Partial order relation on the group of integers

Question

We know that the usual $\leq$ is a partial order relation on the group of integers $\mathbb Z$ and $\mathbb Z$ is a totally ordered with this partial order relation. Is there any other partially order relation exist in $\mathbb Z$ which makes $\mathbb Z$ a partially ordered group (or totally ordered group)?

Answer 1

If $0$ and $1$ are related, then the only possible partial orders that make $\mathbb Z$ into a partially ordered group are the usual order $\le$ and its reverse order $\ge$:

If $0 \prec 1$, then $1 = 0 + 1 \prec 1 + 1 = 2$, and so $n \prec n+1$ for all $n$ by induction.

If $0 \succ 1$, then $n \succ n+1$ for all $n$.

If $0$ and $1$ are unrelated, I don't know right now... I think you need to consider all integers that are related to $0$. Then there'll be several linear orders "starting" at $0$ in both directions. For instance, if $0 \prec n$, then there will be arithmetic progressions of step $n$ starting at $k=0, \dots, n-1$.

Answer 2

Take $P = \{x\in\mathbb Z: x = 0 \text{ or }x\geq 2 \}.$ Then $G = (\mathbb Z, P)$ is a partially ordered group with positive cone $P.$ In this case, the element $3\wedge4$ does not exist in $G.$ If $3\wedge 4 = 3,$ then $3\leq 4, 4 - 3\in P$ but $4 - 3 = 1\notin P.$ This implies $G$ is not orderisomorphic to usual order on the group $\mathbb Z.$