Abelian Groups - László Fuchs [Springer]

170 5 Purity and Basic Subgroups
Theorem 5.12 (Kovács [1]). A subgroup C of a p-group A is contained in a basic
subgroup of A if and only if C is the union of an ascending chain
0 D C 0 C 1 C n :::
of subgroups such that the heights of elements in C n .computed in A/ are bounded
for every n >0.
Proof. If the subgroup C is contained in a basic subgroup B D˚1nD1 B n,where
B n Š˚Z.p n /, then the subgroups C n D C \ .B 1 ˚˚B n / satisfy the stated
condition.
To prove sufficiency, we may assume without loss of generality that C n \p n A D 0
(we may adjoin or delete members in the chain). Consider ascending chains 0 D
G 0 G 1 G n ::: of subgroups of A subject to the following conditions:
C n G n and G n \ p n A D 0 8 n 2 N:
We introduce a partial order in the set of all such chains in the obvious manner: a
chain of the G n is than the chain of the G 0 n if G n G 0 n for each n 2 N. Then Zorn’s
lemma ensures the existence of a maximal chain which we will denote (without
danger of confusion) again by G n .
We claim that the union B D[ 1 nD1 G n for a maximal chain is a basic subgroup
of A. The main step in the proof is to show that G n is p n A-high in A. Ifwe
show this, then we can finish the proof quickly by referring to Khabbaz’s theorem
(Corollary 2.6) thatG n is a summand of A, clearly a maximal p n -bounded one, so
Szele’s theorem (Proposition 5.10) applies.
By way of contradiction, suppose that there is an a 2 A such that, for some
n >0, a … G n ; pa 2 G n satisfies hG n ; ai\p n A D 0: By the maximal choice of the
chain, there is m n C 1 with hG m ; ai \p m A ¤ 0; and it is safe to assume that
m D n C 1. Thus, g nC1 C a D p nC1 c ¤ 0 for some g nC1 2 G nC1 ; c 2 A. Hence
pg nC1 C pa D p nC2 c 2 G nC1 \ p nC1 A D 0 implies pg nC1 D pa 2 G n .Now
g nC1 … G n , for otherwise g nC1 C a D p nC1 c 2hG n ; ai\p n A D 0, a contradiction.
Then G n being maximal in G nC1 with respect to being disjoint from p n A, we argue
that there is a g n 2 G n such that g nC1 C g n D p n d ¤ 0 for some d 2 A. Butthen
a g n D p nC1 c p n d 2hG n ; ai\p n A D 0 implies a D g n 2 G n , a contradiction. ut
Corollary 5.13. Let A be the torsion part of the direct product Q i2I A i of p-groups
A i . Suppose B i D˚1nD1 B in is a basic subgroup of A i where B in Š˚Z.p n / for each
n 2 N, and for each i 2 I. Then
is a basic subgroup of A.
B D M 1
B n with B n D Y B in
nD1 i2I
Proof. For every n >0,wehaveA i D B i1 ˚˚B in ˚ A in where the last summand
has no cyclic summand of order p n . Hence we obtain A D B 1 ˚˚B n ˚A n with