Every abelian group can be embedded in a divisible group I've a proof of the title-statement if I can prove the following:


*

*every cyclic group can be embedded in $\mathbb{C}^{\star}$, the multiplicative group of complex numbers.


Can you suggest me how to prove this?
 A: Do it case by case. First the infinite cyclic groups and then the finite ones.
A: Let's write $G$ as $G=F/R$ which $F$ is free abelian. It leads us to have $F=\sum\mathbb Z\leq\sum\mathbb Q$. Since $$G=F/R=\frac{\sum\mathbb Z}{R}\le\frac{\sum\mathbb Q}{R}$$ and knowing that every quotient group of a divisible group is itself a divisible group so via this way we imbedded $G$ in a divisible groups.
A: Here is a short proof of the whole statement (Following Hilton/Stammbach Homological Algebra exercise 7.2). Let $A$ be an abelian group. There is always an exact sequence $0\to K\to F \to A\to 0$ where $F$ is a free abelian group $F = \bigoplus_I\mathbb Z$. There is an embedding $\bigoplus_I\mathbb Z\hookrightarrow  \prod_I \mathbb Z \hookrightarrow \prod_I\mathbb Q$. There is a unique map $A \to (\prod_I\mathbb Q)/K$ which fits into the diagram

and the snake lemma shows that this map is a monomorphism. The group $\prod_I\mathbb Q/K$ is a quotient of a divisible group and thus divisible. Hence every abelian group can be embedded in a divisible abelian group.
