I'm wondering how to prove that, for every pair of natural numbers $\{m,n\}\in \mathbb{N}$, $m\leq n$, there is a subgroup $G\lhd S_n$ such that $G\cong S_m$. I attempted what i think is a valid proof, that writes like this:
Suppose $\tau:\mathbb{N}\rightarrow\mathbb{N}_{<m}$, and that $G\lhd S_n$ is the subgroup generated by permutations of the form $$\sigma= \pmatrix{1&...&m&m+1&...&n\\\tau(1)&...&\tau(m)&m+1&...&n}$$ Evidently, every permutation $\sigma\in G$ leaves the places $m+1$ to $n$ unchanged. There exists an homomorphism $i:G\rightarrow S_m$ given precisely by $$i(\sigma)=\pmatrix{1&...&m\\\tau(1)&...&\tau(m)}$$ Since every two functions with the same domain, range and output of $\tau$ must be equal, then $i^{-1}:S_m\rightarrow G$ must exist, and hence, $i$ is a bijective homomorphism. Finally, since $\tau$ was chosen arbitrarily, for every $\{\sigma,\pi\}\in S_n$, we can choose $\tau_{\sigma},\tau_{\pi}$ such that: $$i(\sigma\circ\pi)=\pmatrix{1&...&m\\\tau(1)&...&\tau(m)}$$ $$=\pmatrix{1&...&m\\\tau_\sigma(1)&...&\tau_\sigma(m)}\pmatrix{1&...&m\\\tau_\pi(1)&...&\tau_\pi(m)}$$ $$=i(\sigma)\circ i(\pi)$$ Since $i(\sigma\circ\pi)=i(\sigma)\circ i(\pi)$ is a bijective homomorphism, then, it is an isomorphism, and thus $G\cong S_m$.
Is this proof ok? Is the bijection demonstration loose? Any suggestions are welcome.