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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.