# Glauberman's theorem on intersection with the ZJ-subgroup

This article states a fact about the behavior of a finite group relative to a prime number. This fact is true only for odd primes, i.e., it breaks down for the prime two.
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## Statement

Suppose $p$ is an odd prime number, $G$ is a 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 $G$.

## 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)[/itex].
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: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
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: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
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 $L$.