Questions in Geometric Group Theory — Q 12.8

Definitions and mathematical statement

Let $F_n$ be the free group on $n$ generators and let $\operatorname{Out}(F_n)$ be its automorphism group modulo inner automorphisms. A strict total order $<$ on a group $H$ is right invariant if \[ a<b\quad\Longrightarrow\quad ac<bc \qquad(a,b,c\in H). \] Strict total means transitive, irreflexive, and that exactly one of $a<b$, $a=b$, $b<a$ holds. The group is right orderable if such an order exists, with no requirement that the order be computable.

For each integer $n>2$, does $\operatorname{Out}(F_n)$ have a right-orderable subgroup $H$ of finite index? The subgroup and order may depend on the rank, and the index need not satisfy any prescribed bound. Normality of $H$ is not required.

The primary source already records a positive answer at rank three, through virtual residual torsion-free nilpotence and consequent virtual bi-orderability. Thus the remaining general question concerns higher ranks; that recorded case is not presented here as newly open. A bi-invariant order, invariant under both left and right multiplication, is stronger than the requested right order. Torsion-freeness is necessary for orderability but does not itself provide an order, so constructing a torsion-free finite-index subgroup alone is insufficient. The question asks about the abstract subgroup, not an order on free-group words invariant under every outer automorphism.

Short English statement

In higher ranks, does $\operatorname{Out}(F_n)$ have a right-orderable subgroup of finite index?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 1101208 (AMR-010-1208), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Bestvina - Questions in Geometric Group Theory (pdf) (2004), Question 12.8 (PDF page 20). Original source indexed by AMR-010-1208. https://www.math.utah.edu/~bestvina/eprints/questions-updated.pdf.