I was wondering if it's possible to find a noncyclic group with a totally ordered subgroup lattice.
For the finite case I attacked the problem like this:
First, note that $G$ can't be nonabelian (Even in the infinite case), since, if it were, then there would be $a,b\in G$ which don't commute, so $a\notin \langle b \rangle$ and $b\notin \langle a \rangle$ (Since these cyclic groups are abelian).
Now, since $G\neq 1$ is abelian, if $G$ is finite (Or even finitely generated), then $$G\cong H\times K$$ where $H$ is cyclic and nontrivial and $K$ is another abelian group (This follows from the fundamental theorem of finitely generated abelian groups). But then $H\not\leq K$ and if $K\neq 1$, $K\not\leq H$, so if $G$ is not cyclic its lattice subgroup is not totally ordered.
In the infinite case, I can still argue that $G$ is abelian, so a counterexample must be an abelian group which is not finitely generated. How can such a counterexample be found (if possible)?
The question was based on a comment here in which they said that in the finite case the group must be cyclic, but didn't say anything about the infinite case.