# Thompson transitivity theorem

## 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 $q \ne p$.

## 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.

## 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 $A$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 $D$.

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]