# 

```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
&lt;L&gt;=A[p].
Proof. It is clear that &lt;L&gt;  A[p] , since by definition each element of L is of order p.
To prove A[p]  &lt;L&gt;. 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]=&lt;L&gt;.
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 j1 , ..., c k of height  n  1 . If we write m i c i  p n mi a i for i=1, …, j, then we have
p n (b  m1 a 1  ...  mj a j )  m j1c j1  ...  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
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