Let an "order preserving" group be any group $(G,*)$ for which there exists a relation "$>$" that satisfies: $$\text{(A1) }a>b \text{ and }b>c\implies a>c,\forall a,b,c\in G$$$$\text{(A2) }a>b \Leftrightarrow b\not >a,\forall a,b\in G,a\not = b$$$$\text{(A3) }a>b \implies a*c>b*c,\forall a,b,c\in G$$
Some examples: $(\mathbb{Z},+)$ and $(\mathbb{R_{>0}},\cdot)$ are order preserving, but $(\mathbb{R} \backslash \{0\},\cdot)$ is not since $2>1\not \Rightarrow -2>-1$.
Conjecture: All order preserving groups are abelian
Work I've done:
To simplify the question, we can define the "positives" $P=\{a\in G|a>e\}$ (where $e$ is the identity). We can create necessary conditions for $P$ given that $G$ is order preserving:
$$\text{(A1)+(A3) }a*b^{-1}\in P \text{ and }b*c^{-1}\in P\implies a*c^{-1}\in P,\forall a,b,c\in G$$ $$\text{(A2)+(A3) }a*b^{-1}\in P\Leftrightarrow b*a^{-1}\not\in P,\forall a,b\in G,a\not = b$$ Or more simply: $$\text{(B1) }a\in P \text{ and }b\in P\implies a*b\in P,\forall a,b\in G$$ $$\text{(B2) }a\in P\Leftrightarrow a^{-1}\not\in P,\forall a\in G\backslash\{e\}$$
These conditions are are also sufficient, i.e. if there exists a $P\subset G$ that satisfies (B1) and (B2) then $G$ is order preserving. Simply define $a>b\Leftrightarrow a*b^{-1}\in P$ and the properties follow.
Mostly so far I've tried to find counter-examples to the conjecture. I've searched online for any current work on this problem, found that the braid group is supposedly a counter-example, but I cannot find the definition of ">" used nor have I been able to come up with my own (I tried defining it in terms of the leftmost "twist", thinking that the ordering of $a$ and $b$ might be "protected" by the other twists from $c$). I've played around with various non-abelian groups, modified their definitions slightly, etc. but not gotten very far.
edit: Thank you to @stewbasic who found the ordering for the braid group, it's called the Dehornoy order. I still don't know how to prove it works, however.
To prove the conjecture I think a first step would be to show that a version of $(A3)$ where $c$ is multiplied to the left giving $c*a>c*b$ must hold as well.
This ordering is known as the Dehornoy Order on the braid group. Let $\sigma_1,...,\sigma_{n-1}$ be the usual generators of $B_n$. We let $P$ be the collection of elements of $B_n$ such that $x \in P$ if $x$ can be written as a word in $\sigma_1^\pm,...,\sigma_{n-1}^\pm$ and $\exists i \in \{1,...,n-1\}$ where by $\sigma_i$ is present in the word, but for $j<i$, $\sigma_j^{\pm}$ and $\sigma_i^{-1}$ are not present in the word.
Then, we say that $a < b$ if $a^{-1}b \in P$.
Here is the explanation of why this is true. This is a very non-trivial result and I would have never come up with such an ordering on my own.