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  "body": "\n## Statement\n\nGiven $\\theta,$ 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.\n\nThe 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.\n\nSeveral 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.\n\nThe 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:\n\n- $H\\cap C_G(a) \\subseteq \\theta(a)$ for all nonidentity $a \\in A.$\n\nFor example, the subgroups $\\theta(a)$ belong to И as a result of the balance condition of *θ*.\n\nThe 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.**\n\nThus, 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.\n",
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