Abelian Groups - László Fuchs [Springer]

5 Basic Subgroups 171
A n as the torsion part of Q i2I A in. It is clear that A n cannot have any cyclic summand
of order p n . An appeal to Proposition 5.10 concludes the proof.
ut
Quasi-Basis The existence of basic subgroups makes it possible to find a simple
generating set in p-groups (that will be used later on).
Let B denote a basic subgroup in a p-group A. We write
B D˚i2I ha i i and A=B D˚j2J C j where C j Š Z.p 1 /:
If Cj is generated by the cosets c jn .n 2 N/ such that pc j1 D 0; pc jnC1 D c jn
.n 2 N/,
then by the purity of B in A, we can pick a representative c jn 2 A of each c jn such
that o.c jn / D o.c jn /.Thec jn satisfy the relations
pc j1 D 0; pc jnC1 D c jn b jn .n 2 N/
where b jn 2 B is necessarily of order p n . We now consider the set fa i .i 2 I/,
c jn .j 2 J; n 2 N/g, called a quasi-basis of A. It satisfies:
Proposition 5.14 (Fuchs [2]). If fa i ; c jn g is a quasi-basis of the p-group A, then
every x 2 A can be written in the form
x D s 1 a i1 CCs k a ik C t 1 c j1 n 1
CCt m c jm n m
with distinct indices, where s i ; t j 2 Z and every t j is prime to p. This form is unique
for the given quasi-basis in the sense that the terms sa i and tc jn are determined by x.
Proof. Given x 2 A, we first write the coset x D x C B as
x D t 1 c j 1 n 1
CCt m c j m n m
with gcdft j ; pg D1:
Then x .t 1 c j1 n 1
CCt m c jm n m
/ 2 B is equal to a linear combination s 1 a i1 CCs k a ik .
That all this is done with uniquely determined terms is obvious.
ut
A frequently quoted corollary is the following (cp. also Theorem 3.2 in
Chapter 10).
Corollary 5.15 (Kulikov [3]). A basic subgroup B of a reduced p-group A satisfies
jAj jBj @ 0
:
Proof. Using the notation we developed for quasi-basis, we observe that if for
indices j ¤ k, the equality b jn D b kn were true for all n 2 N, then the differences
c jn c kn .n 2 N/ would generate a divisible subgroup in A. Therefore, the cardinality
of a reduced A cannot exceed the cardinality of the set of sequences fb jn g n2N which
is jBj @ 0
:
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