Glauberman's theorem on intersection with the ZJ-subgroup

Statement
Suppose $$p$$ is an odd prime number, $$G$$ is a fact about::p-stable group, and $$P$$ is a $$p$$-Sylow subgroup. Suppose, further, that $$B$$ is a nontrivial normal $$p$$-subgroup of $$G$$. Then, $$B \cap Z(J(P))$$ is also a nontrivial normal $$p$$-subgroup of $$G$$.

Related facts

 * Glauberman-Thompson normal p-complement theorem
 * p-constrained and p-stable implies Glauberman type for odd p
 * Glauberman's replacement theorem

Facts used

 * 1) uses::Characteristic of normal implies normal
 * 2) uses::Normality is strongly intersection-closed
 * 3) uses::Glauberman's replacement theorem
 * 4) uses::Sylow satisfies permuting transfer condition
 * 5) uses::Frattini's argument
 * 6) uses::Any class two normal subgroup whose commutator subgroup is in the ZJ-subgroup normalizes an abelian subgroup of maximum order: This is an immediate corollary of Glauberman's replacement theorem.
 * 7) uses::Normality is centralizer-closed
 * 8) uses::Thompson's replacement theorem

Proof
Given: A $$p$$-stable group $$G$$. A nontrivial normal $$p$$-subgroup $$B$$ of $$G$$. A $$p$$-Sylow subgroup $$P$$ of $$G$$.

To prove: $$B \cap Z(J(P))$$ is a nontrivial normal $$p$$-subgroup of $$G$$.

Proof: We prove this by assuming a counterexample group $$G$$, and a nontrivial normal $$p$$-subgroup $$B$$ of $$G$$ that provides a counterexample of minimum order for that particular prime $$p$$.

For convenience, denote $$Z = Z(J(P))$$.

Note that since $$B$$ is a nontrivial normal $$p$$-subgroup, $$B \le O_p(G)$$, and in particular, $$B \le P$$. Note also that $$B \cap Z$$ is nontrivial, because $$Z$$ contains the center of $$P$$, and is thus normality-large.

First part of the proof: showing that $$B$$ satisfies the conditions for Glauberman's replacement theorem
To prove: $$B$$ is the normal closure of $$B \cap Z(J(P))$$, $$B$$ has class two, $$[B,B] \le Z(J(P))$$.

Proof:


 * 1) $$B$$ is the normal closure of $$B \cap Z$$: Suppose $$B_1$$ is the normal closure of $$B \cap Z$$. Then, $$B \cap Z \le B_1 \le B$$. $$B_1$$ is thus a nontrivial normal $$p$$-subgroup containing $$B \cap Z$$. Thus, $$B_1 \cap Z = B \cap Z$$. By minimality of the order of $$B$$ as a counterexample, we conclude that $$B_1 = B$$.
 * 2) Let $$B' = [B,B]$$ be the commutator subgroup. Then, $$B'$$ is normal in $$G$$: $$B'$$, being the commutator subgroup of $$B$$, is characteristic in $$B$$. By fact (1), we obtain that $$B'$$ is normal in $$G$$.
 * 3) $$Z \cap B'$$ is normal in $$G$$: Since $$B$$ is a $$p$$-group, $$B'$$ is a proper subgroup of $$B$$. Thus, by minimality of $$B'$$ as a counterexample, and the fact that $$B'$$ is normal in $$G$$ by the preceding step, $$Z \cap B'$$ is normal in $$G$$.
 * 4) $$[Z \cap B,B] \le Z \cap B'$$: $$Z$$ and $$B$$ are both normal in $$P$$, so by fact (2), $$Z \cap B$$ is also normal in $$P$$. Thus, $$[Z \cap B,B] \le [Z \cap B, P] \le Z \cap B$$. Combining this with the obvious fact that $$[Z \cap B, B] \le [B,B] = B'$$, we get $$[Z \cap B,B] \le Z \cap B'$$.
 * 5) For any conjugate $$L$$ of $$Z \cap B$$, $$[L,B] \le Z \cap B'$$: Suppose $$L = g(Z \cap B)g^{-1}$$. Since $$B$$ is normal, $$B = gBg^{-1}$$. Thus, $$[L,B] = [g(Z \cap B)g^{-1},gBg^{-1}] = g[Z \cap B,B]g^{-1} \le g(Z \cap B')g^{-1}$$. By step (3), $$Z \cap B'$$ is normal, so $$[L,B] \le Z \cap B'$$.
 * 6) $$B' \le Z$$: Since $$B$$ is generated by all the conjugates of $$Z \cap B$$ (step (1)), $$B'$$ is the normal closure in $$B$$ of the subgroups generated by $$[L,B]$$ for all conjugates $$L$$ of $$Z \cap B$$. But each of these is contained in $$Z \cap B'$$, which is normal in $$G$$ and hence in $$B$$. Thus, $$B' \le Z \cap B'$$, and in particular, $$B' \le Z$$.
 * 7) $$[Z \cap B,B']$$ is trivial: This is because both are subgroups of the abelian group $$Z$$.
 * 8) $$[L,B']$$ is trivial for every conjugate $$L$$ of $$Z \cap B$$ in $$G$$: This follows by reasoning similar to step (5), using that $$B'$$ is normal in $$G$$ (step (2)).
 * 9) $$C_B(B') = B$$, so $$B' \le Z(B)$$ and $$B$$ has class two: By the previous step, $$C_B(B')$$ contains every conjugate of $$Z \cap B$$. By step (1), $$B$$ is generated by these conjugates, so $$C_B(B') = B$$. In particular, $$B' \le Z(B)$$, so $$B$$ has class two.

Second part of the proof: the normal core of normalizer of $$Z \cap B$$ does not contain $$J(P)$$
Given: Let $$L$$ be the largest normal subgroup contained in $$N_G(Z \cap B)$$. In other words, $$L$$ is the normal core of normalizer of $$Z \cap B$$.

To prove: $$L$$ does not contain $$J(P)$$.

Proof: We prove that if $$L$$ contains $$J(P)$$, then $$B \cap Z$$ is normal in $$G$$, a contradiction.


 * 1) $$P \cap L$$ is a $$p$$-Sylow subgroup of $$L$$: This follows from fact (4).
 * 2) $$G = LN_G(P \cap L)$$: This follows from fact (5), the previous step, and the fact that $$L$$ is normal.
 * 3) $$J(P) = J(P \cap L)$$: If $$J(P) \le L$$, then $$J(P) \le P \cap L$$, so by the definition of $$J$$, $$J(P)$$ = J(P \cap L).
 * 4) $$N_G(P \cap L) \le N_G(Z)$$: From the previous step, $$Z = Z(J(P)) = Z(J(P \cap L))$$, so $$Z$$ is characteristic in $$P \cap L$$. Thus, by fact (1), $$Z)$$ is normal in $$N_G(P \cap L)$$, so $$N_G(P \cap L)$$ is contained in the normalizer $$N_G(Z)$$.
 * 5) $$G = LN_G(Z)$$: This follows from steps (2) and (4).
 * 6) $$Z \cap B$$ is normal in $$G$$: $$L$$ normalizes $$Z \cap B$$ by assumption. $$N_G(Z)$$ normalizes $$Z$$, and since $$B$$ is normal, $$N_G(Z)$$ normalizes $$Z \cap B$$. Thus, $$N_G(Z \cap B)$$ contains both $$L$$ and $$N_G(Z)$$. The previous step thus forces $$N_G(Z \cap B) = Z$$.

Meat of the proof
We continue to denote by $$L$$ the normal core of normalizer of $$B \cap Z$$.


 * 1) By fact (6), there exists an abelian subgroup $$A$$ of maximum order in $$P$$ such that $$B$$ normalizes $$A$$. In particular, $$[[B,A],A] = 1$$: In the first part, we established that $$B$$ satisfies the conditions for fact (6): $$B$$ is class two normal and $$[B,B] \le Z(J(P))$$. Thus, fact (6) yields that there exists an abelian subgroup $$A$$ of maximum order in $$P$$ such that $$B$$ normalizes $$A$$. Since $$B$$ normalizes $$A$$, $$[B,A] \le A$$. Further, since $$A$$ is abelian, $$[A,A] = 1$$, so $$[[B,A],A] = 1$$.
 * 2) Let $$C = C_G(B)$$. Then, $$AC/C \le O_p(G/C)$$: Both $$A$$ and $$B$$ are $$p$$-subgroups. Since $$B$$ is normal in $$G$$, so is $$O_{p'}(G)B$$. Further, $$A \le N_G(B)$$, again since $$B$$ is normal, and $$[[B,A],A] = 1$$ as established in the previous step. Thus, $$G$$ being $$p$$-stable yields $$AC_G(B)/C_G(B) \le O_p(N_G(B)/C_G(B))$$. Using $$C = C_G(B)$$ and $$G = N_G(B)$$ yields the result.
 * 3) $$C \le L$$: Since $$C$$ centralizes $$B$$, $$C$$ centralizes and thus normalizes $$B \cap Z$$. Also, by fact (7) and the given datum that $$B$$ is normal, $$C$$ is normal. By the fact that $$L$$ is the unique largest normal subgroup that normalizes $$B \cap Z$$, we get that $$L$$ is normal.
 * 4) $$AL/L \le O_p(G/L)$$: This is a direct consequence of the previous two steps.
 * 5) $$O_p(G/L)$$ is trivial:
 * 6) $$A \le L$$: This is a direct consequence of the previous two steps.
 * 7) $$J(P \cap L)$$ is contained in $$J(P)$$: $$P \cap L$$ is a subgroup of $$P$$ that, by the preceding step, has an abelian subgroup of maximum order in $$P$$. Thus, the maximum possible order of an abelian subgroup in $$P \cap L$$ is the same as in $$P$$. Thus, the subgroups generating $$J(P \cap L)$$ form a subset of the subgroups generating $$J(P)$$, so $$J(P \cap L)$$ is contained in $$J(P)$$.
 * 8) $$B$$ is abelian:
 * 9) There exists an abelian subgroup $$A_1$$ of $$P$$ of maximum order that is not contained in $$P \cap L$$: This follows from the second part of the proof, which showed that $$J(P)$$ is not contained in $$P \cap L$$.
 * 10) $$[[B,A_1],A_1] \ne 1$$: If $$[[B,A_1],A_1]$$ were trivial, then replacing $$A$$ by $$A_1$$ in steps (2)-(6) of the above argument would yields $$A_1 \le L$$ contradicting the previous step.
 * 11) Among the set of possibilities for $$A_1$$, pick $$A_1$$ such that $$A_1 \cap B$$ has maximum order. Then, there exists $$A^*$$ such that $$A^*$$ normalizes $$A_1$$ and $$A_1 \cap B$$ is properly contained in $$A^* \cap B$$: This follows from fact (8) (Thompson's replacement theorem).
 * 12) $$A^* \le L$$: This follows from the maximality of $$A_1 \cap B$$ among the subgroups of maximum order in $$P$$ that are not contained in $$L$$.