Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G = F_n$ be the free group with $n$ generators, $x_1,...,x_n$. Let $[a,b] = a^{-1}b^{-1}ab$, and $G_i$ be the $i^{th}$ group in the lower central series. We know that $G_2/G_3$ is generated by $[x_i,x_j]$ when $i>j$, and that $G_3/G_4$ is generated by (the basic commutators) $[[x_i,x_j],x_k]$ where $i>j$ and $j \leq k$. This means that the element $a = [[x_i,x_j],x_k] \in G_3/G_4$, when $i>j>k$, can be written as a product of the basic commutators of $G_3/G_4$. However, I wasn't able to express $a$ as such a product. Can anyone help me with that?

Thank you very much, Alex

share|cite|improve this question

Use the Hall-Witt identity, that modulo $G_4$ becomes $$ [[x_i, x_j], x_k] \cdot [[x_j, x_k], x_i] \cdot [[x_k, x_i], x_j] = 1, $$ so if $i > j > k$ you have $$ [[x_i, x_j], x_k] = [[x_j, x_k], x_i]^{-1} \cdot [[x_i, x_k], x_j], $$ and now the commutators on the RHS are basic.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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