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 $q\neq p$.

Related facts

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

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 $q\neq p$.

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 $i\neq j$, 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] Since both $Q_{1}$ and $Q_{2}$ are maximal with respect to the property of being $A$-invariant $q$-subgroups, neither can be contained in the other. Thus, their intersection is proper in both.
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] $D$ is $A$-invariant because it is the intersection of two $A$-invariant subgroups. $R_{1}$ is an $A$-invariant subgroup because it is defined purely in terms of the $A$-invariant subgroups $D$ and $Q_{1}$. Similarly for $R_{2}$. Thus, $A$ has a well-defined action on the two quotients.
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 $q\neq p$.	Step (3)	[SHOW MORE] Apply Fact (1), noting that ${\overline {R_{1}}},{\overline {R_{2}}}$ are both finite $q$-groups with $A$ acting on them. Note that the notation of Fact (1) differs from the notation here: the $q$ of Fact (1) is what we call $p$ here, and the $r,s$ of Fact (1) are both equal to what we call $q$ here.
5	$C_{R_{i}}(u)D/D=C_{\overline {R_{i}}}(u)$ for $i=1,2$	Fact (3)	$q\neq p$, so $u$ is acting as a coprime automorphism.	Step (3)	[SHOW MORE] This follows from fact (3), applied to the homomorphism $R_{i}\to R_{i}/D$ and the automorphism given by conjugation by $u$.
6	$T_{i}=C_{R_{i}}(u)$ is not contained in $D$ for $i=1,2$			Steps (4), (5)	[SHOW MORE] The image of $C_{R_{i}}(u)$ modulo $D$ equals $C_{\overline {R_{i}}}(u)$, which is nontrivial.
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] $T_{i}$ by definition is contained in $R_{i}=N_{Q_{i}}(D)$, which in turn is contained in $N_{G}(D)$.

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References

Journal references

• Solvability of groups of odd order by Walter Feit and John Griggs Thompson, Pacific Journal of Mathematics, Volume 13, Page 775 - 1029(Year 1963): This 255-page long paper gives a proof that odd-order implies solvable: any odd-order group (i.e., any finite group whose order is odd) is a solvable group.^{Project Euclid pageMore info}

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}