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forms a basis for G. In other words,
() =+=
() +
( .
rank G
t
s
rank K
rank H
Part (2) follows from (1) by letting G = H ≈ K and letting j and y be the natural
inclusion and projection map, respectively, and the theorem is proved.
B.5.9. Theorem. Let G be a free abelian group with basis {g 1 ,g 2 ,...,g n } . If H is any
group and h 1 , h 2 ,..., h n Œ H, then there exists a unique homomorphism j :G Æ H
such that j(g i ) = h i .
Proof.
Show that any element g of G has a unique representation of the form
gkg kg
=
+
+◊◊◊+
kg
,
k
Œ
Z
11
2 2
nn
i
and define
j gkh kh
() =
+
+ ◊◊◊+
k nn
.
11
22
Free abelian groups behave very similarly to free groups. Compare Theorem B.5.9
with Theorem B.4.25. There is also an analogous commutative diagram to describe
what is going on and we again have a universal factorization property . Alternatively,
Theorem B.5.9 can be paraphrased by saying that, given a free group G and any other
group H, in order to define a homomophism from G to H it suffices to define it on a
basis of G, because any map from a basis of G to H extends uniquely to a homo-
morphism of G. The freeness of G is important because consider
G == {
Z 3
012
,,
}
and define
j 11
() Z .
The map j does not extend to a homomorphism j : Z 3 Æ Z . Of course, although 1 is
a basis for Z 3 , the group Z 3 is not a free group and so there is no contradiction.
Next, let G and H be groups.
Definition.
Let
(
) =
{
}
hom
GH
,
h h G
:
Æ
H is a homomorphism
and if h i Πhom(G,H), then define
hh GH
1
+
:
Æ
2
by
(
h hg hghg rgG
1
+
)( ) =
() +
()
Œ
.
2
1
2
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