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I'm stuck in this problem:

If $A$ is an abelian group which is torsion free, prove that there exists a divisible group $D$ and an injective homomorphism $A \to D$.

Same question also but for $A$ abelian and any torsion group.

Any help is appreciated.

Thanks a lot

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I would advise you to think of accepting answer for your previous questions - otherwise '0% accept rate' may discourage from answering new ones. –  Ilya Oct 25 '11 at 6:46

2 Answers 2

In the torsion-free case you could put together the following facts:

  • Every abelian group $A$ is a $\mathbb{Z}$-module in a natural way.

  • For a $\mathbb{Z}$-module $A$ one can form the localization $A_0:=(\mathbb{Z}\setminus \{0\})^{-1}A$.

  • The natural homomorphism $A\rightarrow A_0$ is injective.

In the case of a torsion group $A$ the relevant facts are these:

  • $A$ is isomorphic to a group of the form $\mathbb{Z}^{(I)}/N$, where $\mathbb{Z}^{(I)}$ is the direct sum over the index set $I$.

  • A factor group of a divisible group is divisible.

  • Define $D:=\mathbb{Q}^{(I)}/N$ as $\mathbb{Z}$-modules.

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Fist for any $A$ abelian group you could consider the free abelian group $\mathbb{Z}^{(A)}$ with basis $\{e_a\}_{a\in A}$ and the function that asigns to each $e_a$ to $a$, so the function extends to a morphism of groups which for construction is an epimorphism, so apply the first theorem of iso morphism to get that $A\cong \mathbb{Z}^{(A)}/N$ where $N$ is the kernel of the previous morphism, then consider the isomorphism in the reverse direction that is $f:A\longrightarrow \mathbb{Z}^{(A)}/N$ and compose it with the inclusion $i:\mathbb{Z}^{(A)}/N\longrightarrow \mathbb{Q}^{(A)}/N$. This with the facts that $\mathbb{Q}$ is divisible, the coproduct of divisibles is divisible and the quotient of divisible is divisible.

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