Let $G_1$ and $G_2$ be finitely presentable groups and let $f : G_1 \rightarrow G_2$ be a surjective homomorphism. Denoting the kth term of the lower central series of $G_i$ by $\gamma_k(G_i)$, assume that $f$ induces an isomorphism $G_1 / \gamma_k(G_1) \rightarrow G_2 / \gamma_k(G_2)$ for all $k \geq 1$. Is $f$ necessarily an isomorphism?
EDIT : I forgot the obvious assumption that the intersection of the lower central series of $G_i$ is trivial for $i=1,2$.
No, there exist 1-relator groups with the same nilpotent factors as the free group: http://caissny.org/pdfs/Parafree%20one-relator%20groups.pdf
Edit I did not notice the assumption that $G_2$ is a factor-group of $G_1$. Then the answer is "yes". Suppose that there is a kernel of $\phi\colon G_1\to G_1/N=G_2$. For some $n$, $N$ is not contained in $\gamma_n(G_1)$. We have $\gamma_n(G_2)=\phi(\gamma_n(G_1))=\phi(N\gamma_n(G_1))$. Hence $G_2/\gamma_n(G_2)$ is a proper homomorphic image of $G_1/\gamma_n(G_1)$. If these two groups ($G_1/\gamma_n(G_1)$ and $G_2/\gamma_n(G_2)$) were isomorphic, you would have a finitely generated non-Hopfian nilpotent group $G_1/\gamma_n(G_1)$ that is impossible since these groups are residually finite.
Please allow me to give some explanation for answer of Mark Sapir for the beginners in order to avoid confusion for notation: Assume that $f$ is not an isomorphism. Since $f$ is surjective homomorphism, then $N=Ker f \ne 1$ and $G_2\cong G_1/N$. Without loss of generality, we can assume that $G_2=G_1/N$. Since the intersection of the lower central series is trivial by assumption, there exists $n$ such that $N $ is not contained in $\gamma_n(G_1)$. Now $\gamma_n(G_2)=\gamma_n(G_1)N/N$, and then $G_2/\gamma_n(G_2) \cong G_1/\gamma_n(G_1)N$, a proper homomorhic image of $G_1/\gamma_n(G_1)$. Then we get a contradiction as Mark Sapir.
