Classification of symmetric groups that are N-groups

This article classifies the members in a particular group family symmetric group that satisfy the group property N-group.

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

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.