Classification of symmetric groups that are N-groups

Statement

Suppose $n$ is a positive integer. The symmetric group $S_n$ is a N-group if and only if $n\in \{0,1,2,3,4,5,6\}$.

Proof

Proof of failure for larger $n$

Consider $n\ge 7$. Describe $S_n$ concretely as the group of all permutations on the set $\{1,2,\dots ,n\}$. Let $x$ be the element $(1,2)$ in $S_n$. The centralizer $C_G(x)$ equals the normalizer $N_G(⟨x⟩)$ and it is the internal direct product $\mathrm{S}\mathrm{y}\mathrm{m}\{1,2\}\times \mathrm{S}\mathrm{y}\mathrm{m}\{3,4,\dots ,n\}$. This is isomorphic to $S_2\times S_{n-2}$. For $n\ge 7$, $n-2\ge 5$, so $S_{n-2}$ is non-solvable (follows from alternating groups are simple for degree $\ge 5$), hence $S_2\times S_{n-2}$ is non-solvable.

Proof of success

For $n\in \{0,1,2,3,4\}$, the whole group is solvable, so it is a N-group.

The case $n=5$ is also direct: the only non-solvable subgroups are the whole group symmetric group:S5 and A5 in S5 (isomorphic to alternating group:A5) and neither of them has a nontrivial solvable normal subgroup.

For the case $n=6$, the only non-solvable subgroups are the whole group symmetric group:S6, the subgroup A6 in S6 (isomorphic to alternating group:A6), and subgroups isomorphic to alternating group:A5 and symmetric group:S5. None of them have a nontrivial solvable normal subgroup. Thus, the group is a N-group.