$S_5$ does not act transitively on $\{1,2,3\}$? I have seen a statement: let $S_n$ act transitively on a set with $m$ elements. Then, $m\leq 2$ or $n \leq m$. I was able to prove it, so I believe it. However, I find this strikingly unintuitive. For instance, $S_5$ contains $S_3$, which acts transitively on $\{1,2,3\}$, so shouldn't $S_5$ also act transitively? Where is the gap in my logic here?
$\textbf{Edit:}$ Can someone give me an example of a non-trivial action of $S_5$ on $\{1,2,3\}$ that is not transitive?
 A: You would need to extend the acton of $S_3$ on $\{1, 2, 3\}$ to an action of $S_5$ on $\{1, 2, 3\}$, i.e. you would need to extend the identity $\mathrm{id} \colon S_3 \to S_3$ to a group epimorphism $p \colon S_5 \to S_3$. This is not possible.
More generally, there exists no group epimorphism $p \colon S_n \to S_m$ if $n \geq 5$ and $n \neq m > 2$ because then $A_n$ is the only proper nontrivial normal subgroup of $S_n$. We would then have $\ker p = \{1\}$, and contradicting $n \neq m$, or $\ker p = A_n$, contradicting $m \neq 2$, or $\ker p = S_n$, contradicting $m \neq 1$.
PS: As we can see from this argument there are precisely four actions of $S_5$ on $\{1,2,3\}$, classified by the group homomorphisms $p \colon S_5 \to S_3$. The case $\ker p = S_5$ corresponds to the trivial action. If $\ker p = A_5$ then $\mathrm{im} \ p \cong S_2$, so we have three such actions, corresponding to the three transpositions of $S_5$: So we fix any of the three elements and let $S_5$ swap the other two where $A_5$ does nothing.
A: Standard argument (above) is through group action and homomorphism. Here is a little different:
suppose $S_5$ acts on $\{1,2,3\}$ transitively. Then $|S_5|=|\mathrm{Orbit}(i)|.|\mathrm{Stab}(i)|=3.|\mathrm{Stab}(i)|$, hence $\mathrm{Stab}(i)$ is subgroup of index $3$ in $S_5$, so its order is $40$. In group of order $40$, Sylow-$5$ subgroup is normal, i.e. normalizer of Sylow-$5$ subgroup in $S_5$ contains $\mathrm{Stab}(i)$. So index of normalizer in $S_5$ of Sylow-$5$ subgroup is $\leq 3$, i.e. number of Sylow-$5$ subgroups in $S_5$ is $\leq 3$, contradiction (it is $6$).
