Thompson transitivity theorem

Statement
Suppose $$G$$ is a finite group and $$p$$ is a prime number. Suppose further that $$G$$ is a fact about::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 $$q \ne p$$.

Necessity of assumptions

 * Analogue of Thompson transitivity theorem fails for abelian subgroups of rank two
 * Analogue of Thompson transitivity theorem fails for groups in which not every p-local subgroup is p-constrained

Opposite facts

 * Normal rank two Sylow subgroup for least prime divisor has normal complement

Corollaries/applications

 * Corollary of Thompson transitivity theorem

Application to groups of particular kinds
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) uses::Corollary of centralizer product theorem for rank at least three
 * 2) uses::Prime power order implies nilpotent, uses::Nilpotent implies normalizer condition
 * 3) uses::Centralizer of coprime automorphism in homomorphic image equals image of centralizer
 * 4) uses::Lemma on containment in p'-core for Thompson transitivity theorem

Proof
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 $$A$$invariant 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 $$D$$.