construct groups using quotient group

Question

If I have a surjective group endomorphism $\phi$ between $G$ and $H$. If H' is subgroup maximal abelian of $H$, let be $J=\phi^{-1}(H')$, I know that $J/ker\phi$ is isomorphic $H'$, but is possible construct one abelian subgroup $J'$ of $J$ that $\phi(J')=H'$? I think this is possible because for each $h \in H'$, I can catch one $a \in J$ such that $\phi(a)=h$, and I do this for all $h \in H'$. But I can not hold this.

Answer 1

Let $G=\langle a,b\rangle$ be the free group on two generators $a,b$, let $H=H'=\Bbb Z\oplus\Bbb Z$, and $\phi$ maps $a,b$ to the standard generators of $H$. The question is whether we can find an abelian subgroup $J'$ of $G$ that maps onto $H$. We can't because two words in $G$ commute iff they are powers of a common element, i.e., a subgroup of $G$ is abelian iff it is cyclic.