Centralizer-large subgroups permute and their product and intersection are centralizer-large
Statement
Suppose is a group of prime power order and are Centralizer-large subgroup (?)s of . Then, , and both and are both centralizer-large subgroups of .
Related facts
- Centrally large subgroups permute and their product is centrally large
- All minimal CL-subgroups have the same commutator subgroup
Facts used
Proof
Given: A group of prime power order, centralizer-large subgroups of .
To prove: , and both and are also centralizer-large subgroups.
Proof: Note that if and only if it is a subgroup, if and only if it is equal to . In general, we have .
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 the value . Then, combining (*) and (**), we see that:
with equality holding in all the inequalities. But and maximize , hence equality must hold. Thus, , . Also, and by definition. Thus, the subgroups and are centralizer-large.
Since is maximized for and , we must have equality. Hence . Moreover, we get that and .
References
Journal references
- A measuring argument for finite groups by Andrew Chermak and Alberto Delgado, Proceedings of the American Mathematical Society, Volume 107,Number 4, Page 907 - 914(Year 1989): ^{Official copy}^{More info}
- Centrally large subgroups of finite p-groups by George Isaac Glauberman, Journal of Algebra, ISSN 00218693, Volume 300,Number 2, Page 480 - 508(Year 2006): ^{Official copy}^{More info}