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Subword Complexity of Profinite Words and Subgroups of Free Profinite Semigroups
- Internat. J. Algebra Comput
, 2003
"... We study free profinite subgroups of free profinite semigroups of the same rank using, as main tools, iterated implicit operators, subword complexity and the associated entropy. ..."
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Cited by 14 (6 self)
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We study free profinite subgroups of free profinite semigroups of the same rank using, as main tools, iterated implicit operators, subword complexity and the associated entropy.
Nonarithmetic superrigid groups: counterexamples to Platonov’s conjecture
- Ann. of Math
"... Margulis showed that “most ” arithmetic groups are superrigid. Platonov conjectured, conversely, that finitely generated linear groups which are superrigid must be of “arithmetic type. ” We construct counterexamples to Platonov’s Conjecture. 1. Platonov’s conjecture that rigid linear groups are arit ..."
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Cited by 6 (1 self)
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Margulis showed that “most ” arithmetic groups are superrigid. Platonov conjectured, conversely, that finitely generated linear groups which are superrigid must be of “arithmetic type. ” We construct counterexamples to Platonov’s Conjecture. 1. Platonov’s conjecture that rigid linear groups are aritmetic (1.1) Representation rigid groups. Let Γ be a finitely generated group. By a representation of Γ we mean a finite dimensional complex representation, i.e. essentially a homomorphism ρ: Γ − → GLn(C), for some n. We call Γ linear if some such ρ is faithful (i.e. injective). We call Γ representation rigid if, in each dimension n ≥ 1, Γ admits only finitely many isomorphism classes of simple (i.e. irreducible) representations. Platonov ([P-R, p. 437]) conjectured that if Γ is representation rigid and linear then Γ is of “arithmetic type ” (see (1.2)(3) below). Our purpose here is to construct counterexamples to this conjecture. In fact our counterexamples are
Groups of intermediate subgroup growth and a problem of Grothendieck
- Grothendieck, Duke Math. J
"... Let f be a function such that for every ε> 0, n log n ≤ f (n) ≤ n εn holds if n is sufficiently large. Suppose that log f (n) / log n is nondecreasing. Using sequences of finite alternating groups, for every such f we construct a 4-generator group Ɣ such that sn(Ɣ), the number of subgroups of index ..."
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Cited by 5 (0 self)
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Let f be a function such that for every ε> 0, n log n ≤ f (n) ≤ n εn holds if n is sufficiently large. Suppose that log f (n) / log n is nondecreasing. Using sequences of finite alternating groups, for every such f we construct a 4-generator group Ɣ such that sn(Ɣ), the number of subgroups of index at most n in Ɣ, grows like f (n). This essentially completes the investigation of the “spectrum ” of possible subgroup growth types and settles several questions posed by Lubotzky, Mann, and Segal. As a by-product we obtain continuously many nonisomorphic 4-generator residually finite groups with isomorphic profinite completions. Our construction also sheds some light on a problem of Grothendieck [Gr]; we obtain an abundance of pairs of finitely generated residually finite groups Ɣ0 < Ɣ, such that the natural map î: ̂Ɣ0 → ̂Ɣ between profinite completions is an isomorphism, but Ɣ0 ̸ ∼ = Ɣ. Let Ɣ be a finitely generated group. Denote by sn(Ɣ) the number of subgroups of
DECISION PROBLEMS AND PROFINITE COMPLETIONS OF GROUPS
, 810
"... Abstract. We consider pairs of finitely presented, residually finite groups P ↩ → Γ for which the induced map of profinite completions ˆ P → ˆ Γ is an isomorphism. We prove that there is no algorithm that, given an arbitrary such pair, can determine whether or not P is isomorphic to Γ. We construct ..."
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Cited by 2 (1 self)
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Abstract. We consider pairs of finitely presented, residually finite groups P ↩ → Γ for which the induced map of profinite completions ˆ P → ˆ Γ is an isomorphism. We prove that there is no algorithm that, given an arbitrary such pair, can determine whether or not P is isomorphic to Γ. We construct pairs for which the conjugacy problem in Γ can be solved in quadratic time but the conjugacy problem in P is unsolvable. Let J be the class of super-perfect groups that have a compact classifying space and no proper subgroups of finite index. We prove that there does not exist an algorithm that, given a finite presentation of a group Γ and a guarantee that Γ ∈ J, can determine whether or not Γ ∼ = {1}. We construct a finitely presented acyclic group H and an integer k such that there is no algorithm that can determine which k-generator subgroups of H are perfect. For Karl Gruenberg, in memoriam
Direct factors of profinite completions and decidability
- J. Group Theory
"... Abstract. We consider finitely presented, residually finite groups G and finitely generated normal subgroups A such that the inclusion A ↩ → G induces an isomorphism from the profinite completion of A to a direct factor of the profinite completion of G. We explain why A need not be a direct factor o ..."
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Cited by 1 (1 self)
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Abstract. We consider finitely presented, residually finite groups G and finitely generated normal subgroups A such that the inclusion A ↩ → G induces an isomorphism from the profinite completion of A to a direct factor of the profinite completion of G. We explain why A need not be a direct factor of a subgroup of finite index in G; indeed G need not have a subgroup of finite index that splits as a non-trivial direct product. We prove that there is no algorithm that can determine whether A is a direct factor of a subgroup of finite index in G. Let G be a finitely generated residually finite group. The inclusion A ↩ → G of any finitely generated subgroup induces a morphism of profinite completions ι: Â → ˆ G. If A is a direct factor of G then ι is injective and we can identify the closure A of ι(A) with Â. In [13] Nikolov and Segal answered a question of Goldstein and Guralnick [11] by showing that the converse of the preceding observation is false: there exist pairs of finitely generated residually finite groups A ↩ → G, with A is normal in G, such that ι: Â → A is an isomorphism,
THE TRUE PROSOLUBLE COMPLETION OF A GROUP EXAMPLES AND OPEN PROBLEMS
, 2005
"... Abstract. The true prosoluble completion P S(Γ) of a group Γ is the inverse limit of the projective system of soluble quotients of Γ. Our purpose is to describe examples and to point out some natural open problems. We answer the analogue of a question of Grothendieck for profinite completions by pro ..."
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Abstract. The true prosoluble completion P S(Γ) of a group Γ is the inverse limit of the projective system of soluble quotients of Γ. Our purpose is to describe examples and to point out some natural open problems. We answer the analogue of a question of Grothendieck for profinite completions by providing examples of pairs of non–isomorphic residually soluble groups with isomorphic true prosoluble completions. We also provide a new class of finitely generated examples which answer the original Grothendieck’s problem. 1. Introduction. 2. Completion with respect to a directed set of normal subgroups. 3. Universal property.
THE TRUE PROSOLUBLE COMPLETION OF A GROUP EXAMPLES AND OPEN PROBLEMS
, 2005
"... Abstract. The true prosoluble completion P S(Γ) of a group Γ is the inverse limit of the projective system of soluble quotients of Γ. Our purpose is to describe examples and to point out some natural open problems. We answer the analogue of a question of Grothendieck for profinite completions by pro ..."
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Abstract. The true prosoluble completion P S(Γ) of a group Γ is the inverse limit of the projective system of soluble quotients of Γ. Our purpose is to describe examples and to point out some natural open problems. We answer the analogue of a question of Grothendieck for profinite completions by providing examples of pairs of non–isomorphic residually soluble groups with isomorphic true prosoluble completions. We also provide a new class of finitely generated examples which answer the original Grothendieck’s problem. 1. Introduction. 2. Completion with respect to a directed set of normal subgroups. 3. Universal property.

