Finite N-group is solvable or almost simple

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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.