# 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.

## 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.