0
$\begingroup$

Let $f$ be a continuous endomorphism of a compact Hausdorff totally disconnected topological group $G$ and let $H$ be a closed normal subgroup of G such that $f(H)\subseteq H$ and with $\mu(H)=0$ where $\mu$ is the normalized Haar measure on $G.$ Is it true that if $f$ is topologically transitive then also the restriction of $f$ to $H$ is topologically transitive (where I consider the subsapce topology on $H$)?

I think that the answer is no but I am unable to find a counterexample.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

No, take $F$ to be any nontrivial finite group, $G=F^\mathbf{Z}$ and $H\simeq H$ the subgroup of constants in $G$. Let $f$ be the shift ($f(u)(n)=u(n+1)$). Then $f$ is topologically transitive but restricts to the identity on $H$, hence not topologically transitive on $H$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you. With $G=F^\mathbb Z,$ is it possible to find a counterexample to my claim with an $H$ also of the form $H=F_2^{\mathbb Z}$? $\endgroup$
    – Nick Belane
    Commented Nov 15 at 16:45
  • $\begingroup$ I don't know. This is another question. $\endgroup$
    – YCor
    Commented Nov 15 at 17:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .