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Lets call an abelian group $G$, to be semi-free (or SF) if every nonzero subgroup of $G$ is isomorphic to $\mathbb{Z}\times H$ for some abelian group $H$.

Is every semi-free group, a free group? If not, does there exist a good characterization of this class of abelian groups?

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    $\begingroup$ For a module over a ring $R$, this means: module $M$ such that each nonzero submodule $N$ has $\mathrm{Hom}(N,R)\neq 0$. A sufficient condition is to be residually free, or equivalently residually-$R$, i.e., embeddable into $R^X$ for some set $X$. Since $\mathbf{Z}^X$ is indeed not free for $X$ infinite (as indicated in Jeremy Rickard's answer), one can alternatively ask whether there is a non-residually-free module with this property. $\endgroup$
    – YCor
    Commented Dec 16, 2023 at 12:16

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The Baer-Specker group $B$, the direct product of countably many copies of $\mathbb{Z}$, is semi-free but not free.

It is semi-free, because for any nonzero element $x\in B$ there is some projection $\pi:B\to\mathbb{Z}$ onto one of the direct factors for which $\pi(x)\neq0$, so if $K$ is any nonzero subgroup of $B$ there is a nonzero homomorphism $\varphi:K\to\mathbb{Z}$, and so $K\cong\ker(\varphi)\oplus\operatorname{im}(\varphi)\cong\ker(\varphi)\oplus\mathbb{Z}$.

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  • $\begingroup$ Could you please give your proof why this group is semi-free? $\endgroup$
    – Mostafa - Free Palestine
    Commented Dec 16, 2023 at 10:52
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    $\begingroup$ @Mostafa I've added a proof. $\endgroup$
    – Jeremy Rickard
    Commented Dec 16, 2023 at 11:03

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