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Prove that ord(a) = ord(bab^(-1))

  1. Mar 30, 2010 #1
    Prove that ord(a) = ord(bab^(-1)) (Edited/Fixed)

    1. The problem statement, all variables and given/known data
    Prove that ord(a) = ord(bab-1)

    2. Relevant equations
    N/A


    3. The attempt at a solution

    Part 1: Show that ord(bab-1) = n
    Suppose that ord(a) = n
    Then an = e.
    I then have to use the associative law to show (bab-1)n = e but that's where I'm having troublem.

    Part 2: Show that ord(a) = n
    Suppose ord(bab-1) = e
    Then (bab-1)n = e
    I then have to use the associative law to show that an = e but I'm stuck there as well.

    Help is greatly appreciated.
     
    Last edited: Mar 30, 2010
  2. jcsd
  3. Mar 30, 2010 #2
    I fixed the many typos I had.
     
  4. Mar 30, 2010 #3

    jbunniii

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    Try writing out the multiplication:

    [tex](bab^{-1})^n = (bab^{-1})(bab^{-1})\cdots (bab^{-1})[/tex] (product of n identical terms)

    and simplifiy.
     
  5. Mar 30, 2010 #4
    So, doing that I get this:

    [tex](bab^{-1})^n = (bab^{-1})(bab^{-1})\cdots (bab^{-1})[/tex] = [tex]ba^nb^{-1} = beb^{-1} = bb^{-1} = e[/tex]

    Is that a valid thing to do?

    On that thought, How would I show [tex]a^n = e[/tex] for part 2?
     
  6. Mar 30, 2010 #5

    jbunniii

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    Yes, it's valid. So this shows that [itex](bab^{-1})^n = e[/itex]. It does NOT necessarily follow that [itex]o(bab^{-1}) = n[/itex], because there could be some smaller exponent [itex]k[/itex] such that [itex](bab^{-1})^k = e[/itex]. You have to do a little more work to prove that this can't happen.

    I'm not sure why you think this is a two-part proof. You want to prove that

    [tex]o(a) = o(bab^{-1})[/tex]

    So let [itex]n = o(a)[/itex], and then prove that [itex]o(bab^{-1}) = n[/itex] and you're done.
     
  7. Mar 30, 2010 #6
    Ok.


    That's how my professor outlined the proof for me. They broke it up into two parts. The conclusion that is to come out of that as well is that doing both parts will prove that both orders are either infinite or both finite and equal. I believe that's why they want it broken up into two parts.
     
  8. Mar 30, 2010 #7

    jbunniii

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    I'm not sure I understand the prof's rationale. Structuring the proof as

    "Part 1: show that [itex]o(bab^{-1}) = n[/itex]
    Part 2: show that [itex]o(a) = n[/itex]"

    begs the question: what is [itex]n[/itex]? I don't see how this helps address the finite vs. infinite cases. More natural would seem:

    "Part 1: Suppose that [itex]o(a) = n < \infty[/itex]. Show that [itex]o(bab^{-1}) = n[/itex]
    Part 2: Suppose that [itex]o(a) = \infty[/itex]. Show that [itex]o(bab^{-1}) = \infty[/itex]"
     
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