Let $(G,+,<)$ be a totally ordered abelian group i.e. $(G,+)$ is an abelian group with partial order $<$ such that for every $a,b\in G$, exactly one of $a=b$ or $a<b$ or $b<a$ holds; and for every $a,b,c\in G$, $a<b \implies a+c<b+c$. Let us call a subgroup $H$ of a totally ordered abelian group $G$ to be isolated if $H\ne G$ and $a\in H, -a<a \implies b\in H, \forall -a<b<a$.
My question is: If $(G,+,<)$ is a totally ordered abelian group with a unique isolated subgroup i.e. the trivial subgroup $\{e\}$ is the only isolated subgroup, then is it true that there is an order preserving isomorphism between $G$ and a subgroup of $(\mathbb R,+)$ with the usual order inherited from real line ?
Best Answer
Accordingly to this page: "Order Preserving Isomorphism", it is sufficient to infer the Archimedean property (for every $a,b\in G$, if $a>0$, then there is a $n\in\mathbb{N}$ such that $na\geq b$).
Let $a>0$ be an element of $G$. Let $H=\{x\in G:\exists n\in\mathbb{N}:-na<x<na\}$. Then $-a<0<a$, so $0\in H$. For $x,y\in H$, there are $m,n\in\mathbb{N}$ such that $-ma<x<ma$ and $-na<y<na$, so $-(m+n)a<x-y<(m+n)a$, so $x-y\in H$. Also, for $x,y\in G$, if $0\leq y\leq x$ and $x\in H$ then there is $n\in\mathbb{N}$ such that $0\leq x<na$, so $0\leq y\leq x<na$, so $y\in H$. Therefore, either $H$ is an isolated subgroup (then $H=\{0\}$) or $H=G$, but $-2a<a<2a$, so $a\in H$, so $H\neq\{0\}$, so $H=G$. Therefore, $G$ is Archimedean.