Proof for conjugate cycles Let $\alpha =(a_1,a_2,...,a_s)$ be a cycle and $\pi$ a permutation in $S_n$. Then $\pi\alpha\pi^{-1}$ is the cycle $(\pi(a_1),...,\pi(a_s))$.
I'm having trouble understanding the proof of this:
Take $\pi(a_i)$. Then $\pi\alpha\pi^{-1}\pi(a_i)=\pi\alpha(a_i)=\pi(a_{i+1\; (\mathrm{mod}\;\mathrm{s})})$, then follows on from here.
But I'm not understanding how this proves the statement, since it feels like it's proving that $\pi\alpha=(\pi(a_1),...,\pi(a_s))$.
 A: Consider the permutations $\beta\colon \{1, \ldots, n\} \to \{1, \ldots, n\}$ and $\gamma\colon \{1, \ldots, n\} \to \{1, \ldots, n\}$ defined by:
$$
\beta(x) = \pi(\alpha(\pi^{-1}(x)))
\qquad\text{and}\qquad
\gamma(x) = \pi(a_{j+1 \pmod s}) \text{, where } \pi(a_j) = x
$$
Note that since $\pi$ is a permutation (and thus bijective), $\gamma$ is well-defined. We want to show that $\beta = \gamma$. That is, we want to show that both functions send the same input to the same output. Indeed, choose any $x \in \{1, \ldots, n\}$ and let $a_j = \pi^{-1}(x)$ so that $x = \pi(a_j)$. Then:
\begin{align*}
\beta(x)
&= \pi(\alpha(\pi^{-1}(x))) \\
&= \pi(\alpha(a_j)) \\
&= \pi(a_{j+1 \pmod s}) \\
&= \gamma(x)
\end{align*}
as desired.
A: Why does it "feel like it's proving that $\pi \alpha = (\pi(a_1),\ldots,\pi(a_s))$"? That would mean that $\pi\alpha(\pi(a_i)) = \pi(a_{i+1})$, which is not the case.
To say that $\pi\alpha\pi^{-1}$ is the cycle $(\pi(a_1),\ldots,\pi(a_s))$ means that $\pi\alpha\pi^{-1}$ maps $\pi(a_i)$ to $\pi(a_{i+1})$, and that is exactly what you've written.
