Show $\pi\alpha\pi^{-1}$ and $\pi\beta\pi^{-1}$ are disjoint given $\alpha$ and $\beta$ are disjoint for an arbitrary permutation $\pi$ Let $\alpha$ and $\beta$ be  disjoint cycles of the same length $s$. Show that for any permutations $\pi$ we have that
$$\pi \alpha \pi^{-1}\quad \text{and }\quad      \pi \beta\pi^{-1} $$
are disjoint as well. 
$\alpha = (a_1a_2\ldots a_s)$ and $\beta= (b_1b_2\ldots b_s)$. Define the sets of elements $A=(a_1,\ldots ,a_s)$ and $B=(b_1,\ldots ,b_s)$. Now it is possible that there are elements that are unchanged by both permutations. Let these elements be the set $C=(c_1 ,\ldots ,c_r)$. Now for each element $x$, if $\pi^{-1}(x) \in A$ we have that
$$\pi \alpha \pi^{-1}(x) \neq x\quad \text{and }\quad      \pi \beta\pi^{-1}(x) = x=e(x) $$
Similarly if $\pi^{-1} \in B$ we find that 
$$\pi \alpha \pi^{-1}(x) =x=e(x)\quad \text{and }\quad      \pi \beta\pi^{-1}(x)\neq x$$
Lastly, if $\pi^{-1}(x) \in C$ we have that 
$$\pi \alpha \pi^{-1}(x) =   \pi \beta\pi^{-1}(x) = x=e(a) $$
I feel like this is leading to a proof of the desired fact... I am not sure how to rigorously finish the argument though. Any help would be greatly appreciated. Thanks!

Hint: if $\alpha = (a_1 \; a_2 \; a_3 \;\cdots\; a_s)$, then $\pi \alpha \pi^{-1} = (\pi(a_1) \; \pi(a_2)\; \cdots \; \pi(a_n))$ (do you see why?)
