N-group not implies solvable or minimal simple

Statement

It is possible for a group (in fact, a finite group) to be a N-group (?) (i.e., a group in which every local subgroup (the normalizer of a nontrivial solvable subgroup) is a solvable group) but to not itself be either a Solvable group (?) or a Minimal simple group (?).

Related facts

Converse

• Minimal simple implies N-group
• Solvable implies N-group

Proof

Further information: symmetric group:S5

Suppose $G$ is $S_{5}$, i.e., symmetric group:S5. We note that $G$ is neither a solvable group nor a minimal simple group, because it has a subgroup $H$ isomorphic to alternating group:A5 as a proper subgroup, and this subgroup is a simple non-abelian group.

To see that $G$ is a N-group, note the following:

1. The only subgroups of $G$ that are not solvable are $G$ and $H$. Of these, $H$ is simple non-abelian.
2. The only nontrivial normal subgroups of $G$ are $G$ and $H$. The only nontrivial normal subgroup of $H$ is $H$ itself.
3. For any nontrivial subgroup $K$ of $G$, we know that $K$ is a nontrivial normal subgroup of $N_{G}(K)$. Combining with Steps (1) and (2), we see that if $N_{G}(K)$ is not solvable, then $K$ must also not be solvable. Thus, the contrapositive is true: if $K$ is a nontrivial solvable group, so is $N_{G}(K)$.