# 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: "everyAis aBor aC."

## Statement

Suppose is a finite N-group, i.e., is a finite group that is also a N-group, i.e., the normalizer of any nontrivial solvable subgroup of is solvable. Then, 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

## Proof

**Given**: A finite N-group .

**To prove**: is either solvable or almost simple.

**Proof**:

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | If has a nontrivial solvable normal subgroup, then is solvable. | By definition, the normalizer of the nontrivial solvable normal subgroup, which is , must be solvable. | |||

2 | If has a subgroup that is simple non-abelian, then is trivial. | Fact (1) | is a N-group | Suppose is nontrivial. Take a non-identity element . Let . Then, . Since is non-solvable, is non-solvable even though is solvable, contradicting the assumption that is a N-group. | |

3 | If has a minimal normal subgroup , then must be either solvable or simple non-abelian. | Facts (2), (3) | is finite | Step (2) | By Facts (2) and (3), if is non-solvable, it is an internal direct product of pairwise isomorphic simple non-abelian subgroups . If , then contains , contradicting Step (2). Hence, . |

4 | is either solvable or almost simple. | is finite | Steps (1), (2), (3) | If is trivial, it is solvable, so assume nontrivial. Let be a minimal normal subgroup of . By Step (3), is either solvable or simple non-abelian. If is solvable, then is solvable by Step (1). If is simple non-abelian, then it is a centralizer-free non-abelian simple normal subgroup by Step (2), hence is almost simple. |