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Let $G$ be a finitely generated group.

For a set of generators $B$ of $G$, $\ell_B(x)$ is the length of the smallest sequence of elements(and inverse of the elements) in $B$, such that the product equals $x$. $\ell_B(X) = \sum_{x\in X} \ell_B(x)$.

We want to find some function $f$, such that any set of generators $B'$ and $b\in B$, we have $\ell_{B'}(b)\leq f(\ell_B(B'))$.

For example, if $G=\mathbb{Z}$, $B=\{1\}$, then we have $$ \ell_{B'}(1)\leq \ell_B(B') $$

Indeed, we can pick two relatively prime elements $u,v\in B'$, and solve for $au+bv = 1$ while minimizing $|a|+|b|$. By Bézout's lemma, this can be bounded by $|u|+|v|\leq \ell_B(B')$.

In particular, I would like to see a result like this for $B=\{a,b\}$ and $G$ is the free group generated by $B$.

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  • $\begingroup$ Can you ask a more specific question? $\endgroup$
    – Qiaochu Yuan
    Mar 5, 2017 at 3:21
  • $\begingroup$ @QiaochuYuan I'm interested in techniques or references that solve this kind of problems. Seeing an example for the free group of rank $2$ is good enough. $\endgroup$
    – Chao Xu
    Mar 5, 2017 at 3:36
  • $\begingroup$ In the case of the free group in $B=\{a,b\}$, isn't $\ell_{B'}(x) \le \max\{\ell_{B'}(a),\ell_{B'}(b)\} \ell_B(x)$ an answer? $\endgroup$
    – Felipe Voloch
    Mar 5, 2017 at 3:56
  • $\begingroup$ @FelipeVoloch that's true, we only need bound on $\ell_{B'}(b)$ for all $b\in B$. I see why I wasn't being clear. I have modified the question. $\endgroup$
    – Chao Xu
    Mar 5, 2017 at 4:39
  • $\begingroup$ Both Profs Leininger and Kapovich in the math dept at UIUC are experts on free-group automorphisms, and could certainly help you with this question. $\endgroup$
    – HJRW
    Mar 5, 2017 at 7:10

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