# Bounding the index $|G' \cap H : H'|$ for $G$ a finitely generated nilpotent group

Let $G$ be an infinite finitely generated torsion-free nilpotent group. Let $H$ be a finite-index subgroup of index $n$. Let $G^\prime$, $H^\prime$ denote the derived groups of $G$ and $H$ respectively.

I want to show that the index $|G^\prime \cap H : H^\prime|$ is bounded by a polynomial in $n = |G:H|$.

I suggest the polynomial $n^{h(G)}$, where $h(G)$ denotes the Hirsch length of $G$ should do the trick, perhaps even $n^{h(G^\prime)}$, but alas I am unable to prove such!

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This one might be more suitable for mathoverflow.. –  Derek Holt Sep 19 '12 at 13:19

This is not possible. Consider the group $G$ generated by $s_i$, $i=1,\dots,m$ and central elements $t_{i,j}$, $1\leq i < j \leq m$, with the relations $s_i^2=t_{i,j}^2=e$, $[s_i,s_j]=t_{i,j}$, $i<j$. Let $H$ be the subgroup generated by $s_i$, $i\geq 2$ and all $t_{i,j}$'s. Then $[G:H]=2$, but $[G':H']=2^{m-1}$, since $G'$ is generated by all $t_{i,j}$'s, while $H'$ is generated by those with $i\neq 1$.
Note also that the question reduces to one about finite groups upon factoring by the normal core of $H$, so one cannot expect the Hirsch length to play a role.