What is the number of elements in the set {x^(13n) : n is a positive integer}?

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Discussion Overview

The discussion revolves around determining the number of elements in the set {x^(13n) : n is a positive integer} within the context of group theory, specifically focusing on cyclic groups and their properties. Participants explore the implications of the order of an element in a cyclic group of order 15 and the conditions under which certain elements can be equal.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant claims that the number of elements in the set {x^(13n) : n is a positive integer} is 3, based on the properties of a cyclic group of order 15.
  • Another participant explains that if the set {x^3, x^5, x^9} has exactly two elements, then one of the elements must be equal to another, leading to the assumption that |x|=3.
  • It is noted that if |x|=3, then |x^(13)| can be calculated using the formula |x^k|=n/gcd(n,k), resulting in |x^(13)|=3.
  • A participant expresses confusion about the assumption of |x|=3 and requests clarification on the formula used and an example of a set that meets the discussed conditions.
  • Further clarification is provided regarding the theorem that relates the order of an element to the divisibility of indices, emphasizing that if |x|=3, then x^3 must equal x^9.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the assumptions made about the order of the element x. While some points are clarified, there remains uncertainty about the initial assumption of |x|=3 and the application of the relevant theorems.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the order of elements in cyclic groups and the specific conditions under which certain elements can be equal. The discussion does not resolve these uncertainties.

yxgao
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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 :
3.

WHy is the answer 3? Thanks!
 
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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].

Doug
 
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?
Thanks!
 
Originally posted by yxgao
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?
Thanks!

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.

Doug
 

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