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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. How many elements are in the set $\{ x^{13n}:n \mbox{ is a positive integer}\}$?

I feel like some form of $\textbf{gcd}$ or $\textbf{lcm}$ is involved, but I could be wrong. Any form of hints will be greatly appreciated.

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Hint: if the set $\{x^3,x^5,x^9\}$ has exactly two elements, one of them must be the identity (why?). Hence, apply Lagrange. – Zen Jun 6 '13 at 2:48
@Zen thanks for the hints. Rather than one of them being the identity, why can't one of them equal another? – math-visitor Jun 6 '13 at 2:50
One of them must equal another - that is your premise no? In fact, the ones that are equal must both be equal to the identity. – Zen Jun 6 '13 at 2:51
@Zen Thank you. Things are not that clear right now but I'll keep thinking about this question. – math-visitor Jun 6 '13 at 2:53
Suppose none of them is the identity. Then you have either $x^3=x^5\implies x^2=e$ which implies you have a subgroup of order 2, which by Lagrange's is impossible. Or you have $x^5=x^9\implies x^4=e$, also impossible. [edited: I had a brain fart] – Zen Jun 6 '13 at 3:10
up vote 1 down vote accepted

Hints: The group is cyclic, and of order $15$. Keep in mind that one of $x^3,x^5,x^9$ is distinct from the other two, so we certainly can't have $x=e$.

If $x^3=x^9,$ then $x^6=e$, but the order of $x$ must divide the order of the group (and also divide $6$, in this case), and so the order of $x$ is $3$.

Show that the other two cases are impossible.

Thus, $x$ has order $3.$ What can we then say about the number of distinct values that powers of $x$ can take? Since $13n$ ranges through all values (modulo $3$) for positive integers $n$, what can we say about the number of values that $x^{13n}$ does take?

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