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If $A$ is an abelian group, then $\mbox{Hom}(A,\mathbb{Z}) \ne 0$ if and only if $A$ has an infinite cyclic direct summand.

The hint is to use

If $F$ is a free abelian group and $g:B \to F$ is a surjective homomorphism from some abelian group $B$ then B = \mbox{ker} g \oplus F', where F' \simeq F.

I guess I am misinterpreting the question. Can't I just take $A = \mathbb{Z}/2\mathbb{Z}$, and define $\phi(z) = \begin{cases} 0 &\mbox{ if } z = 0 \\ 1 &\mbox{ if } z = 1 \end{cases}$

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    yes, you only need the lemm$a$ for one direction; the other follows from the definition of direct sum.2011-05-04

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One direction is easy: if $A\cong \mathbb{Z}\oplus B$, then the projection onto the first component is a nontrivial homomorphism from $A$ to $\mathbb{Z}$. The converse uses the hint directly, since $\mathbb{Z}$ is the free abelian group of rank $1$. It's using the fact that free abelian groups are projective.

To establish the hint, just pick some $a\in A$ such that $g(a)=1$, and show that $\langle a\rangle \cong \mathbb{Z}$, $\langle a\rangle\cap\mathrm{ker}(g)=\{0\}$, and $A = \mathrm{ker}(g)\oplus\langle a\rangle$.

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    Thank you - it was mainly seeing that $g$ was surjective that was throwing me around2011-05-04