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Let $G$ be a group generated by $n$ generators, $x_1,...,x_n$ and let $[a,b] = a^{-1}b^{-1}ab$. Also let $G_i$ be the $i^{th}$ group in the lower central series. Thus, $[a,b]\cdot[b,a]=1$ (which can be though as an identity in $G_2/G_3$) and that the Hall-Witt identity in $G_3/G_4$ becomes: $$ [[a,b],c]\cdot[[b,c],a]\cdot[[c,a],b]=1. $$

Is this can be generalized to $G_n/G_{n+1}$? Explicitly, is the following true in $G_n/G_{n+1}$ for $x_1,\ldots,x_n \in G$ and $\sigma = (1\, 2\, 3\, ...\, n) \in Sym_n$: $$ \prod_{i=0}^{n-1}[[\ldots[\sigma^i(x_1),\sigma^i(x_2)],\sigma^i(x_3)],\ldots,\sigma^i(x_n)] = 1. $$

Thank you!

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No, this formula doesn't work in general.

We will build an explicit counterexample. Let $G$ be the group of unitriangular matrices of order $5$, generated by matrices $$ a = 1 + e_{12},\ b = 1 + e_{23},\ c = 1+e_{34}, d = 1 + e_{45}. $$ Here $1$ is the identity matrix and $e_{ij}$ is a matrix having $1$ at position $(i,j)$ and $0$ in all other entries. Consider this product: $$ g = [[[a,c],b],d] \cdot [[[c,b],d],a] \cdot [[[b,d],a],c] \cdot [[[d,a],c],b]. $$ A closer look shows that $[a,c]=[b,d]=[d,a]=1$, so the product can be considerably simplified: $$ g = [[[c,b],d],a]. $$ Now, a straightforward calculation shows $[[[c,b],d],a] = 1 + e_{45} \neq 1$ (please check this!).

Finally, $G_5= 1$, so $g$ goesn't vanish in $G_4/G_5$, and we are done.

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thanks! Is there any (known) identity that relates the element $[[[a,b],c],d] \in G_4/G_5$ to the other three elements (when $G$ is any group)? – David Apr 16 '13 at 15:18
I doubt it, but I'm not sure. Your question is essentially not about group $G$ but about its associated graded Lie algebra, which is free if $G$ is free. So maybe you can find something useful by looking at free Lie algebras. – Dan Shved Apr 16 '13 at 15:48
Although it is clear that there cannot be any fixed linear relation between these four elements, because we've seen an example when one of them is non-zero and all the others are zero. (I have inadvertently switched to additive terminology here). – Dan Shved Apr 16 '13 at 15:52

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