Finite group having at least two conjugacy classes of involutions has order less than the cube of the maximum of orders of centralizers of involutions

Statement
Suppose $$G$$ is a finite group such that there are at least two conjugacy classes of involutions (non-identity elements of order two) in $$G$$. Then, if $$h$$ is the maximum of the orders of all subgroups of $$G$$ that arise as a fact about::centralizer of involution, we have:

$$|G| < h^3$$.

Similar facts

 * Dihedral trick
 * Brauer-Fowler theorem on existence of subgroup of order greater than the cube root of the group order

Opposite facts

 * Finite group having exactly one conjugacy class of involutions need not have order less than the cube of the order of the centralizer of involution
 * Finite simple non-abelian group has order greater than product of order of proper subgroup and its centralizer
 * Every proper abelian subgroup of a finite simple non-abelian group has order less than its square root

Facts used

 * 1) uses::Involutions are either conjugate or have an involution centralizing both of them
 * 2) uses::Size of conjugacy class equals index of centralizer

Qualitative explanation for appearance of cubes
Fact (1) is the crucial fact, since it says that we either have conjugacy or two-step centralizing. Fact (2) says that the size of the conjugacy class is the reciprocal of the size of the centralizer.

Thus, if we take the centralizer and then take the centralizers of the involutions in there, we will end up with all the involutions that are not in the same conjugacy class. If we have two conjugacy classes, we can play the classes against each other. The two steps of centralization, and the one step of conjugacy, together add up to three steps, explaining the occurrence of the cube.

Proof details
Given: A group $$G$$ with at least two conjugacy classes of involutions. $$h$$ is the maximum possible order of a subgroup arising as the centralizer of an involution of $$G$$.

To prove: $$|G| < h^3$$.

Proof: