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A question in groups and presentation

  1. Nov 18, 2012 #1
    let X(2n) be the group whose presentation is :
    ( x,y l x^n = y^2 = 1 , xy = yx^2 )

    show that if n = 3k then X2n has order 6
    and if (3,n)=1 then x satisfies the additional relation x=1 , in this case , deduce that X(2n) has order 2
    note that :
    x^3 = 1


    ____________
    I tried to slove it and I think that I found the right solution and I will explain the main idea of it because the details is quite long , and I need to know if the main idea is right or not

    the idea is :
    first step ,
    I suppose that S is a set : S= { 1 , x , y , x^2 , xy , yx }
    then I proved that the elements of S are distinct " I mean that if a , b belongs to S then it's nessary that a=\= b "

    then I proved that S = X(2n )
    I did it as follows :
    first I proved that every element of X(2n) is a product of finite elements if S
    then I proved that any product of finite elements of S equals to some elements of S
    the I deduced that every element of X(2n) belongs to S so X(2n) ⊂ S
    but S ⊂ X(2n) because {x,y} generates X(2n) then {x,y} ⊂ X(2n) and every possible product of them also belongs to X(2n) because multipication is closed in groups

    so X(2n) = S

    but
    lSl = the order of S = 6
    so l X(2n) l = 6

    at last , I show that if n is not equal to 3k then elements of S are not distinct so
    l s l = 6 if n = 3k

    is this idea true ??

    the details is long

    so I will omit it ,

    and I wonder about your ideas to slove this problem ??
    I find my one too long

    thank you very much
     
  2. jcsd
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