Cyclic group

[vvaann]

This is a question from the GRE practice book. Let's discuss it.

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^(13*n): n is a positive integer} is

A. 3
B. 5
C. 8
D. 15
E. infinite

Answer key:

SPOILER: A

[Dragonfinity]

Easy.

If G = <a> (that is, a generates G) is a group of order 15, and {x^3, x^5, x^9} has only 2 elements, then x = a^5. So, we get

x^(13*1) = a^(5*13) = a^65 = a^5
x^(13*2) = a^(5*26) = a^130 = a^10
x^(13*3) = a^(5*39) = a^195 = a^0 = e
x^(13*4) = x^(13*3)*x^(13*1) = e*a^5 = a^5

and so on.

[Kennyc]

Although the answer is right, I am pretty sure x can be a^5 or a^10, where a^5 does not equal a^10. To check the order of x^13n, you just need to check a^5 since a^10 is a^5(2). Since both have order 3 the answer is A.