two subgroups of $S_{n}$ and $S_{m}$ If $H\subseteq S_{n}$ and $K\subseteq S_{m}$ how can I then show that I can think of $H\times K$ as it was a subgroup of $S_{m+n}$?

In hopes of getting this off the Unanswered list, here’s a hint expanding on Jyrki’s first comment.
$K$ is a group of permutations of the set $\{1,\dots,m\}$, so each $k\in K$ is a bijection $$k:\{1,\dots,m\}\to\{1,\dots,m\}\;.$$ For each $k\in K$ let $$\hat k:\{n+1,\dots,n+m\}\to\{n+1,\dots,n+m\}:n+i\mapsto n+k(i)\;,$$ and let $\widehat K=\{\hat k:k\in K\}$.

Show that $\widehat K$ is a group of permutations of $\{n+1,\dots,n+m\}$ and is isomorphic to $K$.

For each $\langle h,k\rangle\in H\times K$ let $$g_{\langle h,k\rangle}:S_{n+m}\to S_{n+m}:i\mapsto\begin{cases}
h(i),&\text{if }1\le i\le n\\
\hat k(i),&\text{if }n+1\le i\le n+m\;,
\end{cases}$$
and let $G=\{g_{\langle h,k\rangle}:\langle h,k\rangle\in H\times K\}$.

Show that $G$ is a subgroup of $S_{n+m}$ and is isomorphic to $H\times K$.

