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.

Thank you for your help.


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.

  • $\begingroup$ 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! $\endgroup$ – Hashy Nov 21 '19 at 0:43
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    $\begingroup$ @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: $\langle x\rangle \supset \langle x^2\rangle \supset \langle x^4\rangle \supset\cdots$ $\endgroup$ – Arturo Magidin Nov 21 '19 at 2:56
  • $\begingroup$ (for “ideals” read “subgroups”) $\endgroup$ – Arturo Magidin Nov 21 '19 at 3:04
  • $\begingroup$ It's still unclear to me how every element of the group is finite by the DCC. $\endgroup$ – Hashy Nov 21 '19 at 16:32
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    $\begingroup$ @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. $\endgroup$ – Arturo Magidin Nov 21 '19 at 16:58

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