Results 1 to 3 of 3

Thread: conjugacy question abstract algebra

  1. #1
    Newbie
    Joined
    Dec 2012
    From
    Los Angeles
    Posts
    12

    conjugacy question abstract algebra

    If x~_c y in a group G, show that x and y have infinite order or o(x)=o(y)


    (Note: the relation x~_c y is called conjugacy, and as for any equivalence relation, the set G is partitioned by its equivalence classes)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member Nehushtan's Avatar
    Joined
    Mar 2013
    From
    Europe
    Posts
    42
    Thanks
    12

    Re: Conjugacy question in abstract algebra

    Hint: Show that $\displaystyle \left(gxg^{-1}\right)^n=gx^ng^{-1}$ for all $\displaystyle g\in G$ and $\displaystyle n\in\mathbb Z^+$. It will follow that $\displaystyle \left(gxg^{-1}\right)^n=e\Leftrightarrow x^n=e$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member jakncoke's Avatar
    Joined
    May 2010
    Posts
    387
    Thanks
    80

    Re: conjugacy question abstract algebra

    This stuff becomes easy if you realize $\displaystyle (gxg^{-1})^{n} = gx^ng^{-1} $ for $\displaystyle (gxg^{-1})^{n} = gxg^{-1}gxg^{-1}gxg^{-1}....= gx^ng^{-1}$

    If |x| = $\displaystyle \infty $.
    if |y| = k ( a finite number)
    then $\displaystyle y^k = e $
    so since x ~ y, there exists a $\displaystyle g \in G $ s.t $\displaystyle gxg^{-1} = y $ then $\displaystyle (gxg^{-1})^{k} = y^k = e $
    so $\displaystyle gx^{k}g^{-1} = e $
    or $\displaystyle gx^k = g $
    or $\displaystyle x^k = e $ which cannot be since order of x is $\displaystyle \infty $.

    Thus is x has order $\displaystyle \infty $ so does y.

    If |x| = n (finite)
    then $\displaystyle (gxg^{-1})^{n} = y^n $ so $\displaystyle e = y^n $, so either |y| = n, so |y| is less than n. if |y| = k < n $\displaystyle y^k = (gxg^{-1})^{k} $
    so $\displaystyle e = gx^kg^{-1}$ or $\displaystyle x^k = e $ contradiction so n = k.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. abstract algebra: subgroups question
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Mar 9th 2013, 01:46 PM
  2. Abstract Algebra Identity Element question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 4th 2012, 11:51 PM
  3. Replies: 0
    Last Post: Apr 23rd 2010, 11:37 PM
  4. Abstract Algebra question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Feb 17th 2010, 05:15 AM
  5. [SOLVED] Abstract algebra semigroup question
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: Jul 9th 2008, 09:33 PM

Search Tags


/mathhelpforum @mathhelpforum