6
$\begingroup$

Let $F_X$ be a free group on a countably infinite set $X$. Let $\alpha$ be an automorphism of $F_X$ and $H$ a closed subgroup of $F_X$ in the profinite topology.

Is it possible that $\alpha(H) \lneq H$?

Note that if $\alpha$ is the conjugation by some $r \in F_X$ then the answer is negative.

Improve this question
$\endgroup$
5
  • $\begingroup$ Isn't every finitely generated subgroup of a free group closed in the profinite topology? If so, just take $H=\langle x\rangle$ and make $\alpha$ exchange $x$ with some other free generator... $\endgroup$
    – Arturo Magidin
    Commented Nov 23, 2015 at 18:08
  • $\begingroup$ @ArturoMagidin I am asking whether $\alpha$ can map $H$ to a proper subgroup of itself. $\endgroup$
    – Pablo
    Commented Nov 23, 2015 at 18:15
  • 1
    $\begingroup$ Oh; misinterpreted that. Sorry. $\endgroup$
    – Arturo Magidin
    Commented Nov 23, 2015 at 19:09
  • 6
    $\begingroup$ Choose $X=\mathbf{Z}$, and $\alpha$ be the shift on generators $x_n\mapsto x_{n+1}$, and $H=F_\mathbf{N}$. Clearly $\alpha(H)$ is properly contained in $H$. Also $H$ is closed in the profinite topology, by a simple argument. $\endgroup$
    – YCor
    Commented Nov 23, 2015 at 23:27
  • $\begingroup$ @YCor you are right! $\endgroup$
    – Pablo
    Commented Nov 24, 2015 at 0:43

0

Reset to default

You must log in to answer this question.