I'm reviewing for the math GRE (it's been 8+ years since I took abstract algebra) and came across this question:
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 $\{x^{13n}: n \text{ a positive integer } \}$ is what?
Can someone show me how to approach this problem, and what concepts are in play here?
Best Answer
You have $3$ possibilities:
$$x^3=x^5 \Rightarrow x^2=e \,.$$
In this case since $x^{15}=e$ it follows that $x=e$, which is not possible (since you only get one value in your set).
$$x^5=x^9 \Rightarrow x^4=e \,.$$
Again, this implies that $x=e$, not possible.
$$x^3=x^9 \Rightarrow x^6=e \,.$$
Thus, $x^3=x^{\operatorname{gcd}(6,15)}=e$. This means that $x$ must have order $1$ or $3$, but again $x=e$ is not possible.
Thus, $x$ is an element of order $3$ in your group, and from there it is easy: $x^{13n}=x^{13m} \Leftrightarrow 3|13(m-n) \Leftrightarrow 3|m-n$... So how many distinct values do you get?