Centralizer-large subgroups permute and their product and intersection are centralizer-large

Statement
Suppose $$P$$ is a group of prime power order and $$A,B$$ are fact about::centralizer-large subgroups of $$P$$. Then, $$AB = BA$$, and both $$AB$$ and $$A \cap B$$ are both centralizer-large subgroups of $$P$$.

Related facts

 * Centrally large subgroups permute and their product is centrally large
 * All minimal CL-subgroups have the same commutator subgroup

Facts used

 * 1) uses::Product formula

Proof
Given: A group $$P$$ of prime power order, centralizer-large subgroups $$A,B$$ of $$P$$.

To prove: $$AB = BA$$, and both $$AB$$ and $$A \cap B$$ are also centralizer-large subgroups.

Proof: Note that $$AB = BA$$ if and only if it is a subgroup, if and only if it is equal to $$\langle A, B \rangle$$. In general, we have $$|AB| \le \left| \langle A,B \rangle \right|$$.

By fact (1), we have:

$$\left| \langle A,B \rangle \right| \ge |AB| = \frac{|A||B|}{|A \cap B|} \tag{(*)}$$

and:

$$\left| \langle C_P(A),C_P(B) \rangle \right| \ge |C_P(A)C_P(B)| = \frac{|C_P(A)||C_P(B)|}{|C_P(A) \cap C_P(B)|} \tag{(*)}$$

Also, we have:

$$C_P(C_P(A) \cap C_P(B)) \ge \langle A,B \rangle \tag{(**)}$$

and:

$$C_P(A \cap B) \ge \langle C_P(A),C_P(B)\rangle \tag{(***)}$$.

Denote by $$f_1(D)$$ the value $$|D||C_P(D)|$$. Then, combining (*) and (**), we see that:

$$f_1(A \cap B)f_1(C_P(A) \cap C_P(B)) \ge f_1(A)f_1(B)$$

with equality holding in all the inequalities. But $$A$$ and $$B$$ maximize $$f_1$$, hence equality must hold. Thus, $$AB = BA = C_P(C_P(A) \cap C_P(B))$$, $$C_P(B)C_P(A) = C_P(A)C_P(B) = C_P(A \cap B)$$. Also, $$C_P(AB) = C_P(A) \cap C_P(B)$$ and $$C_P(C_P(A)C_P(B)) = A \cap B$$ by definition. Thus, the subgroups $$A \cap B$$ and $$AB$$ are centralizer-large.

Since $$f_1$$ is maximized for $$A$$ and $$B$$, we must have equality. Hence $$AB = BA$$. Moreover, we get that $$C_P(AB) = C_P(A) \cap C_P(B)$$ and $$C_P(C_P(A)C_P(B)) = A \cap B$$.