TITLE: How does $(12\dots n)$ and $(a\, b)$ generate $S_n$?
QUESTION [7 upvotes]: I know that $S_n$ is generated by a number of things, like all transpositions, all transpositions of the form $(1\,a)$, the transpositions $(1\,2),(2\,3),\dots,(n-1\, n)$, and  just the two elements $(123\dots n),(1\,2)$. 

Suppose $n$ is prime. If you just have $(123\dots n)$ and some arbitrary transposition $(a\, b)$, how does this also generate $S_n$? 

Can you somehow get to $(1\,2)$ or reduce it to some other previous case?

REPLY [3 votes]: Another slightly different way of looking at it for $n = p$ a prime, if you prefer:
Let $\sigma = (12 \ldots p)$. Then $\sigma (ab) \sigma^{-1} = (\sigma(a) \sigma(b)) = (a+1 \ \ b+1)$. Continuing, we generate all transpositions of the form $(a+k\ \ b+k$) (working modulo $p$). Let $t = b - a$, and we see we have all transpositions of the form $(a \ \ a+t)$. Given any $c,d \in \{0 , \ldots , p-1 \}$, there is an $s$ such that $c + st = d \ (\mathrm{mod} \ p)$ (specifically $s = t^{-1}(c-d)$, which always exists as $\mathbb Z / p \mathbb Z$ is a field; this wouldn't work if $n$ wasn't prime). So we have generated all transpositions, and so the whole group.