Let $n \ge 2$, $n \in \mathbb N$ and set $\sigma=(1,2,3,...,n)$ and $\tau=(1,2)$. Show that $\mathrm{Sym}(n)=\left<\sigma,\tau\right>$.
Approach: Induction
Proof:
Base case $n=2$
$\sigma=(1,2)$
$\tau=(1,2)$
$Sym(2)=\{Id_2,(1,2)\}$ $(1,2)=\tau$ and $Id_2=\tau\sigma$
so base case holds
Inductive step assume $Sym(k)=<\sigma,\tau>$ where $\sigma=(1,2,3,4,...,k)$ and $\tau=(1,2)$
Show $Sym(k+1)=<\lambda,\tau>$ where $\lambda={1,2,...,k+1}$ and $\tau=<1,2>$
To prove the inductive step, I was thinking we have to use the fact that every $x\in Sym(k+1)$ can be represented as the product of pairwise disjoint cycles. I think that we can express $\lambda$ in terms of $\sigma$
Your previous question was about transpositions generating $S_n$. So we may try an approach where we attempt to generate all transpositions from the given two permutations.
First, conjugate $(12)$ by $(12\cdots n)$ repeatedly to obtain $(23),(34),\cdots,(n-1\,n)$.
This makes sense once you understand this: $\sigma(a_1\,a_2\,\cdots\,a_k)\sigma^{-1}=(\sigma(a_1)\,\sigma(a_2)\,\cdots\,\sigma(a_k))$.
Second, see if you can conjugate $(i\,i+1)$ by $(j\,j+1)$s systematically to get any $(i\,k)$. For example, conjugating $(12)$ by $(23)$ gives $(13)$, then conjugating that by $(34)$ gives $(14)$.