Thompson transitivity theorem

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Statement

Suppose G is a finite group and p is a prime number. Suppose further that G is a Group in which every p-local subgroup is p-constrained (?): the normalizer of any non-identity p-subgroup of G is a p-constrained group.

Suppose A \in SCN_3(p). In other words, A is maximal among abelian normal subgroups inside some p-Sylow subgroup P (note that normality is in P, not in G) and further, A has rank at least three. In other words, any generating set for A must comprise at least three elements.

Then, C_G(A) permutes transitively under conjugation the set of all maximal A-invariant q-subgroups of G for any prime q \ne p.

Related facts

Necessity of assumptions

Opposite facts

Corollaries/applications

Application to groups of particular kinds

Group property Meaning Is the property stronger than being a group in which every p-local subgroup is p-constrained? Is the conclusion of the Thompson transitivity theorem true even if we drop the rank at least three requirement? If so, why?
finite nilpotent group (includes the case of finite abelian group) direct product of its Sylow subgroups, so all Sylow subgroups are normal Sylow subgroups Yes Yes, because a maximal A-invariant q-subgroup in this case is simply a q-Sylow subgroup, and it is unique because the group is nilpotent).
p-nilpotent group there is a normal p-complement, i.e., the p-Sylow subgroup is a retract Yes Unclear
finite solvable group no simple non-abelian composition factors Yes Unclear
p-solvable group has a chief series where all chief factors are either p-groups or p'-groups Yes Unclear
N-group every local subgroup is a solvable group Yes Unclear
minimal simple group simple non-abelian group in which every proper subgroup is solvable Yes Unclear
group in which every p-local subgroup is p-solvable every p-local subgroup is a p-solvable group Yes Unclear

In particular, we note that the Thompson transitivity theorem applies to minimal simple groups. This is an important ingredient in the classification of finite minimal simple groups and the odd-order theorem.

Facts used

  1. Corollary of centralizer product theorem for rank at least three
  2. Prime power order implies nilpotent, Nilpotent implies normalizer condition
  3. Centralizer of coprime automorphism in homomorphic image equals image of centralizer
  4. Lemma on containment in p'-core for Thompson transitivity theorem

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

Given: A finite group G, a prime p. Every p-local subgroup of G is p-constrained. A is maximal among abelian normal subgroups in some p-Sylow subgroup P. Also, A has rank at least three.

To prove: C_G(A) permutes transitively under conjugation the set of all maximal A-invariant q-subgroups of G for any prime q \ne p.

Proof: First, note that conjugation by any element in C_G(A) sends Ainvariant subgroups to A-invariant subgroups. In particular, it permutes the set of maximal A-invariant q-subgroups under conjugation.

Let S_1, S_2, \dots, S_t be the orbits of maximal A-invariant q-subgroups of G. We want to prove that t = 1. We break the proof into two steps.

The intersection of any two subgroups in distinct orbits is trivial

To prove: If Q_1 \in S_i, Q_2 \in S_j for i \ne j, then Q_1 \cap Q_2 is trivial.

Proof: Suppose not. Among all possible pairs Q_1,Q_2 of subgroups in distinct orbits for which the intersection is nontrivial, pick a pair such that the intersection Q_1 \cap Q_2 has largest possible order. Call the intersection D.

Step no. Assertion/construction Facts used Given data/assumptions used Previous steps used Explanation
1 D is a proper subgroup in both Q_1 and Q_2 Q_1,Q_2 are both maximal A-invariant q-subgroups. [SHOW MORE]
2 R_i = N_{Q_i}(D) properly contains D for i = 1,2 Fact (2) Q_1,Q_2 are q-groups, i.e., groups of prime power order Step (1) Fact-step-given-combination direct
3 Let \overline{R_i} = R_i/D, for i = 1,2. Then, D, R_1, R_2 are A-invariant and hence A acts on \overline{R_1},\overline{R_2}, both of which are q-groups. Q_1, Q_2 are A-invariant Steps (1), (2) [SHOW MORE]
4 There exists a non-identity element u of A such that C_{\overline{R_1}}(u) and C_{\overline{R_2}}(u) are both nontrivial. Fact (1) A is a finite abelian p-group of rank at least three, and q \ne p. Step (3) [SHOW MORE]
5 C_{R_i}(u)D/D = C_{\overline{R_i}}(u) for i = 1,2 Fact (3) q \ne p, so u is acting as a coprime automorphism. Step (3) [SHOW MORE]
6 T_i = C_{R_i}(u) is not contained in D for i = 1,2 Steps (4), (5) [SHOW MORE]
7 H = C_G(\langle u \rangle) is p-constrained By hypothesis, normalizers of non-identity p-subgroups are p-constrained. direct, noting that u \in A, so \langle u \rangle is a p-subgroup.
8 Each T_i is contained in O_{p'}(H). Fact (4) Fact-direct (need to explain).
9 Each T_i is contained in N_G(D). Steps (2), (6) [SHOW MORE]
PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

There is in fact only one orbit

References

Journal references

Textbook references

  • Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 292, Theorem 5.4, Chapter 8 (p-constrained and p-stable groups), Section 5 (The Thompson transitivity theorem), More info