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?


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 $x^ry^s=b$.

  • $\begingroup$ How x and y generate a and b? Sorry,I don't get your point. $\endgroup$ – Zhongmin Jin Nov 14 '13 at 22:37
  • $\begingroup$ This is just standard theory of free abelian groups. I've added some details. $\endgroup$ – Derek Holt Nov 15 '13 at 8:28
  • $\begingroup$ You are right. Thanks. I feel naive to ask the question. Now I am just wondering whether it is possible to add some assumptions to get that uniform bound? @DerekHolt $\endgroup$ – Zhongmin Jin Nov 16 '13 at 6:10

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.

  • $\begingroup$ I think we can talk about the symmetric generators, which is generators together with its inverse. In that case, $r_1=g_1^{-1}$ $|r_1|=1$. I think I need a little of my question. $\endgroup$ – Zhongmin Jin Nov 14 '13 at 22:06
  • $\begingroup$ Are you changing the definition of the word norm? In that case maybe $r_1 = g_1^{|G|/2}$ is a good counterexample. $\endgroup$ – apt1002 Nov 14 '13 at 22:15
  • $\begingroup$ $r_1=g^{\frac{|G|}{2}}$ is not a generators. In my question, we only care about the word norm of generators (elements of generators number exact the minimal. ). We don't care about the word norm of arbitrary group element. The word norm's definition is the same as wiki, link $\endgroup$ – Zhongmin Jin Nov 14 '13 at 22:23
  • $\begingroup$ True! My mistake. But it is fixable. I see you have asked a new question. Let's continue there... $\endgroup$ – apt1002 Nov 14 '13 at 22:30

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