Isomorphism of subgroup of $S_n$. Let $n,k \in \mathbb{N}$ with $1<k<n$. We define a subgroup $H \leq S_n$ as:
$$H=\{\sigma \in S_n | \; \sigma(\{1,...,k\}) \subseteq \{1,...,k\}\}$$
I have to show that $H$ is isomorphic to $S_k\times S_{n-k}$.
My idea is that for all permutations in $H$, we can divide it in two parts, one of them only permutates elements from $\{1,...,k\}$ and the other one permutates only elements from $\{k+1,...,n\}$, so I could write $\sigma \in H$ as $\sigma = \rho\tau, \rho \in S_k$ and $\tau \in S_{n-k}$.
With that, an isomorphism should be easy to define, but I don't really know how can I formalize that idea (or if it's really right). Any hints? Thanks.
 A: Deploying my hint in the comment, let's define $I:=\{1,\dots,k\}$ and $J:=\{k+1,\dots,n\}$:

*

*Injectivity: for every $\sigma,\tau\in S_n$,
\begin{alignat}{2}
&\varphi(\sigma)=\varphi(\tau) &&\Longrightarrow \\
&(\sigma_{|I}=\tau_{|I}) \wedge (f\sigma_{|J}f^{-1}=f\tau_{|J}f^{-1}) &&\Longrightarrow \\
&(\sigma_{|I}=\tau_{|I}) \wedge (\sigma_{|J}=\tau_{|J}) &&\Longrightarrow \\
&\sigma=\tau
\end{alignat}

*Surjectivity: for every $(\psi_1,\psi_2)\in S_k\times S_{n-k}$, define $\sigma\in S_n$ by $\sigma_{|I}:=\psi_1$ and $\sigma_{|J}:=f^{-1}\psi_2f$; then $(\psi_1,\psi_2)=\varphi(\sigma)$.

*Preservation of the operation (composition): for every $\sigma,\tau\in H$ and $i\in I$,
\begin{alignat}{2}
(\sigma\tau)_{|I}(i) &\stackrel{i\in I}{=}  &&(\sigma\tau)(i) \\
&\stackrel{\sigma,\tau\in S_n}{=} &&\sigma(\tau(i)) \\
&\stackrel{\tau\in H}{=} &&\sigma_{|I}(\tau(i)) \\
&\stackrel{i\in I}{=} &&\sigma_{|I}(\tau_{|I}(i)) \\
&\stackrel{\sigma_{|I},\tau_{|I}\in S_k}{=} &&(\sigma_{|I}\tau_{|I})(i) \\
\end{alignat}
whence $(\sigma\tau)_{|I}=\sigma_{|I}\tau_{|I}$. Likewise, $(\sigma\tau)_{|J}=\sigma_{|J}\tau_{|J}$, because $\lambda\in H\iff \lambda\in K:=\{\gamma\in S_n\mid \gamma(J)\subseteq J\}$. Therefore:
\begin{alignat}{1}
\varphi(\sigma\tau) &= ((\sigma\tau)_{|I}, f(\sigma\tau)_{|J}f^{-1}) \\
&= (\sigma_{|I}\tau_{|I}, f\sigma_{|J}\tau_{|J}f^{-1}) \\
&= (\sigma_{|I}\tau_{|I}, (f\sigma_{|J}f^{-1})(f\tau_{|J}f^{-1})) \\
&= (\sigma_{|I}, f\sigma_{|J}f^{-1})(\tau_{|I}, f\tau_{|J}f^{-1}) \\
&= \varphi(\sigma)\varphi(\tau)  
\end{alignat}
and $H \cong S_k\times S_{n-k}$.
