Ordered Groups: Left Multiplication vs Right Multiplication Given that $G$ is a linearly ordered group (bi-ordered). I want to try and understand the difference between the “size” of left multiplication vs right multiplication (which I have written below using additive notation, but this is not meant to imply that $G$ is abelian).
To be more specific, given $\varepsilon \in G^{>0}$, does there always exist some $\delta \in G^{>0}$ such that $x + \varepsilon \geq \delta + x$ (for all $x \in G$)?
To be a bit more qualitative, I considered the function $\phi : G^{>0} \rightarrow G^*$ defined by $\phi(\varepsilon) := \inf\{x + \varepsilon + (-x) : x \in G\}$ (where $G^*$ is the infimum closure of $G$, which is just a set and not necessarily a group).
It is not too difficult to show that $\phi(x) \geq 0$ for all $x > 0$, and $\phi$ is an increasing function. Therefore the question can be reduced to asking:
Can there exist $g \in G^{>0}$ such that $\phi(g) = 0$?
 A: Have you heard of transseries? Those are formal series involving real coefficients, exponentials and logarithms. The ordered field $\mathbb{T}$ of log-exp transseries is equipped with a partial composition law $\circ: \mathbb{T} \times \mathbb{T}^{>\mathbb{R}} \rightarrow \mathbb{T}$ where $\mathbb{T}^{>\mathbb{R}}$ is the set of transseries which are larger than each real number.
This law is such that $G:=(\mathbb{T}^{>\mathbb{R}},\circ,x,<)$ is a linearly bi-ordered group with identity element $x$, which corresponds to the identity function. In $G$, you have a formal version $\operatorname{e}^x$ of the exponential, a formal version $x+1$ of the translation by $1$. And all compositions of those and their inverses. I write $E_n$ fot the $n$-fold composition of $\operatorname{e}^x$, $L_n$ for its formal reciprocal (i.e. inverse in $G$) and $T_1:=x+1$. I also omit the composition symbol between transseries in the sequel.
The order on $G$ can be informally understood as $f<g$ if $f(x)<g(x)$ for sufficiently large $x$, for the corresponding real-valued functions $f,g$ defined on final segments of $\mathbb{R}$. Many transseries do not correspond to functions, at least in a straight forward way, but many do, and in any case the intuition is pretty useful.
So we have our linearly bi-ordered group. I claim that the set of transseries $L_n T_1 E_n$ for $n \in \mathbb{N}$ is coinitial in $G^{>x}$. Indeed, transseries have formal Taylor expansions around every point. In particular, for $n >0$, we have $L_n(T_1 E_n)=x+\rho$ where $\rho$ is a transseries which is positive and smaller than $\frac{1}{E_{n-1}}$ (note that we are taking advantage of the extra field structure on $\mathbb{T}$ but we are still talking about things happenning in $G$).
But the set $\{\frac{1}{E_n}: n \in \mathbb{N}\}$ is coinitial in $\mathbb{T}^{>0}$ so no transseries above $x$ can be smaller than each $L_n T_1 E_n$. This proves that $\phi(T_1)=0$ in $G$.

One could do without transseries and directly think of $L_n T_1 E_n$ as (germs at $+\infty$ of) functions in the group generated by the germs at $+\infty$ of the real-valued functions $\exp$ and $r \mapsto r+1$. But one then requires some model theoretic arguments to justify that this is linearly ordered and that the set of germs $L_n T_1 E_n$ is indeed coinitial. At least I don't see an easy way to prove the result.

As a final comment, one could imagine groups where $\phi(\varepsilon)>0$ for all $\varepsilon$. The idea would be that for groups of functions or formal series such as that which I used, the terms $x+\varepsilon+(-x)$ are very close to $0$ if $\varepsilon$ is close to $0$ and most of all if $x$ is very small. But one could make sure that $G^{>0}$ has no countable coinitial subset whereas $G^{>0}$ has countable cofinality. Picking a cofinal sequence $(y_n)_{n \in \mathbb{N}}$ in $G^{>0}$, the sequence $((-y_n)+\varepsilon+y_n)$ would be the best candidate to reach the infimum for $\phi(\varepsilon)$. But it could not reach $0$ because $G^{>0}$ has no coinitial subsequence.
Constructive examples of such groups will be more exotic and difficult to introduce.
