# 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.

## 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, $L \le C_G(Q)$. 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).