Let $S_n$ be the Permutation group on ${1,...,n}$ , where $n\in \mathbb{N}$, and let $H$ be a subgroup of $S_{n+m}$,where $\sigma\in H$ iff for each $1 \le i \le n \ : \ 1 \le \sigma(i) \le n$ .
Prove that $H$ is isomorphic to $S_n\times S_m$.
My attempt:
Let $\phi(\sigma):H\to S_n\times S_m$ be: $(\sigma_1,\sigma_2)$ where :
$\sigma_1(i):[n]\to [n] , \sigma_1(i) = \sigma(i)$
$\sigma_2(j):[m]\to [m] , \sigma_2(j)=\sigma(j)$
How can I prove that it's a bijection?
Or maybe it isn't at all?
I can't guess what $[m]$ means, but I presume that your $\sigma$ is
\begin{align} \phi(\sigma)=(\sigma |_{\{1,\cdots,n\}} , \sigma|_{\{{1',\cdots,m'\}}})\end{align} where we consider $S_{n+m}$ as the permutation group of $\{1,\cdots,n,1',\cdots,m'\}$.
Since $|H|=|S_n \times S_m |$, showing injectivity will be suffice. This follows
\begin{align} \phi(\sigma)=id &\Leftrightarrow \sigma |_{\{1,\cdots,n\}}=id|_{\{1,\cdots,n\}} \,\;\mathrm{and}\;\, \sigma|_{\{{1',\cdots,m'\}}} =id|_{\{1',\cdots,m'\}}\\ &\Leftrightarrow \sigma=id \end{align} where $id$ is the identity function on $\{1,\cdots,n,1',\cdots,m'\}$.