Criteria for subgroup in Abelian group having a direct complement Let $A$ be an Abelian group (not necessarily finitely generated) and $A' \subset A$ be a subgroup with the following property:
For any $a \in A, n \in \mathbb{Z} \setminus {0}$ the condition $n \cdot a \in A'$ implies $a \in A'$. 
Is it true that $A = A' \oplus A''$ for some $A''$?
 A: No, this condition does not imply $A'$ is a direct summand of $A$.  Here's a very general way of getting counterexamples.  Let $B$ be any torsion-free abelian group which is not free (for instance, $B=\mathbb{Q}$), let $A$ be a free abelian group with a surjection $f:A\to B$, and let $A'=\ker(f)$.  If $A'$ were a direct summand of $A$, then $B$ would be a direct summand of $A$.  But that would imply $B$ is free since $A$ is free, which is a contradiction.  So $A'$ is not a direct summand of $A$.  On the other hand, $A'$ does satisfy your condition, since your condition is equivalent to saying that $A/A'=B$ is torsion-free.
(From the perspective of homological algebra, if $B=A/A'$, you are asking for a torsion free abelian group $B$ and a nontrivial element of $\operatorname{Ext}(B,A')$.  For any non-projective abelian group, you can find an abelian group $A'$ such that $\operatorname{Ext}(B,A')\neq 0$.  Since projective is the same as free for abelian groups, this means you can take $B$ to be any non-free torsion free abelian group.)
Here's a somewhat different counterexample that is a nice example to be aware of.  Take $A=\mathbb{Z}^\mathbb{N}$ and let $A'\subset A$ be the subgroup of functions $\mathbb{N}\to\mathbb{Z}$ with finite support.  Then $A'$ satisfies your condition (since multiplying an element of $A$ by a nonzero integer does not change its support), but $A'$ is not a direct summand of $A$.  Indeed, notice that the function $f(n)=n!$ is divisible by every integer as an element of the quotient group $A/A'$.  If $A'$ were a direct summand of $A$, then $A/A'$ would be isomorphic to a subgroup of $A$.  But no nonzero element of $A$ is divisible by every integer, so this is impossible.
