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) ?
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3$\begingroup$ You mean, other than the two obvious ones (trivial and improper)? $\endgroup$– Arturo MagidinCommented Mar 24, 2015 at 17:15
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1$\begingroup$ You mean, a nontrivial totally ordered group? $\endgroup$– Emil JeřábekCommented Mar 24, 2015 at 17:42
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1$\begingroup$ @HJRW the given definition is that of bi-orderable (specified to $x=y$ it means the order is left-invariant and specified to $z=t$ it means it's right invariant). $\endgroup$– YCorCommented Mar 24, 2015 at 18:26
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$\begingroup$ @YCor, yes, in the light of Dave's answer I realize that now. $\endgroup$– HJRWCommented Mar 25, 2015 at 1:40
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1 Answer
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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 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.)
answered Mar 24, 2015 at 17:51