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A cyclic group of order 15

  1. Dec 20, 2003 #1
    A cyclic group of order 15 has an element x such that the set {x^3, x^5, x^9} has exactly two elements. The number of elements in the set {x^(13n) : n is a positive integer} is :

    WHy is the answer 3? Thanks!!
  2. jcsd
  3. Dec 20, 2003 #2
    Let [tex]x[/tex] be an element of a cyclic group of order 15. If [tex]\{x^3,x^5,x^9\}[/tex] has exactly 2 elements, then one element must be the same as another. If [tex]|x|=3[/tex] then [tex]x^3=x^9[/tex]. If [tex]|x|=3[/tex] then [tex]|x^{13}|=3/gcd(3,13)}=3[/tex]. Hence [tex]|<x^{13}>|=3[/tex].

  4. Dec 20, 2003 #3
    Thanks so much for the explanation!! However I haven't actually studied group theory before so there's some things I still am not sure about. Why did you assume |x| = 3? What formula did [tex]|x^{13}|=3/gcd(3,13)}=3[/tex] come from?
    Can you give an example of a set that meets this condition?
  5. Dec 20, 2003 #4
    Your original post stated that there exists an element [tex]\inline{x}[/tex] such that [tex]\inline{\{x^3,x^5,x^9\}}[/tex] has exactly two elements. Thus I need to find two elements that are the same. There is a theorem that states if [tex]\inline{|x|=n}[/tex] then [tex]\inline{x^i=x^j}[/tex] if and only if [tex]\inline{n}[/tex] divides [tex]\inline{i-j}[/tex]. If [tex]\inline{x^3=x^5}[/tex] then [tex]\inline{|x|=2}[/tex] which is not possible in a cyclic group of order 15 because 2 does not divide 15. Likewise, [tex]\inline{x^5{\neq}x^9}[/tex]. However if [tex]\inline{|x|=3}[/tex] then [tex]\inline{x^3=x^9}[/tex] because 3 divides [tex]\inline{i-j=9-3=6}[/tex]. Also, a cyclic group of order 15 can have an element with order 3.

    There is a theorem that states if [tex]\inline{|x|=n}[/tex] then [tex]\inline{|x^k|=n/gcd(n,k)}[/tex].

    The group [tex]\inline{Z_{15}}[/tex] is the group of integers modulo 15 under addition. Both the element 5 and the element 10 have order 3.

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