Ex. 5 Prove that any group can be embedded in a divisible group.
Proof: Let G be an abelian group. It is known that any abelian group is a
homomorphic image of a free abelian group, that is, there exists a free abelian group,
say F, which is a direct sum of copies of Z and a subgroup K of F such that G is
isomorphic to F K . But Z can be considered as a subgroup of Q ( the group of
rational numbers), so there exists a group F which is a direct sum of copies of Q, and
F can be considered as a subgroup of F , hence F K is a subgroup of F K . If  is the
isomorphism of G onto F K and i is the inclusion map of F K into F K , then the
composition i is a monomorphism of G into F K , that is, G is embedded in F K .
On the other hand, F is divisible ( since it is a direct sum of copies of the divisible
group Q) and F K is divisible ( since it is a homomorphic image of the divisible
group F ). Therefore G is embedded in a divisible group.
Ex. 7 Prove that the following three properties of group are equivalent:
a) G is divisible; b) G is a direct summand of every containing group; c ) If K is a
group and f a homomorphism of a subgroup of K into G, then f can be
extended to a homomorphism of K into G.
Proof: (a  b): By Theorem 2.
(b  c): By Ex. 6, G is divisible, then by Ex. 1 (c ) holds.
(c  a): By Ex. 5, G can be considered as a subgroup of a divisible group, say
D, now by the assumption of ( c ), the identity of G onto itself can be extended to a
homomorphism of D onto G, that is G is a homomorphic image of a divisible group.
Therefore G itself is divisible.