# 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 Maximal among abelian normal subgroups (?) in $P$. Then, $A$ is a Direct factor (?) in $C_G(A)$.

## Related facts

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

Other related facts:

## 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, 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.