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shows that the condition that n i divides n i+1 is important for the uniqueness part of
Theorem B.5.7.
B.5.8. Theorem.
Assume that G, H, and K are finitely generated abelian groups.
(1) If j :K Æ G and y :G Æ H are homomorphisms satisfying
(a) j is injective,
(b) im(j) = ker(y), and
(c) y is surjective,
then
() =
() +
( .
rank G
rank K
rank H
(2) The function rank is “additive”, that is,
(
) =
() +
( .
rank H
K
rank H
rank K
Before proving Theorem B.5.8, we introduce some more notation.
Definition. The subset {g 1 ,g 2 ,...,g n } of an abelian group G is said to form a basis
for G provided that
(1) G = Z g 1 + Z g 2 + ···+ Z g n , and
(2) if k 1 g 1 + k 2 g 2 + ···+ k n g n = 0, for k i Œ Z , then k i g i = 0 for all i.
It is easy to show that {g 1 ,g 2 ,...,g n } is a basis for the group G if and only if
Gg g
=
ZZ
≈◊◊◊≈
Z
n
.
1
2
Thus, by Theorem B.5.7 each finitely generated group has a basis.
Definition. Any abelian group which is isomorphic to G 1 ≈G 2 ≈ ···≈G n , where each
G i is isomorphic to Z , is called a free abelian group .
It follows easily from Theorem B.5.7 and the definitions that the following con-
ditions on a finitely generated abelian group are equivalent:
(1) G is free.
(2) G is torsion-free.
(3) G has a basis consisting of elements of infinite order.
Proof of Theorem B.5.8. To prove part (1), assume without loss of generality that
all three groups G, H, and K are torsion-free. Let {h 1 ,h 2 ,...,h s } and {k 1 ,k 2 ,...,k t } be
bases for H and K, respectively, where s = rank(H) and t = rank(K). Choose elements
g 1 , g 2 ,..., and g s in G such that y(g i ) = h i . It is now easy to show that
{
jj
()()
kk
,
,...,
j
()
kgg g
t
,
,
,...,
}
1
2
1
2
s
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