group theory question regarding orders of elements

Question

Let $G$ be a group with $y\in{G}$ and $n,r\in\mathbb{N}$. If $o(y)=n$, what is $o(y^r)$?

My attempt: Let $$o(y^r)=a,$$ Then we have $$1_G=(y^r)^a=(y)^{ra}.$$ So we have that $$n\mid ra,$$ So either $r$ or $a$ (or both) is a multiple of $n$. I'm not too sure where to go from here or if this is even the most effective approach.

Answer 1

Start from the definition of the order: $a$ is the least positive integer $a$ such that $(y^r)^a=y^{ra}=e$.

Now, if $y^{ra}=e$, $ra$ is a multiple of the order $n$ of $y$. So $a$ is the least positive integer such that $ra$, which is a multiple of $r$, is also a multiple of $n$, i.e. $ra=\operatorname{lcm}(r,n)$.

Can you continue?

Answer 2

Let $\gcd(n,r)=d$. Then the answer is $o(y^r)=\frac{n}{d}$. Let's prove it. First of all, note that $(y^r)^{\frac{n}{d}}=(y^n)^{\frac{r}{d}}=e^{\frac{r}{d}}=e$. So the order of $y^r$ is at most $\frac{n}{d}$.

Now, if $(y^r)^a=e,a\geq 1$ then $y^{ra}=e$ and hence $n|ra$. This implies $\frac{n}{d}|\frac{r}{d}a$. But $\frac{n}{d},\frac{r}{d}$ are coprime and hence $\frac{n}{d}|a$. This proves the order of $y^r$ is at least $\frac{n}{d}$.