Let $n\geq 2$ be an integer, and consider $H=\langle (1 \ 2 \ \cdots \ n), (a \ b)\rangle\subset S_n.$
It is known that $H=S_n$ if and only if $b-a$ and $n$ are coprime.
Question. What is the order of $H$ in general ?
For the moment, I think I have managed to prove the following facts:
Let $d=\gcd(n, b-a)$. Then
- $H$ contains all the transpositions $(k \ k+\frac{n}{d})$
In particular, $H$ contains the subgroup generated by $(1 \ n/d+1 \ 2n/d+1\cdots )$, which is isomorphic to $S_d$, and $d!$ divises the order of $H$.
- $H$ contains all the transpositions $(1 \ 1+ \ell(b-a)),$
and that's pretty much it.
Thoughts.
I wonder if $H$ is isomorphic to a wreath product $S_d\wr S_{n/d}$ or something like that, which would solve the problem.
Of course, if one may show that $H$ is a maximal subgroup of $S_n$, then it would do the trick, since $H$ is transitive and imprimitive if $d>1$.
I woud be extremely surprised if the result was not already known.
Thanks for your help.