For every abelian groups $G$ let $\mathrm{d}G$ be its maximal divisible subgroup. Then $G \mapsto G/\mathrm{d}G$ is a right exact functor $\mathbf{Ab} \to \mathbf{Ab}$.
Let $$ 0 \to G \xrightarrow{i} H \xrightarrow{p} K \to 0 $$ be an exact sequence. I'm having trouble showing for the sequence $$ G/\mathrm{d}G \xrightarrow{i'} H/\mathrm{d}H \xrightarrow{p'} K/\mathrm{d}K \to 0 $$ (where $p' \colon h + \mathrm{d}H \mapsto p(h) + \mathrm{d}K$ ) that $\ker p' \subseteq \operatorname{im} i'$.
So let $h + \mathrm{d}H \in \ker p'$. Then $p(h) \in \mathrm{d}K$. I need to show that there is $h'$ in $\mathrm{d}H$ such that $p(h) = n p(h')$ for some integer $n$ in order to draw the conclusion.
Any help would be appreciated!