Results 1 to 4 of 4

Math Help - Proof Involving Orbits

  1. #1
    Junior Member
    Joined
    Sep 2008
    From
    Indiana
    Posts
    26

    Proof Involving Orbits

    I'm given: Let A be a set and let \sigma \in S_A. For a fixed a \in A, the set
    O_{a,\sigma} = { \sigma^n (a)|n \in Z}
    is the orbit of a under \sigma.

    I have to prove:
    Let a,b \in A and \sigma \in S_A. Show that if O_{a,\sigma} and O_{b,\sigma} have an element in common, then O_{a,\sigma} = O_{b,\sigma}.

    The statement makes sense, I just have no idea what to do when it comes to proving it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by Ryaη View Post
    I'm given: Let A be a set and let \sigma \in S_A. For a fixed a \in A, the set
    O_{a,\sigma} = { \sigma^n (a)|n \in Z}
    is the orbit of a under \sigma.

    I have to prove:
    Let a,b \in A and \sigma \in S_A. Show that if O_{a,\sigma} and O_{b,\sigma} have an element in common, then O_{a,\sigma} = O_{b,\sigma}.

    The statement makes sense, I just have no idea what to do when it comes to proving it.

    You know that orbits of an action of a group on a set are either identical or disjoint (because they in fact are equivalence classes of an equivalence relation)? Well, this is the same: O_a,sygma is the orbit of the element a under the natural action of the group <sygma>,,,and we're done!

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    From
    Indiana
    Posts
    26
    I'm not sure I completely understand.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    Yeah, what tonio says is true, this is just a small case of the theorem which states that the orbits of a group action partition the set upon which it acts.
    In this case, its really easy to prove.

    So suppose they have nontrivial intersection, we show this means the sets are the same. I will only show one containment because the other way is completely symmetric.

    Let x\in O_{a,\sigma} \cap O_{b,\sigma}
    \sigma^i(a)=x=\sigma^j(b) by definition of O_{a,\sigma} and O_{b,\sigma}

    But then just hit this equation by \sigma^{-i} on both sides.
    \sigma^{-1}\sigma^i(a)=\sigma^{-i}\sigma^j(b)
    a=\sigma^{j-i}(b) Note this uses group action definition.

    This shows x\in O_{a,\sigma} \subset O_{b,\sigma}. See why? Now show the other containment
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof about closed orbits of a planar system
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 21st 2011, 01:07 PM
  2. Proof involving GCD
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 11th 2010, 03:15 PM
  3. A proof involving logs
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 26th 2010, 03:23 PM
  4. Proof involving e
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 2nd 2010, 03:44 PM
  5. Proof involving GCD
    Posted in the Discrete Math Forum
    Replies: 11
    Last Post: December 5th 2009, 04:36 PM

Search Tags


/mathhelpforum @mathhelpforum