A subgroup $H$ of a group $G$ is said to be subnormal in $G$ if there exists a subnormal series from $H$ to $G$, that is a finite series \begin{align*} H = H_{0}\trianglelefteq H_{1}\trianglelefteq . . . \trianglelefteq H_{n} = G \end{align*} of subgroups of $G$, each of those is normal in the following one.
Let $n\in\mathbb{N}$ with $n\geq 2$. Find all subnormal subgroups of $S_{n}$.
Firstly, I know that for $n\geq 5$ $A_{n}$ is the only proper non-trivial normal subgroup of $S_{n}$.
For the other cases:
n=1 , $S_{1}$ has no nontrivial normal subgroups.
n=2, $S_{2}=C_{2}$ the unique group on 2 elements, so it has no nontrivial normal subgroups. n=3, $S_{3}$ has one nontrivial proper normal subgroup, the group generated by (1 2 3).
n=4, $S_{4}$ has $A_{4}$ and $V_{4}$ being the only non-trivial proper normal subgroups of $S_{3}$.
From here, I'm not really sure how to apply the known normal subrgroups of $S_{n}$ in order to obtain all the subnormal subgroups.
I know that every abelian group is also a nilpotent group and every subgroup of a nilpotent group is subnormal. However, in this case, $S_{n}$ is only abelian if and only if $n\leq 2$.