By Karel Dekimpe

Ranging from simple wisdom of nilpotent (Lie) teams, an algebraic idea of almost-Bieberbach teams, the basic teams of infra-nilmanifolds, is constructed. those are a traditional generalization of the well-known Bieberbach teams and lots of effects approximately usual Bieberbach teams end up to generalize to the almost-Bieberbach teams. in addition, utilizing affine representations, particular cohomology computations will be performed, or leading to a class of the almost-Bieberbach teams in low dimensions. the idea that of a polynomial constitution, an alternate for the affine constructions that usually fail, is brought.

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**Extra info for Almost-Bieberbach Groups: Affine and Polynomial Structures**

**Sample text**

1 Let r be any polycyclic-by-finite group, then r has a unique maximal finite normal subgroup. Proofi Let E1 C_ E2 C_ E3 C . . Fin C_ E,~+I C_ . . be any ascending chain of finite normal subgroups of F. This chain is necessarilly finite, since every ascending chain of subgroups of r is finite (see [59]). Therefore we can choose a finite normal subgroup H o f t which is maximal among all finite normal subgroups. Now it is easy to see that H is unique, for if K was another such a normal subgroup, then H .

Now it is easy to see that H is unique, for if K was another such a normal subgroup, then H . K would contradict the maximality of H . 2 For a polycyclic-by-fmite group r , we will denote its maximal normal finite subgroup by F ( r ) . 3 In general, F ( r ) is not maximal among all finite subgroups of r . For example, when r = Z :~Z2, where Z2 acts non trivially on Z, one easily sees that F ( r ) = 1, while r has subgroups of order 2. 40 Chapter 3: Algebraic characterizations of AC-groups D e f i n i t i o n 3 .

4 If F is any polycyclic-by-finite group, then C r ( F i t t (Y)) c Fitt (Y). Proof: Y fits in a short exact sequence 1 --~ Fitt (F) ---, Y ~ Q ~ 1 where Q is abelian-by-finite. Let A denote an abelian normal subgroup of finite index in Q. Consider the group F' = p - l ( A ) which is solvable. Then, since F i t t ( r ' ) = F i t t ( r ) , we see that C r , ( F i t t ( r ) ) C_ F i t t ( r ) (using the exercise of [59]). 4 The closure of the Fitting subgroup 43 A = {1}. So, C r ( F i t t ( F ) ) .