Solvability is 2-local for finite groups

Statement

Verbal statement

The following are equivalent for a finite group:

• The group is solvable, i.e., it is a finite solvable group.
• The subgroup generated by any two elements of the group is a solvable group.

Statement with symbols

Let $G$ be a finite group. The following are equivalent for $G$:

• $G$ is solvable.
• For any $a,b\in G$, the subgroup $⟨a,b⟩$ is solvable.

Related facts

• Abelianness is 2-local
• Cyclicity is 2-local for finite groups
• Nilpotency is 2-local for finite groups

Facts used

1. Solvability is subgroup-closed
2. Every finite non-solvable group has a minimal simple group as subquotient
3. Finite minimal simple implies 2-generated
4. Solvability is quotient-closed

Proof

Solvable implies the subgroup generated by any two elements is solvable

This follows from fact (1): any subgroup of a solvable group is solvable.

Subgroup generated by any two elements is solvable implies solvable

Given: A group $G$ that is finite and not solvable.

To prove: There exist elements $a,b\in G$ such that $⟨a,b⟩$ is solvable.

Proof:

1. Since $G$ is not solvable, fact (2) tells us that $G$ has a minimal simple subquotient. In particular, there exist subgroups $M◃N\le G$ such that $N/M$ is a minimal simple group. Let $\alpha :N\to N/M$ be the quotient map.
2. By fact (3), there exist elements $x,y\in N/M$ such that $N/M=⟨x,y⟩$.
3. Let $a={\alpha }^{-1}(x),b={\alpha }^{-1}(y)$. We know that $\alpha (⟨a,b⟩)=⟨\alpha (a),\alpha (b)⟩=S$, which is simple non-Abelian. Thus, $⟨a,b⟩$ has a simple non-Abelian homomorphic image. By fact (4), this forces $⟨a,b$ to not be solvable, completing the proof.