G-equidecomposable and equivalence relation

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SUMMARY

The discussion centers on the equivalence relation A~B for subsets A and B of a set X, where G acts via isometries. The relation is established as an equivalence relation on subsets of X if and only if A and B are G-equidecomposable. Key concepts include the Banach-Schroder theorem and properties of equidecomposability, which involve checking three conditions related to group actions, including the existence of inverses and the associative property of composition.

PREREQUISITES
  • Understanding of group theory concepts, particularly isometries.
  • Familiarity with the Banach-Schroder theorem.
  • Knowledge of equivalence relations and their properties.
  • Basic principles of G-equidecomposability.
NEXT STEPS
  • Study the Banach-Schroder theorem in detail.
  • Learn the definition and properties of G-equidecomposable sets.
  • Explore examples of equivalence relations in group theory.
  • Investigate the implications of group actions on subsets of a set.
USEFUL FOR

Mathematicians, particularly those specializing in group theory, and students seeking to understand the relationship between group actions and equivalence relations.

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G acts via isometries on a set X, and A,B are subsets of X. Prove that the relation A~B is an equivalence relation on subsets of X iff A and B are G-equidecomposable.


I think this has to do with the Banach-Schroder theorem, but am not sure. I know it is a definition in group theory, but am not sure how to prove it since it seem pretty self explanitory to me.
 
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Can someone help?
 
define G-equidecomposable, please,

the result should be easy isnce it just checking 3 things for the property of equidecomposability, whatever that may be, which are generally trivial for groups since groups have inverses, and identity and composition is associative. eg if X~Y and g in G effects this relation, then g^{-1} will (probably) effect the relation Y~X, if X~Y and Y~Z and the relation is because of elements f and g resp. then gf wil mean that X~Y, and X~X because e(X)=X.

note i haven't a clue what equidecompsoability is, but this will be the proof, I'm almost sure of it.
 

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