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 $n\geq 7$. 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 $n\geq 7$, $n-2\geq 5$, so $S_{n-2}$ is non-solvable (follows from alternating groups are simple for degree $\geq 5$), 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.