Graphics Reference
In-Depth Information
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