The word norm bound between two minimal generators set of a group.

Question

Given any finite generated group G, there are many generators set. Let (I called length of G) $n=\min \{m$ | $\{g_1,g_2,\cdots,g_m\}$ is symmetric generators set of G $ \}$ . Symmetric generators set means generators with their inverse inside. Now we focus on generators set with exactly n elements.

The word norm of $g\in G$ respect to a generators set $\{g_1,g_2,\cdots,g_n\}$ is : $|g|=\min\{l: g=g_{i_1}g_{i_2}\cdots g_{i_l} : g_{is} \in \{g_1,g_2,\cdots,g_n\} 1\leq s\leq l \}$

Now suppose $\{g_1,g_2,\cdots,g_n\}$ and $\{r_1,r_2,\cdots,r_n\}$ be any two generators set of G, where n is defined as above.

Does there exist a constant, such that word norm (w.r.t {$g_i$}), $|r_i|<L, for\ \forall 1\leq i\leq n $, where L is independent of choice of generators set ? Can The constant L be uniformly chosen independent with the group, the length of G, or at least the any generators set with minimal length?

Answer 1

Consider a cyclic group, and $r_1 = g_1^{-1}$. Then $|r_1|$ is one less than the size of G, which can be as large as you like.

Answer 2

Let $G$ be a free abelian group of rank 2 generated by $a$ and $b$. Then $G$ is generated by $x = a^ib^j$ and $y=a^kb^l$ when $il-jk = \pm 1$. The lengths of $x$ and $y$ w.r.t $a$ and $b$ are $|i|+|j|$ and $|k|+|l|$, which can be arbitrarily large.

If $il-jk = \pm 1$ then the inverse of the matrix $\left(\begin{array}{cc}i&j\\k&l\end{array}\right)$ has entries in ${\mathbb Z}$.

Let the inverse be $\left(\begin{array}{cc}p&q\\r&s\end{array}\right)$.

Then $x^py^q = a^{ip+kq}b^{jp+lq} = a$ and similarly $x^ry^s=b$.