It is well known, that any finite group of order $n$ is isomorphic to a subgroup of $S_n$. Let’s call a finite group $G$ incompressible iff it is not isomorphic to any subgroup of $S_{|G|-1}$ . Does there exist some sort of classification of incompressible groups?
What I currently know:
Any non-trivial incompressible group has non-trivial center
If the center of a group $G$ is trivial, then it acts faithfully by conjugation on $G \setminus \{e\}$.
If an incompressible group is non-trivially decomposed into a direct product of two its subgroups, it is isomorphic to $C_2 \times C_2$
One can construct a faithful action of $H \times K$ on $H \cup K$. It is defined as $(h, k)h_0 \mapsto hh_0$ and $(h, k)h_0 \mapsto kk_0$ for $h, h_0 \in H$, $k, k_0 \in K$.
$|H| + |K| \geq |H||K|$ iff either one of the groups is trivial, or both of them are isomorphic to $C_2$.
$C_2 \times C_2$ is the only possible group and indeed is not contained in $S_3$.
I also conjecture, that «direct product» in this statement can be replaced with «semidirect product», but do not know how to prove that.
All cyclic $p$-groups are incompressible
If $p$ is prime, then $S_{p^n - 1}$ does not have an element of order $p^n$
$Q_8$ is incompressible
$S_7$ does not contain $Q_8$ as a subgroup