Is every subgroup of a free abelian group a direct summand? My guess is NO, because take $G=\mathbb{Z}$ and $F=2\mathbb{Z}$ is a subgroup but not a direct summand.
 A: That every subgroup of an abelian group $A$ has a direct summand is equivalent to saying that $A$ is semisimple as a $\mathbb{Z}$-module. As you have already noticed $\mathbb{Z}$ itself is not semisimple.
If $A$ is an abelian group such that some element $a \in A$ has infinite order, then $\mathbb{Z}$ is isomorphic to the subgroup generated by $a$. Because submodules of semisimple modules are also semisimple it follows that $A$ cannot be semisimple as a $\mathbb{Z}$-module.
As a special case of this we get that the trivial group is the only free abelian group which is a semisimple $\mathbb{Z}$-module, because each other free abelian group contains an element of infinite order.
A: NO, because take $G=\mathbb{Z}$ and $F=2\mathbb{Z}$ is a subgroup but not a direct summand. It is not a direct summand because if $\mathbb{Z}$ was $\mathbb{Z}=2\mathbb{Z} \oplus H$
Then since the intersection of $2\mathbb{Z}$ and $H$ is trivial, $H$ must consist of all odd numbers plus the zero. But that does not constitute a subgroup.
