ON THE FIXED-POINT SET OF AN AUTOMORPHISM OF A GROUP

Let φ be an automorphism of a group G. Under various finiteness or solubility hypotheses, for example under polycyclicity, the commutator subgroup [G,φ] has finite index in G if the fixed-point set CG(φ) of φ in G is finite, but not conversely, even for polycyclic groups G. Here we consider a stronger, yet natural, notion of what it means for [G,φ] to have ‘finite index’ in G and show that in many situations, including G polycyclic, it is equivalent to CG(φ) being finite. 2010 Mathematics Subject Classification: 20F16, 20E36.

Endimioni and Moravec in [2] prove that if φ is an automorphism of a polycyclic group G with its fixed-point set C G (φ) finite, then G modulo [G, φ] = g −1 • gφ : g ∈ G is also finite (so [G, φ] is large), but the converse is false, even if φ has order 2, C G (φ) = 1 and G is polycyclic and metabelian.These and related results are extended in [9] to, in particular, soluble groups of finite rank.Is there some stronger notion of [G, φ] being large such that in these situations C G (φ) is finite if and only is [G, φ] is large in this sense?We will see below that the answer to this is yes.
Let φ be an automorphism of a group G and define the map γ if G into itself by gγ = [g, φ] = g −1 • gφ.Then ker γ = {g ∈ G : gγ = 1} is C G (φ) and assumptions on C G (φ) should give information about Gγ.The problem is that γ is not usually a homomorphism and Gγ is not usually a subgroup of G.If S is any subset of G say that S has finite index in G if S contains a subgroup of G (normal if you wish) of finite index in G.If Gγ has finite index in G then so does [G, φ], since [G, φ] = Gγ .We shall see that in suitable situations C G (φ) is finite if and only if Gγ has finite index in G, and these situations include polycyclic groups and soluble groups of finite rank.
We start by defining the classes of group we shall be mainly considering.A group G has finite Hirsch number if it has a series of finite length, each factor of which is either locally finite or infinite cyclic, the number of infinite cycles here being an invariant of G, called the Hirsch number of G.For any set π of primes a group G satisfies min-π (min-q if π = {q}) if the set of π-subgroups of G satisfies the minimal condition under inclusion.
An FAR group (that is, a finite-abelian-ranks group, also called an S 0 -group) is a soluble group with finite Hirsch number satisfying min-q for every prime q.See [5] and [6] for discussions and alternative definitions of this class of FAR groups.Note in particular that all soluble groups of finite (Prüfer) rank and hence all polycyclic groups are FAR groups.
Theorem 1.Let G be a finite extension of a soluble FAR group and φ an automorphism of G such that φ p = 1 for some prime p. Define maps γ and ψ of G into itself by The following are equivalent.Here ker ψ = {g ∈ G : gψ = 1}.Again note that ψ is not usually a homomorphism and ker ψ is not usually a subgroup of G; also always Gγψ = {1}, so ker ψ ⊇ Gγ.The main part of the theorem is the equivalence of a) and b).The equivalence of a) and c) just for polycyclicby-finite groups is almost immediate from Corollary C of [1].Further Theorem B of [1] implies that in Theorem 1 above C G (φ) being finite is equivalent to ker(ψ 2 ) having finite index in G.
Sometimes one can deduce that G/[G, φ] is periodic from the periodicity of C G (φ), e.g.see Lemma 1 of [11].If S is a subset of a group G, say that G is periodic modulo S if for each g ∈ G there is a positive integer n with g n ∈ S. Note that this is weaker than requiring S to contain a normal subgroup N of G with G/N periodic, even in the simplest of cases.For example, let G denote the additive group of the integers Z and S = {±ip i : i = 0, 1, 2, . . .}, where p i denotes the (i + 1)-th prime.Trivially G is periodic modulo S, but if s ∈ S with s = 0 it is easy to see that 2s / ∈ S, so S contains no non-zero subgroup of G. Also say G has finite exponent modulo S if for some integer e > 0, g e ∈ S for all g ∈ G.
Theorem 2. Let G be a finite extension of a soluble FAR group and φ an automorphism of G such that φ p = 1 for some prime p. Define maps γ and ψ of G into itself by The following are equivalent.
We can squeeze a little more out of this theorem.
Corollary.Let G be a group that has a local system consisting of finite extensions of soluble FAR groups.Suppose φ is an automorphism of G satisfying φ p = 1 for some prime p.With γ and ψ as in Theorem 2, the following are equivalent.
So, suppose in Theorem 2 (or its corollary) that G is a (torsion-free)by-finite FAR group.If C G (φ) is periodic, then C G (φ) is finite.Hence if G is periodic modulo Gγ (resp.ker ψ), then C G (φ) is finite by Theorem 2 and consequently by Theorem 1 the subset Gγ (resp.ker ψ) has finite index in G. Hence in particular if G is polycyclic-by-finite and G is periodic modulo Gγ or ker ψ, then G has a normal subgroup N with ker ψ ⊇ Gγ ⊇ N and G/N periodic (i.e.finite).This does not hold for soluble FAR groups in general, indeed not even for minimax such groups.
Example 1.There is a metabelian minimax group G of rank 3 with an automorphism φ of order 2 such that C G (φ) is periodic and yet Gγ and ker ψ do not contain normal subgroups N of G such that G/N is periodic.
If φ has infinite order then we can no longer define ψ, but at least we can say something about Gγ.Theorem 3. Let G be a finite extension of a nilpotent FAR group and let φ be an automorphism of G with γ defined as in the previous theorems.
as usual, then the following are equivalent.i) C G (φ) is finite.
ii) Gγ has finite index in G.
iii) ker ψ has finite index in G.
Theorem 4. Let G be a finite extension of a soluble FAR group and φ an automorphism of G with C G (φ) finite.Then with γ as above (Gγ) [3] = {xγ • yγ • zγ : x, y, z ∈ G} has finite index in G.
Theorem 4 strengthens Part ii) of the theorem of [9], which says that G/[G, φ] is finite.Throughout this paper if φ is an automorphism of a group G and m is a positive integer with φ m = 1, we define maps γ and ψ of G into itself by for all g ∈ G. Notice that Gγψ = {1}, so always ker ψ ⊇ Gγ.However Gψγ need not be {1} in general.If m = p is prime we write p for m in the definition of ψ.Also ker(ψ r ) ⊇ ker ψ for every positive integer r.
Obviously the very much simpler case where the group G is abelian is basic to our results.In this case γ and ψ are endomorphisms, so Gγ = [G, φ] and ker ψ are subgroups of G, ker γ = C G (φ) and G/C G (φ) ∼ = Gγ ≤ G. Thus if G is also finitely generated, then C G (φ) is finite if and only if G/[G, φ] is finite.In general if φ has finite order then G has a local system of φ-invariant, finitely generated subgroups and a simple localization argument, together with a special case of Lemma 3 below shows that C G (φ) is periodic if and only if G/[G, φ] is periodic.
For abelian groups in general we cannot replace periodic by finite here and this already just for abelian groups shows the necessity for some sort of rank restrictions at least in Theorem 1.If G is a free abelian group of infinite rank and φ is inversion on G, Let A be the direct product of Prüfer q-groups A q , one for each prime q, and let φ denote the automorphism of A of infinite order that for each q raises each element of A q to its (1 + q)-th power.Then A q γ = (A q ) q = A q and so [A, φ] = Aγ = A. Also C A (φ) is the direct product of cyclic groups of order q, one for each prime q, so C A (φ) is infinite.Clearly A is an FAR group with rank 1 and Hirsch number 0. Consequently the converses of Theorem 3a) and Theorem 4 are both false.
However for special FAR groups it is possible to go further.Following the terminology of [5], a soluble group G is an FATR group (called an S 1 -group in [6]) if it is an FAR group and if the set of primes q, such that G contains an element of order q, is finite.In Proposition 3 below we prove that if φ is an automorphism of a finite extension of a soluble FATR group G with Gγ of finite index in G, then C G (φ) is finite.Thus in particular, if in Theorem 3 we replace the FAR assumption by FATR, then C G (φ) is finite if and only if Gγ has finite index in G. Theorem 4, of course, still leaves unanswered questions.
To prove Theorem 1 we need the following four lemmas.The first three are very elementary, but Lemma 4 is the heart of the proofs of Theorems 1 and 2 and, in parts, of Theorems 3 and 4. Lemma 1 is immediate from [9, Lemma 2].
Lemma 1.Let φ be a fixed-point-free automorphism of the torsion-free abelian group A of finite rank.Then A/Aγ is finite.
and suppose N 1 has finite exponent e.Then (N i+1 ) e ≤ N i for each i and N has exponent dividing e n .If N also satisfies the minimal condition on subgroups, then N is finite.
for each i and the remainder of the lemma follows easily.Lemma 3. Let φ be an automorphism of finite order m of the group G.
If e is such that g e ∈ ker ψ for all g ∈ G, then x em = 1 and C G (φ) has exponent dividing em.Lemma 4. Let φ be an automorphism of finite order m of the nilpotent group G with C = C G (φ) a finite π-group for some finite set π of primes.If m is a π-number and if G has finite Hirsch number and satisfies min-q for all q ∈ π (equivalently min-π here), then Gγ has finite index in G.
Proof: Note first (e.g. by [4, 3.13] and [8, Lemma 4]) that any image of a group with finite Hirsch number and min-q for some prime q also satisfies min-q.Assume by induction on the Hirsch number that the lemma is valid for groups (if any) of smaller Hirsch number than that of G.
Consider first the case where the maximum normal locally finite subgroup T of G is finite.If G has Hirsch number 0 then G = T is finite in this case and the claim is vacuous.Suppose G has positive Hirsch number.Then G is (torsion-free)-by-finite (e.g. by [8,Lemmas 4 and 6]).Also G is periodic and hence finite if its centre ζ 1 (G) is finite (e.g. by Lemma 2), so G has an infinite, φ-invariant, torsion-free central subgroup Z. Then C Z (φ) = C ∩ Z = 1 and (Z : Zγ) is finite by Lemma 1.Since Zγ is infinite and torsion-free, the Hirsch number of G/Zγ is less than that of G. Further by Lemma 1c) of [10] the order of C G/Zγ (φ) divides (Zγ : (Zγ) m )|C|, which is a finite π-number.By induction there exists a normal subgroup N of G of finite index with Zγ ≤ N and N/Zγ lying in By induction there is a normal subgroup N of G of finite index with Z ≤ N and with N/Z contained in (G/Z)γ = Gγ/Z.Then Gγ ⊇ N and Gγ has finite index in G.
We now consider the general case.By hypothesis O π (G) satisfies the minimal condition on subgroups.Hence O π (G) is a Chernikov group and as such has a characteristic subgroup D of finite index that is a direct product of a finite number, say r, of Prüfer q-groups for the various primes q in π.Here we induct on r, the case r = 0 having been covered above.
Assume r > 0. By Lemma 2 the subgroup The proof of the lemma is complete.
Proposition 1.Let φ be an automorphism of the group G with φ p = 1 for some prime p such that C G (φ) is a finite π-group for some set π of primes.Suppose G has finite Hirsch number and satisfies min-q for all primes q ∈ π.Then Gγ has finite index in G.In particular C G (φ) is a π-group for some finite set π of primes.For each prime q by hypothesis every q-subgroup of G is a Chernikov group and Chernikov groups of finite exponent are finite.Therefore C G (φ) is finite.
Example 2. We cannot remove the min-q condition completely from either Theorem 1 or Proposition 1.
Example 18 of [1] constructs a group G with the following properties.G is nilpotent of class 2 and prime exponent q and has an automorphism φ of prime order p = q such that C G (φ) has order q and N ψ = {1} for all subgroups N of G of finite index.Clearly then neither Gγ nor ker ψ has finite index in G. Trivially G has Hirsch number 0 and satisfies min-r for all primes r = q.Thus we do need to restrict the r-subgroups of G at least for those primes r involved in C G (φ). Moreover in this construction the prime p (but not the prime q) can be chosen arbitrarily.
We now come to the proof of Theorem 2. We need some further lemmas.Lemma 5 below is a special case of [8,Lemma 15].Lemma 5. Let φ be an automorphism of the group G with φ p = 1 for some prime p. Suppose A is an abelian φ-invariant normal p -subgroup of G and set C = C G (φ) and K/A = C G/A (φ).Then K = (A ∩ K)C.If C is also periodic, then so is K. Lemma 6.Let φ be an automorphism of the group G with φ p = 1 for some prime p. Suppose A is an abelian φ-invariant normal divisible p-subgroup of G with finite rank r and set The actions of k and φ on A commute, so k and kb −1 normalize A∩C.Then φ stabilizes the series kb Therefore (kb −1 ) ps ∈ C, k ps ∈ AC, and K ps ≤ AC.Finally if C is periodic, then so is AC and therefore so is K. Lemma 7. Let G be a nilpotent group with finite Hirsch number.Then G is periodic modulo one (and hence many) of its finitely generated subgroups.
Proof: The upper central factors , where i runs over the positive integers at most the class of G, and N = X .We prove by induction on the class of G that G is periodic modulo N .
If G is abelian the claim is clear.We may assume by induction that G is periodic modulo N ζ 1 (G).Then if g ∈ G there exists m ≥ 1 with g m ∈ N ζ 1 (G), say g m = yz, where y ∈ N and z ∈ ζ 1 (G).Now ζ 1 (G)/ X 1 is periodic, so there exists n ≥ 1 with z n ∈ X 1 ≤ N .Consequently g mn = y n z n ∈ N .The lemma follows.
Proposition 2. Let G be a soluble-by-finite countable group with finite Hirsch number satisfying min-p for some prime p. Suppose G has an automorphism φ with φ p = 1.Then the following are equivalent.
Proof: Suppose C is periodic.By [4, 3.17 and 3.13] there is a characteristic series where each T i /T i−1 is an abelian p -group, T /T s is a divisible abelian p-group of finite rank, T ∞ /T is finite and T ∞ is the maximum periodic normal subgroup of G. Then Lemmas 5 and 6 yield that C G/T (φ) is periodic.Now G/T is (torsion-free)-by-finite ([8, Lemmas 4 and 6] again).Thus C G/T (φ) is actually finite and therefore (see [1,Theorem D]) G has a φ-invariant normal subgroup H of finite index with T ≤ H and H/T nilpotent.Since we have assumed that G is countable, it follows from Part b) of the Proposition of [11] that H has a nilpotent subgroup K with H periodic modulo K (note we cannot in general choose K normal in H).Since φ has finite order we may choose K to be φ-invariant.Further K is periodic modulo some finitely generated subgroup L of K by Lemma 7. Again we may choose L to be φ-invariant.Since L is finitely generated and nilpotent, so C L (φ) is finite.Consequently, e.g. by Theorem 1, there is a subgroup N of L of finite index with Lγ ⊇ N .Clearly N is finitely generated and nilpotent, G is periodic modulo N and Gγ ⊇ Lγ ⊇ N .Therefore a) implies b).Trivially b) implies c), always c) implies d), and d) implies a) by Lemma 3. The proof is complete.
The Proof of Theorem 2: Finite extensions of soluble FAR groups are always countable and satisfy min-q for every prime q.Thus Theorem 2 follows from Proposition 2.
The Proof of the Corollary: Suppose C G (φ) is periodic.Since φ has finite order, so G is the union of its φ-invariant subgroups X that are finite extensions of FAR groups.Then C X (φ) is periodic and so X is periodic modulo Xγ by Theorem 2. Consequently if x ∈ X then some positive power of x lies in Xγ and so in Gγ.Thus a) implies b), b) implies c), and c) implies a) as in the previous cases.We will frequently apply Lemma 8 when A is central in G.In this case we may always choose B = Aγ.If A is torsion-free of finite rank and C is periodic, then A/Aγ is finite by Lemma 1 and we can choose B of finite index in A.
If C and A/B are both finite, then clearly K/B is also finite.
The proof of the following lemma is nearly a repeat of the proof of Lemma 4. As with Lemma 4 it suffices in Lemma 9 to assume that G is nilpotent of finite Hirsch number and satisfies min-q for some finite set π of primes q.However it is messy to specify π.Apart from the prime divisors of |C| we also need certain primes that depend on the way φ acts on the torsion-free abelian φ-invariant sections of G.It is simpler to assume the FAR condition and hence to assume G satisfies min-q for all primes: q.Lemma 9. Let φ be an automorphism of the nilpotent FAR group G with C = C G (φ) finite.Then Gγ has finite index in G.
Proof: Assume by induction on the Hirsch number that the lemma is valid for groups (if any) of smaller Hirsch number than that of G. Let π denote the set of prime divisors of the order of C.
Consider first the case where the maximum normal locally finite subgroup T of G is finite.If G has Hirsch number 0 then G = T is finite in this case and the claim is vacuous.Suppose G has positive Hirsch number.Then G is (torsion-free)-by-finite (e.g. by [8,Lemmas 4 and 6]).Also G is periodic and hence finite if its centre ζ 1 (G) is finite (e.g. by Lemma 2), so G has an infinite, φ-invariant, torsion-free central subgroup Z. Then C Z (φ) = C ∩ Z = 1 and (Z : Zγ) is finite by Lemma 1.Since Zγ is infinite and torsion-free, the Hirsch number of G/Zγ is less than that of G.
By induction there is a normal subgroup N of G of finite index with Z ≤ N and with N/Z contained in (G/Z)γ = Gγ/Zγ.Then Gγ ⊇ N and Gγ has finite index in G.
We now consider the general case.By hypothesis O π (G) satisfies the minimal condition on subgroups.Hence O π (G) is a Chernikov group and as such has a characteristic subgroup D of finite index that is a direct product of a finite number, say r, of Prüfer q-groups for the various primes q in π.Here we induct on r, the case r = 0 having been covered above.
Assume r > 0. By Lemma 2 the subgroup Proof: We induct on the derived length of G; if G = 1 everything is vacuous.Suppose G satisfies the lemma and let x ∈ G.The abelian group G/G satisfies min-q for every prime q and hence has only finitely many elements of each order.Thus xφ m ∈ xG for some positive integer m, say xφ m = xy where y ∈ G .
By induction yφ n = y for some positive integer n.Also y • yφ m • yφ 2m • • • yφ m(n−1) has finite order, r say.Then Thus x φ is finite and φ has finite orbits in G. Suppose The Proof of Theorem 4: Note first that C S (φ) is finite for every φ-invariant section S of G by [9, Theorem ii)].Let π denote the set of prime divisors of the order of C G (φ). Denote the maximum periodic soluble normal subgroup of G by T 0 and set T = O π (T 0 ).Then T = T γ by Lemma 10.Now T 0 /T is a Chernikov group by [4, 3.17 and 3.13] (of course here π is finite).Hence in the notation of [5] the group G/T is a finite extension of a soluble FATR group and therefore its Fitting subgroup H/T is nilpotent and G/H is abelian-by-finite, see [5, 5.2.2].By Theorem 3 there is a subgroup M/T of H/T of finite index with (H/T )γ ⊇ M/T ; that is, with Hγ • T γ = Hγ • T ⊇ M .Replacing M by M (H:M ) T we may assume that M is characteristic in G. Then G/M is finite-by-abelianby-finite and hence is nilpotent-by-finite.By Theorem 3 again there is a subgroup N/M of G/M of finite index with (G/M )γ ⊇ N/M .Consequently Gγ • M ⊇ N and therefore (Gγ) [3] ⊇ Gγ • Hγ • T γ ⊇ N .The proof is complete.Proposition 3. Let φ be an automorphism of the finite extension G of a soluble FATR group.If Gγ has finite index in G, then C G (φ) is finite.
where each G i /G i−1 is either a torsion-free abelian group of finite rank or is a divisible abelian q-group of finite rank for some prime q and G/G n is finite, cf.[8, Lemma 4].We induct on n and assume that G/G 1 satisfies the proposition.To simplify notation set A = G 1 .Then K/A = C G/A (φ) is finite.
By hypothesis there is a subgroup N of G of finite index contained in Gγ.Let T be a (necessarily finite) transversal of A to K and consider a ∈ A ∩ N .Then a = gγ for some g in G. Necessarily g ∈ K. Hence g = tb for some t ∈ T and b ∈ A. Now tγ ∈ Kγ, which lies in the abelian group A, so a) C G (φ) is finite.b) Gγ has finite index in G. c) ker ψ has finite index in G.
is an infinite elementary abelian 2-group.(The above already shows that if G is a torsion-free abelian group with G/[G, φ] periodic and |φ| finite, then C G (φ) = 1 .)If G is a direct product of infinitely many Prüfer 2-groups and again φ denotes inversion, then |φ| = 2, G/[G, φ] = 1 and C G (φ) is an infinite elementary abelian 2-group.(If G is periodic abelian with C G (φ) = 1 , then G = [G, φ], e.g. by [3, 10.1.1].) by the centrality of z.Therefore Gγ ⊇ N and Gγ has finite index in G. Now we consider the case where O π (G) is finite.We further induct on the least c such that O π (G) ≤ ζ c (G). Obviously c is bounded by the class of G and the case c = 0 is covered by the case completed above.Suppose c ≥ 1 and set Z = O π (G) ∩ ζ 1 (G) = O π (ζ 1 (G)).Then φ acts fixed-point freely on Z, so Z = Zγ, e.g. by [3, 10.1.1].Again C G/Z (φ) is a finite π-group by [10, Lemma 1c)] since by hypothesis no prime divisor of m lies in π , so Z m = Z.Clearly and D/E is a direct product of less than r Prüfer groups.Clearly E m = E and then C G/E (φ) is a finite π-group by [10, Lemma 1c)] again.By induction there is a normal subgroup N of G of finite index with E ≤ N and N/E contained in (G/E)γ.It follows that Gγ ⊇ N and that Gγ has finite index in G.
1 and p ∈ π.Thus either way G satisfies min-p and we may assume that p ∈ π.By Lemma 4 there is a subgroup N of M of finite index with M γ ⊇ N .Clearly N has finite index in G and Gγ ⊇ M γ ⊇ N .The proposition follows.The Proof of Theorem 1: a) implies b) by Proposition 1 and b) implies c) since ker ψ ⊇ Gγ.Finally if c) holds then C G (φ) has finite exponent by Lemma 3.

Lemma 8 .
Let φ be an automorphism of a group G and A and B abelian φ-invariant normal subgroups of G with B ≤ Aγ.Set C = C G (φ) and K/B = C G/B (φ).Then K = C(A ∩ K).If C is finite and B has finite index in A, then K/B is finite.
Further C G/Zγ (φ) is finite by Lemma 8.By induction there exists a normal subgroup N of G of finite index with Zγ ≤ N and N/Zγ lying in(G/Zγ)γ = Gγ • Zγ/Zγ = Gγ/Zγ, since if g ∈ G and z ∈ Z, then g −1 • gφ • z −1 • zφ = (gz) −1 (gz)φ ∈ Gγby the centrality of z.Therefore Gγ ⊇ N and Gγ has finite index in G. Now we consider the case where O π (G) is finite.We further induct on the least c such that O π (G) ≤ ζ c (G). Obviously c is bounded by the class of G and the case c = 0 is covered by the case completed above.Suppose c ≥ 1 and set Z = O π (G) ∩ ζ 1 (G) = O π (ζ 1 (G)).Then φ acts fixed-point freely on Z, so Z = Zγ, e.g. by [3, 10.1.1].Again C G/Z (φ) is a finite π-group by Lemma 8. Clearly and D/E is a direct product of less than r Prüfer groups.Also Eγ is divisible and ker(γ| E ) = C E (φ) is finite, so E ∼ = Eγ and E = Eγ.Hence C G/E (φ) is a finite π-group by Lemma 8 again.By induction there is a normal subgroup N of G of finite index with E ≤ N and N/E contained in (G/E)γ.It follows that Gγ ⊇ N and that Gγ has finite index in G.The proof of the lemma is complete.The Proof of Theorem 3: a) By Fitting's Lemma there is a φ-invariant nilpotent normal subgroup M of G of finite index, which clearly must also be an FAR group.Suppose C G (φ) is finite.Then by Lemma 9 there is a subgroup N of M of finite index with M γ ⊇ N .Clearly then N has finite index in G and Gγ ⊇ M γ ⊇ N .Thus Gγ has finite index in G. b) Now suppose φ has finite order m.If Gγ has finite index in G always so does ker ψ.If ker ψ has finite index in G, then C G (φ) has finite exponent by Lemma 3.But C G (φ) is a finite extension of a nilpotent FAR group.Consequently C G (φ) is finite.Lemma 10.Let φ be an automorphism of the periodic soluble FAR group G. Then every orbit of φ in G is finite.If also C G (φ) = 1 , then G = Gγ.

a
= gγ = g −1 • gφ = b −1 (t −1 • tφ)bφ = tγ • bγ.Therefore t∈T tγ • Aγ ⊇ A ∩ N .But T and (A : A ∩ N ) are finite.Consequently (A : Aγ) is finite.If A is torsion-free of finite rank, then rank A = rank Aγ and C A (φ) = ker(γ| A ) is torsion-free of rank 0. Thus in this case C A (φ) = 1 .Suppose A is a divisible abelian q-group of finite rank.Then A/Aγ is finite and divisible.Hence A = Aγ and so C A (φ) = ker(γ| A ) is finite.Either wayC A (φ) is finite.Trivially C G (φ) ≤ K and C A (φ) = A∩C G (φ). Therefore C G (φ) has order at most |C A (φ)|(K : A), which is finite.Theorem 3 and Proposition 3 immediately yield the following.Corollary.Let φ be an automorphism of the finite extension G of a nilpotent FATR group.The following are equivalent.a) C G (φ) is finite.b) Gγ has finite index in G.The Proof of Example 1: For any prime p let C be a Prüfer p-group and x the automorphism of C given by c x = c 1+p for all c ∈ C. Then x has infinite order.Let A = C 1 × C 2 be the direct product of two copies of C and y the automorphism of A that acts as x on C 1 and as x −1 on C 2 .The split extension G = y A of A by y is metabelian, minimax and of rank 3. Now G has an automorphism φ of order 2 given by yφ = y −1 and (c 1 d 2 )φ = d 1 c 2 (here c, d ∈ C with c i and d i denoting the corresponding elements of C i ).Then φ acts fixed-point freely on G/A and therefore C G (φ) ≤ A, which is periodic.Always ker ψ ⊇ Gγ.Suppose N is a normal subgroup of G with G/N periodic and ker ψ ⊇ N .Then y i ∈ N for some positive integer i and hence c −11 y i c 1 ∈ N for every c ∈ C.But c −1 1 y i c 1 = y i c −1 1 [c −1 1 , y i ]c 1 = y i [c −1 , x i ] 1 .For any i ≥ 1 we can pick c ∈ C such that [c −1 , x i ] = 1.Then for this c we have (c −1 1 y i c 1 )ψ = (y i [c −1 , x i ] 1 ) • y −i [c −1 , x i ] 2 ) = 1.Thus no such N exists.