Orbit-like maps on linearly ordered groups

[Natsugou]

I want to extend the indexes of \(a\) of a recurrence relation \(a_{n+1} = f(a_n)\) to arbitrary linearly ordered groups.
For a map \(a: \mathbb{Z} \to X, n \mapsto a_n\), there exists \(f: X \to X\) such that \(a_{n+1} = f(a_n)\) if and only if \(a_m = a_n \implies a_{m+k} = a_{n+k}\) for any positive integer \(k\).
When \(a: \mathbb{Z} \to X\) satisfies the above condition, the periods of \(a\), \(P(a) = \{m-n \mid a_m = a_n\}\) is a subgroup of \(\mathbb{Z}\).

To generalize to arbitrary linearly ordered groups, I will introduce some definitions.
For a map \(a\) defined on a linearly ordered group \(G\),
\(a\) is peep if \(a(g) = a(h) \implies a(fg) = a(fh)\) for all \(f > 1\).
\(a\) is bib if \(a(g) = a(h) \implies a(gf) = a(hf)\) for all \(f > 1\).
\(a\) is peep bib if \(a\) is peep and bib.
I define periods of \(a\) as
\begin{align}
  P(a) &= \{g^{-1}h \mid a(g) = a(h)\} \\
  &= \{p \in G \mid \exists g, h \in G, p = g^{-1}h\: \text{and}\: a(g) = a(h)\} \\
  &= \{p \in G \mid \exists g \in G, a(g) = a(gp)\}.
\end{align}
A subgroup \(H\) of \(G\) is said to be fee if \(g^{-1}Hg \subset H\) for all \(g > 1\).

Let \(G\) be a bi-ordered group and \(a\) be a map defined on \(G\). (Edit: \(G\) need not to be bi-ordered, just need to be linearly ordered.) Then
\begin{equation}
  \begin{array}{ccccc}
  & & a\: \text{is}\: \textit{peep bib} & \implies & a\: \text{is}\: \textit{peep} \\
  & & \Downarrow & & \Downarrow \\
  P(a)\: \text{is normal} & \implies & P(a)\: \text{is}\: \textit{fee} & \implies & P(a)\: \text{is a subgroup} \\
  \end{array}
\end{equation}
and the converses of these are not true.

For example, let \(F\) be the free group generated by \(a, b\) and bi-ordred in the way of Definition 3 of https://doi.org/10.4153/CJM-2003-034-2
Then the subgroup \(\langle a \rangle\) is not fee.
The fee subgroup \(\langle b^{F_+} \rangle\) is not normal (shown in the attached pdf
  momo (4).pdf (Size: 223.12 KB / Downloads: 134) ).
We define a map \(a\) on \(\mathbb{Z}\) as
\begin{equation}
  \dots, -7, -6, -5, 22, -3, 22, -1, 0, 1, 0, 1, 0, 1, 0, 1, \dots
\end{equation}
Then \(a\) is neither peep nor bib, and \(P(a) = 2\mathbb{Z}\).

[Natsugou]

This is a proof of the above diagram. We define \(g \sim_a h\) as \(a(g) = a(h)\).

If \(G\) is linearly ordered and \(a\) on \(G\) is peep, then \(\forall p, q \in P(a)\), \(\exists g \in G\), \(g \sim_a gp \sim_a gq\).
\(\because\) \(\forall p, q \in P(a)\), \(\exists g, h \in G\), \(g \sim_a gp\) and \(h \sim_a hq\). Since \(G\) is linearly ordered, \(gh^{-1} \geq 1\) or \(hg^{-1} \geq 1\) holds. From \(a\) is peep, \(g \sim_a gq\) or \(h \sim_a hp\) holds.

If \(G\) is bi-ordered and \(a\) on \(G\) is peep, then \(P(a)\) is a subgroup.
\(\because\) It is trivial that \(1 \in P(a)\) and \(P(a)\) is closed under inverses.
(i) Where \(pq \geq 1\): For all \(p, q \in P(a)\), we see \(q^{-1} \in P(a)\). As shown above, there exist \(g \in G\) such that \(g \sim_a gp \sim_a gq^{-1}\). Since \(G\) is bi-ordered, \(gpqg^{-1} \geq 1\). By using peep property, \(gpq \sim_a gpqp \sim_a gp\). Therefore \(g \sim_a gpq\).
(ii) Where \(pq < 1\): Then \(q^{-1}p^{-1} > 1\). By the same argument as in (i), we can indicate \(g \sim_a gq^{-1}p^{-1}\). Since \(P(a)\) is closed under inverses, \(pq \in P(a)\).

If \(P(a)\) is a subgroup of \(G\) and \(a\) is bib, then \(P(a)\) is fee.
\(\because\) For any \(p \in P(a)\) there exists \(g, h \in G\) such that \(p = g^{-1}h\) and \(g \sim_a h\). Since \(a\) is bib, \(gf \sim_a hf\). Thus \(f^{-1}pf = f^{-1}g^{-1}hf \in P(a)\).

Moreover, if \(G\) is linearly ordered and \(a\) on \(G\) is peep bib, then \(P(a)\) is fee.
\(\because\) (i) Where \(pq \geq 1\): For all \(p, q \in P(a)\), we see \(p^{-1} \in P(a)\). As shown above, there exist \(g \in G\) such that \(g \sim_a gp^{-1} \sim_a gq\). By using bib property and adding \(pq \geq 1\), we get \(gpq \sim_a gq \sim_a gqpq\). Therefore \(g \sim_a gpq\).
(ii) Where \(pq < 1\): Then \(q^{-1}p^{-1} > 1\). By the same argument as in (i), we can indicate \(g \sim_a gq^{-1}p^{-1}\). Since \(P(a)\) is closed under inverses, \(pq \in P(a)\). We have \(P(a)\) is a subgroup and \(a\) is bib, thus \(P(a)\) is fee.

Edit: Later, I noticed that if  \(G\) is linearly ordered and \(a\) on \(G\) is peep, then \(P(a)\) is a subgroup.
\(\because\) As shown above, there exist \(g \in G\) such that \(g \sim_a gp \sim_a gq^{-1}\). Since \(G\) is linearly ordered, \(gpqg^{-1} \geq 1\) or \(gq^{-1}p^{-1}g^{-1} \geq 1\) holds.
(i) Where  \(gpqg^{-1} \geq 1\):  By using peep property, \(gpq \sim_a gpqp \sim_a gp\). Therefore \(g \sim_a gpq\).
(ii) Where  \(gq^{-1}p^{-1}g^{-1} \geq 1\): Similarly, we get \(gq^{-1}p^{-1} \sim_a gq^{-1} \sim_a gq^{-1}p^{-1}q^{-1}\), and so \(g \sim_a  gq^{-1}p^{-1}\).  Since \(P(a)\) is closed under inverses, \(pq \in P(a)\).