Talk:Signalizer functor

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According to Kurzweil–Stellmacher page 304 line 1, A should not be cyclic. If A is cyclic, then W can be a p′-subgroup with all the right properties, but И need not have a unique maximal element (and W need not be one of the maximal elements). Basically, if A is cyclic, then it can act fixed-point-freely on elements of И, so that they trivially satisfy the balance condition.

An explicit example is A the alternating group on {1,2,3}, G the symmetric group on {1,2,3,4,5}, and θ(a) = 1. Clearly θ is the signalizer functor defined by W = 1, but И contains 2 Klein 4-groups acting regularly on {1,2,3,4} and {1,2,3,5}. In particular, И has 2 distinct maximal elements, so is not complete under the И definition. JackSchmidt (talk) 05:05, 4 January 2012 (UTC)Reply

Jack- Of course your example is nice, but the solvable sig. functor thm says that A must have at least three generators (not one) before И is guaranteed to have a unique maximal element. On pages 310-311 is an example where A has rank 2, yet θ is still not complete. Echocampfire (talk) 15:16, 6 January 2012 (UTC)Reply

The text of the introduction references a few named theorems, e.g. the "Solvable Signalizer Functor Theorem." For theorems that have specific names but are not named after authors, is the proper formatting to capitalize all the nouns in the name? Or would "solvable signalizer functor theorem" be more appropriate? MoonriseCleric (talk) 16:29, 11 August 2024 (UTC)Reply