Prove that the only way $S_3$ can act transitively on a set of order 3 is by giving the permutations of the set

Question

I've seen it often stated that if $S_3$ acts trasitively on a three element set, then it must just permute that set. I.e., we can label the elements of the set $S$, as $s_1, s_2, s_3$ so that for any $\pi \in S_3$, $\pi \ast s_i = s_{\pi(i)}$. I tried to give a formal proof that this must be true, but I haven't been able to get anywhere. Any help would be appreciated!

Answer 1

Yes, it is true. I will formulate the problem in the following way:

If $S_3$ acts on a set $X$ with three elements, then we have a permutation representation: $\varphi: S_3\to S_3$. Your question is equivalent to: if the action is transitive, then $\varphi$ is an isomorphism.

If it is not an isomorphism, the image is a proper subgroup of $S_3$. Transitivty of the action says the image must be $\{1,(123),(132)\}$, so $|\ker \varphi|=6/3=2$. But $S_3$ has no normal subgroup of order $2$, so the image cannot be proper. $\varphi$ must be an isomorphism. The proof is completed.

Answer 2

If $S_3$ acts transitively on a set $X=\{x,y,z\}$ the stabilizer of any of them must have two elements, thus is generated by a transposition.

Obviously the order two subgroups generated by transposition meet trivially one with another, thus a transposition fixes one element and interchanges the other two.

Now you conclude using the fact that $S_3$ is generated by transpositions.

(It may help renaming $x_1$ the element fixed by $(2\ 3)$, $x_2$ the element fixed by $(1\ 3)$, and $x_3$ the element fixed by $(1\ 2)$)