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.
View similar facts

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 $G$.

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. Characteristic of normal implies normal
2. Normality is strongly intersection-closed
3. Glauberman's replacement theorem
4. Sylow satisfies permuting transfer condition
5. Frattini's argument
6. 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. Normality is centralizer-closed
8. 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, $B\leq O_{p}(G)$, and in particular, $B\leq P$. 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, $[B,B]\leq Z(J(P))$.

Proof:

1. $B$ is the normal closure of $B\cap Z$: Suppose $B_{1}$ is the normal closure of $B\cap Z$. Then, $B\cap Z\leq B_{1}\leq B$. $B_{1}$ 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. $[Z\cap B,B]\leq Z\cap B'$: $Z$ and $B$ are both normal in $P$, so by fact (2), $Z\cap B$ is also normal in $P$. Thus, $[Z\cap B,B]\leq [Z\cap B,P]\leq Z\cap B$. Combining this with the obvious fact that $[Z\cap B,B]\leq [B,B]=B'$, we get $[Z\cap B,B]\leq Z\cap B'$.
5. For any conjugate $L$ of $Z\cap B$, $[L,B]\leq Z\cap B'$: Suppose $L=g(Z\cap B)g^{-1}$. Since $B$ is normal, $B=gBg^{-1}$. Thus, $[L,B]=[g(Z\cap B)g^{-1},gBg^{-1}]=g[Z\cap B,B]g^{-1}\leq g(Z\cap B')g^{-1}$. By step (3), $Z\cap B'$ is normal, so $[L,B]\leq Z\cap B'$.
6. $B'\leq Z$: 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, $B'\leq Z\cap B'$, and in particular, $B'\leq Z$.
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 $B'\leq Z(B)$ 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, $B'\leq Z(B)$, 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 $J(P)\leq L$, then $J(P)\leq P\cap L$, so by the definition of $J$, $J(P)$ = J(P \cap L)</math>.
4. $N_{G}(P\cap L)\leq N_{G}(Z)$: 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 $[B,B]\leq Z(J(P))$. Thus, fact (6) yields that there exists an abelian subgroup $A$ of maximum order in $P$ such that $B$ normalizes $A$. Since $B$ normalizes $A$, $[B,A]\leq A$. Further, since $A$ is abelian, $[A,A]=1$, so $[[B,A],A]=1$.
2. Let $C=C_{G}(B)$. Then, $AC/C\leq O_{p}(G/C)$: Both $A$ and $B$ are $p$-subgroups. Since $B$ is normal in $G$, so is $O_{p'}(G)B$. Further, $A\leq N_{G}(B)$, again since $B$ is normal, and $[[B,A],A]=1$ as established in the previous step. Thus, $G$ being $p$-stable yields $AC_{G}(B)/C_{G}(B)\leq O_{p}(N_{G}(B)/C_{G}(B))$. Using $C=C_{G}(B)$ and $G=N_{G}(B)$ yields the result.
3. $C\leq L$: 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. $AL/L\leq O_{p}(G/L)$: 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]
   One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.
6. $A\leq L$: 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]
   One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.
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. $[[B,A_{1}],A_{1}]\neq 1$: If $[[B,A_{1}],A_{1}]$ were trivial, then replacing $A$ by $A_{1}$ in steps (2)-(6) of the above argument would yields $A_{1}\leq L$ 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. $A^{*}\leq L$: This follows from the maximality of $A_{1}\cap B$ among the subgroups of maximum order in $P$ that are not contained in $L$.

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.

References

Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 278, Theorem 2.10, Chapter 8 (p-constrained and p-stable groups), Section 8.2 (Glauberman's theorem), ^{More info}