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Bijection between Orbit and Stabilizer

  1. Oct 16, 2009 #1
    So I know this is the orbit-stabilizer theorem. I saw it in Hungerford's Algebra (but without that name).

    So we want to form a bijection between the right cosets of the stabilizers and the orbit. Could I define the bijection as this:

    f: gG/Gx--->gx

    Where H=G/Gx

    f(hx)=gx h in H

    ^ Is that what the function is suppose to look like? I'm really stuck on understanding the proof since it doesn't show me this function but it guarantees that it is a bijection. I'm not sure if there is a better way to word this question except, perhaps, what is the function between the right cosets and the orbit.
     
    Last edited: Oct 16, 2009
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  3. Oct 17, 2009 #2

    Fredrik

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    Have a look at theorem 3 on page 5 of this article. Is the function that the author defines in the proof what you're looking for?
     
  4. Oct 17, 2009 #3
    Sort of, but I'm getting kind of thrown off by his notation.

    So the function that is defined is sending f: g*H---->Orb(x) where H=G/Gx.

    Is this just saying that f(gh)=gx or f(g)=gx because H is everything that doesn't move x so its not really worth mentioning.
     
    Last edited: Oct 17, 2009
  5. Oct 17, 2009 #4

    Fredrik

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    He's saying that if [itex]y\in\mathcal O_x[/itex], then there exists [itex]g\in G[/itex] such that [itex]y=gx[/itex]. Then he defines [itex]f(y)=gK_x[/itex], where [itex]K_x[/itex] is the stabilizer subgroup for x. It's not obvious that this f is well-defined, because there could exist a [itex]g'\neq g[/itex] such that [itex]y=gx=g'x[/itex]. To prove that the above actually defines a function, we must prove that [itex]gx=g'x[/itex] implies [itex]gK_x=g'K_x[/itex].

    Here's the proof: Suppose that [itex]z\in gK_x[/itex]. Then there's a [itex]k\in K_x[/itex] such that [itex]z=gk=g'g'^{-1}gk[/itex]. This is a member of [itex]g'K_x[/itex] if [itex]g'^{-1}gk\in K_x[/itex], and it's not hard to see that it is:

    [tex]g'^{-1}gkx=g'^{-1}gx=g'^{-1}g'x=x[/tex]

    This proves that [itex]gK_x\subset g'K_x[/itex], and we can of course repeat the argument with g and g' swapped.

    So f is a well-defined function from the orbit [itex]\mathcal O_x[/itex] into the set of right cosets [itex]G/K_x[/itex]. The equation that defines f can also be written as [itex]f(gx)=gK_x[/itex].
     
    Last edited: Oct 17, 2009
  6. Oct 17, 2009 #5
    The very last function you defined is from the orbit to the stabilizer (I know its a bijection but I just want to make sure).
     
  7. Oct 17, 2009 #6

    Fredrik

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    No, [itex]gK_x=\{gk|k\in K_x\}[/itex] is a right coset, so

    [tex]f:\mathcal O_x\rightarrow G/K_x[/tex]​

    if we use the notation [itex]G/K_x[/itex] for the set of right cosets.
     
  8. Oct 18, 2009 #7
    Great, it all makes sense. However, how does this bijection prove that O*Gx:K=G. To me it just proves they are equal, but that can't always be the case.
     
  9. Oct 20, 2009 #8

    Fredrik

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    The theorem says that the action restricted to [itex]\matcal O_x[/itex] is equivalent (in a technical sense) to the canonical action on [itex]G/K_x[/itex]. This splits the problem of understanding a group action into two pieces: 1. Find the orbits. 2. Study the action on [itex]G/K_x[/itex], for each different [itex]K_x[/itex].
     
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