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 $P$.

Related facts

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

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 $|AB|\leq \left|\langle A,B\rangle \right|$.

By fact (1), we have:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left| \langle A,B \rangle \right| \ge |AB| = \frac{|A||B|}{|A \cap B|} \tag{(*)}}

and:

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

and:

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

$f_{1}(A\cap B)f_{1}(C_{P}(A)\cap C_{P}(B))\geq f_{1}(A)f_{1}(B)$

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

• 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 copyMore 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 copyMore info}