## Every nontrivial subgroup of a solvable group contains normal abelian subgroups

Prove that if is a nontrivial subgroup of a solvable group then there is a nontrivial subgroup which is abelian and normal in .

We proceed by induction on the width of – that is, the number of prime factors of (including multiplicity).

If has width 1, then is simple and no such exists.

If has width 2, say , and is nontrivial, then without loss of generality .

Lemma: Let be a nonempty partially ordered set such that every chain has bounded length. Then there exists a minimal element . Proof: Let be a chain of maximal length , where . If an element exists such that , then is a chain of length , a contradiction. So is -minimal in .

Consider the set . Note that is nonempty since . Note that because is solvable, has a finite composition series, so that by the Jordan-Holder Theorem, all composition series for have the same finite length . Now every chain in can be extended to a composition series for , and thus has length bounded by . By the lemma, then, there exists a -minimal element . By Lemma 5 to this previous theorem, is abelian.

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## Comments

“Note that because G is solvable, G has a finite composition series.” Can this be right? The problem doesn’t assume G is finite, and Z is a solvable group that doesn’t have a composition series.

Ah, this proof implicitly assumes that is finite. I’ll have to mark this for later.

Thanks!