Let $(\Lambda, \leq ,+)$ be a totally ordered abelian group. Say such an abelian group is simple if it has no nontrivial quotients (the only quotients are $0$ and itself).

One might wish to understand totally ordered abelian groups in terms of the simple ones. We may wish to embed a totally ordered abelian group $\Lambda$ as $\Lambda \rightarrow \prod_{i \in I} \Lambda_i$, where the $\Lambda_i$ are simple and the product is given some kind of dictionary order.

My question is, is there a terminal simple totally ordered abelian group? I would suspect it is $(\mathbb{R}, \leq, +)$, but I cannot seem to show this.

Note: by simple, I mean there are no subgroups $S \subset \Lambda$ such that $[a, b] \subset S$ for each $a, b \in S$.

Lemma 1: A simple totally ordered abelian group $\Lambda$ is Archimedian.

By contrapositive, suppose that $\Lambda$ is not archimedian, and take $a, b \in \Lambda$ such that $an < b$ for each $n$. Then consider the order closure of the subgroup $\langle a \rangle$, i.e. $$ \{ x \in \Lambda : x \in [-na, na] \text{ for some } n \in \mathbb{N}_{\geq 0} \} $$ This is a subgroup of $\Lambda$ which does not include $b$. Hence it is a proper subgroup, so that $\Lambda$ is not simple.

Lemma 2: The terminal simple totally ordered group $\Lambda$ has all suprema and infima.

This is clear enough, as we can embedd any simple totally ordered abelian group $\Lambda$ into its completion $\tilde{\Lambda}$ (where we addded in suprema and infima). Note: we add in the supremum of any set bounded above.

Edit: it was pointed out in the comments that we need to show that the completion $\tilde{\Lambda}$ is simple. Arguing by contrapositive, suppose that the completion is not simple, and take the preimage of a proper nontrivial order closed subgroup $T \subset \tilde{\Lambda}$, and call this preimage $S$. $T$ has a nonzero element $t > 0$ in it, and therefore contains $[-t, t]$. The preimage $S$ of $T$ under $\Lambda \rightarrow \tilde{\Lambda}$ therefore contains at least one nonzero element, as there must be some subset $X \subset \Lambda$ such that $\text{sup}(X) = t$, and such a set must contain an element of $[-t, t]$, which would then be contained in $S$. Hence $S \neq \{ 0 \}$, so that $S = \Lambda$ since $\Lambda$ is simple. Since $T$ is order closed, it must contain all of the elements of the order closure $f(S)$, which is all of $\tilde{\Lambda}$.

Hence we can assume that are are looking for a complete archimedian simple totally ordered abelian group. I would really imagine it is $\mathbb{R}$ at this point, but I cannot manage to show it.

Let's try to construct a map from $\Lambda$ to $\mathbb{R}$. Take $a \in \Lambda$ not equal to $0$ and define $f(a) = 1$. Define $f(na) = n$. Now, for each $m$ and each $b$ such that $mb = na$, define $f(b) = n/m$. Define $f$ on suprema by setting $f( \text{sup} S ) = \text{sup} f(S)$. Now we need to show that this is well defined and defined on all of $\Lambda$.

For positive $b \in \Lambda$, let $\rho(n)$ be the minimal integer such that $\rho_b(n) b \geq n a$. Suppose $b$ and $b'$ are such that $\text{lim}_{n \rightarrow \infty} n/ \rho_b(n) = \text{lim}_{n \rightarrow \infty} n/ \rho_{b'}(n)$. Without loss of generality, $b < b'$. Then we can show that $N(b - b') < a$ for each $N$, violating the archimedian property. This shows that the map constructed is injective.

Well definedness is probably a bit tedious, but I suspect it is true.

  • $\begingroup$ Why should the completion be simple ? By the way, when you say no nontrivial quotients, do you mean quotients of linearly ordered abelian groups or of abelian groups ? $\endgroup$ Dec 6, 2020 at 18:53
  • $\begingroup$ I mean a linearly ordered quotient. $\endgroup$ Dec 6, 2020 at 18:54
  • $\begingroup$ Also, you will have to weaken your request: there can be no terminal such object, as $a\mapsto na$ for $n\in\mathbb N_{>0}$ is always a self map different from the identity (as $A$ must be torsion free). Perhaps you can ask for weakly terminal, i.e. there always exists a map, but no need for it to be unique $\endgroup$ Dec 6, 2020 at 19:57
  • $\begingroup$ @MaximeRamzi Adding a designated point will fix that (and requiring the morphisms to preserve that point). But perhaps there is a better way. $\endgroup$ Dec 6, 2020 at 19:59
  • $\begingroup$ @MaximeRamzi I like your suggestion better, weakly terminal... and also something to prevent it from being merely $\{ 0 \}$. $\endgroup$ Dec 6, 2020 at 20:01

1 Answer 1


Let $A$ be a simple linearly ordered complete divisible archimedean abelian group.

It is in particular a $\mathbb Q$-vector space. For any $a\in A$, we have an embedding $\mathbb Qa\to A$ which automatically extends to $\mathbb R\cong \mathbb Ra\to A$, as $A$ is complete.

I claim that this map must be an isomorphism. Clearly it must be injective as $\mathbb R$ is simple. Moreover, if $b\in A$, $b\leq na$ for some $n$, and so we can actually consider $\{r\in \mathbb R, ra < b\}$ and this has an upper bound, so it has a least upper bound, $r_b$.

Now for any $r< r_b$ we have $(r_b-r)a > r_ba - b$, so for any $n\in \mathbb N$, $a> n(r_ba-b)$. By the archimedean character of $A$, it follows that $r_ba - b \leq 0$, i.e. $r_ba \leq b$ as well.

But now one can also similarly prove $r_ba \geq b$, and so $r_ba = b$

(a simpler way to phrase surjectivity would be to use Dedekind reals, but let me stick to this formulation in case you don't know these).

The claim follows, and so by embedding your linearly ordered group into its rationalization and then the completion of that, you get that $\mathbb R$ is weakly terminal among simple linearly ordered abelian groups.

  • $\begingroup$ Do you think the divisibility assumption is necessary? $\endgroup$ Dec 6, 2020 at 20:20
  • $\begingroup$ What do you mean ? In the last paragraph I explain how to reduce to the divisible case for an arbitrary group $\endgroup$ Dec 6, 2020 at 20:21
  • $\begingroup$ Oh right, my bad! $\endgroup$ Dec 6, 2020 at 20:22
  • $\begingroup$ my greatest thanks for your answer, this is perfect! $\endgroup$ Dec 6, 2020 at 20:23

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