Solvable signalizer functor theorem

Statement

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 weaker versions of the theorem were proven before Glauberman's proof was published. Gorenstein proved it under the stronger assumption that $A$ had rank at least 5. David Goldschmidt proved it 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.

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:

• $H\cap C_G(a) \subseteq \theta(a)$ for all nonidentity $a \in A.$

For example, the subgroups $\theta(a)$ belong to И as a result of the balance condition of θ.

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.

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.