Consider a group $G$ which is generated by three element $a,\ b,\ c$, i.e. $G= (a)\ast (b)\ast (c)$ and $(a)=(b)=(c)=\mathbb{Z}$.
Assume that $N$ is a ${\bf smallest\ normal\ subgroup}$ containing $c^3b^{-1}$. Here my question is whether $G/N$ contains a group $\mathbb{Z}^3$ as a subgroup.
Proof : $c,\ c^2b^{-1}$ commute in $G/N$.
That is $G/N$ contains a subgroup isomorphic to $\mathbb{Z}^2$. But I can not prove whether $G/N$ contains $\mathbb{Z}^3$. How can we prove this ? Thank you in advance.
$G/N=\langle a,b,c\mid c^3b^{-1}\rangle\cong\langle a,c\rangle$ since $b=c^3$ in the quotient.
The Nielsen-Schreier theorem says that every subgroup of a free group is free. Therefore, the largest-rank abelian subgroup of $G/N$ is isomorphic to $\mathbb Z$.
$G/N$ does not contain a $\mathbb{Z}^2$. In your example, $c^2b^{-1}$ is equal to $c^{-1}$ in the quotient, so the reason $c^2b^{-1}$ and $c$ represent commuting elements in $G/N$ is that they are in the same $\mathbb{Z}$ subgroup.