TITLE: What's a non-abelian totally ordered group?
QUESTION [32 upvotes]: Because I have heard the phrase "totally ordered abelian group", I imagine there should be non-abelian ones.  By this I mean a group with a total ordering (not to be confused with a well-ordering) which is "bi-translation invariant": a < b should imply cad < cbd.
Does anyone know any examples?  
Totally ordered abelian groups are easy to come up with: any direct product of subgroups of the reals, with the lexicographic ordering, will do.  Knowing some non-abelian ones would help reveal what aspects of totally ordered abelian groups really depend on them being abelian...
Edit: Via Andy Putman's answer below, I found this great summary of results about ordered and bi-ordered groups (i.e. groups with bi-translation invariant orderings) on Dale Rolfsen's site:
Lecture notes on Ordered Groups and Topology
He shows numerous examples of non-abelian bi-orderable groups, including a bi-ordering (bi-translation invariant ordering) on the free group with two generators.  As well, he mentions, due to Rhemtulla, that a left-orderable group is abelian iff every left-ordering is a bi-ordering, which I think really highlights the relationship between ordering and abelianity.

REPLY [4 votes]: Thompson's group $F$ is totally ordered.  See here for a description of all possible bi-orderings.  Indeed, all diagram groups are totally orderable (see here).  The pure braided Thompson group $BF$ also bi-orderable (see here).