Let $S_k$ be the set of permutations of $(1,...,k)$ and $S(\overline{1,l}),S(\overline{l+1,k}),S(l,k-l) \subset S_k$ be the following subsets \begin{align} S(\overline{1,l}) &= \{ \sigma \in S_k\ |\ \sigma(l+1) = l+1,...,\sigma k=k\} \\ S(\overline{l+1,k}) &= \{ \sigma \in S_k\ |\ \sigma(1) = 1,...,\sigma l=l\} \end{align} and $$ S(l,k-l) = \{\sigma \in S_k\ |\ \sigma:(1,...,k) \mapsto (i_1,...,i_l,j_1,...,j_{k-l}),\\i_1<...<i_l \land j_1<...<j_{k-l}\} $$ Then any $\sigma \in S_k$ can be written as $\sigma = \pi\kappa\rho$ with $\kappa\in S(\overline{1,l}),\rho \in S(\overline{l+1,k})$ and $\pi \in S(l,k-l)$, and therefore $S(l,k-l) \times S(\overline{1,l}) \times S(\overline{l+1,k}) \cong S_k$ under $(\pi,\kappa,\rho) \mapsto \pi\kappa\rho$.
A different way to achieve the same idea is $\sigma = \pi_{\kappa,\rho}$, where $\kappa \in S_l, \rho \in S_{k-l}$ and (as above) $\pi \in S(l,k-l)$, where $$ \pi_{\kappa,\rho}:(1,\dots,k) \mapsto (i_{\kappa1},\dots,i_{\kappa l},j_{\rho 1},\dots,j_{\rho(k-l)}) $$ and therefore $S(l,k-l) \times S_l \times S_{k-l} \cong S_k$ under $(\pi,\kappa,\rho) \mapsto \pi_{\kappa,\rho}$.
Question I: Is this valid (both ways) ? and if it is the case, where can I find a reference with proof ?
Question II: Is it possible to generalise this (written here in the second way) such that one factors the same group as $S_k = S(k_1,\dots,k_r)\times S_{k_1}\times,...,\times S_{k_r}$, where $\sum_j k_j = k$ in such a way that any element $S_k \ni \sigma = \pi_{\kappa_1,\dots,\kappa_r}$, where $\pi \in S(k_1,\dots,k_r)$ i.e. \begin{align} \pi:(1,...,k) &\mapsto (i^{(1)}_1,...,i^{(1)}_{k_1},...,i^{(r)}_1,...,i^{(r)}_{k_r}),& &\begin{cases} i^{(1)}_1<...<i^{(1)}_{k_1},\\ ...,\\ i^{(r)}_1<...<i^{(r)}_{k_r} \end{cases} \end{align} and $\kappa_j \in S_{k_j}$ and $$ \pi_{\kappa_1,\dots,\kappa_r}:(1,...,k) \mapsto (i^{(1)}_{\kappa_1 1},...,i^{(1)}_{\kappa_1 k_1},...,i^{(r)}_{\kappa_r 1},...,i^{(r)}_{\kappa_r k_r}) $$ ?
(necessary but not sufficient test) If we test the cardinatlities we get $|S_k|=k!$, and \begin{align} |S_l|&=l!, & |S_{k-l}|& =(k-l)! & |S(l,k-l)|& =\frac{k!}{l!(k-l)!} \end{align} where the product $|S(l,k-l) \times S_l \times S_{k-l}| = k! =|S_k|$.
Added:
The idea is that one considers the combination $(l,k-l)$, i.e. divide $(1,...,k)$ into two disjoint sets (one has $l$ element) disregarding the order within the single set, consider a shuffle if you like, and then arrange these two sets by $S(\overline{1,l})$ and $S(\overline{l+1,k})$ to get $\sigma$. An argument that may complete the test into a proof is to show that these proucts $\pi\kappa\rho$ are distinct for different triples and so the map \begin{align} A:& S(l,k-l) \times S(\overline{1,l}) \times S(\overline{l+1,k}) \to S_k, & (\pi,\kappa,\rho) &\mapsto \pi\kappa\rho \end{align} is a bijection.