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, \(pq \in P(a)\).
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, \(pq \in P(a)\).