Ex.1 Let G be a group, H a subgroup, D a divisible group. Let f be a homomorphism
of G into D. Show that f can be extended to a homomorphism of G into D .
Proof: First we will need the following lemma.
Lemma. If G, H, D, and f are as assumed above and if x  G \ H , A is the subgroup
of G generated by H and x, then f can be extended to a homomorphism of A into D.
Proof of the lemma: By assumption A  h  nx h  H, n  . There are two cases,
Case1. nx  H, for some n   , let m be the smallest positive integer such that
mx  H , let f (mx )  d . Since D is divisible, then d  md1 for some d 1  D . Define
a mapping g from A into D by g(h  nx )  f (h )  nd 1 , an easy check shows that g is a
well defined homomorphism of A into D, and it is an extension of f.
Case2. nx  H for all 0  n   , then A  H  x , let d 1 be any fixed element in D
and define g from A into D by g(h  nx )  f (h )  nd 1 , since A is a direct sum, and
hence the representation of any element in the form h+nx is unique, g is well defined.
On the other hand g is a homomorphism ,( since by definition of g
g((h 1  n 1 x )  (h 2  n 2 x ))  g((h 1  h 2 )  (n 1  n 2 ) x )  f (h 1  h 2 )  (n 1  n 2 )d 1 , and
since f is a homomorphism, = f (h 1 )  n 1d 1  f (h 2 )  n 2 d 1 ,again by definition of g, =
g(h 1  n 1 x )  g(h 2  n 2 x ) ). Note also, h  nx  H if and only if n=0, hence g(h)=f(h)
for each h in H, that is, g is an extension of f.
Now we continue the proof of the exercise. Let  be the family of all pairs (C i , h i )
where C i is a subgroup of G containing H and h i is a homomorphism of C i into D
extension of f.  is not empty, since the pair ( H, f ) is an element of it.  can be
ordered in this way; (C i , h i )  ( C j , h j ) if C i is subset of C j and h j is an extension of
h i . If  is a chain in  , then the union of all C i 's in  is a subgroup of G, say C,
containing H and h : C  D , defined in the following way: x  C means x  C i for
some C i in  , we define h ( x )  h i ( x ) . It is not difficult to verify that h is a
homomorphism ( since each h i is a homomorphism) and it is an extension of each h i ,
hence of f. Therefore (C, h) is an upper bound of  in  , thus, Zorn's lemma can be
applied on  and a maximal element exists, say ( K, k) . It is claimed that K=G and
k is the desired extension, if not (that is K  G) assume that x  G \ K , then by the
above lemma k can be extended to the subgroup generated by K and x , which
contradicts the maximality of ( K, k) . This ends the proof.
Ex.2 Let G be a group, H a subgroup, and suppose that the identity homomorphism
of H onto itself can be extended to a homomorphism of G onto H. Prove that H is a
direct summand of G.
Proof: Let g be a homomorphism of G onto H such that g(h)=h for each h in H (by
the hypothesis) , let K = Ker (g). We claim that G  H  K .
1. To prove that G= H+K, let x  G , then g(x)=h (for some h in H) and g(h)=h (by
hypothesis) , hence g(x)=g(h), which implies x-h  K, so x=h+k for some k in K ,
that is , G= H+K.
2. To prove H  K  0 , let x  H  K , then g(x)=x and g(x)=0.
Therefore G  H  K .
Ex.3 Combine exercises 1 and 2 to get another proof of Theorem 2.
Proof: Let H be a divisible subgroup of the abelian group G. Then by Ex.1, the
identity homomorphism of H can be extended to homomorphism of G into H (in fact
onto H, since the identity is onto). Hence by Ex. 2, H is a direct summand of G.