# Action of permutation group on set of numbers is transitive

I would appreciate if someone could please tell their opinion about my proof. I think the proof makes sense, but I don't know if it's rigorous enough.

Theorem:

Let $S_n$ be a symmetric group of order $n!$, acting on the set $X=\{1,...,n\}$ of order $n$ by $\sigma \cdot a=\sigma(a)$. Then this action is transitive.

Proof:

Take two arbitrary elements $m,k\in X$. Every $\sigma \in S_n$ is a bijection. Since $S_n$ contains $n!$ unique bijections, there must exist a $\sigma$ for every $m$, such that $\sigma(m)=k$ for any $k$. Thus the action is transitive.

It's not clear by your armgument, why there must exists a $\sigma$ for each $m$ such that $\sigma(m)=k$. On the other hand, there's no need for a counting argument. Given, $1\le m<k\le n$, let $\sigma$ be the transposition $(mk)$. That is, the bijection which switches $m$ and $k$ and leaves every other number fixed.
Alternatively to the other answer, you can use the orbit-stabilizer theorem. For any $$i\in X$$, $$\operatorname{Stab}(i)=\{\sigma\in S_n\mid \sigma(i)=i\}$$. By the standard argument used to show that $$|S_n|=n!$$, you get also $$\left|\operatorname{Stab}(i)\right|=(n-1)!$$ and hence $$|O(i)|=$$ $$\frac{|S_n|}{\left|\operatorname{Stab}(i)\right|}=$$ $$\frac{n!}{(n-1)!}=$$ $$n$$. Therefore, $$O(i)=X$$, namely the action has one orbit, only.