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

Bludov, V. V.; Glass, A. M. W.

doi:10.1112/plms/pdp008

Cited by 24 publications

(44 citation statements)

References 22 publications

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“…As shown in [9], these necessary and sufficient conditions (for a free product of right-orderable groups with amalgamated subgroup to be right orderable) provide the following results where the presentation is as a group.…”

Section: Decision Problems In Right-orderable Groups and -Groupsmentioning

confidence: 89%

“…To see this one crucially uses ultraproducts of right orders (see [9]). Using "normal" families of right orders on the factors and constructing groups of order-preserving permutations of totally ordered sets (sic), one can provide precise necessary and sufficient conditions for a free product of right-orderable groups with amalgamated subgroup to be right-orderable (mutatis mutandis with right-ordered in place of right-orderable) [9]. These ideas extend the non-permutation ideas in [4].…”

Section: Amalgamation In Groups and -Groupsmentioning

confidence: 99%

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Automorphism groups of totally ordered sets: A retrospective survey

Bludov

Droste

Glass

2011

Mathematica Slovaca

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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.

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Section: Decision Problems In Right-orderable Groups and -Groupsmentioning

confidence: 89%

Section: Amalgamation In Groups and -Groupsmentioning

confidence: 99%

Automorphism groups of totally ordered sets: A retrospective survey

Bludov

Droste

Glass

2011

Mathematica Slovaca

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“…Also, the ideas of ultraproduct of orders and of D-order will be used, and the reader is referred to [1] for the definition. Again from [1], a set R of right orders on a group G is called U-closed if it is closed under ultraproducts of orders from R, and D-invariant if it is closed under taking D-orders determined by orders in R. Finally, R is called A-invariant if it is normal, U-closed and D-invariant.…”

Section: Definitionmentioning

confidence: 99%

“…The next lemma is the main point at which input from [1] is needed; the proof of Theorem A given there establishes the lemma in the case that Y has one unoriented edge, and permits an inductive argument. …”

Section: The Case Of a Tree Of Groupsmentioning

confidence: 99%