Results 1 to 9 of 9

Math Help - Permutation Groups

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    224

    Permutation Groups



    I'm sure there is a quick and straightforward method for answering this question. But I can't find it in my textbook.

    Here's what I know:

    \sigma \tau in the array notation would be:

    \begin{bmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 &8&9&10&11\\6 & 4 & 11 & 9 & 1 & 7 & 3 & 5 & 8 &2&10\end{bmatrix}

    and can be written as a product of disjoint cycles:

    (1,6,7,3,11,10,2,4,9,8,5)

    Since it's only one cycle of length 11, the order is lcm(11)=11

    I'd be thankful if anyone could show me what method to use to find (\sigma \tau)^{9000} and (\sigma \tau)^{-21}.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    By what you said about the order, (\sigma\tau)^{11}=1. So, to start with, you should work out 9000 and -21 modulo 11. Can you see why you would want to do this?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    224
    Quote Originally Posted by Swlabr View Post
    By what you said about the order, (\sigma\tau)^{11}=1. So, to start with, you should work out 9000 and -21 modulo 11. Can you see why you would want to do this?
    No, what should do I need to do with it? 9000 mod 11 = 2 and -21 mod 11 = 1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    16
    Since [LaTeX ERROR: Convert failed] , you should now break down the powers 9000 and -21 in terms of 11. I think that will be enough of a hint to solve this thing!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2009
    Posts
    224
    Quote Originally Posted by Vlasev View Post
    Since [LaTeX ERROR: Convert failed] , you should now break down the powers 9000 and -21 in terms of 11. I think that will be enough of a hint to solve this thing!
    Okay, so

    \sigma \tau^{-21} = \sigma \tau^{1} = (1,6,7,3,11,10,2,4,9,8,5)

    Is that right?! And

    \sigma \tau^{9000} = \sigma \tau^{2} = (1,6,7,3,11,10,2,4,9,8,5)^2

    But I don't think this is enough. Should I bring every element to the power of 2 then modulo 11?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    16
    If I understand correctly what you are asking, this is how it's done. (k is some integer)

    [LaTeX ERROR: Convert failed]

    And then just calculate the square. It's similar for the other one!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Dec 2009
    Posts
    224
    Well this is what I mean:

    (\sigma\tau)^2 = (1,6,7,3,11,10,2,4,9,8,5)^2 = (1,36,49,9,121,100,4,16,81,64,25)

    Now writing those numbers modulo 11 we have:

    (1,3,5,9,0,1,4,5,4,9,3)

    Since the 4 and 9 have been repeated twice, the one cycle doesn't make any sense! Do you see the problem? I tried to break this up into new disjoint cycles but that wouldn't work either...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    16
    Ohh, I see, it's just a matter of multiplying permutations!. Just convert the permutation into its array notation and multiply it by itself the same way you are multiplying different permutations!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,311
    Thanks
    1286
    Quote Originally Posted by demode View Post
    Well this is what I mean:

    (\sigma\tau)^2 = (1,6,7,3,11,10,2,4,9,8,5)^2 = (1,36,49,9,121,100,4,16,81,64,25)
    No, that is not what (\sigma\tau)^2 means! (\sigma\tau)^2 means the permutation \sigma\tau done twice. \tau maps 1 into 6 and then 6 into 7 so (\sigma\tau)^2 maps 1 into 7. \tau maps 7 into 3 and then 3 into 11 so (\sigma\tau)^2 maps 7 into 11.

    In cycle notation, (\sigma\tau)^{9000}= (\sigma\tau)^2 starts "(1, 7, 11, ..."


    Now writing those numbers modulo 11 we have:

    (1,3,5,9,0,1,4,5,4,9,3)

    Since the 4 and 9 have been repeated twice, the one cycle doesn't make any sense! Do you see the problem? I tried to break this up into new disjoint cycles but that wouldn't work either...[/QUOTE]
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Permutation of groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 25th 2011, 01:13 AM
  2. Permutation Groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 3rd 2010, 04:04 PM
  3. Permutation groups
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: November 25th 2008, 12:56 AM
  4. Permutation Groups
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 4th 2008, 05:14 PM
  5. Permutation Groups...
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 24th 2007, 08:55 PM

Search Tags


/mathhelpforum @mathhelpforum