A cyclic group of order 15

[yxgao]

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!!

[Mathechyst]

Let x be an element of a cyclic group of order 15. If \{x^3,x^5,x^9\} has exactly 2 elements, then one element must be the same as another. If |x|=3 then x^3=x^9. If |x|=3 then |x^{13}|=3/gcd(3,13)}=3. Hence |<x^{13}>|=3.

Doug

[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 |x^{13}|=3/gcd(3,13)}=3 come from?
Can you give an example of a set that meets this condition?
Thanks!

[Mathechyst]

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 |x^{13}|=3/gcd(3,13)}=3 come from? Can you give an example of a set that meets this condition?
Thanks!

Your original post stated that there exists an element \inline{x} such that \inline{\{x^3,x^5,x^9\}} has exactly two elements. Thus I need to find two elements that are the same. There is a theorem that states if \inline{|x|=n} then \inline{x^i=x^j} if and only if \inline{n} divides \inline{i-j}. If \inline{x^3=x^5} then \inline{|x|=2} which is not possible in a cyclic group of order 15 because 2 does not divide 15. Likewise, \inline{x^5{\neq}x^9}. However if \inline{|x|=3} then \inline{x^3=x^9} because 3 divides \inline{i-j=9-3=6}. Also, a cyclic group of order 15 can have an element with order 3.

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

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

Doug