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Group action and equivalence relation

  1. Dec 16, 2012 #1
    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. jcsd
  3. Dec 17, 2012 #2
    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.
  4. Dec 17, 2012 #3
    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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