Frobenius' normal p-complement theorem

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) uses::Index is multiplicative
 * 2) uses::Second isomorphism theorem
 * 3) uses::Retract implies conjugacy-closed
 * 4) uses::Conjugacy-closed and Sylow implies retract

(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).