Results 1 to 10 of 10

Math Help - Permutations and (Set Theory??) I think.

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    23

    Permutations and (Set Theory??) I think.

    Basically I'm trying to revise for my thursday exam and it's going terribly. I am struggling with these two questions....

    Let X, Y be the permutations:...


    X = (1 2 3 4 5 6 7) and  Y = (1234)(24756)
    ----(1 5 7 4 6 2 3)

    Write X, Y, X^3, XY as products of disjoint cycles and determine orders plus signs!

    I really need a 'dummies' guide so to speak to this one. And X is written exactly like that with the numbers above and below.

    __________________________________________________ ___________

    The next question is:

    Let f: Z5 -> Z5 be given by f(x) = x^2 - draw the diagram of f and determine wether its injective or surjective. Justify.

    part ii) Give an example of a three element subset A -> Z5 such that the restriction of f to A is injective.

    part iii) give an example of subsets B,C such that f takes each element of B to B and the resulting map B->C is surjective.


    These are the two questions I am currently stuck on, if anyone can offer advice to either one I reserve alot of thanks for you. I'm pretty sure I might need help on other questions but not until I have figured these out. Thanks alot in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by Monkens View Post
    Basically I'm trying to revise for my thursday exam and it's going terribly. I am struggling with these two questions....

    Let X, Y be the permutations:...


    X = (1 2 3 4 5 6 7) and  Y = (1234)(24756)
    ----(1 5 7 4 6 2 3)

    Write X, Y, X^3, XY as products of disjoint cycles and determine orders plus signs!

    I really need a 'dummies' guide so to speak to this one. And X is written exactly like that with the numbers above and below.

    x = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\1&5&7&4  &6&2&3\end{array}\right)

    y = (1234)(24756)

    in x we have
    1\rightarrow 1 end of the cycle
    2\rightarrow 5\rightarrow 6 \rightarrow 2 end of the cycle
    3 \rightarrow7 \rightarrow3
    4 \rightarrow4

    so x = (256)(37)

    for y first want to write y

    y = (1234)(24756)

    (1234) = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\2&3&4&1  &5&6&7 \end{array}\right)

    (24756) =  \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\1&4&3&7  &6&2&5\end{array}\right)

    (1234)(24756)= \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\2&3&4&1  &5&6&7 \end{array}\right)\left(\begin{array}{ccccccc}1&2&  3&4&5&6&7\\1&4&3&7&6&2&5\end{array}\right)  = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\4&3&7&1  &6&2&5\end{array}\right)

    write each cycle then to find the product begin with the image of 1

    y = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\4&3&7&1  &6&2&5\end{array}\right)

    write it like x, in y we have

    1\rightarrow 4 \rightarrow 1 end
    2 \rightarrow 3 \rightarrow 7 \rightarrow 5 \rightarrow 6 \rightarrow 2 end

    so
    y=(14)(23756)

    the order of the cycle is the number of the entries in it for (23756) has the order 5
    (14) has the order 2
    the order of y is
    l.c.m(2,5) = 10, l.c.m least common multiple
    that means y^{10} = I
    if y is not written as a disjoint cycles what I did is not true so be careful first write it as a disjoint cycles then find the order

    or you can find y^2,y^3,y^4,... until you face the identity
    I = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\1&2&3&4  &5&6&7\end{array}\right)

    what is the order of x this for you

    now
    xy = (256)(37)(14)(23756)

    xy = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\1&5&7&4  &6&2&3\\4&6&5&1&2&3&7\end{array}\right) as I did before the second row is x the third is the product of xy remove the second row

    xy = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\4&6&5&1  &2&3&7\end{array}\right)

    leave the rest for you

    after I solved it more than 3 times and I see the book that I have and I solved the problems in it and my way it true
    Last edited by Amer; June 2nd 2010 at 01:51 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by Monkens View Post

    The next question is:

    Let f: Z5 -> Z5 be given by f(x) = x^2 - draw the diagram of f and determine wether its injective or surjective. Justify.

    part ii) Give an example of a three element subset A -> Z5 such that the restriction of f to A is injective.

    part iii) give an example of subsets B,C such that f takes each element of B to B and the resulting map B->C is surjective.


    These are the two questions I am currently stuck on, if anyone can offer advice to either one I reserve alot of thanks for you. I'm pretty sure I might need help on other questions but not until I have figured these out. Thanks alot in advance.
    z_5  = \{0,1,2,3,4\}

    f(0) = 0 \;\;, f(1) = 1 \;\;, f(2) =4 \;\;, f(3) =9= 4 \;\;, f(4) =16=1

    f: z_5 \rightarrow z_5
    f(x) = x^2

    it is not injective since f(2)=f(3) is not one-one function, we have same f(1)=f(4) it is enough to find two distinct elements in the domain which have the same image just one pair is enough

    it is not surjective since 3 is in the range but it is not an image same for 2 it is enough to find one element in the range which is not an image

    ii) f: A \rightarrow z_5
    f(x) = x^2
    you can read about restriction here

    choose A such that the image of elements of A covers all the elements in the Images
    you can choose A
    A= \{0,1,2\} or A = \{0,1,3\} or A = \{0,2,4\} or A = \{0,3,4\} anyone of this is correct

    iii)you can choose B anyone of the above, and C is the images {0,1,4}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2010
    Posts
    23
    Thanks very much Amer,

    there is one thing I am a little confused about though..

    Y = (1234)(24756)

    for that, your solution is different to that which is in the solutions book.

    They got Y = (12)(34756)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by Monkens View Post
    Thanks very much Amer,

    there is one thing I am a little confused about though..

    Y = (1234)(24756)

    for that, your solution is different to that which is in the solutions book.

    They got Y = (12)(34756)
    I do not think so


    (1234) = \left(\begin{array}{ccccccc} 1&2&3&4&5&6&7\\2&3&4&1&5&6&7 \end{array}\right) call it f

    (24756) = \left(\begin{array}{ccccccc} 1&2&3&4&5&6&7\\1&4&3&7&6&2&5 \end{array}\right) call it g

    g(f(1)) = g(2) = 4
    g(f(2)) = g(3) = 3
    g(f(3)) = g(4) = 7
    g(f(4)) = g(1) = 1
    g(f(5)) = g(5) = 6
    g(f(6)) = g(6) = 2
    g(f(7)) = g(7) = 5

    (1234)(24756) = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\4&3&7&1  &6&2&5 \end{array}\right)= (14)(23756)

    but

    Y = (12)(34756)

    (12)(34756) = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\2&1&4&7  &6&3&5\end{array}\right) \ne \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\4&3&7&1  &6&2&5 \end{array}\right)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2010
    Posts
    23
    Here is one of the questions I have to do, I really can't get to their solutions from your methods, so It must be something I've said wrong or asked you to do something different.. sorry for wasting your time but could you take a look at this one?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    I do not know sorry I do not have another way to solve it ...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2010
    Posts
    23
    It's ok thanks for your help, The solutions they got for the above problem are...

    pi = (1253)(457)

    p = (146275)

    pi^2 = (16)(23)(475)

    p*pi = (17634)

    I uploaded it so you could see if you were doing what you thought I was asking? As I may have asked incorrectly..

    I can get pi, but nothing else...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    sorry for begin late the difference is simple

    p=(1375)(31462)

    (1375) = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\3&2&7&4  &1&6&5\end{array}\right)

    (31462) = \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\4&3&1&6  &5&2&7\end{array}\right)

    now

    (1375)(31462)= \left(\begin{array}{ccccccc}1&2&3&4&5&6&7\\3&2&7&4  &1&6&5\end{array}\right)\left(\begin{array}{cccccc  c}1&2&3&4&5&6&7\\4&3&1&6&5&2&7\end{array}\right)

    just read it from right to left,
    we will have

    1\rightarrow 4 \rightarrow 4

    2\rightarrow 3 \rightarrow  7

    3\rightarrow 1 \rightarrow  3

    4\rightarrow  6 \rightarrow  6

    5\rightarrow  5 \rightarrow  1

    6\rightarrow  2 \rightarrow  2

    7 \rightarrow  7 \rightarrow  5

    I hope I helped you, best wishes in your exam
    Last edited by Amer; June 2nd 2010 at 02:11 PM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    May 2010
    Posts
    23
    Thanks, that has worked!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Textbooks on Galois Theory and Algebraic Number Theory
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: July 8th 2011, 07:09 PM
  2. Permutations (I think...)?
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: April 2nd 2011, 11:11 AM
  3. Permutations Help
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: March 9th 2010, 08:30 PM
  4. Group Theory - Sylow Theory and simple groups
    Posted in the Advanced Algebra Forum
    Replies: 16
    Last Post: May 16th 2009, 12:10 PM
  5. Problems relating Theory of Automata (Computer Theory)
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: October 17th 2007, 10:52 AM

Search Tags


/mathhelpforum @mathhelpforum