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Questions on the symmetric group

  1. Nov 21, 2012 #1
    first ,
    if p is prime , show that an element has order p in Sn iff it's cycle decomosition is a product of commuting p-cycles

    my solution is very diffrent about the one in the book and I don't know if my strategy is right

    my proof
    ______
    let T is an element of Sn
    and the cycle decomposition of T = t1 t2 t3 .... tm
    t1 , t2 , ... , tm are disjoint cycle
    so
    ti tj = tj ti for all i,j
    T^p = ( t1 t2 .... tm )^p = (t1)^p (t2)^p .... (tm)^p
    now , ti , tj are 2 disjoint cycle so if ti or tj are not equal to identity permutation then ,
    ti tj =\= e

    now if T^p = 1
    then
    (t1)^p (t2)^p .... (tm)^p = 1
    but (t1)^p , (t2)^p , ... , (tm)^p are disjoint
    so if (ti)^p =\= 1 for all i then
    (t1)^p (t2)^p .... (tm)^p =\= 1
    but (t1)^p (t2)^p .... (tm)^p = 1
    so it's nessary that
    (ti)^p = 1 for all i
    so
    l ti l = p
    and ti 's are commuting because they are disjoint

    so we proved that order of T = p where p is prime then it's cycle decomposition has the same order p
    now
    every t1 is a p-cycle

    is this right ?

    the otherway is easy to prove

    plz help


    in this question we can't use the rule which says @ order of an element in Sn is the LCD of the lenghts of the cycles in its cycle decomposition because this rule didn't come in the book and I search for a proof with out it
     
  2. jcsd
  3. Nov 22, 2012 #2
    here is the question on more clear way

    prove that :

    676955475.png
     
  4. Nov 22, 2012 #3

    haruspex

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    On the face of it, that's a rather different question (though no doubt they're equivalent in some way). So if the second post is the real question to be answered, please repost your proof in a way that relates more directly to it.
     
  5. Nov 23, 2012 #4
    sorry , a mistake was made !
    my question is the first one ,
    the second one I work on it

    so , does my attempt on the first question is right or not ?!
     
  6. Nov 23, 2012 #5

    haruspex

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    Not immediately. Tp = ((t1) (t2) ... (tm))p = 1. You need to use the fact that (t1) , (t2) , ... , (tm) are disjoint to show they commute.
     
  7. Nov 23, 2012 #6
    yes , I said that t1 , t2 , ... , tm are disjoint beacuse t's are the cycle decomposition
    so they are disjoint so they are commuting with each other !
     
  8. Nov 23, 2012 #7

    haruspex

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    Yes, you said that later, but not at the point where you made use of it.
     
  9. Nov 24, 2012 #8
    ok :) thank you
     
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