Glauberman's theorem on intersection with the ZJ-subgroup

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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.

Related facts

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 \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)</math>.
  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.
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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