Isomorphism of the Baumslag-Solitar groups

We present a method for computing the number of epimorphisms from a finitely-presented group G to a finite solvable group \Gamma, which generalizes a formula of G\"aschutz. Key to this approach are the degree 1 and 2 cohomology groups of G, with certain twisted coefficients. As an application, we count low-index subgroups of G. We also investigate the finite solvable quotients of the Baumslag-Solitar groups, the Baumslag parafree groups, and the Artin braid groups.Comment: 30 pages; accepted for publication in the Journal of Algebr

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“…The groups BS(m, n) have been classified by Moldavanski [27]. They are in bijection with the set of unordered pairs (m, n) with 0 < m |n|.…”

Section: Baumslag-solitar Groupsmentioning

confidence: 99%

Counting homomorphisms onto finite solvable groups

Matei

Suciu

2005

Journal of Algebra

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“…Using , it is easy to show that the isomorphism type of BS.m; n/ determines m and n (normalized by 1 Ä m Ä jnj) see Moldavanskiȋ [27]. In most cases, m and n are determined by m=n (given by ) and jm nj (given by abelianizing).…”

Section: Remarksmentioning

confidence: 99%

On the automorphism group of generalized Baumslag–Solitar groups

Levitt

2007

Geom. Topol.

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A generalized Baumslag-Solitar group (GBS group) is a finitely generated group G which acts on a tree with all edge and vertex stabilizers infinite cyclic. We show that Out(G) either contains non-abelian free groups or is virtually nilpotent of class ≤ 2. It has torsion only at finitely many primes.One may decide algorithmically whether Out(G) is virtually nilpotent or not. If it is, one may decide whether it is virtually abelian, or finitely generated. The isomorphism problem is solvable among GBS groups with Out(G) virtually nilpotent.If G is unimodular (virtually F n × Z), then Out(G) is commensurable with a semi-direct product Z k ⋊ Out(H) with H virtually free. Contents1. Introduction and statement of results 2. Basic facts about GBS groups 3. The automorphism group of a GBS tree 4. Unimodular groups 5. The deformation space 6. Free subgroups in Out(G) 7. Groups with Out(G) ⊃ F 2 8. Further results References Given Γ, let T be the associated Bass-Serre tree, which we call a GBS tree. Let Out T (G) ⊂ Out(G) be the subgroup leaving T invariant. Most elements of Out T (G) may be viewed as "twists" (see Section 3). Algebraic rigidity implies Out T (G) = Out(G), but in general Out T (G) is smaller.Theorem 1.1. Let G be a GBS group, represented by a labelled graph Γ, and let T be the Bass-Serre tree. Define k as the first Betti number b of Γ if G has a nontrivial center, as b − 1 if the center is trivial.(1) The torsion-free rank of the abelianization of(3) Up to commensurability within Out(G), the subgroup Out T (G) does not depend on Γ.Conversely, any subgroup of Out(G) commensurable with a subgroup of Out T (G) is contained in Out T ′ (G) for some GBS tree T ′ [4].For G = BS(m, n), one has k = 0 if m = n, and k = 1 if m = n. For G = BS(2, 4) with the presentation (1 p ), the group Out T (G) has order 2 p+1 .The converse is also true (see Theorem 8.5). Unimodular groups.A GBS group G is unimodular if xy p x −1 = y q with y = 1 implies |p| = |q|, or equivalently if G is virtually F n × Z (with F n a free group of rank n). The group G then has a normal infinite cyclic subgroup with virtually free quotient, and we show:where k is as above, H is virtually free, and Out 0 has finite index in Out. Since Out(H) is VF [19], we get:Corollary 1.4. Out(G) is virtually torsion-free and VF (it has a finite index subgroup admitting a finite classifying space).Groups with no nontrivial integral modulus. Now consider groups G which do not contain a solvable Baumslag-Solitar group BS(1, n) with n ≥ 2 (there is an equivalent characterization in terms of the modular homomorphism ∆ : G → Q * , see Section 2).Given any GBS group G, the group Out(G) acts on the space P D of all GBS trees (see Section 5), with stabilizers virtually Z k by Theorem 1.1. Clay [3] proved that the space P D is contractible (see also [16]), and Forester [12] proved that the quotient is a finite complex if G does not contain BS(1, n) with n ≥ 2. This gives: 3

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“…So in that case, BS(n, m) is nonamenable. In [Mo91] it was proven that BS(n 1 , m 1 ) ∼ = BS(n 2 , m 2 ) if and only if {n 1 , m 1 } = {εn 2 , εm 2 } for some ε ∈ {−1, 1}. So all nonamenable Baumslag-Solitar groups are up to isomorphism of the form BS(n, m) for some 2 ≤ n ≤ |m|.…”

Section: Baumslag-solitar Groups and Hnn Extensionsmentioning

confidence: 99%

“…These groups were introduced by Baumslag and Solitar in [BS62] to provide the first examples of two generator non-Hopfian groups with a single defining relation. In [Mo91], it was shown that BS(n 1 , m 1 ) ∼ = BS(n 2 , m 2 ) if and only if {n 1 , m 1 } = {εn 2 , εm 2 } for some ε ∈ {−1, 1}. Hence we may always assume that 1 ≤ n ≤ |m|.…”

Section: Introductionmentioning

confidence: 99%

A rigidity result for crossed products of actions of Baumslag–Solitar groups

Meesschaert

2015

Int. J. Math.

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Let BS(n 1 , m 1 ) X 1 and BS(n 2 , m 2 ) X 2 be two ergodic essentially free probability measure preserving actions of nonamenable Baumslag-Solitar groups whose canonical almost normal abelian subgroups act aperiodically. We prove that an isomorphism between the corresponding crossed product II 1 factors forces BS(n 1 , m 1 ) ∼ = BS(n 2 , m 2 ) when |n 1 | = |m 1 | and BS(n 1 , m 1 ) ∼ = BS(n 2 , ±m 2 ) when |n 1 | = |m 1 |. This improves an orbit equivalence rigidity result obtained by Houdayer and Raum in [HR13].

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Isomorphism of the Baumslag-Solitar groups

Cited by 24 publications

References 2 publications

Counting homomorphisms onto finite solvable groups

Counting homomorphisms onto finite solvable groups

On the automorphism group of generalized Baumslag–Solitar groups

A rigidity result for crossed products of actions of Baumslag–Solitar groups

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