# Cyclic group of order 15

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?

• two of the elements have to be equal... And you know that in a group of order $15$ you have $x^{15}=e$... From there you can figure which two elements are equal... – N. S. Nov 7 '11 at 15:05
• Why don't you just write down the sets $\{x^3,x^5,x^9\}$ for all $15$ elements in your group and see what you get? – Rasmus Nov 7 '11 at 15:07

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?

If an element $a$ generates the whole group, then consider $x=a^5$. Then you've got $x^3=e=\text{the identity}$ and $x^9=e$.

The order of $x^{13}$ is the same as the order of $x$ because $\gcd(13,15)=1$. As others have already mentioned, $x$ has order 3. Hence the set has 3 elements.

• Shouldn't it be $\gcd(13,3)$ since the order of $x$ is $3$? – Hrit Roy Jan 3 '18 at 20:05