By Glenn O. E.
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Extra resources for Covariants of Binary Modular Groups (1919)(en)(4s)
Example text
Conversely, suppose F is almost torsion free. It is easy to see that Fitt (F) has to be torsion free, since the elements of finite order of Fitt (F) 44 Chapter 3: Algebraic characterizations of A C-groups form a characteristic subgroup of Fitt (F). So we can consider the exact sequence l~Fitt(F)~P~Q---, 1 where Fitt ( r ) can play the role of N in the discussion above. First we show that no element q of infinite order in Q can be lifted to q E E in such a way that the group generated by F i t t ( P ) and q is nilpotent.
Be a torsion free filtration of F. Then, there exists a representation p : F --~ TI(R K) which is of canonical type with respect to F . Moreover, any two such representations p, p~ : F -~ 7-t(R K) are conjugate to each other inside 7-l(RK). Also, for each canonical type representation p : F -+ 7-l(R K) we have: 1. F acts properly discontinuously on ]~K, via p and with compact quotient. 2. her p =the maximal finite normal subgroup of F, so p is effective if] F is almost torsion free. 3. W / c F : p ( 7 ) : S K -~ R K : ~ ~ (h:(~), h ~ ( x ) , .
Of course, Fitt (F) is also almost torsion free. Because F ( F i t t (F)) is a characteristic subgroup of Fitt (F) and so a normal subgroup of F, which implies that F ( F i t t (F)) = 1. We summarize this in the following theorem. 3 Let F be any polycyclic-by-finite group, then F is almost torsion free F is (almost-crystalIographic)-by-crystallographic. Moreover, in case F is almost torsion free, the intended almost-crystallographic subgroup can be taken as Fitt (F). In [59] one can find, as an exercise, the following property: If F is a solvable group, then C r ( F i t t (F)) C_ Fitt (F).