Transposition $(a\ b)$ with $\gcd(b-a,n) =1$ and cycle $(12\dots n)$ generate $S_n$ Prove that if $1≤a<b≤n$ with $\gcd(b−a,n)=1$, then the transposition $(a\ b)$ and the cycle $(1 2…n)$ generate $S_n$. 
 A: Building on my comment, we let $1<m<n$ be coprime, and take $(m,n),(1,2,…,n)\in S_n$. Note that
$$(m,n)(1,2,…,n) = (1,2,\ldots,m-1,n)(m,m+1,\ldots,n-1).$$
Now, since $m,n-m$ are also coprime, taking $r,s\in\mathbb{Z}$ such that $rm+s(n-m)=1$ we have
$$\sigma:=\left((m,n)(1,2,…,n)\right)^{rm} = (1,2,\ldots,m-1,n)^{rm}(m,m+1,\ldots,n-1)^{1-s(n-m)} = (m,m+1,\ldots,n-1),$$
and similarly
$$\tau:=\left((m,n)(1,2,…,n)\right)^{s(n-m)} = (1,2,\ldots,m-1,n).$$
For all $a\in\{m,m+1,\ldots,n-1\},b\in\{1,2,\ldots,m-1,n\}$, we then have
$$(m,n)^{\sigma^{a-m}\tau^{b\pmod n}} = (a,b),$$
and for all $a_1,a_2\in\{m,m+1,\ldots,n-1\},b_1,b_2\in\{1,2,\ldots,m-1,n\}$ we have
$$(a_1,n)^{(a_2,n)}=(a_1,a_2);\quad (b_1,m)^{(b_2,m)}=(b_1,b_2),$$
thus generating every transposition (and therefore every permutation) in $S_n$.
The case of $m=1$ is more easily handled: $(1,n)(1,2,…,n)=(1,2,…,n-1)$, so by conjugation we can get any transposition $(k,n)$, and therefore any other transposition as well.
As this question is of real interest to me, I'd appreciate feedback in case I made some silly mistake. As an aside, I'm still interested in the question I posed above: if $\gcd(m,n)=d>1$, which subgroup of $S_n$ would $(m,n),(1,2,…,n)$ generate?
