It is well known that a cyclically ordered set may have more than one cut.
I am trying to prove that a cyclically ordered group cannot have more than one cut compatible with the group operation. Is this a correct statement?
A cut of a cyclically ordered set is a linear order $<$ such that
- $a < b < c \implies [a, b, c]$
for any elements $a$, $b$, $c$ of the set.
Since $[a, b, c] = [b, c, a] = [c, a, b]$, then
- $[a, b, c] \iff a < b < c$ or $b < c < a$ or $c < b < a$
for a cut of a cyclically ordered set.
A cut of a cyclically ordered group is compatible with the group operation ($+$) iff:
- $a < b \implies a + x < b + x$ and $x + a < x + b$ for any elements $a$, $b$, $x$ of the group.
A cyclic order of a group is linear if there is a cut compatible with the group operation.
My attempt to prove the main statement:
Let $G(+)$ be a group with a cyclic order $C$ on it.
- There is no compatible cut if there is an apex $x$ in $C$ (Apex of a cyclically ordered group):
- if $x < 0$, then $x + x < 0 + x$, $0 < x$, contradiction;
- if $0 < x$, then $0 + x < x + x$, $x < 0$, contradiction.
Let's assume there is a cut $L$ compatible with group operation ($+$).
- An element $a$ is positive in $C$ iff $a$ is positive in $L$ (Positive and negative elements of a cyclically ordered group):
- if $0 < a$ in $L$, then $-a < 0 < a$ in $L$, $[-a, 0, a]$ in $C$;
- if $[-a, 0, a]$ in $C$, then $-a < 0 < a$ or $0 < a < -a$ or $a < -a < 0$ in $L$,
cases $0 < a < -a$ and $a < -a < 0$ are not compatible with the group operation.
- An element $a$ is negative in $C$ iff $a$ is negative in $L$:
- if $a < 0$ in $L$, then $a < 0 < -a$ in $L$, $[a, 0, -a]$ in $C$;
- if $[a, 0, -a]$ in $C$, then $a < 0 < -a$ or $0 < -a < a$ or $-a < a < 0$ in $L$,
cases $0 < -a < a$ and $-a < a < 0$ are not compatible with the group operation.
- If two elements $a$ and $b$ are positive in $C$, then $a < b$ in $L$ iff $[0, a, b]$ in $C$:
- $[-a, 0, a] \iff 0 < a$;
- $[-b, 0, b] \iff 0 < b$;
- $0 < a < b \iff [0, a, b]$.
- If two elements $a$ and $b$ are negative in $C$, then $a < b$ in $L$ iff $[a, b, 0]$ in $C$:
- $[a, 0, -a] \iff a < 0$,
- $[b, 0, -b] \iff b < 0$,
- $a < b < 0 \iff [a, b, 0]$.
Thus, $L$ (if exists) is uniquely defined by the following rule:
$a < b$ for any $a ≠ b$ iff any of the following statements is true:
- $a$ is not positive and $b$ is not negative in $C$;
- both $a$ and $b$ are positive or negative in $C$, and $[0, a, b]$.
Using this fact, we can introduce the natural cut for an arbitrary cyclically ordered group:
$a < b$ for any $a \ne b$ iff any of the following statements is true:
- $b$ is an apex;
- $a$ is negative or $0$, and $b$ is not negative;
- both $a$ and $b$ are positive or negative, and $[0, a, b]$.
Properties of the natural cut:
- The linear order of a linearly ordered group is the natural cut of its induced cyclic order;
- A cyclic order on a group is linear if and only if its natural cut is compatible with the group operation.