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

Prove that if $H$ is a nontrivial subgroup of a solvable group $G$ then there is a nontrivial subgroup $A \leq H$ which is abelian and normal in $G$.

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

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

If $G$ has width 2, say $|G| = pq$, and $H \leq G$ is nontrivial, then without loss of generality $|H| = p$.

Lemma: Let $(\mathcal{P},\leq)$ be a nonempty partially ordered set such that every chain has bounded length. Then there exists a minimal element $m \in \mathcal{P}$. Proof: Let $\{C_i\}_{i=1}^n \subseteq \mathcal{P}$ be a chain of maximal length $n$, where $C_i \leq C_{i+1}$. If an element $C_0 \in \mathcal{P}$ exists such that $C_0 \leq C_1$, then $\{ C_i \}_{i=0}^n$ is a chain of length $n+1$, a contradiction. So $C_1$ is $\leq$-minimal in $\mathcal{P}$. $\square$

Consider the set $\mathcal{H} = \{ A \vartriangleleft G \ |\ A \leq H, A \neq 1 \}$. Note that $\mathcal{H}$ is nonempty since $H \in \mathcal{H}$. Note that because $G$ is solvable, $G$ has a finite composition series, so that by the Jordan-Holder Theorem, all composition series for $G$ have the same finite length $n$. Now every chain in $\mathcal{H}$ can be extended to a composition series for $G$, and thus has length bounded by $n$. By the lemma, then, there exists a $\leq$-minimal element $A \in \mathcal{H}$. By Lemma 5 to this previous theorem, $A$ is abelian.

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