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Abstract algebra problem concern

  1. Mar 22, 2005 #1
    My prof. assigned this problem as the only one for HW a few days back, and for some reason the answer seems too obvious. What subtleties could I possibly be missing?

    Let G be a group of permutations in a set S. If [tex]x, y \in S[/tex] and [tex] y \in orb_g(x)[/tex], then [tex]orb_g(y) = orb_g(x)[/tex]

    Still, I am unsure how to put it in words. I'll edit this post as I come up with ideas but I think I might need a kick in the right direction, because even as I'm writing this I'm starting to realize it's not entirely trivial.
     
  2. jcsd
  3. Mar 22, 2005 #2

    Hurkyl

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    I think it's obvious too. :smile:

    I would just make sure I did the algebra in painstaking detail, so that I wasn't worried about any of the steps.
     
  4. Mar 22, 2005 #3
    It's good to hear I wasn't overthinking it then! Here is my rough answer that I'll fine-tune later but I'd like to hear opinions on it.

    I define G as follows: [tex]G = \{\phi_e, \phi_1, \phi^{-1}_1, \phi_2, \phi^{-1}_2, ...\}[/tex]

    If [tex]y \in orb_G(x)[/tex], then [tex]\phi(x) = y[/tex] for some [tex]\phi \in G[/tex]. Obviously since G is a group, [tex]\phi^{-1} \in G, \phi^{-1}(y) = x[/tex]. That being said, am I on the right track at least? I know it's not complete, but it shouldn't be hard to wrap up.
     
    Last edited by a moderator: Mar 22, 2005
  5. Mar 22, 2005 #4
    Ok now I'm stumped. I proved that [tex]x \in orb_G(y)[/tex] by [tex]y \in orb_G(x)[/tex] but I can't seem to tie an arbitrary element of [tex]orb_g(x)[/tex] to [tex]orb_g(y)[/tex], ie I can't determine [tex]\phi_i(x) = x_i = \phi_j(y)[/tex]. Any guidance?

    Edit: I think I got the missing key. Since G is a group, any arbitrary [tex]\phi_i, \phi_j \in G[/tex], [tex]\phi_i\phi_j[/tex] must also lie in G. So for an arbitary [tex]x_i \in orb_G(x) | \phi_i(x) = x_i[/tex], if [tex]\phi(x) = y, \phi \in G[/tex] then [tex] \phi_i(\phi^{-1}(y)) = x_i[/tex] and by the closed nature of G, [tex]\phi_i(\phi^{-1}(y)) \in G[/tex]
     
    Last edited by a moderator: Mar 22, 2005
  6. Mar 23, 2005 #5

    Hurkyl

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    Stylistic note: there's no reason to write an enumeration of the elements of G. Furthermore, it is misleading, or even wrong! For example, the group might not be countable, or maybe [itex]\phi_1 = \phi_1^{-1}[/itex]!

    Your edit makes me think you have the right idea... now clean it up! Make it look like:

    Let [itex]z \in \mathrm{Orb}_G(x)[/itex]
    ...
    Therefore [itex]z \in \mathrm{Orb}_G(y)[/itex]

    And vice versa. (Or, do the whole thing with if and only if deductions, so you don't have to do the vice versa)
     
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