A totally ordered group is a group equipped with a compatible total order, that is, $x\leq y$ and $z\leq t$ imply $x+z\leq y+t$ for all $x,y,z,t$ in the group.
Is it true that every totally ordered group has a nontrivial normal subgroup (not necessarily convex) ?



Normal subgroup of a totally ordered group
No. Thompson's group F is bi-orderable (or "totally ordered", in your terminology), and its commutator subgroup [F,F] is simple. So [F,F] is an (infinite) bi-orderable simple group. See [this paper](http://arxiv.org/abs/0808.1688) of Navas and Rivas for a discussion of all the bi-invariant orderings of F, and a reference to a proof that [F,F] is simple.


On the other hand, if a bi-orderable group is finitely generated (and nontrivial), then its abelianization is infinite, so the group certainly has nontrivial, proper, normal subgroups. This fact is in standard textbooks on orderable groups. (See, for example, Theorem 2.3.1 of the book of Kopytov and Medvedev on *Right-Ordered Groups*.)


