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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.

  1. 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 ($+$).

  1. 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.
  2. 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.
  3. If two elements $a$ and $b$ are positive in $C$, then $a < b$ in $L$ iff $[0, a, b]$ in $C$:
    • $[-a, 0, a] \iff 0 < a$;
    • $[-b, 0, b] \iff 0 < b$;
    • $0 < a < b \iff [0, a, b]$.
  4. 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.
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  • $\begingroup$ compatible cyclically ordered means when you multiply all group elements by an element, you rotate them in their cyclical ordered structure? If you cut the circle and straightens it, then it could be that a<b in the linear order and ax>bx? So then it is not possible to find compatible cut? $\endgroup$ – mathpadawan Apr 29 at 17:59
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    $\begingroup$ @mathpadawan: If $a < b$ and $ax > bx$, it means the cut is not compatible. But it does not mean there is no other cut compatible with the operation. I am trying to say that the only possible compatible cut exists between positive and negative elements on apex side of the circle. A cyclic order is compatible iff $[a, b, c] \implies [a x, bx, cx]$ and $[xa, xb, xc]$ $\endgroup$ – Alex C Apr 29 at 18:36

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