|
Let $G$ be a finitely generated group and let $\gamma_i$ be the $i$th group in the lower central series. Could you help me to prove that for every $n$ and for every $i$ with $1\leq i\leq n$, $\gamma_i/\gamma_n$ is finitely generated? The hint that I have is to use the identities:
|
||||
|
|
|
First, prove inductively that $\gamma_n/\gamma_{n+1}$ is finitely generated. Let $g_1,\ldots,g_k$ be a generating set for $G$. We can now proceed by induction on $m$. The result is true if $m=1$, since $\gamma_1/\gamma_2 = G^{\rm ab}$ is finitely generated by $\overline{g_1},\ldots,\overline{g_k}$. Assume now that $\gamma_m/\gamma_{m+1}$ is finitely generated, and let $c_1,\ldots,c_r$ be elements of $\gamma_m$ that generate modulo $\gamma_{m+1}$. Since $\gamma_{m+1} = [\gamma_m,G]$, it is generated by elements of the form $[x,g]$ with $x\in\gamma_m$, $g\in G$. Use the second of your identities that any $[x,g]$ is congruent, modulo $\gamma_{m+2}$, to a product of commutators of the form $[c_i,g]$ and their inverses. Then use the first identity to show that any commutator of the form $[c_i,g]$ can be written, modulo $\gamma_{m+2}$, as a product of commutators of the form $[c_i,g_j]$ and their inverses. Conclude that $\gamma_{m+1}/\gamma_{m+2}$ is finitely generated. Then you can use the fact that each of $\gamma_i/\gamma_{i+1}$, $\gamma_{i+1}/\gamma_{i+2},\ldots,\gamma_{n-1}/\gamma_n$ are finitely generated to conclude that $\gamma_i/\gamma_{n}$ is finitely generated. (If you want to be really ambitious, there is an onto map from $G^{\rm ab}\otimes G^{\rm ab}\otimes\cdots\otimes G^{\rm ab}$ ($n$ factors) to $\gamma_n/\gamma_{n+1}$ via $a_1\otimes\cdots\otimes a_n\mapsto [a_1,a_2,\ldots,a_n]$). |
|||||
|
