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 ?

  • $\begingroup$ Another name for isolated subgroup is convex subgroup. $\endgroup$ Aug 1, 2018 at 23:25
  • $\begingroup$ @DanielKawai : ok ... so ? $\endgroup$
    – user521337
    Aug 2, 2018 at 0:21
  • $\begingroup$ 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. $\endgroup$ Aug 2, 2018 at 1:33
  • $\begingroup$ 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. $\endgroup$ Aug 2, 2018 at 1:38

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.