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This question elaborates on a previous one of mine.

Let $F$ be a free group of finite rank $r>1$.

Question. Is it possible to choose some basis for $F$ which refines the lower central series and in which the expression for every $g\in F$ is finite?

Explanation. The Hall basis of commutators $C=\{c_1,c_2,\ldots\}$ for $F$ (as defined in the collecting process of M. Hall) can be used to express every element $g\in F$ in the form $g=c_1^{n_1}c_2^{n_2}\ldots c_{t}^{n_t}g_k$ with $g_k\in \gamma_k$, the $k$-th term of the lower central series for $F$. This expression (for some $g\in F$) may grow infinitely as $k\to\infty$. Yet, sometimes it is possible to change the basis $C$ to make this expression finite (for a fixed $g\in F$).

Example. If $F=\langle a,b \rangle$ and $g=(ab)^3$ then


We may choose a new basis $C_1=\{a,b,[b,a],g_3,\ldots\}$ in which the expression for $g$ becomes finite. (Note that $g_3$ may be included in a basis, because its expression in $C$ is


with coprime exponents of the terms of weight $3$.) So, the question is whether this sort of "basis correction" can be made for all $g\in F$ uniformly.

Remark. By a basis I mean a set $c_1,c_2,...$ of elements of $F$ whose images (in the appropriate terms) form a basis of the free abelian group

$ \gamma_1/\gamma_2\oplus\gamma_2/\gamma_3\oplus\ldots$

This should explain the phrase "refines the lower central series".

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In what sense is the first expression for $g$ «not finite»? – Mariano Suárez-Alvarez Apr 6 '11 at 18:11
The first expression for $g$ may not be a complete expansion in the basis $C$, because $g_k$ is in general not in $C$. So, this is only a partial expansion modulo $\gamma_k$. Expressing $g_k$ in the basis, may bring about a nontrivial $g_{k+1}$, then $g_{k+2}$, and so on. In this sense the expression is not finite. – Anvita Apr 6 '11 at 19:36

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