Frobenius' 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

Statement

Let $G$ be a finite group and $p$ be a prime number. Then, the following are equivalent:

1. $G$ has a $p$-Sylow subgroup $P$ that is conjugacy-closed in $G$: any two elements of $P$ that are conjugate in $G$ are conjugate in $P$.
2. There is a normal p-complement in $G$: a normal subgroup whose index is a power of $p$ and whose order is relatively prime to $p$. In other words, any $p$-Sylow subgroup is a retract of $G$.
3. For every non-identity $p$-subgroup $Q$ of $G$, the subgroup $N_{G}(Q)$ has a normal p-complement
4. For every non-identity $p$-subgroup $Q$, the quotient $N_{G}(Q)/C_{G}(Q)$ is a $p$-group.

Related facts

• Conjugacy-closed and Sylow implies retract: This is the equivalence of (1) and (2) above.
• Conjugacy-closed Abelian Sylow implies retract: This is a weaker version of the equivalence of (1) and (2) above, that has a more direct proof.
• Equivalence of definitions of Sylow direct factor: This states that a conjugacy-closed normal Sylow subgroup is a direct factor. Note that the statement is again a weakening of the fact that any conjugacy-closed Sylow subgroup is a retract, and it again has a more direct proof.
• Burnside's normal p-complement theorem: A corollary of the fact that any conjugacy-closed Abelian Sylow subgroup is a retract.

Facts used

1. Index is multiplicative
2. Second isomorphism theorem
3. Retract implies conjugacy-closed
4. Conjugacy-closed and Sylow implies retract

Proof

(2) implies (3)

(Note: This implication uses nothing about the special nature of normalizers of $p$-subgroups, and works for all subgroups).

Given: A finite group $G$, a prime $p$, a $p$-Sylow subgroup $P$ of $G$. A normal subgroup $M$ of $G$ such that $MP=G$ and $M\cap P$ is trivial.

To prove: If $Q$ is a non-identity $p$-subgroup of $G$, then $N_{G}(Q)$ has a normal $p$-complement.

Proof: Let $N=N_{G}(Q)$. By the second isomorphism theorem, we have:

$NM/M\cong N/(N\cap M)$.

Since $[NM:M]$ divides $[G:M]$, and $[G:M]$ equals the order of $P$, $[NM:M]$ is a power of $p$. Thus, so is the order of the right side, $[N:N\cap M]$.

Thus, $N\cap M$ is a normal subgroup of $N$ whose order is relatively prime to $p$ (since it is also a subgroup of $M$) and whose index is a power of $p$. Thus, it is a normal $p$-complement in $N$.

(3) implies (4)

Given: A finite group $G$, a prime $p$, a $p$-Sylow subgroup $P$ of $G$. For any non-identity $p$-subgroup $Q$, $N_{G}(Q)$ has a normal $p$-complement.

To prove: If $Q$ is a non-identity $p$-subgroup of $G$, then $N_{G}(Q)/C_{G}(Q)$ is a $p$-group.

Proof: Proof: Let $N=N_{G}(Q)$. Let $L$ be a normal $p$-complement in $N$.

Now, $Q$ and $L$ are both normal subgroups of $N$ and since the order of $L$ is relatively prime to $p$, $Q\cap L$ is trivial. Thus, $L\leq C_{G}(Q)$. Thus, by fact (1):

$[N:L]=[N:C_{G}(Q)][C_{G}(Q):L]$.

Since the left side is a power of $p$, so are both terms of the right side. In particular, $N_{G}(Q)/C_{G}(Q)$ is a $p$-group.

(2) implies (1)

This follows from fact (3).

(1) implies (2)

This follows from fact (4).

References

Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page Theorem 4.5, Chapter 7 (Fusion, transfer and p-factor groups), "Theorem" can not be assigned to a declared number type with value 4.5."Theorem" can not be assigned to a declared number type with value 4.5."Theorem" can not be assigned to a declared number type with value 4.5.^{More info}