9
$\begingroup$

Let $A$ be a (finitely generated) free group and $G$ be a (finitely generated) free $A$-group - that is, a group with an action of $A$, which is free in the category of groups with an $A$-action. Equivalently, $G$ is isomorphic to the free product of copies of a (finitely generated) free group $H$ (with a free basis $(h_1,...h_n)$) indexed by the elements of $A$, the $A$-action being the obvious one.

I feel that the automorphism group of $G$ (as a group with an $A$-action) should be generated by the automorphisms of $H$ and the automorphisms of the following form: $h_i\mapsto\negmedspace^ah_i$ (where $a\in A$ and $i\in\{1,\dots,n\}$ are fixed), $h_j\mapsto h_j$ for $j\neq i$.

Possibly this is well-known and written somewhere. Unfortunately, I did not find anything about this by searching on the web or in classical books of group theory.

Do you have references, or ideas to approach the problem in the most efficient way?

Edit: I began to look carefully at Nielsen reduction arguments to try and adapt them to this situation with action, unfortunately it does not seem to work, at least simply. Maybe we have to be quite more clever in the orderings involved in the induction... or the result that I suggested is false?

Many thanks in advance.

$\endgroup$
  • $\begingroup$ I began to look carefully at Nielsen reduction arguments to try and adapt them to this situation with action, unfortunately it does not seem to work, at least simply. Maybe we have to be quite more clever in the orderings involved in the induction... or the result that I suggested is false? $\endgroup$ – Aurélien Djament Jul 30 '15 at 6:58
4
$\begingroup$

Finally, the answer is positive. See §3.1 in this preprint: https://hal.archives-ouvertes.fr/hal-01214646 that I have just posted. This gives also my motivation to study this question.

$\endgroup$
  • 3
    $\begingroup$ The link doesn't work. $\endgroup$ – few_reps Oct 13 '15 at 8:00
  • $\begingroup$ for me the link works. $\endgroup$ – YCor Oct 13 '15 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.