I asked this question : Composition of permutation to generate all permutations earlier but I didn't phrase it well, here is my new question.
My question is a little bit different I'm looking for a set of permutation $\sigma_1$, ... $\sigma_p$ such that any composition (in this given order) $\sigma_1\circ\sigma_2\circ\cdots\circ\sigma_p\circ\sigma_1\circ\sigma_2\circ\cdots$ will generate all the permutation of a given list.
According to you answer I get that this sequence can be compose only of $\sigma_1=(1,2)$ and $\sigma_2=(1,2,...n)$, but this might not be optimal (because the sequence might be longer.)
Now my question is, what is the minimum sequence of $\sigma_1$ and $\sigma_p$ such that it generates all permutations of a set of $n$ element.
in my example with 1,2,3 above.
if I take $\sigma_1=(1,2)$ and $\sigma_2=(1,3)$
the sequence repeating sequence $\sigma_1\circ\sigma_2$ will generate all permutations :
id gives 1 2 3
$\sigma_1$ gives 2 1 3
$\sigma_1\circ\sigma_2$ gives 3 1 2
$\sigma_1\circ\sigma_2\circ\sigma_1$ gives 1 3 2
$\sigma_1\circ\sigma_2\circ\sigma_1\circ\sigma_2$ gives 2 3 1
$\sigma_1\circ\sigma_2\circ\sigma_1\circ\sigma_2\circ\sigma_1$ gives 3 2 1
I can phrase it this way :
Let $S_n$ be the symmetric group on $n$ elements. You are interested in subsets $\{a_0,a_1,\ldots,a_k,b\}$ where $a_0$ is the identity, such that for any $y∈S_n$, you can find $0≤j≤k$ and and natural number $M$ such that $y=bMaj$. And you want to find the minimum $k$. In other words, you are looking for the number of cosets in $S_n$ relative to the subgroup generated by $b$.
Thanks in advance.