Suppose that $\alpha \in S_{n}$, where $n > 2$. Explain why $\alpha(2\enspace 3)\alpha^{-1}=(\alpha(2)\enspace \alpha (3))$.
I know that we should approach this by looking at $\alpha(2\enspace 3)\alpha^{-1}(j)$ for $j$ taking the values $$ j \in \left\{1, \ldots, n \right\}\setminus\left\{\alpha(2)\alpha(3)\right\},\quad j = \alpha(2),\quad j=\alpha(3). $$ However I'm not sure how to prove the final result. Any help is appreciated.
Try cases:
Case 1: $\alpha^{-1}$ permutes $2$ to $i=2$
Case 2: $\alpha^{-1}$ permutes $2$ to $i=3$
Case 3: $\alpha^{-1}$ permutes $2$ to $i\neq 2,3$
In each case, what does $(2\ 3)$ to $i$ to, what does $\alpha$ permutes the result to.