Assume $A$ and $B$ are infinite abelian groups, $B$ is a subgroup of $A$. Is it true that if every homomorphism from $B$ to $\mathbb Z$ can be extended to a homomorphism from $A$ to $\mathbb Z$, then for all $a\in B$ and all natural number $n$, if there is $c\in A$ such that $nc=a$ in $A$, then there is $x\in B$ such that $nx=a$ in $B$?


Thank you for your answer.



Subgroups with some closed property
No.


counterexample: $A={\mathbb Q}/{\mathbb Z}$. 
$B=$ subgroup generated by any nonzero element of $A$. There are no nonzero homomorphisms from $B$ to ${\mathbb Z}$ since $B$ is torsion. 


edit: torsion free counter example: 


$A={\mathbb Q}$ and $B=\{\frac{p}{q} \in A \ | \ q=2^n \ \text{for some } n\in {\mathbb N} \}$
Again the problem is there are no nonzero homomorphisms from $B\to {\mathbb Z}$ since $B$ is 2-divisible. 


