Prove that $S_n$ acts transitively on the set $A=\{1,2,3,\ldots,n\}$ Im trying to prove that $S_n$ acts transitively on the set $A=\{1,2,3,\ldots,n\}$.
There is a similar question posted here:$S_n$ acting transitively on $\{1, 2, \dots, n\}$
But it doesn't really answer my question as it focuses on another point.
I'm confused as to how i can prove this.
Transitivity implies that if a certain permutation maps $a\Rightarrow b$, and $b \Rightarrow c$ then $a \Rightarrow c$ is implied, if i am correct.
But doesn't this simply follow from the definition of permuting an element in a set?
Edit: Thanks guys. It seems I'm a bit tired :)
 A: By definition, a group $G$ acts transitively on a set $X$ if for any $x,y\in X$ there is some $g\in G$ such that $g.x=y$. In other words, a transitive action is an action with one orbit.
Now, in your example it is easy. For $i,j\in\{1,2,...,n\}$ you can simply take the permutation $\sigma=(ij)$, it satisfies $\sigma.i=\sigma(i)=j$.
A: You have conflated "transitive action of a group" with "transitive relation".
For proving the action of $S_n$ transitive, it is sufficient to notice that $S_n$ contains the transposition $(x\ y)$ for every $x,y \in A$.  (You may want to write a few more words...)
A: $G$ acts transitively on $S$ if for all $s, t \in S$, there exists $g \in G$ s.t. $g \circ s = t$.
The thing you said follows directly from the definition of a group action: if $g \circ a = b$ and $h \circ b = c$ then $c = h \circ b = h \circ (g \circ a) = (hg) \circ a$.
Proving transitivity of $S_n$ is straightforward then. Let $i,j \in \{1,...,n\}$. Define $\sigma \in S_n$ by $\sigma(i) = j, \sigma(j) = i$ and $\sigma(k) = k$ for $k \ne i,j$.
A: The stabilizer of $a\in A$ is $\operatorname{Stab}(a)=\{\sigma\in S_n\mid \sigma(a)=a\}$, and hence $|\operatorname{Stab}(a)|=(n-1)!$ for every $a\in A$. Therefore, by the Orbit-Stabilizer Theorem, $|O(a)|=\frac{n!}{(n-1)!}=n$ for every $a\in A$, and hence there's one orbit only, and the action is transitive.
