# Totally ordered abelian group with a unique " isolated subgroup"

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 ?

• Another name for isolated subgroup is convex subgroup. Aug 1, 2018 at 23:25
• @DanielKawai : ok ... so ? Aug 2, 2018 at 0:21
• YES. This is true. I can give a long but entirely elementary proof. The idea is to fix some $a_0\in G$ with $a_0>e.$ For $e<x\in G$ and $n\in \Bbb N$ let $f(x,n)\in \Bbb N$ such that $nx\leq a_0f(x,n)<(n+1)x.$ Prove that $\psi (x)=\lim_{n\to \infty}n^{-1}f(x,n)$ exists. Let $\psi (y)=-\psi (-y)$ for $y<e$ and $\psi (e)=0$. Now prove that $\psi$ is the desired order-preserving group-isomorphism. There may be a brief sophisticated proof that I don't know about. Aug 2, 2018 at 1:33
• The long proof that I mentioned in my previous comment does not require that G is Abelian, provided that we also have $a<b\implies c+a<c+b$. So we have the result that any fully ordered group with no non-trivial convex sub-groups is Abelian. Aug 2, 2018 at 1:38

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.