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Let $G$ be a finitely generated group, $H$ a subgroup of $G$ of index $n$, with $a_i$ a set of coset representatives and $$G=\displaystyle\bigcup_{i=1}^nH{a_i}.$$ Let $\phi:H\rightarrow G$ be a homomorphism. We define a map $\psi:G\rightarrow G$ as follows. For $g\in H{a_i}$, let $\psi(g)=\phi(ga_i^{-1})$. Then for each $g\in G$, we can define a sequence $s(g)=(i_0i_1\dots i_k\dots)$ with $i_k\in\left\{1,2,\ldots,n\right\}$ such that for $k=0,1,\dots$ we have $\psi^k(g)\in H{a_{i_k}}$.

Let $\mathcal{S}=\left\{s(g)|g\in G\right\}$ be the set of all sequences which can be induced by some $g\in G$.

Question: Does $\mathcal{S}$ only contain preperiodic sequence? That is to say, for each $(i_0i_1\dots i_k)\in\mathcal{S}$,there exist $N$ and $T$ such that $i_{k+T}=i_k$ for all $k\geq N$.

It seems that the answer is positive if $G$ is a Fuchsian group, but I am not sure.

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  • $\begingroup$ For $G = \mathbb{Z}$ one can get a Collatz-like sequence. Take $H = 2\mathbb{Z}$ and $a_0=0$, $a_1=1$. Let $\phi(2r) = 3r$. If $m$ is odd then $\psi^a(2^a m) = 3^a m$, $\psi^{a+1}(2^a m) = 3(3^a m -1)$ and $s(2^am)$ starts $(0,\ldots,0,1,\ldots)$ with $a$ initial $0$s. Hence $s(r)$ is preperiodic only if there is a constant $c$ such that the highest power of $2$ dividing any $\psi^t(r)$ is at most $2^c$. I can't see any reason to expect this. For example, the sequence $s(5)$ begins $10100111001011001010110010110101111100101111111010\ldots$ $\endgroup$ – Mark Wildon Mar 27 '15 at 11:37
  • $\begingroup$ Could you say why you expect this to be true in the Fuchsian case (which if the comment above is any indication, would be interesting)? Did you compute things or is there a deeper reason? Also, it wouldn't hurt to have an idea where this 'weird' map $\psi$ comes from. $\endgroup$ – Jean Raimbault Mar 27 '15 at 11:58

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