# 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 Centralizer-large subgroup (?)s of $P$. Then, $AB = BA$, and both $AB$ and $A \cap B$ are both centralizer-large subgroups of $P$.

## Facts used

1. 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:

$|A||B|}{|A \cap B|} \tag{(*)$

and:

$|C_P(A)||C_P(B)|}{|C_P(A) \cap C_P(B)|} \tag{(*)$

Also, we have:

$(**)$

and:

$(***)$ .

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 $C_P(C_P(A)C_P(B)) = A \cap B$.