N-group not implies solvable or minimal simple

Statement
It is possible for a group (in fact, a finite group) to be a fact about::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 fact about::solvable group or a fact about::minimal simple group.

Converse

 * Minimal simple implies N-group
 * Solvable implies N-group

Proof
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 $$N_G(K)$$.