# Noncyclic (infinite) group with totally ordered subgroup lattice

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.

• A noncyclic group with a totally ordered subgroup lattice is a Prufer group: en.wikipedia.org/wiki/Pr%C3%BCfer_group – Keith Kearnes Dec 20 '17 at 4:06
• @KeithKearnes I had read a bit about the locally cyclic groups and a result of Ore in which it said that a group was locally cyclic iff the subgroup lattice was distributive, but didn't see what happened when it was totally ordered. But now that you show me, I was just looking for the Prüfer Groups. – Nell Dec 20 '17 at 4:10

## 2 Answers

This was already said, but here's a proof as well that the only such groups are $C_{p^n}$ (cyclic of order $p^n$, $p$ prime) and $C_{p^\infty}$ (Prüfer group, which is the inductive limit of $C_{p^n}$ for inclusions, and also realizable as the group of $p$-power roots of unity).

Let $G$ be a group with totally order set of subgroups. Then for any two $x,y$, there's one inclusion between $\langle x\rangle$ and $\langle y\rangle$ and hence $x,y$ commute. So $G$ is abelian. Also, $G$ is indecomposable (not decomposable as nontrivial direct product).

First assume that $G$ is finitely generated. Being abelian, finitely generated, and indecomposable, it is cyclic of infinite or $p$-power order ($p$ prime). The first case (infinite cyclic) is excluded because its subgroups don't form a chain. (Exercise: check this without appealing to the structure theorem for finitely generated abelian groups.)

If $G$ is not finitely generated, all its finitely generated subgroups have the same property and hence are cyclic of $p$-power order, necessarily for a single $p$. It easily follows that $G$ is a Prüfer group.

Consider the multiplicative group of all power of $p$ roots of unity, for some fixed prime $p$.

• Wow. Does this really work? Is there some easy way to classify its subgroups? – Randall Dec 20 '17 at 3:59
• Hmmm . . . Maybe not. – quasi Dec 20 '17 at 4:00
• Try the new one. – quasi Dec 20 '17 at 4:12