I need help understanding hint in the answer to this question: $r$-cycle to a power $k$ is also an $r$-cycle if and only if $\gcd(k, r) = 1$. The third comment also seems to be the same hint.
The hint also appears here: Powers of $m$-cycles are also $m$-cycles
To be clear, the hint seems to be that if $\gcd(k,r)=1$, then $\langle \sigma^k \rangle=\langle \sigma \rangle$.
I know why the hint is true. It is because for any group $G$, if $|x|=r$ then $|x^k|=\frac{r}{\gcd(r,k)}$. Indeed, this fact immediately shows that if $\gcd(k,r)>1$, then $\sigma^k$ is not an $r$-cycle since it has order strictly less than $r$ and $r$-cycles have order $r$.
Conversely, suppose $\gcd(k,r)=1$. What I don't understand is how knowing $\langle \sigma^k \rangle=\langle \sigma \rangle$ helps show that $\sigma^k$ is an $r$-cycle? I've already seen the solution where one shows that $\sigma^k(i)=i+k\pmod r$, but this hint I'm not understanding seems to lead to different solution which somehow avoids this computation.
Look at the contrapositive. If $\sigma^k$ is not an $r$-cycle, then it is the product of shorter cycles, and there are an $i\in\operatorname{dom}\sigma$ and an integer $\ell$ such that $2\le\ell<r$ and $(\sigma^k)^\ell(i)=i$. But then the identity permutation and $(\sigma^k)^\ell$ are distinct members of $\langle\sigma^k\rangle$ sending $i$ to $i$, while $\langle\sigma\rangle$ clearly does not have a non-identity permutation that fixes $i$. Thus, $\langle\sigma^k\rangle\ne\langle\sigma\rangle$.