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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

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1 Answer 1

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.

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