I'm currently reading a book on the representations of the symmetric group of $n$ letters $S_n$ ("The Symmetric Group" by Bruce Sagan), in which the following group is defined as groundwork in order to describe the irreducible representations of $S_n$: Let $\lambda$ be a partition of $n$ (i.e. $\lambda =(\lambda _i)_{i=1}^k$ such that $\lambda _i\geq \lambda _{i+1}$ and $\displaystyle \sum _i\lambda _i=n$.) we define de Young subgroup $S_\lambda$ of $S_n$ as$$S_\lambda :=S_{\{1,\ldots ,\lambda _1\}}\times S_{\{\lambda _1+1,\ldots ,\lambda _1+\lambda _2\}}\times \cdots \times S_{\{n-\lambda _k+1,\ldots ,n\}} \qquad... (1)$$The book asserts (without proof) that there exists a subgroup $H$ of $S_n$ which is isomorphic to $S_\lambda$ but I don't see why.
My thoughts on the question are that, in general, if $H $ and $K$ are subgroups of a group $G$ then $HK = KH$ iff $HK$ is subgroup of $G$. I guess this last result can be extended for a finite arbitrary family of subgroups of a group $G$ and, because $\{1,\ldots ,\lambda _1\}, \{\lambda_1 +1,\ldots ,\lambda _1 + \lambda_2\}, \dots, \{ n - \lambda_k + 1, \dots, \lambda_k \}$ are all disjoint thus $S_{\{1,\ldots ,\lambda _1\}}S_{\{\lambda _1+1,\ldots ,\lambda _1+\lambda _2\}}\cdots S_{\{n-\lambda _k+1,\ldots ,n\}}$ is a subgroup of $S_n$ in which the terms of the product are mutually disjoint and normal in such subgroup and thus itself is isomorphic to the direct product stated in (1)