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I want to show that if $A,B$ are abelian groups then $\operatorname{Hom}_R(A,B)$ is torsion free when $A$ is divisible and $\operatorname{Hom}_R(A,B)$ is divisible when $A$ is torsion free and divisible.

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    Please always make the *body* of your questions self-contained: the question has to be complete even if one cannot see the title. – 2017-01-04
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    What is $R${}{}? – 2017-01-04
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    @MarianoSuárez-Álvarez This wasn't mentioned, can't it be an arbitrary ring? – 2017-01-04
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    If $R$ is to be anything but the integers, you need that $A$ and $B$ be $R$-modules, not just abelian groups for the question to make sense. – 2017-01-04
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    @MarianoSuárez-Álvarez You're right. – 2017-01-04

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I assume that $R$ should be $\mathbb{Z}$?

If $\text{Hom}_\mathbb{Z}(A,B)$ is not torsion free, then for some nonzero homomorphism $\varphi:A\to B$ there is a positive integer $n$ so that $n\varphi=0$. But for any $a\in A$, there is some $a'\in A$ with $a=na'$. So $\varphi(a)=n\varphi(a')=0$. Therefore $\varphi=0$, giving a contradiction.

If also $A$ is torsion free, then for any positive integer $n$ and any $a\in A$ there is a unique $a'\in A$ with $a=na'$ and $\theta:a\mapsto a'$ is a homomorphism $A\to A$. So for any homomorphism $\varphi:A\to B$, $\varphi=n\varphi\theta$, and so $\text{Hom}_\mathbb{Z}(A,B)$ is divisible.

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    The "for any homomorphism" should be "for *some* nonzero homomorphism", no? – 2017-01-04
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    @MarianoSuárez-Álvarez Quite right. I'll edit. – 2017-01-04