Signalizer functor

In mathematics, a signalizer functor gives the intersections of a potential subgroup of a finite group with the centralizers of nontrivial elements of an abelian group. The signalizer functor theorem gives conditions under which a signalizer functor comes from a subgroup. The idea is to try to construct a $$p'$$-subgroup of a finite group $$G$$, which has a good chance of being normal in $$G$$, by taking as generators certain $$p'$$-subgroups of the centralizers of nonidentity elements in one or several given noncyclic elementary abelian $$p$$-subgroups of $$G.$$ The technique has origins in the Feit–Thompson theorem, and was subsequently developed by many people including Daniel Gorenstein who defined signalizer functors, George Glauberman who proved the Solvable Signalizer Functor Theorem for solvable groups, and Patrick McBride who proved it for all groups. This theorem is needed to prove the so-called "dichotomy" stating that a given nonabelian finite simple group either has local characteristic two, or is of component type. It thus plays a major role in the classification of finite simple groups.

Definition
Let A be a noncyclic elementary abelian p-subgroup of the finite group G. An A-signalizer functor on G or simply a signalizer functor when A and G are clear is a mapping θ from the set of nonidentity elements of A to the set of A-invariant p′-subgroups of G satisfying the following properties:
 * For every nonidentity $$a\in A$$, the group $$\theta(a) $$ is contained in $$C_G(a).$$
 * For every nonidentity $$a, b\in A$$, we have $$\theta(a) \cap C_G(b) \subseteq \theta(b).$$

The second condition above is called the balance condition. If the subgroups $$\theta(a)$$ are all solvable, then the signalizer functor $$\theta$$ itself is said to be solvable.

Solvable signalizer functor theorem
Given $$\theta,$$ certain additional, relatively mild, assumptions allow one to prove that the subgroup $$W= \langle \theta(a) \mid a \in A, a \neq 1\rangle$$ of $$G$$ generated by the subgroups $$\theta(a)$$ is in fact a $$p'$$-subgroup. The Solvable Signalizer Functor Theorem proved by Glauberman states that this will be the case if $$\theta$$ is solvable and $$A$$ has at least three generators. The theorem also states that under these assumptions, $$W$$ itself will be solvable.

Several earlier versions of the theorem were proven: Gorenstein proved this under the stronger assumption that $$A$$ had rank at least 5. David Goldschmidt proved this under the assumption that $$A$$ had rank at least 4 or was a 2-group of rank at least 3. Helmut Bender gave a simple proof for 2-groups using the ZJ theorem, and Paul Flavell gave a proof in a similar spirit for all primes. Glauberman gave the definitive result for solvable signalizer functors. Using the classification of finite simple groups, McBride showed that $$W$$ is a $$p'$$-group without the assumption that $$\theta$$ is solvable.

Completeness
The terminology of completeness is often used in discussions of signalizer functors. Let $$\theta$$ be a signalizer functor as above, and consider the set И of all $$A$$-invariant $$p'$$-subgroups $$H$$ of $$G$$ satisfying the following condition: For example, the subgroups $$\theta(a)$$ belong to И by the balance condition. The signalizer functor $$\theta$$ is said to be complete if И has a unique maximal element when ordered by containment. In this case, the unique maximal element can be shown to coincide with $$W$$ above, and $$W$$ is called the completion of $$\theta$$. If $$\theta$$ is complete, and $$W$$ turns out to be solvable, then $$\theta$$ is said to be solvably complete.
 * $$H\cap C_G(a) \subseteq \theta(a)$$ for all nonidentity $$a \in A.$$

Thus, the Solvable Signalizer Functor Theorem can be rephrased by saying that if $$A$$ has at least three generators, then every solvable $$A$$-signalizer functor on $$G$$ is solvably complete.

Examples of signalizer functors
The easiest way to obtain a signalizer functor is to start with an $$A$$-invariant $$p'$$-subgroup $$M$$ of $$G,$$ and define $$\theta(a) = M\cap C_G(a)$$ for all nonidentity $$a \in A.$$ In practice, however, one begins with $$\theta$$ and uses it to construct the $$A$$-invariant $$p'$$-group.

The simplest signalizer functor used in practice is this:

$$\theta(a) = O_{p'}(C_G(a)).$$

A few words of caution are needed here. First, note that $$\theta(a)$$ as defined above is indeed an $$A$$-invariant $$p'$$-subgroup of $$G$$ because $$A$$ is abelian. However, some additional assumptions are needed to show that this $$\theta$$ satisfies the balance condition. One sufficient criterion is that for each nonidentity $$a \in A,$$ the group $$C_G(a)$$ is solvable (or $$p$$-solvable or even $$p$$-constrained). Verifying the balance condition for this $$\theta$$ under this assumption requires a famous lemma, known as Thompson's $$P\times Q$$-lemma.

Coprime action
To obtain a better understanding of signalizer functors, it is essential to know the following general fact about finite groups: $$X = \langle C_X(E_0) \mid E_0 \subseteq E, \text{ and } E/E_0 \text{ cyclic }\rangle$$
 * Let $$E$$ be an abelian noncyclic group acting on the finite group $$X.$$ Assume that the orders of $$E$$ and $$X$$ are relatively prime. Then

To prove this fact, one uses the Schur–Zassenhaus theorem to show that for each prime $$q$$ dividing the order of $$X,$$ the group $$X$$ has an $$E$$-invariant Sylow $$q$$-subgroup. This reduces to the case where $$X$$ is a $$q$$-group. Then an argument by induction on the order of $$X$$ reduces the statement further to the case where $$X$$ is elementary abelian with $$E$$ acting irreducibly. This forces the group $$E/C_E(X)$$ to be cyclic, and the result follows. See either of the books or  for details.

This is used in both the proof and applications of the Solvable Signalizer Functor Theorem. To begin, notice that it quickly implies the claim that if $$\theta$$ is complete, then its completion is the group $$W$$ defined above.

Normal completion
The completion of a signalizer functor has a "good chance" of being normal in $$G,$$ according to the top of the article. Here, the coprime action fact will be used to motivate this claim. Let $$\theta$$ be a complete $$A$$-signalizer functor on $$G$$

Let $$B$$ be a noncyclic subgroup of $$A.$$ Then the coprime action fact together with the balance condition imply that $$W= \langle \theta(a) \mid a \in A, a \neq 1\rangle = \langle \theta(b) \mid b \in  B, b \neq 1\rangle $$.

To see this, observe that because $$\theta(a)$$ is B-invariant, we have

$$\theta(a) = \langle \theta(a) \cap C_G(b) \mid b \in B, b \neq 1\rangle \subseteq \langle \theta(b) \mid b \in B, b \neq 1\rangle.$$

The equality above uses the coprime action fact, and the containment uses the balance condition. Now, it is often the case that $$\theta$$ satisfies an "equivariance" condition, namely that for each $$g \in G$$ and nonidentity $$a \in A$$

$$\theta(a^g) = \theta(a)^g. \, $$

The superscript denotes conjugation by $$g.$$ For example, the mapping $$a \mapsto O_{p'}(C_G(a))$$, which is often a signalizer functor, satisfies this condition. If $$\theta$$ satisfies equivariance, then the normalizer of $$B$$ will normalize $$W.$$ It follows that if $$G$$ is generated by the normalizers of the noncyclic subgroups of $$A,$$ then the completion of $$\theta$$ (i.e. W) is normal in $$G.$$