Let $G$ be a cyclic group of order $15$. Suppose $x\in G$ and exactly two of $x^3,x^5$, and $x^9$ are equal. Find $|x^{12}|$ and $|x^{14}|$.
I have ruled out $x^3=x^5$ because then $|x|=2$. If $|x|=2$ then $x^3=x$ and $x^9= (x^3)^3=(x)^3=x^3$ and we know $x^9$ can't equal $x^3$ if $x^3=x^5$ (because exactly $2$ are equal). So we know $|x|$ is not $2$
I cannot figure out how to narrow it down from here. I can't figure out whether $x^3=x^9$ or $x^5=x^9.$
Any hints?
Now $x^5$ cannot equal $x^9$ because then $|x|$ would then divide $4$ but cannot be 1 [why], so $|x|$ would have to be even, but as $|G|$ is odd this is impossible.
So $x^3=x^9$ which implies that $|x|$ divides 6. So $x^6$ is the identity element. What about $x^{12} =(x^6)^2$ then?
Now, $|x|$ divides 6 but cannot be 1 nor even. So $|x|$ must be 3. Then $x^{14} = x^{-1}$. How does $|x|$ and $|x^{-1}|$ compare?
Alternative: Write $x=a^c$ where $a$ is a generator of $G$; there is such an $a$ as $G$ is cyclic. Then if $x^3=x^5$ then $a^c = a^{2c}$. Thus $c \equiv_{15} 2c$ which implies that $c$ itself must be 0 mod 15 which would imply $x^3=x^5=x^9$.
Likewise $x^5$ cannot equal $x^9$.
So $x^3=x^9$ which gives $x^6$ is the identity which gives $x^{12}$ is the identity.
Can you find the precise value of $|x|$ and then $|x^{-1}|$? Hint: $c$ must be either 5 or 10 mod 15, which implies that $14c$ must be either 5 or 10 mod 15. What is the smallest number then to multiply $14c$ to get a number that is divisible by 15?