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Given a finitely generate free group $G$ with generators $s_1, \dots, s_n$. Let $(b_1, \dots, b_n)$ denote a basis of this free group. A elementary transformation of this basis creates a new basis $(b'_1, \dots, b'_n)$ by one of the following to operations:

  1. For some $1 \leq i \leq n$ replace $b_i$ with $b_i^{-1}$.
  2. For some $1 \leq i, j \leq n$ with $i \neq j$ replace $b_i$ by either $b_j \cdot b_i$, $b_j^{-1} \cdot b_i$, $b_i \cdot b_j$ or $b_i \cdot b_j^{-1}$, i.e. multiply $b_i$ from the left or the right by $b_j$ or its inverse.
  3. Exchange $b_i$ and $b_j$.

Two bases of $G$ are called equivalent if one can get the second in a finite number of elementary transformations from the first.

Question: Are all bases equivalent? If not, how many equivalence classes do exist? (finite, infinite?)

Examples: $(s_1, s_2)$ is equivalent to $(s_1^{-1}, s_1^{-1} s_2 s_1)$.

lalam
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    There is a known generating set for the automorphism group of a free group (see https://en.wikipedia.org/wiki/Automorphism_group_of_a_free_group) and your list of moves is missing some generators… so my guess is your list is not enough to go from any basis to another one. For example, going from $(s_1,s_2)$ to $(s_2,s_1)$ seems unlikely. – Mariano Suárez-Álvarez Dec 29 '17 at 16:07
  • I added the missing step, which I forget. Thanks. That is exactly what I was looking for. – lalam Dec 29 '17 at 16:15
  • Ah, i you add that then that's it. You should add an answer explaining this. – Mariano Suárez-Álvarez Dec 29 '17 at 16:17

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