

*G*. The construction is complete.

Part e). Since *H* is abelian, *G* ^{′} = [*H*,*α*] = *H* *γ* = *G* *γ*, where as above *γ* maps *g* to [*g*,*α*]. Therefore *α* is anticentral in *G*. □

### *Remark 1*

Let H be a polycyclic group with a fpf automorphism *α* of order 2. Then *A* = Fitt*H* is abelian and inverted by *α* (Part a) of Theorem 1, (*H* : *A*) is finite ([8, Theorem 1]) and *A* = *C* _{ H }(*A*) ([6, 2.17]). If we choose *p* = 3 in the construction of the proof of Part c) of Theorem 1, our counter example has A free abelian of rank 2 and (*H* : *A*) = 3. We also have the example with A free abelian of rank 3 and with (*H* : *A*) = 2. These are the smallest possible examples (with *α* not anticentral in G of course), as the following implies.

### **Lemma 6**

*With the above notation, if either* A *has torsion-free rank*(*=* *Hirsch number*) *at most*1*, or* A *has torsion-free rank 2 and*(*H* : *A*) = 2*,* *then* *α* *is anticentral in* *G* =< *α* > *H* *.*

### *Proof*

Here *A* = *B* × *T*, where B is free abelian of rank at most 2 and T is finite abelian. If H is finite or abelian the claim clearly holds (Lemma 1 and Part e) of Theorem 1), so assume otherwise.

Suppose first that A is free abelian of rank 2 and (*H* : *A*) = 2. Now *H* =< *h* > *A*, where *h* ^{2} ∈ *A* and *h* *α* = *h* *b* for some b in A. If *a* ∈ *A*, then [*a*,*α*] = *a* ^{− 2} and [*h* *a*,*α*] = *b* *a* ^{− 2}. Thus *G* *γ* = *H* *γ* = *A* ^{2} ∪ *b* *A* ^{2} =< *b* > *A* ^{2}. If *b* = *a* ^{2} for some a in A, then (*h* *a*)*α* = *h* *b* *a* ^{− 1} = *h* *a*, contradicting *α* fpf. Hence *b*∉*A* ^{2}.

Now *h* ^{− 2} = (*h* ^{2})*α* = (*h* *α*)^{2} = *h* *b* *h* *b* = *h* ^{2} *b* ^{ h } *b*. Thus *b* ^{ h } *b* = *h* ^{− 4} ∈ *A* ^{2}, so *b* ^{ h } *A* ^{2} = *b* *A* ^{2} and *G* *γ*/*A* ^{2} lies in the centre *ζ* _{1}(*H*/*A* ^{2}) of *H*/*A* ^{2}. Clearly |*H*/*A* ^{2}| = 8. If *H*/*A* ^{2} is abelian, then *H* ^{′}≤ *A* ^{2} ≤ *G* *γ*. If not, then *H* ^{′} *A* ^{2}/*A* ^{2} = *ζ* _{1}(*H*/*A* ^{2}), which has order 2. But *b*∉*A* ^{2}; hence *H* ^{′} *A* ^{2} = *G* *γ*/*A* ^{2}. Consequently in either case *H* ^{′}≤ *G* *γ*, so *G* ^{′} = *H* ^{′} *G* *γ* = *G* *γ* and *α* is anticentral in G.

Now assume that A is infinite cyclic. Since *A* = *C* _{ H }(*A*) and H is not abelian, we have (*H* : *A*) = 2. We can now apply the arguments of the previous case. Alternatively apply the previous case directly to *K* = *H*× < *k* >, where k has infinite order and the action of *α* is extended to K by setting *k* *α* = *k* ^{− 1}. The desired conclusion follows from this and the obvious isomorphism *G*≅ < *α* > *K*/ < *k* >.

Finally, consider the general case where *A* = *B* × *T*,*B* and T being as above. Now T has odd order since *α* is fpf, so *α* is fpf on *H*/*T* by [9, Lemma 13 c)]. Thus *G* *γ* *T*/*T* = *G* ^{′} *T*/*T* ≤ *A*/*T* by the previous two cases. If *t* ∈ *T*, then t has odd order and hence *t* = *s* ^{− 2} for some *s* ∈ *T*. Then [*s*,*α*] = *t* and (*g* *s*)*γ* = *g* *γ* ⋅ *t* for any g in G. Consequently, *G* ^{′} *T* = *G* ^{′} and *G* *γ* ⋅ *T* = *G* *γ*. Therefore, *α* is anticentral in G. □

## Notes

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