Show $A=\{(k\:\: k+1), 1≤ k<n\}$ generates $B=\{(1\:\:x), 1<x≤ n\},$ where $B$ is a minimal generating sets for $S_n$.
So I want to show $(k\:\:k+1)$ generates $(1\:\:x)$. I want to use induction for this but I'm not sure how to put it together because I'm a bit confused on how generating sets work for transpositions and $S_n$. For this, I think I need just $n-1$ transpositions.
I have this identity and I'm not sure if it helps: $(a\:b) = (1\:a)(1\:b)(1\:a)$
Here's what I'm thinking so far, no idea if I'm on the right track or if it's right so far:
$A=\{(k\:\:k+1), k=1,\ldots,n-1\} =$ {$(1\: 2),(2\: 3),(3\: 4),\ldots,(n-2\:\:\:\:n-1)$}
Let $n\ge 2$, so base case is $n=2$...
I'm not really sure if it's $n\ge2$, I thought it might be $n\ge3$ maybe?
Let's proof the claim with induction on $x$ for $1<x\le n$. because $x>1$ the base case is $x=2$, in this case $(1\;x)=(1\;2)\in A\subset\langle A\rangle$ as desired. now let us assume the claim is true for $1\le x-1<n$ and prove it for $x$. we will use the fact that
$$ (1\;x)=(x-1\;\;x)(1\;\;x-1)(x-1\;\;x) $$
this equality holds because we can compute for $1,x,x-1$ an get $$ \begin{align} (x-1\;\;x)(1\;\;x-1)(x-1\;\;x)(x)=&(x-1\;\;x)(1\;\;x-1)(x-1)=(x-1\;\;x)(1)=1\\ (x-1\;\;x)(1\;\;x-1)(x-1\;\;x)(x-1)=&(x-1\;\;x)(1\;\;x-1)(x)=(x-1\;\;x)(x)=x-1\\ (x-1\;\;x)(1\;\;x-1)(x-1\;\;x)(1)=&(x-1\;\;x)(1\;\;x-1)(1)=(x-1\;\;x)(x-1)=x \end{align} $$ and every other element is fixed by this permutation. because $(x-1\;\;x)\in A$ and by the induction hypothesis $(1\;\;x-1)\in \langle A\rangle$ we will conclude that $(1\;\;x)\in \langle A\rangle$ as wanted.