This question appears to be new here according to this and this.
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.76 ibid. and it's stated to be a problem from a 2008 GRE practice exam.
I would like to prove this result using the tools given in the textbook prior to this exercise. (A free copy of the book is available online.)
The Question:
If $x$ is an element of a cyclic group of order $15$ and exactly two of $x^3$, $x^5$, and $x^9$ are equal, determine $\lvert x^{13}\rvert$.
My Attempt:
My first concern is whether it's the order of the group, let's call it $G$, or of the element, called $x$, that is $15$. Nonetheless, here are some thoughts, since $x^{15}=e$ either way:
If $x^3=x^5$, then $x^2=e$, but then either $x=e$ or $\lvert x\rvert=2$; in the former case, $\lvert x^{13}\rvert=\lvert e\rvert=0$, but in the latter, $$\begin{align}0&=\lvert x^{15}\rvert\\ &=\lvert (x^2)^7x\rvert \\ &=\lvert x\rvert\\ &=2;\end{align}$$ thus $\lvert x^{13}\rvert=0$. (EDIT: But $x\neq e$.)
If $x^3=x^9$, then $x^6=e$, so $e=x^{15}=(x^6)^2x^3=x^3$, which means $x=e$ or $\lvert x\rvert=3$; thus $\lvert x^{13}\rvert=0$ as before or $$\boxed{\begin{align}\lvert x^{13}\rvert &=\lvert (x^3)^4x\rvert \\ &=\lvert x\rvert\\ &=3.\end{align}}$$ (EDIT: But $x\neq e$.)
If $x^5=x^9$, then $x^4=e$, but now $e=x^{15}=(x^4)^3x^3=x^3$, so that $x^3=x^4$ implies $x=e$; hence $\lvert x^{13}\rvert=0$. (EDIT: But $x\neq e$.)
[EDIT 2: The boxed bit is the only answer.]
Is this correct? What would be a better way of determining $\lvert x^{13}\rvert$?
Please help :)
Since $x^3,\,x^5,\,x^9$ aren't all equal, $|x|$ is positive and not $2$. Instead, it divides $4$ or $6$ (and thus $12$) and also $15$ but not $2$, so $x^{13}=x\implies|x^{13}|=|x|=3$.