# G-equidecomposable and equivalence relation

1. Aug 5, 2005

### JSG31883

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.

2. Aug 6, 2005

### JSG31883

Can someone help????

3. Aug 7, 2005

### matt grime

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.