Nilpotent-by-periodic commutator subgroups of automorphisms

Wehrfritz, B. A. F.

doi:10.1007/s13398-012-0073-7

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“…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.…”

Section: Now G/t Is (Torsion-free)-by-finite ([8 Lemmas 4 and 6] Agamentioning

confidence: 99%

“…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.…”

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On the fixed-point set of an automorphism of a group

Wehrfritz¹

2013

Publicacions Matemàtiques

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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 C G (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 C G (φ) being finite. Mathematics Subject Classification: 20F16, 20E36.Key words: automorphism, fixed-point set, soluble group.Endimioni and Moravec in [2] prove that if φ is an automorphism of a polycyclic group G with its fixed-point setis 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

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Section: Now G/t Is (Torsion-free)-by-finite ([8 Lemmas 4 and 6] Agamentioning

confidence: 99%

mentioning

confidence: 98%