By Leonid Kurdachenko, Javier Otal, Igor Ya Subbotin

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“The conception of modules over workforce earrings RG for endless teams G over arbitrary jewelry R is a really vast and complicated box of study with various scattered effects. … because a number of the effects seem for the 1st time in a e-book it may be prompt warmly to any professional during this box, but additionally for graduate scholars who're offered the wonderful thing about the interaction of the theories of teams, jewelry and representations.” (G. Kowol, Monatshefte für Mathematik, Vol. 152 (4), December, 2007)

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In particular, any direct product of ﬁnite groups and abelian groups is an F C-group. Moreover, let G = Crλ∈Λ Gλ be the Cartesian product. Put Cdrλ∈Λ Gλ = {(gλ )λ∈Λ | gλ ∈ ζ(Gλ ) for all λ ∈ Λ\Λg where Λg is ﬁnite}. The group Cdrλ∈Λ Gλ is called the central direct product of the groups Gλ , λ ∈ Λ. Clearly, a central direct product of ﬁnite and abelian groups is an F C-group. Central-by-ﬁnite groups and groups with ﬁnite derived subgroups (ﬁnite-by-abelian groups) deliver other examples of F C-groups.

Since R is commutative, J = AnnR (a) is a two-sided ideal of R, and then R/J becomes an artinian ring. It follows that R/J is also noetherian (see R. 44]). Hence aR is simultaneously artinian and noetherian, and therefore it has a ﬁnite composition series. In particular, the upper socular series of aR is ﬁnite. 9 that the socle height of A is at most ω. Since A is artinian, A itself is the last term of the upper socle series. Hence A has an ascending series of submodules 0 = A0 ≤ A1 ≤ · · · ≤ Am = A whose factors are simple.

46 Chapter 5. Reduction to subgroups of ﬁnite index Proof. Suppose the contrary: that is, A does not satisfy Min-RH. Then the family M of all RG-submodules B of A that does not satisfy Min-RH is not empty. Since A is an artinian RG-module, M has a minimal element M . Let S be the set of all non-empty ﬁnite subsets X of G with the following property: If {Cn | n ∈ N } is a strictly inﬁnite descending chain of RH-submodules of M , then M = Cn X for all n ∈ N. Let T be a transversal to H in G. Then G = HT .