Finite N-group is solvable or almost simple

This article gives the statement, and possibly proof, of a statement where the conclusion of the statement involves a disjunction (OR) of two possibilities. The prototypical form is: "every A is a B or a C."

Statement

Suppose $G$ is a finite N-group, i.e., $G$ is a finite group that is also a N-group, i.e., the normalizer of any nontrivial solvable subgroup of $G$ is solvable. Then, $G$ is either a solvable group (or equivalently, a finite solvable group) or an almost simple group.

The definition of almost simple that we will use here is: a group is almost simple if it has a centralizer-free non-abelian simple normal subgroup.

Facts used

1. Solvability is subgroup-closed
2. Minimal normal implies characteristically simple
3. Equivalence of definitions of finite characteristically simple group

Proof

Given: A finite N-group $G$.

To prove: $G$ is either solvable or almost simple.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	If $G$ has a nontrivial solvable normal subgroup, then $G$ is solvable.				By definition, the normalizer of the nontrivial solvable normal subgroup, which is $G$, must be solvable.
2	If $G$ has a subgroup $S$ that is simple non-abelian, then $C_{G}(S)$ is trivial.	Fact (1)	$G$ is a N-group		Suppose $C_{G}(S)$ is nontrivial. Take a non-identity element $x\in C_{G}(S)$. Let $Q=\langle x\rangle$. Then, $S\subseteq C_{G}(Q)\subseteq N_{G}(Q)$. Since $S$ is non-solvable, $N_{G}(Q)$ is non-solvable even though $Q$ is solvable, contradicting the assumption that $G$ is a N-group.
3	If $G$ has a minimal normal subgroup $H$, then $H$ must be either solvable or simple non-abelian.	Facts (2), (3)	$G$ is finite	Step (2)	By Facts (2) and (3), if $H$ is non-solvable, it is an internal direct product of pairwise isomorphic simple non-abelian subgroups $T_{1},T_{2},\dots ,T_{n}$. If $n>1$, then $C_{G}(T_{1})$ contains $T_{2}$, contradicting Step (2). Hence, $n=1$.
4	$G$ is either solvable or almost simple.		$G$ is finite	Steps (1), (2), (3)	If $G$ is trivial, it is solvable, so assume $G$ nontrivial. Let $H$ be a minimal normal subgroup of $G$. By Step (3), $H$ is either solvable or simple non-abelian. If $H$ is solvable, then $G$ is solvable by Step (1). If $H$ is simple non-abelian, then it is a centralizer-free non-abelian simple normal subgroup by Step (2), hence $G$ is almost simple.