Results 1 to 5 of 5

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

  1. #1
    Junior Member
    Joined
    Feb 2010
    Posts
    35

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

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

    Here's what I have so far:

    Part 1: Show that ord(bab^{-1}) = n

    Suppose that ord (a) = n,

    Then a^n = e

    The problem I'm having is using the Associative Law to show that (bab^{-1})^n is also e.

    Part 2: Show that ord (a) = n.

    Suppose that ord(bab^{-1}) = n

    Then (bab^{-1})^n = e

    The problem I'm having at this part is using the Associative Law to show that a^n = e.

    Help is appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Iceflash234 View Post
    Prove that  ord(a) = ord(bab^{-1})

    Here's what I have so far:

    Part 1: Show that ord(bab^{-1}) = n

    Suppose that ord (a) = n,

    Then a^n = e

    The problem I'm having is using the Associative Law to show that (bab^{-1})^n is also e.

    Part 2: Show that ord (a) = n.

    Suppose that ord(bab^{-1}) = n

    Then (bab^{-1})^n = e

    The problem I'm having at this part is using the Associative Law to show that a^n = e.

    Help is appreciated.
    Well, clearly (bab^{-1})=(bab^{-1}(bab^{-1})\cdots=ba^nb^{-1}=beb^{-1}=bb^{-1}=e so |bab^{-1}|\leqslant n but what would happen if that inequality was strict?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Iceflash234 View Post
    Prove that  ord(a) = ord(bab^{-1})

    Here's what I have so far:

    Part 1: Show that ord(bab^{-1}) = n

    Suppose that ord (a) = n,

    Then a^n = e

    The problem I'm having is using the Associative Law to show that (bab^{-1})^n is also e.

    Part 2: Show that ord (a) = n.

    Suppose that ord(bab^{-1}) = n

    Then (bab^{-1})^n = e

    The problem I'm having at this part is using the Associative Law to show that a^n = e.

    Help is appreciated.

    Hint: \forall\,\,n\in\mathbb{Z}\,,\,\,(bab^{-1})^n=ba^nb^{-1}

    And remember that it is NOT enough to show x^k=e to conclude that ord(x)=k: you must also show this k is the minimal natural number fulfilling that.


    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2010
    Posts
    35
    Thanks! That helped.

    I see clearly how to show bab^{-1} = e.

    I'm not sure though how to show a^{n} = e for the second part.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    Posts
    68
    Quote Originally Posted by Iceflash234 View Post
    Prove that  ord(a) = ord(bab^{-1})

    Here's what I have so far:

    Part 1: Show that ord(bab^{-1}) = n

    Suppose that ord (a) = n,

    Then a^n = e

    The problem I'm having is using the Associative Law to show that (bab^{-1})^n is also e.

    Part 2: Show that ord (a) = n.

    Suppose that ord(bab^{-1}) = n

    Then (bab^{-1})^n = e

    The problem I'm having at this part is using the Associative Law to show that a^n = e.

    Help is appreciated.
    You are way to confused there. What you need to show that is that  a^n=e\,\Leftrightarrow\,(bab^{-1})^n=e (which will imply that \mathrm{ord}(a)=\mathrm{ord}(bab^{-1})). To do this, you need to use tonioís hint (which can be easily proved by induction).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove a(AB)=(aA)B=A(aB) ..
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 29th 2010, 04:14 AM
  2. Prove: f is one-to-one iff f is onto
    Posted in the Discrete Math Forum
    Replies: 12
    Last Post: June 25th 2010, 10:02 AM
  3. Replies: 2
    Last Post: August 28th 2009, 02:59 AM
  4. Please prove
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 7th 2009, 01:58 PM
  5. Prove this .
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: February 18th 2009, 04:09 AM

Search Tags


/mathhelpforum @mathhelpforum