# If G is a solvable group whose partially ordered set of subgroups satisfies both the ascending and descending chain conditions, then G is finite.

Suppose $$G$$ is a solvable group whose partially ordered set of subgroups satisfies both the ascending and descending chain conditions.

GOAL: Show that $$G$$ is finite.

• Since $$G$$ is solvable, there exists a finite collection of normal subgroups in $$G$$ such that $$1 = N_0 \le N_1 \le \ldots \le N_n=G$$ and the factor groups are all abelian.
• Let $$P$$ be the poset of subgroups of $$G$$. Since $$P$$ satisfies both the ACC and DCC, every ascending chain and descending chain in $$P$$ is 'eventually constant'.
• We know $$P$$ satisfies the ACC $$\iff$$ every nonempty subset of $$P$$ has a maximal element. Likewise, $$P$$ satisfies the DCC $$\iff$$ every nonempty subset of $$P$$ has a minimal element.

Establish the following facts:

• The result will follow if you can prove it for abelian groups.

• Since $$G$$ has ACC on subgroups, it follows that $$G$$ is finitely generated.

• Since $$G$$ has DCC on subgroups, it follows that $$G$$ is torsion.

• Torsion finitely generated abelian groups are finite.

For an example of a nonsolvable group that satisfies the condition but is not finite, look at Tarski monsters.

• Okay, it suffices to show this when G is an abelian group. So now G being abelian and having ACC, we have G is noetherian. Hence every subgroup of G is finitely generated (and thus G is finitely generated?). Likewise, G is artinian. Explain how this implies torsion (we've not discussed torsion groups in particular). Thank you! – Hashy Nov 21 '19 at 0:43
• @Hashy: Noetherian means every subgroup is finitely generated; in particular, $G$ itself, which is a subgroup of itself, is finitely generated (consider the collection of finitely generated subgroups, which must have a maximal element). As to the artinian part, that’s true but you don’t need that much: the infinite cyclic group has infinite descending chains of ideals: $\langle x\rangle \supset \langle x^2\rangle \supset \langle x^4\rangle \supset\cdots$ – Arturo Magidin Nov 21 '19 at 2:56
• (for “ideals” read “subgroups”) – Arturo Magidin Nov 21 '19 at 3:04
• It's still unclear to me how every element of the group is finite by the DCC. – Hashy Nov 21 '19 at 16:32
• @Hashy: An element of infinite order generates an infinite cyclic group. Infinite cyclic groups do not satisfy DCC. Alternatively: pick an element $x\neq e$, and consider the set of all nontrivial subgroups of $\langle x\rangle$. By DCC, it has a minimal element. That minimal element must be a subgroup of prime order (the only nontrivial subgroups that have no proper subgroups are the cyclic groups of prime order). That means the order of $x$ must be finite. – Arturo Magidin Nov 21 '19 at 16:58