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Group action and equivalence relation 
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#1
Dec1612, 03:40 PM

P: 94

Given a group [itex]G[/itex] acting on a set [itex]X[/itex] we get an equivalence relation [itex]R[/itex] on [itex]X[/itex] by [itex]xRy[/itex] iff [itex]x[/itex] is in the orbit of [itex]y[/itex].
My question is, does some form of "reciprocal" always work in the following sense: given a set [itex]X[/itex] with an equivalence relation [itex]R[/itex] defined on it, does it always exist some group [itex]G[/itex] with some action over [itex]X[/itex] such that the set of its orbits coincide with the equivalence classes? I have thoght it, and concluded that for finite sets and groups, the cardinal of [itex]G[/itex] has to be a múltiple of the cardinal of every orbit, but I can't see if it is always possible to construct such group with such an action. Thanks in advance for any help! 


#2
Dec1712, 09:31 AM

P: 350

An equivalence relation is the same as a partition of the set into a disjoint union of subsets (the equivalence classes). Let G be the group of bijections from X to itself. Let H be the subset of G which leaves the equivalence classes invariant. Then H is a subgroup, and it acts in the way you want.



#3
Dec1712, 01:42 PM

P: 94

Hi Vargo,
Thanks for your reply. I think I can see your point. By the subset of G wich leaves the equivalence classes invariant, I think you mean the maximal one with that property (as the trivial susbset of G obviously leaves the classes invariant) Anyway it's interesting that any equivalence relation can be thought as the result of a group action, so every time I see a quotient space of any kind I can think as the result of some group acting by "gluing" some elements together. 


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