# Maps between groups inducing isomorphisms between all nilpotent quotients

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 $i=1,2$.

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## 2 Answers

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.

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Is there a surjection from the free group to that 1-relator group inducing this? (or the other way around, but that's my guess for the direction of the map). –  Louis Mar 14 '12 at 1:11
I did not notice this assumption. No, the 1-related group has one more generator than the free group. –  Mark Sapir Mar 14 '12 at 1:17
Great, thanks!! –  Louis Mar 14 '12 at 1:51
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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.

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