# $\mbox{Hom}(A,\mathbb{Z}) \ne 0$ if and only if $A$ has an infinite cyclic direct summand

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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You could try, but then what would $\phi(2z)$ be? – Qiaochu Yuan May 4 '11 at 3:25
@Qiaochu - but our domain is only $\mathbb{Z}/2\mathbb{Z}$? – Juan S May 4 '11 at 3:28
Yes, but $\phi$ is a homomorphism, so $\phi(2z) = \phi(z) + \phi(z)$. – Qiaochu Yuan May 4 '11 at 3:31
@Qiaochu - OK. So the proof (in one direction at least) is going to be something similar - I can't end up with a homomorphism uneless there is a direct summand of $\mathbb{Z}$ somewhere in there? (I think I only need the lemma for the reverse implication?) – Juan S May 4 '11 at 3:40
yes, you only need the lemma for one direction; the other follows from the definition of direct sum. – Qiaochu Yuan May 4 '11 at 3:46

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 $A = \mathrm{ker}(g)\oplus\langle a\rangle$.
Thank you - it was mainly seeing that $g$ was surjective that was throwing me around – Juan S May 4 '11 at 5:04