Maximal among abelian normal subgroups of Sylow subgroup implies direct factor of centralizer

Statement
Suppose $$G$$ is a finite group and $$p$$ is a prime number. Suppose $$P$$ is a $$p$$-Sylow subgroup of $$G$$, and $$A$$ is fact about::maximal among abelian normal subgroups in $$P$$. Then, $$A$$ is a fact about::direct factor in $$C_G(A)$$.

Related facts
Category:Normal p-complement theorems lists many theorems of a related nature.

Other related facts:


 * Hall and central factor implies direct factor
 * Pi-separable and pi'-core-free implies pi-core is self-centralizing

Facts used

 * 1) uses::Maximal among abelian normal implies self-centralizing in nilpotent
 * 2) uses::Product formula
 * 3) uses::Burnside's normal p-complement theorem

Proof
Given: A finite group $$G$$, a prime $$p$$, a $$p$$-Sylow subgroup $$P$$ of $$G$$. A subgroup $$A$$ that is maximal among abelian normal subgroups of $$G$$.

To prove: $$A$$ is normal in $$C_G(A)$$. Further, there exists a subgroup $$B$$ of $$G$$ such that $$AB = C_G(A)$$ and $$A \cap B$$ is trivial.

Proof: Clearly, A</math, being abelian, is in the center of $$C_G(A)$$, hence normal in $$C_G(A)$$.


 * 1) $$C_G(A) \cap P = C_P(A) = A$$: This follows from fact (1).
 * 2) $$PC_G(A)$$ is a group: Since $$P$$ normalizes $$A$$, $$P$$ also normalizes $$C_G(A)$$. Hence, $$PC_G(A)$$ is a group.
 * 3) $$A$$ is a Sylow subgroup of $$C_G(A)$$: Consider the intersection $$C_G(A) \cap P$$. By the product formula (fact (2)), $$[C_G(A):(C_G(A) \cap P)] = [PC_G(A):P]$$. The right side is relatively prime to $$p$$, hence so is the left side. By step (1), $$C_G(A) \cap P = A$$, so $$[C_G(A):A]$$ is coprime to $$p$$. Thus, $$A$$ is a $$p$$-Sylow subgroup of $$C_G(A)$$.
 * 4) $$A$$ has a normal complement, say $$B$$, in $$C_G(A)$$: By definition, $$A$$ is in the center of $$C_G(A)$$, so it is in the center of its normalizer in $$C_G(A)$$, so by fact (3), $$A$$ has a normal complement.

(4) completes the proof.