A left order on a (discrete) group $G$ is a total order on $G$ satisfying $\forall g,h,k \in G: g < h \implies kg < kh$. A right order is defined symmetrically, and a biorder is an order that is both at once. An order is scattered if it has no densely ordered subset of cardinality at least two, where a set $S$ is densely ordered if $a, c \in S \wedge a \neq c \implies \exists b \in S: a < b < c$.

I'm interested in order types of left orders (and biorders, why not). It's easy to show that left orders on the integers $\mathbb{Z}$ all have order type $\mathbb{Z}$, and on $\mathbb{Z}^2$ there are two constructions of left orders, and as far as I can tell one gives order type $\mathbb{Z}^2$ in lex order (so scattered), and the other orders are dense (so not scattered).

Let $F_2$ be the free nonabelian group on two generators.

Does $F_2$ admit a scattered left order?

Does $F_2$ admit a scattered biorder?

I had an awesome application for this, but I broke it already. Now I'm just curious. I don't actually know what you get from the Magnus embedding $a \mapsto 1+a$, I admit I was too lazy (or scared?) to even give it real thought, and I did not notice a statement in the literature.

More generally, one may ask:

What are the order-types of left orders on $F_2$?

What are the order-types of biorders on $F_2$?

More generally, I'm interested in information on order types of orders on any torsion-free groups, there's plenty of literature on orders but I haven't seen much about order types.

  • $\begingroup$ Weaker than your first question, but $F_2$ acts faithfully on a scattered left order. Indeed, for each quotient by a term of the lower central series one get such a (non-faithful) action, and then one can concatenate. $\endgroup$
    – YCor
    Jan 21, 2020 at 19:20
  • $\begingroup$ Does everything? Is there a simple torsion-free non-orderable group? $\endgroup$
    – Ville Salo
    Jan 21, 2020 at 20:42
  • $\begingroup$ Erm wait, dyadic rationals. $\endgroup$
    – Ville Salo
    Jan 21, 2020 at 20:45
  • $\begingroup$ Don't understand your last 2 questions. You're using "simple" in the meaning "non-complicated"? Anyway every torsion-free abelian group is orderable. Nevertheless I agree that for $p\ge 2$ $\mathbf{Z}[1/p]$ (and hence $BS(1,p)$) cannot act faithfully on any scattered totally ordered space. $\endgroup$
    – YCor
    Jan 21, 2020 at 23:07
  • $\begingroup$ I mean simple in the usual sense. With the first question I meant, does everything act on a scattered left order? If yes, then your observation is not very interesting. With the second question, I figured if you have a simple torsion-free non-orderable group, then that will be a counterexample. But then I realized $\mathbb{Z}[1/2]$ is an easier example. I agree you can also f.g.ify it. $\endgroup$
    – Ville Salo
    Jan 22, 2020 at 5:14

1 Answer 1


Let us first clarify the relationship between scattered, discrete, and dense orders. The last two notions are standard in the theory of ordered groups.

An order (left or two-sided) is discrete if there exists a smallest (necessarily unique) positive element. If such an element does not exist, the order is called dense.

We claim that no dense order (left or two-sided) is scattered. Moreover, we claim that, if the order is dense, then the entire group is a densely ordered set. Indeed, if $a < b$ then $e < a^{-1}b$ and, since the order is dense, there exists $c$ such that $e < c < a^{-1}b$ and we obtain $a < ac < b$.

On the other hand, discrete orders are not necessarily scattered. For instance the lexicographic bi-order on $\mathbb{Z} \times F_2$, with any bi-order on $F_2$, is discrete with smallest element $(1,e)$, but there are densely ordered subsets, namely the copy of $F_2$.

Going back to the posed question, since free groups of rank > 1 do not admit discrete bi-orders (in fact, no centerless group admits a discrete bi-order) they do not admit scattered bi-orders.

On the other hand, there are discrete left-orders on free groups of rank > 1, and, therefore, the question is more interesting there.

The claim that centerless groups do not admit discrete bi-orders follows from Theorem 2.1 in the paper cited below, while the claim that free groups admit discrete orders (that are, moreover Conradian) is Corollary 3.6 in the same paper.

Linnell, Peter A.; Rhemtulla, Akbar; Rolfsen, Dale P. O., Discretely ordered groups, Algebra Number Theory 3, No. 7, 797-807 (2009). ZBL1229.06008.

  • $\begingroup$ Very nice! Can you give a reference for the last two paragraphs? $\endgroup$
    – Ville Salo
    Jun 8, 2022 at 17:42
  • $\begingroup$ I believe you also give a complete answer to the fourth question, as there is a unique dense countable total order without endpoints. $\endgroup$
    – Ville Salo
    Jun 8, 2022 at 18:00
  • $\begingroup$ Added a reference. $\endgroup$ Jun 8, 2022 at 18:10
  • $\begingroup$ By the way, as I recall, the Magnus embedding order is one of those that can be obtained from the lower central series. $\endgroup$ Jun 8, 2022 at 20:34

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