
Symmetry Group Proof
Let $\displaystyle \alpha$ be an mcycle permutation. Prove that $\displaystyle \alpha^j$ is mcycle if and only if $\displaystyle gcd(m,j) = 1$.
I have an idea for both ways but I am getting stuck on concluding either case.
If I assume the gcd is 1 then I am able to reduce down to $\displaystyle (\alpha^j)^s = \alpha$ where s satisfies: $\displaystyle sj+tm=1$. But I struggle concluding that $\displaystyle \alpha^j$ is mcycle.
If I assume $\displaystyle \alpha^j$ is mcycle and that $\displaystyle km \and\ kj$. If I then raise $\displaystyle \alpha^k$ I should be able to expand the product with respect to k and conclude that k must be 1. But I fail to see a systematic way to write out $\displaystyle \alpha^k$.