Burnside's normal p-complement theorem

This article gives the statement, and possibly proof, of a normal p-complement theorem: necessary and/or sufficient conditions for the existence of a Normal p-complement (?). In other words, it gives necessary and/or sufficient conditions for a given finite group to be a P-nilpotent group (?) for some prime number

$p$

.
View other normal p-complement theorems

Name

This result is termed Burnside's normal p-complement theorem and is also sometimes termed Burnside's transfer theorem.

Statement

Statement with symbols

Suppose $p$ is a prime, $G$ is a finite group, and $P$ is a $p$-Sylow subgroup (?) of $G$. Further, suppose $P$ is a Central subgroup of normalizer (?): if $H=N_G(P)$ is its normalizer, and $Z(H)$ is the center of $H$, then $P\le Z(H)$.

Then, $P$ is a Retract (?) of $G$, i.e., there exists a normal p-complement in $G$: a normal subgroup $N$ such that $NP=G$ and $N\cap P$ is trivial.

Related facts

• Conjugacy-closed Abelian Sylow implies retract
• Frobenius' normal p-complement theorem
• Thompson's normal p-complement theorem

Facts used

1. Center of Sylow sugbroup is conjugacy-determined in normalizer: If $P$ is a Sylow subgroup of $G$, then two elements of $Z(P)$ are conjugate in $G$ if and only if they are conjugate in $N_G(P)$.
2. Conjugacy-closed abelian Sylow implies retract: If $P$ is an Sylow subgroup of $G$ such that no two distinct elements of $P$ are conjugate in $G$, then $P$ is a retract of $G$. (Note that the proof of this relies in turn on the focal subgroup theorem).
3. Grün's first theorem on the focal subgroup

Proof

Proof using a rather weak fusion result

Given: $G$ a finite group, $P$ a $p$-Sylow subgroup such that $P\le Z(N_G(P))$.

To prove: $P$ is a retract of $G$: it possesses a normal $p$-complement.

Proof:

1. No two distinct elements of $P$ are conjugate in $G$: Since $P$ is Abelian, $Z(P)=P$, so fact (1) tells us that two elements of $P$ are conjugate in $G$ if and only if they are conjugate in $N_G(P)$. Since $P\le Z(N_G(P))$, no two distinct elements of $P$ are conjugate in $N_G(P)$, and hence no two distinct elements of $P$ are conjugate in $G$.
2. $P$ has a normal complement: This follows from fact (2), since the previous step shows that the conditions for it are satisfied.

Proof using a stronger fusion result

This proof uses fact (3).

See also

• For other theorems called "Burnside's theorem", see Burnside's theorem for disambiguation.