Show that for any $k$-cycle $(a_1 a_2...a_k) \in S_n$, and for any permutation $\pi \in S_n$, $\pi(a_1...a_k)\pi^{-1}=(\pi(a_1)..\pi(a_k))$

Question

Show that for any $k$-cycle $(a_1 a_2...a_k) \in S_n$ where $S_n$ is the set of all permutations of n elements and $(a_1...a_k)$ represents a cycle, like $a_1$ goes to $a_2$ and ... and $a_k$ goes to $a_1$, and for any permutation $\pi \in S_n$, $\pi(a_1...a_k)\pi^{-1}=(\pi(a_1)..\pi(a_k))$.

I thought about this logically, like if an element is not moved in the cycle, then the permutation $\pi$ and $\pi^{-1}$ will just move it back to its original location. But I'm not really sure why the final elements from these permutations are in the form $\pi(a_i)$ instead of having a $\pi^{-1}$ in it somewhere.

Answer 1

In your question, why is: $\pi(a_1...a_k)\pi^{-1}=(\pi(a_1)..\pi(a_k))$, notice what happens if you apply the left side, or the right side, to the number $\pi(a_n)$. In both cases, the result is $\pi(a_{n+1})$ (where $a_{k+1}$ denotes $a_1$).