If $\alpha,\beta\in S_n$ are disjoint, are $\alpha^m, \beta^n$ disjoint for any $m,n\in\Bbb{N}$?

Question

If $\alpha,\beta\in S_n$ are disjoint, are $\alpha^m, \beta^n$ disjoint for any $m,n\in\Bbb{N}$?

I believe that the answer is true but I am not sure how to get a proper proof for this statement.

I have proved one property that is

Let $\alpha=(i_1 i_2\dots i_r)$ and $\beta=(j_1j_2\dots j_s)$. Then $\alpha$ and $\beta$ are disjoint iff $\{i_1,\dots,i_r\}\cap \{j_1,\dots,j_s\}=\varnothing$.

Let $A=\{i_1,\dots,i_r\}$ and $B=\{j_1,\dots,j_s\}$ be the set of elements moved by each $\alpha$ and $\beta$ respectively. Write each $\alpha$ and $\beta$ as a product of disjoint cycles. Since $\alpha$ and $\beta$ are disjoint, their cycles are disjoint too. By the above property, $A\cap B=\varnothing$. Note that the elements moved by $\alpha^m$ and $\beta^n$ will be subset of $A$ and $B$, so the elements moved by both $\alpha^m$ and $\beta^n$ will be subset of $A\cap B$ which is empty set.

Answer 1

We say an element $x$ is moved by a permutation if $\alpha(x)\neq x$. Denote the set of elements moved by $\alpha$ as $\mu(\alpha)$.

Recall that two permutations are said to be disjoint if $\mu(\alpha)\cap \mu(\beta)=\varnothing$.

It is easy to show that $\mu(\alpha^n) \subseteq \mu(\alpha)$.

So clearly if $\mu(\alpha)\cap \mu (\beta)=\varnothing$ then $\mu(\alpha ^n ) \cap \mu (\beta ^m)=\varnothing$