A cyclic group of order 15

1. Dec 20, 2003

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!!

2. Dec 20, 2003

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

3. Dec 20, 2003

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!

4. Dec 20, 2003

Mathechyst

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