I apologize for the lengthy preamble, I have no idea how much of what I am relying on is common knowledge (or even correct to be honest). Let $G=\langle S\mid R\rangle$ be a finitely presented group of rank $n$, and define $\sigma:\mathbb{F}_n\to \mathbb{Z}$ to be the sum of the exponents of an element of $\mathbb{F}_n,$ with the usual basis for $\mathbb{F}_n$. We see $\sigma$ is well defined, since $\mathbb{F}_n$ is a group of equivalence classes on $n$ letters (and $n$ "inverse" letters) where $x\sim y$ if and only if $x=ww'$ and $y=waa^{-1}w'$ where $w,w'$ are strings on $n$ letters and $a$ is a letter. Now, suppose $\sigma(r)=0$ for all $r\in R.$ Then $\sigma$ may be defined similarly on $G$, since $G$ is a group of equivalence classes of elements of $G$ where $x\sim y$ if and only if $x=ww'$ and $y=wpw',$ where either $p=ss^{-1}$ for $s\in S$ or $p\in R.$

Examples of groups of this type include Right Angled Artin Groups, Free Groups (since $R=\{\}$ in this case), and Braid Groups.

My question is this: are all of these groups totally orderable?

  • 1
    $\begingroup$ Your $\sigma$ is only well-defined if you fix a basis. For example, the free group of rank $2$ can be generated by either $x$ and $y$; or by $xy$ and $y$. The sum of exponents of $xy^2$ relative to the first basis is $3$, but the sum of exponents relative to the second basis is $2$, since $xy^2 = (xy)y$ is a product of just two basis elements, raised to the first power. So you need to specify your free basis for $\sigma$ to work. $\endgroup$ Jan 14, 2020 at 7:24
  • $\begingroup$ The title is misleading: I think you mean "with relators whose exponents sum to $0$". $\endgroup$
    – YCor
    Jan 14, 2020 at 19:54
  • $\begingroup$ You should define "totally orderable". It sometimes means: admits a left-invariant total order; sometimes, admits a bi-invariant total order, which is strictly stronger. Anyway this doesn't matter since the answer shows that you can even arrange the group to be non-torsion-free, say with $\langle x,y|[x,y]^2\rangle$. $\endgroup$
    – YCor
    Jan 14, 2020 at 19:55

1 Answer 1


Any group with a presentation of the form $\langle \mathbf{x}\mid R^n\rangle$, with $R\in [F(\mathbf{x}), F(\mathbf{x})]$ and $n>1$, satisfies the exponent-sum condition and contains torsion elements$^{[1]}$. However, these groups are not totally-ordered as any totally-ordered group is torsion-free.

The issue with your idea is that it is simply saying that the group $G$ surjects onto $\mathbb{Z}^{|S|}$, which is a "global" property of the group. For orderability you need "local" properties, so properties of (finitely generated) subgroups. In the above example, although the group itself maps onto $\mathbb{Z}$, the subgroup $\langle R\rangle$ does not.

A condition which does imply orderability is local indicability. A group is locally indicable if every finitely generated subgroup subject into $\mathbb{Z}$. These groups were first studied by Higman in his PhD thesis (regarding zero divisors of group rings), and they admit a total order (although not necessarily a two-sided order). It is a theorem of Brodskii, and independently Howie$^{[2]}$, that torsion-free one-relator groups are locally indicable (so groups $\langle \mathbf{x}\mid R\rangle$ where $R\neq S^n$ for any $n>1$, and no restriction on $R$ only being taken from the derived subgroup!).

For the specific groups you mention: free groups, and more generally RAAGS, are bi-orderable$^{[3]}$, while the Braid groups $B_n$ are not for $n\geq3$$^{[4]}$. However, the braid groups $B_n$ are all right-orderable, $n\geq1$$^{[5]}$.

[1] Theorem 4.12 of the book Wilhelm Magnus, Abraham Karrass, and Donald Solitar. Combinatorial group theory. Dover Publications, 1976.

[2] Howie, James. "On locally indicable groups." Mathematische Zeitschrift 180.4 (1982): 445-461.

[3] Duchamp, Gérard, and Jean-Yves Thibon. "Simple Orderings for Free Partially Commutative Groups." IJAC 2.3 (1992): 351-356.

[4] Neuwirth, Lo Po. "The status of some problems related to knot groups." Topology conference. Springer, Berlin, Heidelberg, 1974.

[5] Dehornoy, Patrick. "Braid groups and left distributive operations." Transactions of the American Mathematical Society 345.1 (1994): 115-150.

  • $\begingroup$ Surely both $B_1$ and $B_2$ are right-orderable, for silly reasons $\endgroup$ Jan 15, 2020 at 15:35
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    $\begingroup$ @RyleeLyman Yes, they are also bi-orderable! The $n\geq3$ was attached to the wrong statement. I've correct it now. Thanks. $\endgroup$
    – user1729
    Jan 15, 2020 at 16:14

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