Conjugacy in lattice-ordered groups and right ordered groups

Abstract

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.

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

“…For example, instead of Corollary 5.3, we easily obtain the following. Corollary 6.7 (Bludov and Glass [5]). If G is a right-ordered group and f, g ∈ G \ {1}, then G, t | f t = g is right orderable with a right order extending that on G.…”

“…Although the class of right-orderable groups fails to be closed under free products with amalgamation in general, amalgamation is possible in the class of right-orderable groups when either H 1 is a cyclic group [5], or when all the following conditions pertain: h i > 1 ⇒ g −1 i h i g i > 1 for all h i ∈ H i , g i ∈ G i (i = 1, 2) and h 1 > 1 if and only if h ϕ 1 > 1 (h 1 ∈ H 1 ), where ϕ is the amalgamating isomorphism (see [6]). Indeed, the techniques lead to similar results for lattice-ordered groups.…”

Automorphism groups of totally ordered sets: A retrospective survey

“…This left open whether one could find such a G # in which, additionally, all elements incomparable to the identity could also be made conjugate. This was recently achieved in [7] using an elaboration of Pierce's original construction.…”

Graph manifolds, left-orderability and amalgamation

“…Both HNN-extensions and free products with amalgamation present considerable difficulty when attempting to determine necessary and sufficient conditions for them to be left-orderable, and related problems remained open for many years (see for example [17,Problem 15.34]). However with recent work, left-orderability of free products with amalgamation and graphs of groups is finally well understood [1,2,6].…”