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 $$n \in \{ 0,1,2,3,4,5,6 \}$$.

Also, the symmetric group on an infinite set is never a N-group. Similarly, the finitary symmetric group on an infinite set is never a N-group.

Related facts

 * Classification of alternating groups that are N-groups

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 $$N_G(\langle x \rangle)$$ and it is the internal direct product $$\operatorname{Sym} \{ 1,2 \} \times \operatorname{Sym} \{ 3,4,\dots,n \}$$. 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.

A similar example works for the symmetric group on an infinite set.

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.