Talk:Signalizer functor

A should not be cyclic
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)


 * 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)