Page 2 of 2 FirstFirst 12
Results 16 to 19 of 19

Math Help - Permutations and Abelian Group proof.

  1. #16
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    \beta=(2,3,9)(6,9,10)=\left(\begin{array}{cccccccc  cc}1&2&3&4&5&6&7&8&9&10\\1&3&9&4&5&6&7&8&2&10\\1&3  &10&4&5&9&7&8&2&6\end{array}\right)


    \beta=\left(\begin{array}{cccccccccc}1&2&3&4&5&6&7  &8&9&10\\1&3&10&4&5&9&7&8&2&6\end{array}\right)=(2  ,3,10,6,9)







    Follow Math Help Forum on Facebook and Google+

  2. #17
    Newbie
    Joined
    Dec 2009
    Posts
    22
    Quote Originally Posted by Amer View Post
    \beta=(2,3,9)(6,9,10)=\left(\begin{array}{cccccccc  cc}1&2&3&4&5&6&7&8&9&10\\1&3&9&4&5&6&7&8&2&10\\1&3  &10&4&5&9&7&8&2&6\end{array}\right)


    \beta=\left(\begin{array}{cccccccccc}1&2&3&4&5&6&7  &8&9&10\\1&3&10&4&5&9&7&8&2&6\end{array}\right)=(2  ,3,10,6,9)







    Ah, I get it now! Thanks very much!

    So now, with Alpha*Eta = Beta,

    that means Eta = Alpha^(-1) * Beta, how do you go about computing that?
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Newbie
    Joined
    Dec 2009
    Posts
    22
    After some calculations, would I be correct to state that:

    Alpha^(-1) = Alpha

    That's what I found. There what I need to compute is:

    Eta = AlphaBeta
    Follow Math Help Forum on Facebook and Google+

  4. #19
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by MickQ View Post
    After some calculations, would I be correct to state that:

    Alpha^(-1) = Alpha

    That's what I found. There what I need to compute is:

    Eta = AlphaBeta
    ok it is easy now to solve it

    \eta = \alpha \beta

    \eta = (1,2)(10,8)(2,3,10,6,9) simplify it
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Abelian Group Proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 25th 2010, 02:44 PM
  2. Is the subgroup of an abelian group always abelian?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 6th 2009, 11:38 PM
  3. Abelian Group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 16th 2009, 07:14 PM
  4. Group Theory: Proof on abelian and isomorphic groups.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 20th 2009, 06:46 PM
  5. Abelian Group proof
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: March 9th 2009, 09:17 PM

Search Tags


/mathhelpforum @mathhelpforum