$\beta^k $ is a cycle ⟺ gcd($k$ , o($\beta$))=$1$

Question

GIVEN: $\beta$ is a cycle and belongs to $S_{n}$, then $($$\beta$$)^k $ is a cycle iff $(k,o($$\beta$$))=1 $

let o($\beta$)=m and $ $$\beta$$=(a_{1} a_{2} .... a_{m}) $

While trying to prove the converse, I assumed that $(k,m)=1$

Then, since, $o($$\beta$$^k)$ $= m/gcd(m,k) $ = $m$

Does this mean $\beta^k$ is also a cycle?

I'm stuck here (and in general), can you please give me a hint on how to prove the statement assuming that $\beta^k$ is a cycle first?

Answer 1

Consider $\beta=(12)(34)$ in $S_4$, its order is $2$ and $\beta$ is not a cycle, $\beta^3=\beta$. But $\gcd(2,3)=1$.