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."
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.
- Solvability is subgroup-closed
- Minimal normal implies characteristically simple
- Equivalence of definitions of finite characteristically simple group
Given: A finite N-group .
To prove: is either solvable or almost simple.
|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.|