Characterization of minimal counterexamples to a characteristic p-functor controlling normal p-complements

Statement
Suppose $$p$$ is a prime number and $$W$$ is a characteristic p-functor that does not control normal p-complements. Suppose $$G$$ is a finite group of the least possible order where such a phenomenon is realized: $$G$$ has a nontrivial $$p$$-Sylow subgroup $$P$$ such that $$N_G(W(P))$$ has a normal p-complement (i.e., is a p-nilpotent group) but $$G$$ does not (i.e., $$G$$ is not a p-nilpotent group). Then, the following are true:


 * 1) $$\! O_{p'}(G)$$ is trivial: In other words, $$G$$ does not possess any nontrivial normal subgroup of order relatively prime to $$p$$.
 * 2) $$\! O_p(G)$$ is nontrivial but not equal to $$P$$.
 * 3) $$\! G/O_p(G)$$ possesses a normal $$p$$-complement. In particular, $$\! G = O_{p,p',p}(G)$$.

Facts used

 * 1) uses::Frobenius' normal p-complement theorem
 * 2) uses::Characteristic of normal implies normal

Proof of (1)
Given: A finite group $$G$$ that is of smallest order with respect to the property that there exists a nontrivial $$p$$-Sylow subgroup $$P$$ such that $$N_G(W(P))$$ possesses a normal $$p$$-complement but $$G$$ does not.

To prove: $$\! O_{p'}(G)$$ is trivial.

Proof strategy: Suppose $$O_{p'}(G)$$ were nontrivial. We will then derive a contradiction from the fact that $$G/O_{p'}(G)$$, being of smaller order, is not a counterexample.

Proof details: Consider the quotient map $$\varphi:G \to G/O_{p'}(G)$$.

Proof of (2)
Given: A finite group $$G$$ that is of smallest order with respect to the property that there exists a nontrivial $$p$$-Sylow subgroup $$P$$ such that $$N_G(W(P))$$ possesses a normal $$p$$-complement but $$G$$ does not.

To prove: $$O_p(G)$$ is a proper nontrivial subgroup of $$P$$.

Proof:


 * 1) By fact (1), there exists a non-identity $$p$$-subgroup $$Q$$ of $$G$$ such that $$N = N_G(Q)$$ does not have a normal $$p$$-complement. Among such $$Q$$, pick a $$Q$$ such that the order of the Sylow $$p$$-subgroup of $$N_G(Q)$$ is maximal. By conjugation, we can further assume that $$P$$ contains the Sylow subgroup of $$N_G(Q)$$, so this Sylow subgroup is $$P \cap N$$.
 * 2) $$P \le N$$: Suppose not. Then set $$R = P \cap N$$, and set $$L = N_N(W(R))), M = N_G(W(R)))$$.
 * 3) * $$P \cap M$$ contains $$N_P(R)$$: Since $$W(R)$$ is a characteristic subgroup of $$R$$, $$W(R)$$ is normal in $$N_P(R)$$ (by fact (2)). Thus, $$N_P(R) \le M$$. In particular, $$P \cap M$$ contains $$N_P(R)$$.
 * 4) * $$P \cap M$$ properly contains $$R$$: Since $$R$$ is proper in $$P$$, $$R$$ is proper in $$N_P(R)$$. Thus, $$R$$ is properly contained in $$P \cap M$$ by the previous step.
 * 5) * $$M$$ has a normal $$p$$-complement: We have found a subgroup $$R$$ such that $$R \le P$$ and the $$p$$-Sylow subgroup of $$N_G(R)$$ is strictly bigger in size that that of $$N_G(Q)$$. By the maximality assumption, $$M = N_G(R)$$ must have a normal $$p$$-complement.
 * 6) * $$L$$ has a normal $$p$$-complement: This follows from the previous statement, and the observation that $$L \le M$$.
 * 7) * $$N$$ has a normal $$p$$-complement: If $$N = G$$, then $$P \le N$$, which we're assuming is false. Thus, $$N$$ is proper in $$G$$. By the previous step, $$N$$ satisfies the conditions for the induction hypothesis, so $$N$$ has a normal $$p$$-complement.
 * 8) * The desired contradiction is achieved: The desired contradiction is achieved because by our assumption, $$N = N_G(Q)$$ does not possess a normal $$p$$-complement.
 * 9) $$N = G$$ and $$O_p(G)$$ is nontrivial. Without loss of generality, $$Q = O_p(G)$$: If $$N$$ were a proper subgroup of $$G$$, we'd have $$P \le N \le G$$ and it would possess a normal $$p$$-complement by step (2). This contradicts the assumption that $$N = N_G(Q)$$ does not have a normal $$p$$-complement. Further, since $$N_G(Q) = G$$ and $$Q$$ is nontrivial, $$Q$$ is a nontrivial normal $$p$$-subgroup. Thus, $$O_p(G)$$ is nontrivial, and we can assume without loss of generality that $$Q = O_p(G)$$.
 * 10) If $$O_p(G) = P$$, it has a normal $$p$$-complement. Thus, we may assume that $$Q = O_p(G)$$ is a proper subgroup of $$P$$: By fact (2) again, $$W(P)$$ is normal in $$G$$. Thus, $$G = N_G(W(P))$$ and the assumption that $$N_G(W(P))$$ has a normal $$p$$-complement yields that $$G$$ has a normal $$p$$-complement.

Proof of (3)
Given: $$G$$ is a group of smallest order with respect to the following: $$G$$ has a nontrivial $$p$$-Sylow subgroup such that $$N_G(W(P))$$ has a normal $$p$$-complement but $$G$$ does not.

To prove: $$G = O_{p,p',p}(G)$$.

Proof: We've already established in part (2) that $$Q = O_p(G)$$ is a proper nontrivial subgroup of $$P$$. We proceed from there.

Let $$\varphi:G \to G/Q$$ be the quotient map. Then $$\varphi(P) = P/Q$$ is $$p$$-Sylow in $$\varphi(G)$$. Let $$M_1 = \varphi^{-1}(W(P/Q))$$ and $$N_1 = \varphi^{-1}(N_{G/Q}(W(P/Q))))$$.


 * 1) $$N_1 = N_G(M_1)$$: This follows from some form of the lattice isomorphism theorem, and the fact that both subgroups contain the kernel of $$\varphi$$.
 * 2) If $$P \le T \le G$$, and $$T$$ is proper in $$G$$, $$P$$ has a normal complement in $$T$$:
 * 3) * If $$P \le T \le G$$, then $$P$$ is a $$p$$-Sylow subgroup of $$T$$, because Sylow satisfies intermediate subgroup condition. Further, $$N_T(W(P))$$ is a subgroup of $$N_G(W(P))$$ containing $$P$$. In particular, since $$P$$ has a normal complement in $$N_G(W(P))$$, it has a normal complement in $$N_T(W(P))$$, because retract satisfies intermediate subgroup condition.
 * 4) * In particular, if we assume the induction hypothesis holds for $$T$$, then we deduce that $$P$$ has a normal complement in $$T$$.
 * 5) $$N_1$$ has a normal $$p$$-complement: This is a direct consequence of the previous step, and the observation that since $$\varphi(P) \le \varphi(N_1)$$, $$P \le N_1$$.
 * 6) $$\varphi(N_1)$$ has a normal $$p$$-complement: This follows from the preceding step.
 * 7) $$\varphi(G)$$ has a normal $$p$$-complement: From the previous step, we see that $$N_{G/Q}(W(P/Q))$$ has a normal $$p$$-complement. $$\varphi(G) = G/Q$$, having order strictly smaller than that of $$G$$, satisfies the property that $$W$$ controls normal $$p$$-complements in it. Thus, $$\varphi(G)$$ has a normal $$p$$-complement.
 * 8) $$G = O_{p,p',p}(G)$$: From the previous step, $$\varphi(G) = O_{p',p}(\varphi(G))$$, yielding that $$G = O_{p,p',p}(G)$$.