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The centralizer of an infinite-order element in a hyperbolic group is virtually cyclic
Statement
Let be a hyperbolic group and let have infinite order. Then its centralizer
contains a cyclic subgroup of finite index.
Facts & Assumptions
Given: A hyperbolic group and an infinite-order element .
There is such that every geodesic segment from to lies in the -neighborhood of the powers of .
A geodesic quadrilateral in a -hyperbolic Cayley graph is -thin, and every metric ball in a locally finite Cayley graph is finite.
Infinite-order elements are undistorted (Infinite-order elements of hyperbolic groups are undistorted).
Proof
By [L1] and [A1], the powers of form a quasi-axis: geodesics joining distant powers stay uniformly close to the power orbit.
Let and choose so large that the two long sides of the quadrilateral with vertices have points outside the -neighborhoods of its short sides. By [A2], some point on is within of . Using [A1] on both long sides gives integers with . Since commutes with , this says that the coset has a representative of word length at most .
The ball of radius is finite by [A2], so step 2.1 leaves only finitely many cosets of in . Hence has finite index in , and the centralizer is virtually cyclic.
Depends on
Used by
Dependency tree · two levels
11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Clara Löh, Geometric Group Theory, Section 6.5.2 (standard reference, not scraped)