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Theorem17.1 [Fuchs]: A p-group A is a direct sum of cyclic groups if and only if A is the union of an ascending chain of subgroups (1) A1 A 2 ... A n ... , A n A , n 1 Such that the heights of elements 0 of A n are less than a finite bound k n . Proof: Necessity. If A is a direct sum of cyclic groups, then in a decomposition, collect the cyclic summands of the same order p n , for every n, and denote their direct sum by B n . Clearly, A n B1 B2 ... Bn satisfy the conditions with k n =n. Sufficiency. For the proof of the sufficiency we need the following lemmas. Lemma1. Assume that a p-group A is the union of an ascending chain of subgroups (1) A1 A 2 ... A n ... , A n A , Such that the heights of elements 0 n 1 of A n are less than a finite bound k n . Then there exists an ascending chain G1 G 2 ... G n ... such that (i) A n G n , (ii) G n p n A 0 and (iii) G n is maximal with respect to condition (ii). For every n. Proof. There is no loss of generality if we assume that k n = n, that is, A n p n A 0 For every n (we may adjoin 0’s to the beginning of the chain A1 A 2 ... A n ... and repeat some terms A n [a finite number of times] to reach the above assumption). Consider all chains C1 C2 ... Cn ... of subgroups of A such that (i) A n C n and (ii) C n p n A 0 for all n. Define : the chain of C n less than or equal to the chain of B n if Cn Bn for each n. Then the set of all chains with (i) and (ii) is inductive ( satisfying the conditions of Zorn’s lemma) and Zorn’s lemma applies to conclude the existence of a chain G1 G 2 ... G n ... satisfying (i) and (ii) which is maximal in our present sense. Lemma2. Assume that G1 G 2 ... G n ... is an ascending chain of subgroups of a p-group A such that (i) A n G n , (ii) G n p n A 0 and (iii) G n is maximal with respect to condition (ii). For every n. For every n select a maximal independent set L n of elements in the subgroup G n [p] p n 1 A , and define L L n . Then n <L>=A[p]. Proof. It is clear that <L> A[p] , since by definition each element of L is of order p. To prove A[p] <L>. Let a A[p] ( note that a A G n ). If a G1 , then n 1 a G[p] a p A , that is , a L1 L . Assume that a G r implies a L is true, and let b G r 1[p] \ G r . Then b G r implies G r , b p r A 0 ( since G r is maximal with the condition G r p r A 0 ) , let 0 g kb c p r A , where g G r and k=1 may be assumed (for otherwise we multiply by s with ks 1 mod p). Now c G r 1 and h (c) r (c p r A) , thus pc G r 1 p r 1A 0 (by (ii)), hence o(c)=p and so c G r 1[p] c p r A which implies c L r 1 L . On the other hand g G r and pg=pc-pb=0 implies g A[ p] , by the induction assumption g L , hence b g c L . 0 Therefore A[p]=<L>. Lemma3. Let A be a group having the conditions of Lemma1, and let G1 G 2 ... G n ... be the ascending chain of Lemma1 . Let L be the set of Lemma2 . For every c i L with mi h(c i ) choose an a i A satisfying p mi a i c i . Then A ..., a i , ... a i A . Proof. Assume we have proved that every a A of order p n belongs to A ( note that, for n=1 the above assumption is true, since in Lemma2 we have proved A[p] L and L A (by definition of A )). Let b A be of order p n 1 with n 1 . Then p n b A[p] L (by Lemma2). So p n b m1c1 ... m k c k with some c i L . Let c1 , ... , c j be of height n and c j1 , ..., c k of height n 1 . If we write m i c i p n mi a i for i=1, …, j, then we have p n (b m1 a 1 ... mj a j ) m j1c j1 ... m k c k G n 1 . Condition (ii) of Lemma1 implies b m1 a 1 ... m j a j is of order p n , thus in A , and consequently, b A . Proof of the theorem(sufficiency) . Since A is a direct sum of cyclic groups, by Lemmas1 ,2 and 3 A= A is a direct sum of cyclic groups