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

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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.

Related facts

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:

Failed to parse (unknown function "\tag"): \left| \langle A,B \rangle \right| \ge |AB| = \frac{|A||B|}{|A \cap B|} \tag{(*)}

and:

Failed to parse (unknown function "\tag"): \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:

Failed to parse (unknown function "\tag"): C_P(C_P(A) \cap C_P(B)) \ge \langle A,B \rangle \tag{(**)}

and:

Failed to parse (unknown function "\tag"): 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.

References

Journal references