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290 U. Bunke, T. Schick, M. Spitzweck, and A. ThomFor the last statement we consider finite abelian groups as topological groups withthe discrete topology. Note that the homomorphisms G ! F.i/ are surjective for alli 2 I . We call the system F 2 Ab I an inverse system.Lemma 4.35 ([HR63]). The following assertions on a topological abelian group Gare equivalent:1. G is compact and totally disconnected.2. Every neighbourhood U G of the identity contains a compact subgroup Ksuch that G=K is a finite abelian group.3. G is profinite.Lemma 4.36. Let G be a profinite abelian group and n 2 Z. We define the groups K; Qas the kernel and cokernel of the multiplication map by n, i.e. by the exact sequenceThen K and Q are again profinite.0 ! K ! G n ! G ! Q ! 0: (34)Proof. We write G WD lim j 2J G j for an inverse system .G j / j 2J of finite abelian groups.We define the system of finite subgroups .K j / j 2J by the sequences0 ! K j ! G jn! Gj :Since taking kernels commutes with limits the natural projections K ! K j , j 2 Jrepresent K as the limit K Š lim j 2J K j .Since cokernels do not commute with limits we will use a different argument for Q.Since G is compact and the multiplication by n is continuous, nG G is a closedsubgroup. Therefore the group theoretic quotient Q is a topological group in thequotient topology.A quotient of a profinite group is again profinite [HM98], Exercise E.1.13. Here isa solution of this exercise, using the following general structural result about compactabelian groups.Lemma 4.37 ([HR63]). If H is a compact abelian group, then for every open neighbourhood1 2 U H there exists a compact subgroup C U such that H=C ŠT a F for some a 2 N 0 and a finite group F .Since G is compact, its quotient Q is compact, too. This lemma in particularimplies that Q is the limit of the system of these quotients Q=C . In our case, since Gis profinite, it can not have T a as a quotient, i.e given a surjectionG ! Q ! T a Fwe conclude that a D 0. Hence we can write Q as a limit of an inverse system of finitequotients. This implies that Q is profinite.