TITLE: When do $(i \enspace j)$ and $(1 \enspace 2 \ldots n)$ generate $\mathfrak{S}_n$?
QUESTION [0 upvotes]: Let $n$ be an integer greater than $1$, and let $i, j\in \{1, \ldots ,n\}$ with $i<j$. The permutations $(i \enspace j)$ and $(1 \enspace 2 \ldots n)$ generate the whole symmetric group $\mathfrak{S}_n$ if and only if $j-i$ is prime to $n$.  

I have been facing lots of troubles trying to prove this statement. I couldn't show the direct nor the converse implications.  
For the converve implication, of course I know very well the case where $i=1$ and $j=2$, which is a rather classic exercise. The thing is, if I denote $\tau = (1 \enspace 2)$ and $\sigma = (1 \enspace 2 \ldots n)$, then we have $\sigma^{k-1}\tau\sigma^{-(k-1)} = (k\enspace k+1)$ for $k\in \{1,\ldots n-1\}$. It is then not very difficult to decompose any transposition as a product of transpositions of the form $(k \enspace k+1)$.
I tried to copy this proof for the general case where $j-i$ is prime to $n$. Note that in this situation, if I denote $\sigma'= \sigma^{j-i}$, then $\sigma'$ is still an $n$-cycle and has the form $(i \enspace j \ldots)$. I thought this could be useful because it is similar to the $i=1, j=2$ case, however the fact that $j-i$ may be different from one is giving me troubles.  
For the direct implication, I have no clue about how to proceed. Well, if we manage to somehow compute the order of $\sigma^{j-i}$ and show that it is $n$, then $\operatorname{gcd}(j-i,n)=1$ as wanted. This may be a pist, but I really can't find how to proceed.  
Could someone please give me the solution, a hint or a reference for this exercise ? 
Thank you very much.

REPLY [0 votes]: To show that if $\gcd(j-i,n)=g>1$ implies that $(i\,j)$ and $\sigma$ generate a proper subgroup, consider the subgroup $H$ they generate, and prove that $H$
is imprimitive, fixing a nontrivial partition having $\{1,g+1,2g+1,\ldots, n-g+1\}$ as one of its parts.
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