Let $\alpha = (\alpha_1 \, \alpha_2 \, \ldots \, \alpha_s)$ be a cycle, for positive integers $\alpha_1 , \alpha_2 , \ldots , \alpha_s$. Let $\pi$ be any permutation. Show that $\pi \alpha \pi^{-1}$ is the cycle $(\pi(\alpha_1) \, \pi(\alpha_2) \, \ldots \, \pi(\alpha_s))$.
I started by choosing a specific $\alpha$ and $\pi$, and tried finding $\pi \alpha \pi^{-1}$ to give myself some idea of what to do but have had no luck. Tips?
Define $\alpha_{s + 1} := \alpha_1$ and consider that $\alpha$ only moves elements $\alpha_\ell$.
So as what does an element $\beta$ reach $\alpha$ through $\pi^{-1}$? Or asked another way: when is an element $\beta$ transformed by $\pi^{-1}$ into an $\alpha_\ell$?
Consider $(\pi^{-1}(\beta) = \alpha_\ell) \equiv (\beta = \pi(a_\ell))$ and so $(\pi\circ \alpha\circ \pi^{-1})(\pi(\alpha_\ell)) = (\pi\circ \alpha)(a_\ell) = \pi(a_{\ell + 1})$. Thus every element $\pi(\alpha_\ell)$ is mapped to $\pi(a_{\ell + 1})$. Now, just write this down in cycle notation and you have your result.
Do you get the idea? Perhaps my exposition is to sparse?
