# Lower central series

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:

$[a,bc]=[a,b][a,c][[c,a],b]$

$[ab,c]=[b,c][[c,b],a][a,c]$

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 $a_1\otimes\cdots\otimes a_n\mapsto [a_1,a_2,\ldots,a_n]$).

• Oops! Thanks for catching my erroneous (now deleted) other answer. (Which, incidentally, falsely claimed that the $\gamma_i$ themselves were finitely generated.) – Cam McLeman Feb 15 '12 at 5:03
• @CamMcLeman: No problem: it's a common (and very tempting) error. – Arturo Magidin Feb 15 '12 at 5:15