Solvability is 2-local for finite groups

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 $$\langle a,b \rangle$$ 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) uses::Solvability is subgroup-closed
 * 2) uses::Every finite non-solvable group has a minimal simple group as subquotient
 * 3) uses::Finite minimal simple implies 2-generated
 * 4) uses::Solvability is quotient-closed

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 $$\langle a,b \rangle$$ 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 $$M \triangleleft N \le G$$ 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 $$N/M = \langle x,y \rangle$$.
 * 3) Let $$a = \alpha^{-1}(x), b = \alpha^{-1}(y)$$. We know that $$\alpha(\langle a,b \rangle) = \langle \alpha(a), \alpha(b)\rangle = S$$, which is simple non-Abelian. Thus, $$\langle a,b\rangle$$ has a simple non-Abelian homomorphic image. By fact (4), this forces $$\langle a,b$$ to not be solvable, completing the proof.