Finitely generated lattice-ordered groups with soluble word problem

Glass, A. M. W.

doi:10.1515/jgt.2008.001

Cited by 6 publications

(5 citation statements)

References 15 publications

Supporting

Mentioning

Contrasting

Order By: Relevance

“…Ì ÓÖ Ñ Jº ( [29], cf. [11]) A finitely generated -group has soluble word problem iff it can be -embedded in a simple -group which can be -embedded in a finitely presented -group.…”

Section: ì óö ñ Iº ([28] Cf [38]) a Finitely Generated -Group Can Bmentioning

confidence: 99%

Automorphism groups of totally ordered sets: A retrospective survey

Bludov

Droste

Glass

2011

Mathematica Slovaca

View full text Add to dashboard Cite

ABSTRACT. In 1963, W. Charles Holland proved that every lattice-ordered group can be embedded in the lattice-ordered group of all order-preserving permutations of a totally ordered set. In this article we examine the context and proof of this result and survey some of the many consequences of the ideas involved in this important theorem. GenesisIn 1957, P. M. Cohn noted that Cayley's (right regular) Representation Theorem applies immediately to right-ordered groups [12]. That is, let G be a group with a total order that is preserved by multiplication on the right. Embed G in the group A(G) := Aut(G, ≤) of all order-preserving permutations of the set (G, ≤) via the right regular representation; i.e., f (gϑ) = fg (f, g ∈ G). Now A(G) can be right totally ordered as follows: let ≺ be any well-ordering of the set G and define a < b iff α 0 a < α 0 b, where α 0 is the least element of G (under ≺) of the support of ba −1 . If we choose ≺ so that its least element is the identity of G, then the right regular embedding preserves the group operation and the total order:What can one do if the order on G is not total, even if multiplication on the left also preserves the partial order? The easiest situation would be to 2010 M a t h e m a t i c s S u b j e c t C l a s s i f i c a t i o n: Primary 06F15, 20F60, 20B27, 20F10. K e y w o r d s: totally ordered set, permutation group, representation, primitive permutation group, amalgamation, free product with amalgamated subgroup, right-orderable group, normal subgroups, Bergman property, outer automorphism groups.

show abstract

“…Ì ÓÖ Ñ Jº ( [29], cf. [11]) A finitely generated -group has soluble word problem iff it can be -embedded in a simple -group which can be -embedded in a finitely presented -group.…”

Section: ì óö ñ Iº ([28] Cf [38]) a Finitely Generated -Group Can Bmentioning

confidence: 99%

Automorphism groups of totally ordered sets: A retrospective survey

Bludov

Droste

Glass

2011

Mathematica Slovaca

View full text Add to dashboard Cite

show abstract

“…We do not know if the groups B(T ) are lattice-orderable; they are certainly not if there are elements g, h which are not conjugate with g 2 = h 2 . Indeed, Theorems D and E have lattice-ordered group analogues (see [12,13]). A finitely presented lattice-ordered group with insoluble group word problem was constructed in [14].…”

Section: Concluding Remarks and Questionsmentioning

confidence: 99%

Word problems, embeddings, and free products of right-ordered groups with amalgamated subgroup

Bludov¹,

Glass²

2009

Proceedings of the London Mathematical Society

View full text Add to dashboard Cite

Dedicated to Graham Higman, in memoriam, with gratitude for his research. AbstractWe use permutation groups to give necessary and sufficient conditions for the free product of right ordered groups with amalgamated subgroup to be right orderable. We obtain several consequences answering previously posed problems and also prove the right orderable analogues of the Higman Embedding Theorem and the Boone-Higman Theorem.--------------AMS Classification: 06F15, 20F60, 20B27, 20F10.

show abstract

“…Proof: In [10] A. M. W. Glass proved that if G is any finitely generated lattice-ordered group, then G has soluble word problem if and only if it can be embedded in an -simple lattice-ordered group S that can be embedded in a finitely presented lattice-ordered group. The lattice-ordered group S had the property that any two strictly positive elements of S were conjugate.…”

Section: Proof Of Theorem Amentioning

confidence: 99%

“…However, if G has soluble word problem and we use G(0) of the proof of Corollary 2.1, then G(0) is finitely generated and has soluble word problem. Instead of using Pierce's result, we apply the proof of Theorem A (together with the aside) to G(0) in the construction in [10] (substituting P ∪ N ∪ I in place of P ) and obtain that S has exactly 4 conjugacy classes; it is simple as a group by the proof of Corollary 2.1. / / 3 Deduction of Theorems B and C Proof of Theorem B: Let j = 1, 2 and G j be a right orderable group with non-trivial cyclic subgroup C j .…”

Section: Proof Of Theorem Amentioning

confidence: 99%

Conjugacy in lattice-ordered groups and right ordered groups

Bludov¹,

Glass²

2008

Journal of Group Theory

View full text Add to dashboard Cite

We prove that if G j are right orderable groups with non-trivial cyclic subgroups C j (j = 1, 2), then the (group) free product of G 1 and G 2 with C 1 and C 2 amalgamated is right orderable. We will deduce this from our main theorem that every lattice-ordered group can be embedded in one in which there are precisely four conjugacy classes. As a consequence of our method, we also show that every right ordered group can be embedded in a right ordered group in which any two non-identity elements are conjugate.--------------AMS Classification: 06F15, 20F60, 20B27, 20F10.

show abstract

Finitely generated lattice-ordered groups with soluble word problem

Cited by 6 publications

References 15 publications

Automorphism groups of totally ordered sets: A retrospective survey

Automorphism groups of totally ordered sets: A retrospective survey

Word problems, embeddings, and free products of right-ordered groups with amalgamated subgroup

Conjugacy in lattice-ordered groups and right ordered groups

Contact Info

Product

Resources

About